Chapter 4 – Chemical Composition - University of …chemweb/chem1000/docs...4.3 The percent composition of calcium in calcium carbonate is set up as: % Ca = 33 grams Ca 9.88g Ca

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Chapter 4 – Chemical Composition

4.1 (a) mole; (b) Avogadro’s number; (c) empirical formula; (d) solute; (e) molarity; (f) concentrated solution

4.2 (a) molar mass; (b) percent composition by mass; (c) solvent; (d) concentration; (e) dilute solution; (f)

dilution

4.3 The percent composition of calcium in calcium carbonate is set up as:

% Ca = 3 3

grams Ca 9.88 g Ca100% 100%

grams CaCO 24.7 g CaCO = 40.0 % Ca

4.4 The percent composition of oxygen in calcium carbonate is set up as:

% O = 3 3

grams O 1.80 g O100% 100%

grams CaCO 3.75 g CaCO = 48.0 % O

The percent composition of carbon in calcium carbonate is set up as:

% C = 3 3

grams O 0.450 g O100% 100%

grams CaCO 3.75 g CaCO = 12.0 % C

4.5 Mass percent is defined as the mass of the part divided by the mass of the whole times 100%. It does not

matter what mass units are used, just so long as they are the same for the part and the whole. That allows

the units to cancel properly.

Mass of PartMass % = 100%

Mass of Whole

30.0 μgMass %C = 100%

50.0 μg = 60.0% C

4.6 Mass percent is defined as the mass of the part divided by the mass of the whole times 100%. It does not

matter what mass units are used, just so long as they are the same for the part and the whole. That allows

the units to cancel properly.


Mass of Whole

3.54 mgMass %N = 100%

20.0 mg = 17.7% N

4.7 To calculate the actual mass of a substance in a sample, we rearrange the percent equation in the following

way:


Mass of Whole becomes

Mass % Mass of WholeMass of Part =

100%

Mass of lithium = 1.2 g 18.8%

100%

= 0.226 g lithium

4.8 To calculate the actual mass of a substance in a sample, we rearrange the percent equation in the following

way:


Mass of Whole becomes

Mass % Mass of WholeMass of Part =

100%

42

Mass of silicon = 348 g 46.7%

100%

= 163 g silicon

4.9 Chemical formulas must have whole number subscripts. (a) H2S; (b) A ratio of 1.5 oxygen atoms to 1

nitrogen atom means that the number of each atom needs to be doubled to give whole-number coefficients

for both atoms. Doubling the values works because it converts 1.5 to 3 (a whole number). This gives a

ratio of 3 oxygen atoms to 2 nitrogen atoms (note that both quantities are doubled). The chemical formula

is N2O3. (c) A ratio of one-half calcium ion to one chloride ion means that the number of each ion needs to

be doubled to give whole-number coefficients for both ions. This gives a ratio of 1 calcium ion to 2

chloride ions. The chemical formula for calcium chloride is CaCl2.

4.10 Chemical formulas must have whole number subscripts. (a) PH3; (b) A ratio of 2.5 oxygen atoms to 1 atom

means that the number of each atom needs to be doubled to give whole-number coefficients for both atoms.

This works because it converts 2.5 to 5 (whole number). This gives a ratio of 5 oxygen atoms to 2 nitrogen

atoms. The chemical formula is N2O5. (c) A ratio of 1/3 aluminum ion to one chlorate ion implies that the

number of each ion must be tripled to give whole-number coefficients for both atoms. This gives a ration

of one aluminum ion to three chlorate ions. The chemical formula is Al(ClO3)3.

4.11 The formulas are derived by counting the atoms of each element in the figures. (a) H2SO4; (b) SCl4; (c)

C2H4

4.12 The formulas are derived by counting the atoms of each element in the figures. (a) H2SO3; (b) SCl6; (c)

C2H6

4.13 Each molecule pictured contains 1 carbon atom and 2 oxygen atoms. The formula is CO2.

4.14 Each molecule pictured shows 2 nitrogen and four hydrogen atoms. The formula is N2H4.

4.15 (a) The formula unit for an ionic compound is described by its formula, which shows the ratio of ions in

lowest possible whole numbers. Since the formula of sodium chloride is NaCl, one formula unit is

NaCl. From the image, you can see a one-to-one correspondence of sodium ions and chloride ions in

the nearest atoms (you should be able to count 14 chloride and 14 sodium ions). This also gives the

formula NaCl (1:1 ratio of Na+ and Cl

).

(b) For molecular compounds consisting of discrete molecules, the formula unit is the same as its

molecular formula. In this case there are two Cl atoms in each molecule, so the formula unit is Cl2.

(c) For methane, another molecular compound, each molecule contains one carbon atom and four

hydrogens so the molecular formula and the formula unit is CH4.

(d) The image of silicon dioxide is more difficult to analyze because it is not composed of discrete

molecules. As with ionic compounds, the formula unit is a ratio with lowest possible whole numbers.

On average each silicon atom shares four oxygen atoms with its neighbors, so the formula unit is one

silicon and two oxygens (½ × 4 oxygens); SiO2. However, as with ionic compounds and molecules the

formula unit can be derived from its name, silicon dioxide.

4.16 (a) The formula unit for an ionic compound is described by its formula, which shows the ratio of ions in

lowest possible whole numbers. Since the formula of cesium chloride is CsCl, one formula unit is

CsCl. From the image, you can see a one-to-one correspondence of cesium ions and chloride ions.

This also gives the formula CsCl (1:1 ratio of Cs+ and Cl

).

(b) For molecular compounds consisting of discrete molecules, the formula unit is the same as its

molecular formula. In this case there are two oxygen atoms in each molecule, so the formula unit is

O2.

(c) For sulfur dioxide, there is one sulfur and two oxygens in each molecule, so the formula is SO2.

(d) For sodium, a metal, the formula unit is simply Na.

43

4.17 One mole of any object contains 6.022 1023

of those objects. One mole of NH3 contains 6.022 1023

NH3 molecules. One-half mole of NH3 contains half of that:

3 3Molecules NH = 0.5 mol NH23

3

3

6.022×10 NH molecules

1 mole NH 23

3 3.0×10 NH molecules

Since there is one nitrogen atom for each NH3, we can expect that there are 3.0 1023

N atoms as well. We

can express that as follows:

N Atoms = 2333.0 10 NH

3

1 N atom

1 NH = 3.0 10

23 N atoms

In a similar fashion we can see that there are three hydrogen atoms in each NH3.

H atoms = 2333.0 10 NH

3

3 H atoms

1 NH = 9.0 10

23 H atoms

4.18 One mole of any object contains 6.022 1023

of those objects. One mole of SO2 molecules contains

6.022 1023

SO2. If we have 1.75 moles then we have 1.75 times that number:

2 2Molecules SO = 1.75 mol SO23

2

2

6.022×10 SO molecules

1 mole SO 24

2 1.05×10 SO molecules

Since there is one sulfur for each sulfur dioxide, we can expect that there are 1.05 1024

S atoms as well.

We can express that as follows:

S Atoms = 2421.05 10 SO

2

1 S atom

1 SO = 1.05 10

24 S atoms

The molecular formula also shows us that there are two oxygen atoms in each SO2 molecule, so there are

twice as many O atoms than SO2 molecules:

O Atoms = 2421.0538 10 SO

2

2 O atom

1 SO = 2.11 10

24 O atoms

4.19 One mole of Cu2S contains 6.022 1023

Cu2S formula units. One-half mole of Cu2S contains half that

value:

2 2Formula units of Cu S = 0.50 mol Cu S23

2

2

6.022×10 Cu S formula units

1 mole Cu S 23

2 3.0×10 Cu S formula units

4.20 One mole of CuSO4 contains 6.022 1023

CuSO4 formula units. 1.5 moles of CuSO4 contains 1.5 times

that many formula units:

4 4Formula units of CuSO = 1.50 mol CuSO23

4

4

6.022×10 CuSO formula units

1 mole CuSO 23

4 9.0×10 CuSO formula units

4.21 To calculate the number of S atoms in 0.2 mol of SO2, calculate the number of molecules in 0.2 mol and

then multiply by the number of S atoms in each SO2 molecule (1 S per molecule) as mapped below:

Map: AN formula ratioMoles Number of molecules Number of atoms

Atoms S = 0.2 mol236.022 10 molecule

mol

1S atom

1 molecule = 1 10

23 S atoms

44

4.22 To calculate the number of O atoms in 0.2 mol of SO2, calculate the number of molecules in 0.2 mol and

then multiply by the number of O atoms in each SO2 molecule (2 O atoms per molecule) as mapped below:

Map: AN formula ratioMoles Number of molecules Number of atoms

Atoms O = 0.2 mol236.022 10 molecule

mol

2 O atom

1 molecule = 2 10

23 O atoms

4.23 To calculate the number of calcium ions in 1 mol of CaCl2, calculate the number of formula units in 1 mol

and then multiply by the number of calcium ions per formula unit (1 Ca2+

per CaCl2) as mapped below:

Map: AN formula ratioMoles Number of formula units Number of ions

Ca2+

ions = 1 mol

236.022 10 formula unit

mol

21 Ca

1 formula unit

= 6 1023

Ca2+

ions

4.24 To calculate the number of chloride ions in 2 mol of CaCl2, calculate the number of formula units in 2 mol

and then multiply by the number of chloride ions per formula unit (2 Cl per CaCl2) as mapped below:

Map: AN formula ratioMoles Number of formula units Number of ions

Cl ions = 2 mol

236.022 10 formula unit

mol

2 Cl

1 formula unit

= 2 1024

Cl ions

4.25 The molar mass is the sum of the masses of the component elements in the chemical formula. The masses

for one mole of each element are found on the periodic table.

NaCl

Mass of 1 mol of Na = 1 mol 22.99 g/mol = 22.99 g

Mass of 1 mol of Cl = 1 mol 35.45 g/mol = 35.45 g

Mass of 1 mol of NaCl = 58.44 g

The molar mass is 58.44 g/mol

Cl2



CH4

Mass of 1 mol of C = 1 mol 12.01 g/mol = 12.01 g

Mass of 4 mol of H = 4 mol 1.008 g/mol = 4.03 g

Mass of 1 mol of CH4 = 16.04 g


SiO2

Mass of 1 mol of Si= 1 mol 28.09 g/mol = 28.09 g

Mass of 2 mol of O= 2 mol 16.00 g/mol = 32.00 g

Mass of 1 mol of SiO2 = 60.09 g


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CsCl

Mass of 1 mol of Cs = 1 mol 132.9 g/mol = 132.9 g


Mass of 1 mol of CsCl = 168.4 g


O2

Mass of 2 mol of O= 2 mol 16.00 g/mol = 32.00 g


SO2

Mass of 1 mol of S = 1 mol 32.07 g/mol = 32.07 g

Mass of 2 mol of O = 2 mol 32.00 g/mol = 32.00 g

Mass of 1 mol of SO2 = 64.07 g


Na





(a) Hg2Cl2

Mass of 2 mol of Hg = 2 mol 200.6 g/mol = 401.2 g


Mass of 1 mol of Hg2Cl2 472.1 g

Molar mass = 472.1 g/mol

(b) CaSO4•2H2O Note that the mass of 1 water molecule was calculated separately as 18.016 g/mol.

Mass of 1 mol of Ca = 1 mol 40.08 g/mol = 40.08 g



Mass of 2 mol of H2O = 2 mol 18.016 g/mol = 36.032 g

Mass of 1 mol of CaSO4•2H2O = 172.17 g


(c) Cl2O5



Mass of 1 mol of Cl2O5 150.90 g


(d) NaHSO4





Mass of 1 mol of NaHSO4 = 120.06 g


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(a) K2SO4

Mass of 2 mol of K = 2 mol 39.10 g/mol = 78.20 g



Mass of 1 mol of K2SO4 174.26 g


(b) NiCl2•6H2O; Note that the mass of 1 water molecule was calculated separately as 18.016 g/mol (four

significant figures).

Mass of 1 mol of Ni = 1 mol 58.69 g/mol = 58.69 g


Mass of 6 mol of H2O = 6 mol 18.016 g/mol = 108.108 g

Mass of 1 mol of NiCl2•6H2O = 237.69 g


(c) C2H4Cl2




Mass of 1 mol of C2H4Cl2 98.95 g


(d) Mg(NO3)2

Mass of 1 mol of Mg = 1 mol 24.31 g/mol = 24.31 g

Mass of 2 mol of N = 1 mol 14.01 g/mol = 28.02 g


Mass of 1 mol of Mg(NO3)2 = 148.33 g




(a) I2

Mass of 2 mol of I = 2 mol 126.9 g/mol = 253.8 g


(b) CrCl3

Mass of 1 mol of Cr = 1 mol 52.00 g/mol = 52.00 g


Mass of 1 mol of CrCl3 = 158.4 g


(c) C4H8



Mass of 1 mol of C4H8 56.10 g




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(a) P4

Mass of 4 mol of P = 4 mol 30.97 g/mol = 123.9 g


(b) CrO2Cl2

Mass of 1 mol of Cr = 1 mol 52.00 g/mol = 52.00 g



Mass of 1 mol of CrO2Cl2 = 154.90 g


(c) CaF2

Mass of 1 mol of Ca = 1 mol 40.08 g/mol = 40.08 g

Mass of 2 mol of F = 2 mol 19.00 g/mol = 38.00 g

Mass of 1 mol of CaF2 78.08 g


4.31 To measure out a useful number of atoms by counting would not be possible because atoms are too small

for us to see and manipulate. If you could count 1000 water molecules per second for slightly over 20

billion years, you would have counted the molecules in one small drop of water (about 0.020 mL). Instead,

because we know the mass of one mole of any substance, we can carefully weigh out an amount of any

substance and determine the number of atoms our sample contains.

4.32 One mole is 6.022 1023

of any object. Scientists use the mole because atoms, ions, and molecules are too

small to work with individually. Large collections of these units are more reasonable to deal with.

4.33 The molar mass and atomic mass have the same numerical value but different units. If the molar mass of

LiCl is 42.39 g/mol, the mass of one formula unit of LiCl is 42.39 amu.

4.34 The molar mass and atomic mass have the same numerical value but different units. If the mass of one

formula unit of CuCl2 is 134.5 amu, then the mass of 1.00 mol of CuCl2 is 134.5 g (i.e. the molar mass is

134.5 g/mol).

4.35 The units for molar mass are grams per mole:

Molar mass = g

mol

This means that if you have the numbers of grams and moles, you can calculate the molar mass by dividing

the number of grams by the number of moles of substance this mass represents. In this case, you are given

grams (12.0) and molecules of substance.

Map: ANNumber of molecules Moles

Molecules = 232.01 10 molecules

23

mole

6.022 10 molecules

= 0.334 mol

Molar mass = 12.0 g

0.334 mol = 35.9 g/mol

4.36 In this problem you are asked to convert from molecules to mass. You can make this conversion by first

converting from molecules to moles; then use the molar mass to calculate mass of the sample.

Map: ANNumber of molecules Moles Mass in grams

MM

48

Mass = 233.01 10 moleculesmole

236.022 10 molecules

98.09 g

mole = 49.0 g

4.37 The conversion from mass to number of moles uses the following problem solving map (MM is the molar

mass):

Map: Mass in grams MolesMM

(a) The molar mass of KHCO3 is 100.12 g/mol (calculated by adding the masses of the component

elements of the chemical formula).

Moles KHCO3 = 310.0 g KHCO 3

3

1 mol KHCO

100.12 g KHCO = 0.0999 mol KHCO3

(b) The molar mass of H2S is 34.08 g/mol.

Moles H2S = 210.0 g H S 2

2

1 mol H S

34.08 g H S = 0.293 mol H2S

(c) The molar mass of Se is 78.96.

Moles Se = 10.0 g Se1 mol Se

78.96 g Se = 0.127 mol Se

(d) The molar mass of MgSO4 is 120.37 g/mol.

Moles MgSO4 = 410.0 g MgSO 4

4

1 mol MgSO

120.37 g MgSO = 0.0831 mol MgSO4


mass):


(a) The molar mass of SO2 is 64.06 g/mol.

Moles SO2 = 2100.0 g SO 2

2

1 mol SO

64.06 g SO = 1.561 mol SO2

(b) The molar mass of Na2SO4 is 142.05 g/mol.

Moles Na2SO4 = 2 4100.0 g Na SO 2 4

2 4

1 mol Na SO

142.05 g Na SO = 0.7040 mol Na2SO4

(c) The molar mass of BaSO4 is 233.4 g/mol.

Moles BaSO4 = 4100.0 g BaSO 4

4

1 mol BaSO

233.4 g BaSO = 0.4284 mol BaSO4

(d) The molar mass of KAl(SO4)2 is 258.2 g/mol.

49

Moles KAl(SO4)2 = 4 2100.0 g KAl(SO ) 4 2

4 2

1 mol KAl(SO )

258.2 g KAl(SO ) = 0.3873 mol KAl(SO4)2


mass). In these problems, if the mass is not given in grams, it must first be converted using an appropriate

problem solving map:


(a) The molar mass of NaCl is 58.44 g/mol.

Moles NaCl = 32.5 g NaCl1 mol NaCl

58.44 g NaCl = 0.556 mol NaCl

(b) The molar mass of C9H8O4 is 180.15 g/mol. The mass is converted to grams using 1 mg = 103

g.

Moles C9H8O4 = 9 8 4250.0 mg C H O

39 8 410 g C H O

9 8 41 mg C H O

9 8 4

9 8 4

1 mol C H O

180.15 g C H O = 1.388 10

3 mol C9H8O4

(c) The molar mass of CaCO3 is 100.09 g/mol. The mass is converted to grams using 1 kg = 103 g.

Moles CaCO3 = 373.4 kg CaCO

3310 g CaCO

31 kg CaCO

3

3

1 mol CaCO

100.09 g CaCO = 733 mol CaCO3

(d) The molar mass of CuS is 95.61 g/mol. The mass is converted to grams using 1 g = 106

g.

Moles CuS = 5.47 μg CuS

610 g CuS

1 μg CuS

1 mol CuS

95.61 g CuS = 5.72 10

–8 mol CuS


mass). In these problems, if the mass is not given in grams, it must first be converted using an appropriate

problem solving map:


(a) The molar mass of K2SO4 is 174.26 g/mol.

Moles K2SO4 = 2 472.2 g K SO 2 4

2 4

1 mol K SO

174.26 g K SO = 0.414 mol K2SO4

(b) The molar mass of C18H21NO4 is 315.36 g/mol. The mass is converted to grams using 1 mg = 103

g.

Moles C18H21NO4 = 18 21 4160.0 mg C H NO

318 21 410 g C H NO

18 21 41 mg C H NO

18 21 4

18 21 4

1 mol C H NO

315.36 g C H NO

= 5.074 104

mol C18H21NO4

(c) The molar mass of Fe3O4 is 231.55 g/mol. The mass is converted to grams using 1 kg = 103 g.

410

Moles Fe3O4 = 3 42.82 kg Fe O

33 410 g Fe O

3 41 kg Fe O

3 4

3 4

1 mol Fe O

231.55 g Fe O = 12.2 mol Fe3O4

(d) The molar mass of C12H22O11 is 342.30 g/mol. The mass is converted to grams using 1 g = 106

g.

Moles C12H22O11 = 12 22 115.00 μg C H O

612 22 1110 g C H O

12 22 111 μg C H O

12 22 11

12 22 11

1 mol C H O

342.30 g C H O

= 1.46 10–8

mol C12H22O11

4.41 If the molar mass of a substance is relatively small, it will take more moles of that substance to equal 1

gram than it would take of a substance with a larger molar mass. This means that carbon, which has the

smallest molar mass of the substances given, contains the most moles of atoms in a 1.0-g sample.

4.42 If the molar mass of a substance is relatively large, it will take fewer moles of that substance to equal 5.0 g

than it would take of a substance with a smaller molar mass. This means that silver, which has the highest

molar mass of the substances given, contains the least moles of atoms in a 5.0-g sample.

4.43 To convert moles to grams we use the following problem solving map (MM = molar mass)

Map: Moles Mass in gramsMM

(a) The molar mass of Ba(OH)2 is 171.32 g/mol.

Mass Ba(OH)2 = 42.50 mol BaSO 4

4

171.32 g BaSO

1 mol BaSO = 428 g Ba(OH)2

(b) The molar mass of Cl2 is 70.90 g/mol.

Mass Cl2 = 22.50 mol Cl 2

2

70.90 g Cl

1 mol Cl = 177 g Cl2

(c) The molar mass of K2SO4 is 174.26 g/mol.

Mass K2SO4 = 2 4 2.50 mol K SO 2 4

2 4

174.26 g K SO

1 mol K SO = 436 g K2SO4

(c) The molar mass of PF3 is 87.97 g/mol.

Mass K2SO4 = 32.50 mol PF 3

3

87.97 g PF

1 mol PF = 220. g K2SO4

4.44 To convert moles to grams we use the following problem solving map (MM = molar mass)


(a) The molar mass of I2 is 253.80 g/mol.

Mass I2 = 20.750 mol I 2

2

253.80 g I

1 mol I = 190. g I2

(b) The molar mass of Mg(NO3)2 is 148.33 g/mol.

Mass Mg(NO3)2 = 3 20.750 mol Mg NO

3 2

3 2

148.33 g Mg NO

1 mol Mg NO = 111 g Mg(NO3)2

(c) The molar mass of SiO2 is 60.09 g/mol.

Mass SiO2 = 20.750 mol SiO 2

2

60.09 g SiO

1 mol SiO = 45.1 g SiO2

411

(c) The molar mass of Na3PO4 is 163.94 g/mol.

Mass Na3PO4= 3 4 0.750 mol Na PO 3 4

3 4

163.94 g Na PO

1 mol Na PO = 123 g Na3PO4

4.45 The conversion of 2.7 moles of Zn(CH3CO2)2 to an equivalent number of grams requires the molar mass

(MM). We use the following problem solving map:


The molar mass of Zn(CH3CO2)2 is 183.47 g.

Grams Zn(CH3CO2)2 = 3 2 22.7 mol Zn(CH CO ) 3 2 2

3 2 2

183.47 g Zn(CH CO )

mol Zn(CH CO ) = 5.0 10

2 g Zn(CH3CO2)2

4.46 The conversion of 3.4 moles of Cu(HCO3)2 to an equivalent number of grams requires the molar mass

(MM). We use the following problem solving map:


The molar mass of Cu(HCO3)2 is 185.59 g.

Grams Cu(HCO3)2 = 3 23.4 mol Cu(HCO ) 3 2

3 2

185.59 g Cu(HCO )

mol Cu(HCO ) = 6.3 10

2 g Cu(HCO3)2

4.47 (a) The conversion of 30.0 g NH3 to an equivalent number of moles requires the molar mass (MM). We

use the following problem solving map:


The molar mass of NH3 is 17.03 g/mol.

Moles NH3 = 330.0 g NH 3

3

mol NH

17.03 g NH = 1.76 mol NH3

(b) To calculate the number of NH3 molecules in the sample, we use Avogadro’s number (NA = 6.022

1023

) and the following problem solving map:

Map: AN Moles Number of molecules

Molecules NH3 = 31.7612 mol NH23

3

6.022 10 molecules

mol NH

= 1.06 10

24 molecules NH3

(c) To calculate the number of nitrogen atoms in the sample, we look at the ratio of the elements in the

chemical formula. The ratio for nitrogen in ammonia is 1 atom N/1 molecule NH3. Therefore, the

number of N atoms is the same as the number of NH3 molecules, 1.06 1024

N atoms.

(d) From the chemical formula, we know that there are 3 moles of H for each mole of NH3. This is the

formula ratio needed to calculate the moles of H in the sample of NH3:

Map: 3

formula ratioMoles NH Moles H

412

Moles H = 31.7612 mol NH3

3 mol H

1 mol NH = 5.28 mol H

4.48 (a) The conversion of 15.0 g Na2CO3 to an equivalent number of moles requires the molar mass (MM). We

use the following problem solving map:


The molar mass of Na2CO3 is 105.99 g/mol.

Moles Na2CO3 = 2 315.0 g Na CO 2 3

2 3

mol Na CO

105.99 g Na CO = 0.142 mol Na2CO3

(b) To calculate the number of formula units in the sample, we use Avogadro’s number (NA = 6.022

1023

) and the following problem solving map:

Map: AN Moles Number of formula units

Formula units Na2CO3 = 2 30.14152 mol Na CO23

2 3

6.022 10 formula units

mol Na CO

= 8.52 1022

formula units Na2CO3

(c) To calculate the number of sodium ions in the sample, we look at the ratio of the elements in the

chemical formula. The ratio for sodium in Na2CO3 is 2 Na+ ions/1 formula unit.

Map: formula ratio

Number of formula units Number of Na ions

Na+ ions = 22

2 38.5223 10 formula units Na CO2 3

2 Na ions

formula units Na CO

= 1.70 1023

Na+ ions

(d) To calculate the number of moles carbonate ions in the sample, we look at the ratio of the carbonate

ions to formula units of Na2CO3. This ratio is 1 mol CO32

per mol of Na2CO3. Therefore, the number

of carbonate ions is the same as the number of moles of Na2CO3: 0.142 mol CO32

.

4.49 The number of atoms per mole of substance depends on the number of atoms in the chemical formula.

H2SO4 has the most atoms per mole because it has more atoms per molecule. Na has the least atoms per

mole.

4.50 The number of atoms per mole of substance depends on the number of atoms in the chemical formula.

C2H6 has the most atoms per mole of substance, and Fe has the least atoms per mole of substance.

4.51 To calculate the number of molecules in 0.050 g of water, we need to find the appropriate conversion

factors. This can be done using the following conversion map:

Map: ANMass in grams Moles Number of molecules

MM

Note that the units of molar mass (g/mol) serve as a connection between mass and moles, and the units of

Avogadro’s number (molecules/mole) serve as a connection between molecules and moles.

The molar mass of water is 18.02 g/mol.

413

Molecules H2O = 20.050 g H O2mol H O

218.02 g H O

232

2

6.022 10 molecules H O

mol H O

= 1.7 1021

molecules H2O

4.52 To calculate the number of SiO2 formula units in 1 grain of sand (7.7 104

g) we need to find the

appropriate conversion factors. Assuming that sand is 100% SiO2, we use the following conversion map:

Map: ANMass in grams Moles Number of formula units

MM

Note that the units of molar mass (g/mol) serve as a connection between mass and moles, and the units of

Avogadro’s number (formula units/mole) serve as a connection between formula units and moles.

The molar mass of SiO2 is 60.09 g/mol.

Formula units SiO2 = 427.7 10 g SiO

2mol SiO

260.09 g SiO

232

2

6.022 10 formula units SiO

mol SiO

= 7.7 1018

formula units SiO2

4.53 To calculate the number of formula units we use the following conversion map:


MM

(a) Br2 = 159.80 g/mol

Formula units Br2 = 2250.0 g Br2mol Br

2159.80 g Br

232

2

6.022 10 formula units Br

mol Br

= 9.421 1023

formula units Br2

(b) MgCl2 = 95.21 g/mol

Formula units MgCl2 = 2250.0 g MgCl2mol MgCl

2159.80 g MgCl

232

2

6.022 10 formula units MgCl

mol MgCl

= 1.581 1024

formula units MgCl2

(c) H2O = 18.02 g/mol

Formula units H2O = 2250.0 g H O2mol H O

218.02 g H O

232

2

6.022 10 formula units H O

mol H O

= 8.355 1024

formula units H2O

(d) Fe = 55.85 g/mol

Formula units Fe = 250.0 g Femol Fe

58.85 g Fe

236.022 10 formula units Fe

mol Fe

= 2.696 1024

formula units Fe

4.54 To calculate the number of formula units we use the following conversion map:


MM

(a) Cu = 63.55 g/mol

414

Formula units Cu = 375.0 g Cumol Cu

63.55 g Cu

236.022 10 formula units Cu

mol Cu

= 3.554 1024

formula units Cu

(b) NaBr = 102.89 g/mol

Formula units NaBr = 375.0 g NaBrmol NaBr

102.89 g NaBr

236.022 10 formula units NaBr

mol NaBr

= 2.195 1024

formula units NaBr

(c) SO2 = 64.06 g/mol

Formula units SO2 = 2375.0 g SO2mol SO

264.06 g SO

232

2

6.022 10 formula units SO

mol SO

= 3.525 1024

formula units SO2

(d) NH4Cl = 53.49 g/mol

Formula units NH4Cl = 4375.0 g NH Cl4mol NH Cl

453.49 g NH Cl

234

4

6.022 10 formula units NH Cl

mol NH Cl

= 4.222 1024

formula units NH4Cl

4.55 To calculate the number of atoms or ions of each element in 140.0 g of each substance, the best strategy is

to first calculate the number of formula units. Once you have determined this, use the formula ratios to

calculate the number of atoms or ions. In part (b), for example, you will calculate the formula units of

Ca(NO3)2 and then use formula ratios to calculate the number of Ca2+

and NO3– ions. Two problem solving

maps are applied:

Number of formula units:

ANMass in grams Moles Number of formula units

MM

Number of ions or atoms: formula ratio

Number of formula units ions or atoms

(a) Since there we are only looking for atoms of one element, we can combine the two problem solving

maps. The appropriate conversions are: H2 = 2.016 g/mol; 2 H atom = 1 H2 molecule

Atoms = 2140.0 g H2mol H

22.016 g H

2326.022 10 H

2mol H 2

2 H atom

H = 8.364 10

25 H atoms

(b) Ca(NO3)2 = 164.10 g/mol

Formula units Ca(NO3)2 =

3 2140.0 g Ca NO

3 2mol Ca NO

3 2

164.10 Ca NO

233 2

3 2

6.022 10 Ca NO

mol Ca NO

= 5.138 10

23 Ca(NO3)2

Ca2+

ions: 1 Ca2+

ion = 1 Ca(NO3)2

Ca2+

ions = 233 2

5.138 10 Ca NO

2

3 2

1 Ca

1Ca NO

= 5.138 1023

Ca2+

ions

NO3 ions: 2 NO3

ions = 1 Ca(NO3)2

415

NO3 ions = 23

3 25.138 10 Ca NO

3

3 2

2 NO

1Ca NO

= 1.028 1024

NO3 ions

(c) N2O2 = 60.02 g/mol

Formula units N2O2 = 2 2140.0 g N O2 2mol N O

2 260.02 g N O

232 2

2 2

6.022 10 N O

mol N O

= 1.405 10

24 N2O2

N atoms = 242 21.405 10 N O

2 2

2 N

1N O = 2.809 10

24 N atoms

O atoms = 242 21.405 10 N O

2 2

2 O

1N O = 2.809 10

24 O atoms

(d) K2SO4 = 174.26 g/mol

Formula units = 2 4140.0 g K SO2 4mol K SO

2 4174.26 g K SO

232 4

2 4

6.022 10 K SO

mol K SO

= 4.838 10

23 K2SO4

K+ ions: 2 K

+ ions = 1 K2SO4

K+ ions = 23

2 44.838 10 K SO2 4

2 K

1K SO

= 9.676 1023

K+ ions

SO42

ions: 1 SO42

ion = 1 K2SO4

SO42

ions = 232 44.838 10 K SO

24

2 4

1SO

1K SO

= 4.838 1023

SO42

ions

4.56 To calculate the number of atoms or ions of each element in 140.0 g of each substance, the best strategy is

to first calculate the number of formula units. Once you have determined this, use the formula ratios to

calculate the number of atoms or ions. In part (a), for example, you will calculate the formula units of

BaSO4 and then use formula ratios to calculate the number of Ba2+

and SO42–

ions. Two problem solving

maps are applied:

Number of formula units:

ANMass in grams Moles Number of formula units

MM

Number of ions or atoms: formula ratio

Number of formula units ions or atoms

(a) BaSO4 = 233.36g/mol

Formula units = 4140.0 g BaSO4mol BaSO

4233.36 g BaSO

234

4

6.022 10 BaSO

mol BaSO

= 3.613 10

23 BaSO4

Ba2+

ions = 2343.613 10 BaSO

2

4

1 Ba

1BaSO

= 3.613 1023

Ba2+

ions

SO42–

ions = 2343.613 10 BaSO

24

4

1SO

1BaSO

= 3.613 1023

SO42–

ions

Total ions = (3.613 1023

Ba2+

ions) + (3.613 1023

SO42–

ions) = 7.225 1023

ions

You can also calulate the number of sulfur and oxygen atoms:

416

S atoms = 2343.613 10 BaSO

4

1S

1BaSO = 3.613 10

23 S atoms

O atoms = 2343.613 10 BaSO

4

4 O

1BaSO = 1.445 10

24 O atoms

(b) Mg3(PO4)2 = 262.87 g/mol;

Formula units = 3 4 2140.0 g Mg PO

3 4 21mol Mg PO

3 4 2

262.87 g Mg PO

233 4 2

3 4 2

6.022 10 Mg PO

mol Mg PO

= 3.207 1023

Mg3(PO4)2

Mg2+

ions = 233 4 2

3.207 10 Mg PO

2

3 4 2

3 Mg

1Mg PO

= 9.622 1023

Mg2+

ions

PO43–

ions = 233 4 23.207 10 Mg (PO )

34

3 4 2

2 PO

1Mg (PO )

= 6.414 1023

PO43–

ions

Total ions = (9.622 1023

Mg2+

ions) + (6.414 1023

PO43–

ions) = 1.604 1024

ions

You can also calulate the number of phosphorus and oxygen atoms:

P atoms = 233 4 2

3.207 10 Mg PO 3 4 2

2 P

1Mg PO = 6.414 10

23 P atoms

O atoms = 233 4 2

3.207 10 Mg PO 3 4 2

8 O

1Mg PO = 2.566 10

24 O atoms

(c) Since we are only looking for one type of atom, we can combine the two problem solving maps. The

appropriate conversions are: O2 = 32.00 g/mol; 2 O atoms = 1 O2 formula unit

O atoms = 2140.0 g O2mol O

232.00 g O

2326.022 10 O

2mol O 2

2 O atom

O = 5.269 10

24 O atoms

(d) KBr = 119.00 g/mol;

Formula units = 140.0 g KBrmol KBr

119.00 g KBr

236.022 10 KBr

mol KBr

= 7.085 10

23 KBr

K+ ions = 237.085 10 KBr

1 K

1KBr

= 7.085 1023

K+ ions

Br ions = 237.085 10 KBr

1 Br

1KBr

= 7.085 1023

Br ions

Total ions = (7.085 1023

K+ ions + (7.085 10

23 Br

ions) = 1.417 10

24 ions

4.57 To calculate the mass of 6.4 10

22 molecules of SO2 we use the following conversion map:


MM

The molar mass of SO2 is 64.06 g/mol.

417

Mass SO2 = 2226.4 10 molecules SO

21 mol SO

2326.022 10 moleculesSO

2

2

64.06 g SO

1 mol SO = 6.8 g SO2

4.58 To calculate the mass of 1.8 1021

molecules of H2SO4 we use the following conversion map:


MM

The molar mass of H2SO4 is 98.08 g/mol.

Mass H2SO4 = 212 41.8 10 molecules H SO

2 41 mol H SO

232 46.022 10 molecules H SO

2 4

2 4

98.08 g H SO

1 mol H SO

= 0.29 g H2SO4

4.59 The substance that has the most nitrogen atoms per 25.0-g sample will also have the highest number of

moles of nitrogen atoms. We use the following conversion map:

Map: formula ratio

Mass in grams Moles Moles of N atomsMM

The molar mass of NH3 is 17.03 g/mol and there is 1 mol N atoms per mol NH3.

Moles N atoms = 325.0 g NHmol

3

3

NH

17.03 g NH 3

1 mol N atoms

mol NH = 1.47 mol N atoms

The molar mass of NH4Cl is 53.49 g/mol and there is 1 mol N atoms per mole NH4Cl.

Moles N atoms = 425.0 g NH Clmol

4

4

NH Cl

53.49 g NH Cl 4

1 mol N atoms

mol NH Cl = 0.467 mol N atoms

The molar mass of NO2 is 46.01 g/mol and there is 1 mol N atoms per mole NO2.

Moles N atoms = 225.0 g NOmol

2

2

NO

46.01 g NO 2

1 mol N atoms

mol NO = 0.543 mol N atoms

The molar mass of N2O3 is 76.02 g/mol and there are 2 mol N atoms per mole N2O3.

Moles N atoms = 2 325.0 g N Omol

2 3

2 3

N O

76.02 g N O 2 3

2 mol N atoms

mol N O = 0.658 mol N atoms

Because NH3 has the highest number of moles of nitrogen atoms, we know that it has the most nitrogen

atoms per 25.0-g sample.

4.60 The substance that has the most chlorine atoms per 100.0-g sample will also have the highest number of

moles of chlorine atoms. We use the following conversion map:

Map: formula ratio

Mass in grams Moles Moles of Cl atomsMM

The molar mass of NaCl is 58.44 g/mol and there is 1 mol Cl atoms per mol NaCl.

418

Moles Cl atoms = 100.0 g NaClmol

NaCl

58.44 g NaCl

1 mol Cl atoms

mol NaCl = 1.711 mol Cl atoms

The molar mass of PCl3 is 137.3 g/mol and there are 3 mol Cl atoms per mole PCl3.

Moles Cl atoms = 3100.0 g PClmol

3

3

PCl

137.3 g PCl 3

3 mol Cl atoms

mol PCl = 2.185 mol Cl atoms

The molar mass of CaCl2 is 110.98 g/mol and there are 2 mol Cl atoms per mole CaCl2.

Moles Cl atoms = 2100.0 g CaClmol

2

2

CaCl

110.98 g CaCl 2

2 mol Cl atoms

mol CaCl = 1.802 mol Cl atoms

The molar mass of HClO2 is 68.46 g/mol and there is 1 mol Cl atoms per mole HClO2.

Moles Cl atoms = 2100.0 g HClOmol

2

2

HClO

68.46 g HClO 2

1 mol Cl atoms

mol HClO = 1.461 mol Cl atoms

Because PCl3 has the highest number of moles of chlorine atoms, we know that it has the most chlorine

atoms per 100.0-g sample.

4.61 No. Molecules of the same substance have the same percent compositions.

4.62 Percent composition can be used to compare the composition of different substances. Copper ore, for

example, is analyzed for percent copper composition, and fertilizers are analyzed for their percent nitrogen

content. Percent composition can also be used to determine the empirical formula of a compound.

4.63 The empirical formula shows the relative amounts of each atom in a compound, expressed as small whole

numbers. The molecular formula shows the exact numbers of each atom present in one molecule of that

compound.

4.64 Chemical formulas are much more compact and give the same information (i.e. the relative composition of

substances). For example, we could report that water is 88.81% oxygen and 11.19% hydrogen by mass or

simply report that its formula is H2O.

4.65 The empirical and molecular formulas are different if the subscripts in the molecular formula are all

divisible by a common factor other than 1. For example, in the formula H2O2, both subscripts are divisible

by 2, so the empirical formula (HO) is different than the molecular formula. Of the substances listed, those

with different empirical and molecular formulas are:

Molecular Empirical

H2O2 HO

N2O4 NO2

4.66 The empirical and molecular formulas are the same if the subscripts in the molecular formula are not

divisible by a common factor other than 1. For example, in the formula N2O3, the subscripts have no

common factors other than 1, so the empirical formula is the same as the molecular formula. Of the

substances listed, those with the same empirical and molecular formulas are N2O3 and NaCl.


divisible by a common factor other than 1. Often, it is desirable to simplify the molecular formula in order

419

to determine if the molecular and empirical formulas are different. For part (d) HO2CC4H8CO2H can be

simplified to C6H10O4.

Molecular Common Empirical

Formula Factor Formula

(a) P4O10 2 P2O5

(b) Cl2O5 none same as molecular

(c) PbCl4 none same as molecular

(d) C6H10O4 2 C3H5O2


divisible by a common factor other than 1.



(a) As4O6 2 As2O3

(b) H2S2 2 HS

(c) CaCl2 none same as molecular

(d) C3H6 3 CH2





(a) C6H4Cl2 2 C3H2Cl

(b) C6H5Cl none same as molecular

(c) N2O5 none same as molecular





(a) N2O4 2 NO2

(b) H2C2O4 2 HCO2

(c) C2H4O2* 2 CH2O

*It is easier to find the empirical formula of CH3CO2H if the molecular formula is simplified to C2H4O2.

4.71 NO2 and N2O4 have the identical empirical formulas (NO2). The empirical formulas of the other

compounds are different than NO2.



N2O none N2O

NO none NO

NO2 none NO2

N2O3 none N2O3

N2O4 2 NO2

N2O5 none N2O5

4.72 C2H4 and C3H6 have identical empirical formulas (CH2). The empirical formulas of the other compounds

are different than CH2.



CH4 none CH4

C2H4 2 CH2

C3H6 3 CH2

C4H12 4 CH3

C6H6 6 CH

420

4.73 The empirical formula shows the whole-number ratio of moles of each element in a compound. To

determine the empirical formula of a compound, we calculate the number of moles of each element in the

sample, and then determine the mole ratios. When presented with percent composition data, it is easiest to

assume that we have exactly 100 grams of the substance.

(a) The percent composition of our sample is 72.36% Fe and 27.64%O. If we have a 100-gram sample, it

will contain 72.36 g Fe and 27.64 g O. We can convert these masses to the equivalent number of

moles as follows:

Moles Fe = 72.36 g Fe1 mol Fe

55.85 g Fe = 1.296 mol Fe

Moles O = 27.64 g O1 mol O

16.00 g O = 1.728 mol O

Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 1.296 mol

Fe).

moles Fe 1.296 mol Fe 1 mol Fe


moles O 1.728 mol O 1.333 mol O


Notice that one of the ratios is not a whole number. Since the subscripts of chemical formulas must be

whole numbers (i.e. you can’t have fractions of atoms), we multiply each ratio by 3 to convert the ratio

into a whole number. We find that the empirical formula has 4 moles of oxygen for every 3 moles of

iron. The empirical formula is Fe3O4.

(b) The percent composition of our sample is 58.53% C, 4.09% H, 11.38% N, and 25.99% O. If we have a

100 gram sample, it will contain 58.53 g C, 4.09 g H, 11.38 g N, and 25.99 g O. We can convert these

masses to the equivalent number of moles as follows:

Moles C = 58.53 g C1 mol C

12.01 g C = 4.873 mol C

Moles H = 4.09 g H1 mol H

1.008 g H = 4.06 mol H

Moles N = 11.38 g N1 mol N

14.01 g N = 0.8123 mol N


16.00 g O = 1.624 mol O

Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 0.8123

mol N).

moles C 4.873 mol C 6.000 mol C

moles N 0.8123 mol N 1 mol N

421

moles H 4.06 mol H 5.00 mol H






The empirical formula has 6 moles of carbon, 5 moles of hydrogen, and 2 moles of oxygen for every

1 mole of nitrogen. The empirical formula is C6H5NO2.

4.74 The empirical formula shows the whole-number ratio of moles of each element in a compound. To

determine empirical formula of a compound, we calculate the number of moles of each element in the

sample and then determine the mole ratios. When presented with percent composition data, it is easiest to

assume that we have exactly 100 grams of the substance.

(a) The percent composition of our sample is 85.62% C and 14.38%H. If we have a 100 gram sample, it

will contain 85.62 g C and 14.38 g H. We can convert these masses to the equivalent number of moles

as follows:


12.01 g C = 7.129 mol C


1.008 g H = 14.27 mol H


C).

moles C 7.129 mol C 1 mol C




Rounding the mole ratios to nearest whole number we find that for each mole of carbon there are 2

moles of hydrogen. The empirical formula is CH2.

(b) The percent composition of our sample is 63.15% C, 5.30% H, and 31.55% O. If we have a 100-gram

sample, it will contain 63.15 g C, 5.30 g H, and 31.55 g O. We convert these masses to the equivalent

number of moles as follows:


12.01 g C = 5.258 mol C


1.008 g H = 5.26 mol H

422


16.00 g O = 1.972 mol O


O).


moles O 1.972 mol O 1 mol O





The ratios are not all whole numbers and must be converted to whole numbers by multiplying by an

appropriate factor. In this case, the fractional portion (0.667) represents 2/3, so we should be able to

convert these values to whole numbers by multiplying each ratio by 3. Thus for every 3 moles of

oxygen we have 8 moles of carbon and 8 moles of hydrogen. The empirical formula is C8H8O3.

4.75 The percent composition of our sample is 73.19% C, 19.49% O, and 7.37% H. If we have a 100-gram

sample, it will contain 73.19 g C, 19.49 g O, and 7.37 g H. To determine the empirical formula, we convert

the masses of each element to the equivalent number of moles, and then determine the relative number of

moles of each element in the substance.


12.01 g C = 6.094 mol C


16.00 g O = 1.218 mol O


1.008 g H = 7.31 mol H

Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 1.218 mol O).







Rounding the mole ratios to the nearest whole numbers, we find that chemical formula has 5 moles of

carbon, 6 moles of hydrogen, and 1 mole of oxygen. The empirical formula for eugenol is C5H6O.

4.76 The percent composition of our sample is 52.66% Ca, 12.30% Si, and 35.04% O. If we have a 100 gram

sample, then it will contain 52.66 g Ca, 12.30 g Si, and 35.04 g O. To determine the empirical formula, we

423

convert the masses of each element to the equivalent number of moles, and then determine the relative

number of moles of each element in the compound.

Moles Ca = 52.66 g Ca1 mol Ca

40.08 g Ca = 1.314 mol Ca

Moles Si = 12.30 g Si1 mol Si

28.09 g Si = 0.4379 mol Si


16.00 g O = 2.190 mol O

Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 0.4379 mol Si).

moles Ca 1.314 mol Ca 3.001 mol Ca

moles Si 0.4379 mol Si 1 mol Si





Rounding the mole ratios to the nearest whole number, we find that chemical formula has 3 moles of

calcium and 5 moles of oxygen for every 1 mole of silicon. The empirical formula is Ca3SiO5.

4.77 The percent composition of our sample is 37.01% C, 2.22% H, 18.50% N, and 42.27% O. Assuming a 100

gram sample, it will contain 37.01 g C, 2.22 g H, 18.50 g N, and 42.27 g O. To determine the empirical

formula, we convert the masses of each element to the equivalent number of moles, and then determine the

relative number of moles of each element in the substance.


12.01 g C = 3.082 mol C


1.008 g H = 2.202 mol H


14.01 g N = 1.320 mol N


16.00 g O = 2.642 mol O

Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 1.320 mol N).



424







Notice that two of the ratios are not whole numbers. The fractional portions (0.333 and 0.668) represent

one-third (1/3) and two-thirds (2/3). We can convert these values to whole numbers by multiplying each

ratio by 3. The empirical formula has 7 moles of carbon, 5 moles of hydrogen, and 6 moles of oxygen for

every 3 moles of nitrogen. The empirical formula is C7H5N3O6.

4.78 The percent composition of strychnine is 75.42% C, 6.63% H, 8.38% N, and 9.57% O. A 100-gram sample

will contain 75.42 g C, 6.63 g H, 8.38 g N, and 9.57 g O. To determine the empirical formula, we convert

the masses of each element to the equivalent number of moles, and determine the relative number of moles

of each element in the substance.


12.01 g C = 6.280 mol C


1.008 g H = 6.58 mol H


14.01 g N = 0.598 mol N


16.00 g O = 0.598 mol O

Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 0.598 mol O or

0.598 mol N).









Notice that ratio of carbon to oxygen is not a whole number. The fractional portion (0.5) represents one-

half (1/2). We can convert this value to a whole number by multiplying each ratio by 2. The empirical

425

formula for strychnine has 21 moles of carbon, 22 moles of hydrogen, 2 moles of oxygen, and 2 moles of

nitrogen. The empirical formula is C21H22N2O2.

4.79 Convert the percentages to moles of each element, assuming that we have a 100.0 g sample:

1 mol Cmol C = 45.42 g C = 3.782 mol

12.01 g C

1 mol Hmol H = 2.720 g H = 2.698 mol

1.008 g H

1 mol Omol O = 51.86 g C = 3.241 mol

16.00 g O

Divide each by the smallest number:

mol C 3.782 mol = 1.402

mol H 2.698 mol

mol H 2.698 mol = 1.000

mol H 2.698 mol

mol O 3.241 mol = 1.201

mol H 2.698 mol

Although we did not get whole numbers for the relative amounts of carbon and oxygen, these amounts can

be converted to whole numbers by multiplying by 5:

1.402 mol C × 5 = 7.010 mol C = 7 mol C

1.000 mol H × 5 = 5.000 mol H = 5 mol H

1.201 mol O × 5 = 6.005 mol O = 6 mol O

Now that we have whole number element amounts, we can write an empirical formula: C7H5O6.

4.80 Convert the percentages to moles of each element, assuming that we have a 100.0 g sample:

1 mol Pmol P = 35.57 g P = 1.149 mol

30.97 g P

1 mol Smol S = 64.43 g S = 2.010 mol

32.06 g S

Divide by the smallest number:

mol S 2.010 mol = 1.749

mol P 1.149 mol

mol P 1.149 mol = 1.000

mol P 1.149 mol

Although we did not get a whole number for the relative amount of sulfur, these amounts can be converted

to whole numbers by multiplying by 4:

1.749 mol S × 4 = 6.996 mol S = 7 mol S

1.000 mol P × 4 = 4.000 mol P = 4 mol P

Now that we have whole number element amounts, we can write an empirical formula: P4S7.

4.81 The percent composition by mass and the molar mass are needed to determine the molecular formula.

4.82 A molecular formula represents the exact numbers of atoms of each element in one molecule of the

compound while an empirical formula only represents the whole-number molar ratios of each element in

the compound.

4.83 The ratio of the molar mass to the mass calculated from the empirical formula gives the information

necessary to determine the molecular formula. The mass of the empirical formula, CH2O is:

426




Mass of 1 mol of CH2O 30.03 g

The ratio of the molar mass to empirical formula mass is:

Molar mass ratio = 90 g / mol

30.03 g / mol = 3

Multiplying each of the subscripts in CH2O by three, we obtain the molecular formula C3H6O3.

4.84 The ratio of the molar mass to the mass calculated from the empirical formula gives the information

necessary to determine the molecular formula. The mass of the empirical formula, NO2 is:



Mass of 1 mol of NO2 46.01 g

The ratio of the molar mass to empirical formula mass is:


46.01 g / mol = 2.0

Multiplying each of the subscripts in NO2 by 2, we obtain the molecular formula N2O4.

4.85 The strategy for determining the molecular formula from percent composition and molar mass involves two

key steps. First, determine the empirical formula and the molar mass of the empirical formula from the

percent composition data. Next, use the ratio of the molar mass of the compound to the empirical formula

mass to determine the molecular formula.

The percent composition of our sample is 40.00% C, 6.72% H, and 53.29% O. A 100-gram sample of the

compound will contain 40.00 g C, 6.72 g H, and 53.29 g O. To determine the empirical formula, we

convert the masses of each element to the equivalent number of moles, and then determine the relative

number of moles of each element in the compound.


12.01 g C = 3.331 mol C


1.008 g H = 6.67 mol H


16.00 g O = 3.331 mol O

Next divide the number of moles of each element by the smallest of the molar amounts (i.e. 3.331 mol C or

3.331 mol O).





427



The empirical formula has 2 moles of hydrogen for each mole of oxygen and carbon. The empirical

formula is CH2O. The mass of the empirical formula is calculated as:




Mass of 1 mol of CH2O 30.03 g

The ratio of the molar mass to the empirical formula molar mass is:


30.03 g / mol = 6.0

Multiplying the subscripts in CH2O by 6, we obtain the molecular formula C6H12O6.

4.86 The strategy for determining the molecular formula from percent composition and molar mass involves two

key steps. First, determine the empirical formula and the molar mass of the empirical formula from the

percent composition data. Next, use the ratio of the molar mass of the compound to the empirical formula

mass to determine the molecular formula.

The percent composition of our sample is 50.7% C, 9.9% H, and 39.4% N. A 100-gram sample of the

compound will contain 50.7 g C, 9.9 g H, and 39.4 g N. To determine the empirical formula, we convert

the masses of each element to the equivalent numbers of moles, and then determine the relative number of

moles of each element in the compound.


12.01 g C = 4.22 mol C


1.008 g H = 9.8 mol H


14.01 g N = 2.81 mol N

Next divide the number of moles of each element by the smallest of the molar amounts (i.e.2.81 mol N).







Notice that two of the ratios are not whole numbers. The fractional portions (0.5) represent one-half (1/2).

Therefore, we can multiply the ratios by 2 to produce whole numbers. The empirical formula contains 3

moles carbon and 7 moles hydrogen for every 2 moles of nitrogen. The empirical formula is C3H7N2.

428

The mass of the empirical formula is calculated as:




Mass of 1 mol of C3H7N2 71.11 g

The ratio of the molar mass to the empirical formula mass:


71.11 g / mol = 2.00

Multiplying the subscripts in C3H7N2 by 2, we obtain the molecular formula: C6H14N2.

4.87 To calculate the percent composition of a compound from the chemical formula, assume a sample size of 1

mole. Calculate the mass of each element in 1 mole of the compound, divide by the molar mass of the

compound and multiply by 100.

(a) Percent composition of SO2



Mass of 1 mol of SO2 = 64.06 g

% S = 32.06 g

100%64.06 g

= 50.05% S

% O = 32.00 g

100%64.06 g

= 49.95% O

(b) Percent composition of CuCl2

Mass of 1 mol of Cu = 1 mol 63.55 g/mol = 63.55 g


Mass of 1 mol of CuCl2 134.45 g

% Cu = 63.55 g

100%134.45 g

= 47.27% Cu

% Cl = 70.90 g

100%134.45 g

= 52.73% Cl

(c) Percent composition of Na3PO4


Mass of 1 mol of P = 1 mol 30.97 g/mol = 30.97 g


Mass of 1 mol of Na3PO4 163.94 g

% Na = 68.97 g

100%163.94 g

= 42.07% Na

429

% P = 30.97 g

100%163.94 g

= 18.89% P

% O = 64.00 g

100%163.94 g

= 39.04% O

(d) Percent composition of Mg(NO3)2

Mass of 1 mol of Mg = 1 mol 24.31 g/mol = 24.31 g



Mass of 1 mol of Mg(NO3)2 148.33 g

% Mg = 24.31 g

100%148.33 g

= 16.39 % Mg

% N = 28.02 g

100%148.33 g

= 18.89 % N

% O = 96.00 g

100%148.33 g

= 64.72 % O

4.88 To calculate the percent nitrogen of a compound from the chemical formula, assume a sample size of 1

mole. Calculate the mass of nitrogen in 1 mole of the compound, divide by the molar mass of the

compound and multiply by 100.

(a) Percent nitrogen in NaNO3



Mass of 3 mol of O = 3 mol 16.00 g/mol = 48.00g

Mass of 1 mol of NaNO3 85.00 g

% N = 14.01 g

100%85.00 g

= 16.48% N

(b) Percent nitrogen in NH4Cl




Mass of 1 mol of NH4Cl 53.49 g

% N = 14.01 g

100%53.49 g

= 26.19% N

(c) Percent nitrogen in N2H4



Mass of 1 mol of N2H4 32.05 g

% N = 28.02 g

100%32.05 g

= 87.43% N

(d) Percent nitrogen in N2O

430



Mass of 1 mol of N2O 44.02 g

% N = 28.02 g

100%44.02 g

= 63.65% N

4.89 A comparison of the percent composition of each mineral shows that cuprite, CuO, has the highest

percentage of copper (79.89% Cu) with chalcocite, Cu2S, an extremely close second at 79.84% Cu.

(a) Cu2S



Mass of 1 mol of Cu2S = 159.2 g

%Cu = 127.1 g

100%159.2 g

= 79.84% Cu

(b) Cu2(CO3)(OH)2





Mass of 1 mol of Cu2(CO3)(OH)2 221.1 g

%Cu = 127.1 g

100%221.1 g

= 57.49% Cu

(c) CuO



Mass of 1 mol of CuO 79.55 g

%Cu = 63.55 g

100%79.55 g

= 79.89% Cu

(d) Cu3(CO3)2(OH)2





Mass of 1 mol of Cu3(CO3)2(OH)2 344.7 g

%Cu = 190.7 g

100%344.7 g

= 55.32% Cu

4.90 A comparison of the percent composition of each mineral shows that wustite, FeO, has the highest

percentage of iron (77.73% Fe).

FeO

Mass of 1 mol of Fe = 1 mol 55.85 g/mol = 55.85 g

431


Mass of 1 mol of FeO 71.85 g

%Fe = 55.85 g

100%71.85 g

= 77.73% Fe

Fe2O3



Mass of 1 mol of Fe2O3 159.7 g

%Fe = 111.7 g

100%159.7 g

= 69.94% Fe

Fe3O4



Mass of 1 mol of Fe3O4 231.55 g

%Fe = 167.55 g

100%231.55 g

= 72.36%

FeCO3




Mass of 1 mol of FeCO3 115.86 g

%Fe = 55.85 g

100%115.86 g

= 48.20% Fe

4.91 A solution is a homogenous mixture of two or more substances. Some common solutions: clear drinks

(coffee and tea), window cleaner, soapy water, tap water, air, brass (a homogenous mixture of copper and

zinc).

4.92 The solute is the substance that is dissolved and is usually present in lesser amount. The solvent is the

substance doing the dissolving and is usually present in greater amount. In the case of solutions formed

from solids and liquids, however, the solid is considered the solute and the liquid is the solvent. Most

products available in the grocery store contain many different solutes.

Solution Solute

Soda Sugar (C12H22O11)

Soy Sauce Salt (NaCl)

Vinegar Acetic acid (CH3COOH)

Window Cleaner Ammonia (NH3)

Bleach Sodium hypochlorite (NaClO)

4.93 Water is the solvent because it is present in the larger amount. Calcium chloride, CaCl2, is the solute

because it is present in the smaller amount. Pure calcium chloride is a solid, but when it is added to water,

it dissociates into its ions as it is dissolving (as shown in the figure).

432

4.94 Water is the solvent because it is present in the larger amount. Sodium chloride is the solute because it is

present in the smaller amount. Pure sodium chloride is a solid, but when it is added to water, it dissociates

into its ions as it is dissolving (as shown in the figure).

4.95 Dilute and concentrated are relative terms. A concentrated solution has a relatively high solute

concentration while a dilute solution has a lower solute concentration.

4.96 Brine is a very concentrated solution of sodium chloride and is used to pickle different kinds of foods.

Saline eye drops are also solutions of sodium chloride, but they are very dilute. There is a much lower

concentration of sodium chloride in eye drops than there is in brine. Liquid pesticides can also be dilute or

concentrated. If you purchase a bottle that is “ready to use,” it is a dilute solution; if the instructions say to

mix with water, then the solution is concentrated and must be diluted before use. Liquid detergent would

also be a good example of a concentrated solution. You add it to water to make it into a dilute solution

before you use it.

4.97 Concentration describes the relationship between the quantities of solute and solvent in a solution.

Molarity (number of moles of solute per liter of solution) and percent concentration (number of grams of

solute per gram of solution) are quantitative ways of expressing concentration.

4.98 Molarity is defined as the number of moles of solute dissolved in one liter of solution.

mol moles of soluteMolarity

L liters of solution

4.99 Solution A is more concentrated. The volume shown in both figures is the same, but solution A shows

more solute particles.

4.100 Solution A is more dilute. The volume shown in both figures is the same, but solution A has fewer solute

particles.

4.101 The molarity of a solution is calculated from the number of moles of solute and volume of solution

(measured in liters). The volumes are given, but the solute quantities are given in grams. In each case, we

must convert the mass of solute into moles using the molar mass. The overall process can be described by

the following map:

Map: volume

Grams of solute Moles of solute Molarity MM

(a) CH3CO2H (60.05 g/mol)

Moles of CH3CO2H = 3 2122 g CH CO H 3 2

3 2

1 mol CH CO H

60.05 g CH CO H = 2.03 mol CH3CO2H

Molarity = 3 22.03 mol CH CO H

1.00 L solution= 2.03 M CH3CO2H

(b) C12H22O11 (342.3 g/mol)

Moles of C12H22O11 = 12 22 11185 g C H O 12 22 11

12 22 11

1 mol C H O

342.3 g C H O = 0.540 mol C12H22O11

Molarity = 12 22 110.540 mol C H O

1.00 L solution= 0.540 M C12H22O11

(c) HCl (36.46 g/mol)

Moles of HCl = 70.0 g HCl1 mol HCl

36.46 g HCl = 1.92 mol HCl

433

Molarity = 1.92 mol HCl

0.600 L solution= 3.20 M HCl

(d) This problem is slightly different than the others because the volume is not given in liters.

KOH (56.11 g/mol)

Moles of KOH = 45.0 g KOH1 mol KOH

56.11 g KOH = 0.802 mol KOH

Convert the volume to liters:

Volume (L) = 250.0 mL1 L

1000 mL = 0.2500 L

Molarity = 0.802 mol KOH

0.2500 L solution= 3.21 M KOH

4.102 The molarity of a solution is calculated from the number of moles of solute and volume of solution

(measured in liters). To calculate molarity we must convert the mass of solute into moles using the molar

mass, and convert the volumes from milliliters to liters. The process can be described by the following

map:

Map: volume

Grams of solute Moles of solute Molarity MM

(a) HNO3 (63.02 g/mol)

Volume (L) = 255 mL1 L

1000 mL = 0.255 L

Moles of HNO3 = 36.30 g HNO 3

3

1 mol HNO

63.02 g HNO = 0.100 mol HNO3

Molarity = 30.100 mol HNO

0.255 L solution= 0.392 M HNO3

(b) H2SO4 (98.08 g/mol)


1000 mL = 0.125 L

Moles of H2SO4 = 2 449.0 g H SO 2 4

2 4

1 mol H SO

98.08 g H SO = 0.500 mol H2SO4

Molarity = 30.500 mol HNO

0.125 L solution= 4.00 M H2SO4

(c) KOH (56.11 g/mol)


1000 mL = 0.525 L

Moles of KOH = 2.80 g KOH1 mol KOH

56.11 g KOH = 0.0499 mol KOH

434

Molarity = 0.0499 mol KOH

0.525 L solution= .0951 M KOH

(d) Ca(OH)2 (74.10 g/mol)

Volume (L) = 200.0 mL1 L

1000 mL = 0.2000 L

Moles of Ca(OH)2 = 7.40 g KOH1 mol KOH

74.10 g KOH = 0.0999 mol Ca(OH)2

Molarity = 20.0999 mol Ca(OH)

0.2200 L solution= 0.499 M Ca(OH)2

4.103 To calculate moles of substance, it helps to remember that molarity is the conversion between moles and

volume (L). Since you are given molarity and volume, you can calculate moles of a substance. The moles

of ions are calculated using the formula ratios of ions per formula unit.

Volume Moles of soluteM

Moles Na2SO4 = 2 4150.0 mL Na SO

32 410 L Na SO

2 41mL Na SO

2 4

2 4

0.124 mol Na SO

L Na SO = 0.0186 mol Na2SO4

2 42 4

2 mol Na 1 mol Na SOMoles of Na SO mol Na

Mol Na+ = 2 40.0186 mol Na SO

2 4

2 mol Na

1mol Na SO

= 0.0372 mol Na+

2

24 2 42 4 4

1 mol SO 1 mol Na SOMoles of Na SO mol SO

Mol SO42

= 2 40.0186 mol Na SO2

4

22 4

1 mol SO

1mol Na SO

= 0.0186 mol SO4

2

4.104 To calculate moles of substance, it helps to remember that molarity is the conversion between moles and

volume (L). Since you are given molarity and volume, you can calculate moles of a substance. The moles

of ions are calculated using the formula ratios of ions per formula unit.

Volume Moles of soluteM

Moles Mg(NO3)2 = 3 2225.0 mL Mg(NO )

33 210 L Mg(NO )

3 21mL Mg(NO )

3 2

3 2

1.20 mol Mg(NO )

L Mg(NO )

= 0.270 mol Mg(NO3)2

2

23 23 2

1 mol Mg 1 mol Mg(NO ) Moles of Mg(NO ) mol Mg

Mol Mg2+

= 3 20.270 mol Mg(NO )2

3 2

1 mol Mg

1mol Mg(NO )

= 0.270 mol Mg2+

3 3 23 2 3

2 mol NO 1 mol Mg(NO ) Moles of Mg(NO ) mol NO

435

Mol NO3 = 3 20.270 mol Mg(NO ) 3

3 2

2 mol NO

1mol Mg(NO )

= 0.540

4.105 Multiplying the volume of solution (in liters) by the molarity gives you the moles of solute:

Map: molarity

Volume of solution Moles of solute

Make sure to convert the volume to liters. Then use the molar mass of the substance to convert the moles

of solute to mass:

Map: Moles of solute Grams of soluteMM

(a) 250.0 mL of 1.50 M KCl (MM = 74.55 g/mol)

Volume of solution = 250.0 mL1 L

1000 mL = 0.2500 L

Moles KCl = 0.2500 L1.50 mol KCl

1 L = 0.375 mol KCl

Grams KCl = 0.375 mol KCl74.55 g KCl

1 mol KCl = 28.0 g KCl

(b) 250.0 mL of 2.05 M Na2SO4 (MM = 142.04 g/mol)


1000 mL = 0.2500 L

Moles Na2SO4 = 0.2500 L 2 42.05 mol Na SO

1L = 0.512 mol Na2SO4

Grams Na2SO4 = 2 40.5125 mol Na SO 2 4

2 4

142.04 g Na SO

1 mol Na SO = 72.8 g Na2SO4

4.106 Multiplying the volume of solution (in liters) by the molarity gives you the moles of solute:

Map: molarity

Volume of solution Moles of solute

Make sure to convert the volume to liters. Then use the molar mass of the substance to convert the moles

of solute to mass:

Map: Moles of solute Grams of soluteMM

(a) 150.0 mL of 0.245 M CaCl2 (MM = 110.98 g/mol)


1000 mL = 0.1500 L

Moles CaCl2 = 0.1500 L 20.245 mol CaCl

L = 0.0368 mol CaCl2

Grams CaCl2= 20.0368 mol CaCl 2

2

110.98 g CaCl

1 mol CaCl = 4.08 g CaCl2

(b) 1450 mL of 0.00187 M H2SO4 (MM = 98.08 g/mol)

436

Volume of solution = 1450 mL1 L

1000 mL = 1.45 L

Moles H2SO4 = 1.45 L 2 40.00187 mol H SO

1 L = 0.00271 mol H2SO4

Grams H2SO4 = 2 40.00271mol H SO 2 4

2 4

98.08 g H SO

1 mol H SO = 0.266 g H2SO4

4.107 We convert moles of solute to volume of solution using molarity:

Map: molarity

Moles of solute Volume of solution

Make sure you check to determine that your answer makes sense. If the number of moles of solute you

need is greater than the molarity of the solution, you will need more than one liter of solution. Conversely,

if the number of moles of solute you need is smaller than the molarity of the solution, you will need less

than one liter of solution.

(a) Volume of solution = 30.250 mol AlCl3

1 L

0.250 mol AlCl = 1.00 L

(b) Volume of solution = 0.250 mol HCl1 L

3.00 mol HCl = 0.0833 L

4.108 We convert moles of solute to volume using molarity:

Map: molarity

Moles of solute Volume of solution

Make sure you check to determine that your answer makes sense. If the number of moles of solute you

need is greater than the molarity of the solution, you will need more than one liter of solution. Conversely,

if the number of moles of solute you need is smaller than the molarity of the solution, you will need less

than one liter of solution.

(a) Volume of solution = 2 40.250 mol H SO2 4

1 L

1.50 mol H SO = 0.167 L

(b) Volume of solution = 0.250 mol NaCl1 L

0.750 mol NaCl = 0.333 L

4.109 Count the copper ions in the image before and after the addition of water. Before the addition of water,

there are 10 copper ions. After the addition of water, there are two copper ions shown in the same volume.

When diluting solutions, the concentration is inversely proportional to volume. Since the concentration

decreased by a factor of five (i.e. 2/10 = 1/5) the volume must have increased by five (i.e.10/2 = 5).

Starting with a volume of 10.0 mL, the final volume is 50.0 mL. 40.0 mL of water was added to the

original solution.

4.110 Count the H2S molecules in the image before and after the addition of water. Before the addition of water,

there are 10 H2S molecules. After the addition of water, there are six H2S molecules shown in the same

volume. When diluting solutions, the concentration is inversely proportional to volume. Since the

concentration decreased by a ratio of 6/10, the volume must have increased by the inverse proportion

(10/6). Starting with a volume of 50.0 mL, the final volume is 83.3 mL. 33.3 mL of water was added to

the original solution.

437

4.111 The relationship between the concentration and molarity of the dilute and concentrated solutions is given

as:

con con dil dilM V M V

To calculate the volume added, you need to first determine the volume of the dilute solution, and then

calculate the increase in volume.

con condil

dil

M V

VM

To help you solve this problem it is important to identify the appropriate variables:

Mcon = 0.1074 M

Vcon = 935.0 mL

Mdil = 0.1000 M

Since the volume of the concentrated solution is given in milliliters, we first convert the volume to units of

liters:

Vcon = 935.0 mL1 L

1000 mL = 0.9350 L

Vdil = 0.1074 M 0.9350 L

0.1000 M = 1.004 L

The final volume required is 1.004 L. A volume of 0.0692 L should be added. You should note that this

answer is reasonable since the volume of the dilute solution should always be larger than the more

concentrated solution.


as:


To calculate the volume added, you need to first determine the volume of the dilute solution, and then

calculate the increase in volume.

con condil

dil

M V

VM

To help you solve this problem correctly it is important to identify the appropriate variables in the problem:

Mcon = 3.0 M

Vcon = 25 mL

Mdil = 0.055M

Since the volume of the concentrated solution is given in milliliters, we first convert the volume to units of

liters:

Vcon = 25 mL1 L

1000 mL = 0.025 L

Vdil = 3.0 M 0.025 L

0.055 M = 1.4 L

The volume of the dilute solution is 1.4 L. Note that this answer is reasonable because the volume of the

dilute solution should always be larger than the more concentrated solution.

438


as:


To calculate the molarity of the dilute solution, we solve the equation for Mdil.

con condil

dil

M V

MV

(a) To help you solve this problem correctly it is important to identify the appropriate variables in the

problem:

Mcon = 0.1832 M

Vcon = 24.75 mL

Vdil = 250.0 mL

At this point, there are two ways to calculate the molarity of the dilute solution. Since the volumes are

given in milliliters, it is not necessary to convert the units to liters. Notice how the mL units cancel in

the following calculation:

Mdil = 0.1832 24.75 mLM

250.0 mL = 0.01814 M

Alternatively, you could convert the volumes to units of liters and carry out the same calculation:

Mdil = 0.1832 0.02475 LM

0.2500 L = 0.01814 M

(b) Mcon = 1.187 M

Vcon = 125 mL = 0.125 L (The volumes need to be converted to the same units.)

Vdil = 0.500 L

Mdil = 1.187 0.125 LM

0.500 L = 0.297 M

(c) Mcon = 0.2010 M

Vcon = 10.00 mL

Vdil = 50.00 mL

Mdil = 0.2010 10.00 mLM

50.00 mL = 0.04020 M


as:


To calculate the molarity of the dilute solution, we solve the equation for Mdil.

con condil

dil

M V

MV

(a) To help you solve this problem correctly it is important to identify the appropriate variables in the

problem:

439

Mcon = 6.00 M

Vcon = 35.45 mL

Vdil = 150.0 mL

At this point, there are two ways to calculate the molarity of the dilute solution. Since the volumes are

given in milliliters, it is not necessary to convert the units to liters. Notice how the mL units cancel in

the following calculation:

Mdil = 6.00 35.45 mLM

150.0 mL = 1.42 M

Alternatively, you could convert the volumes to units of liters and carry out the same calculation:

Mdil = 6.00 0.03545 LM

0.1500 L = 1.42 M

(b) Mcon = 0.00102 M

Vcon = 250.0 mL

Vdil = 500.0 mL

Mdil = 0.00102 250.0 mLM

500.0 mL = 5.10 10

4 M

(c) Mcon = 0.8045 M

Vcon = 5.00 mL

Vdil = 250.0 mL

Mdil = 0.8045 5.00 mLM

250.0 mL = 0.0161 M

4.115 To calculate the mass of 0.100 mol of Cu(OH)2 we use the molar mass:


Cu(OH)2




Mass of 1 mol of Cu(OH)2 97.57 g

Mass Cu(OH)2 = 20.100 mol Cu(OH)2

97.57 g

1 mol Cu(OH) = 9.76 g Cu(OH)2

4.116 (a) To calculate the number of moles of V2O5 in 52.5 g of V2O5, we use the molar mass and the mole ratio

of vanadium in the compound:

Map: formula ratio

Mass in grams Moles Moles VMM

Molar Mass of V2O5

Mass of 2 mol of V = 2 mol 50.94 g/mol = 101.88 g


Mass of 1 mol of V2O5 181.88 g

440

Moles V= 2 552.5 g V O2 51 mol V O

2 5181.88 g V O 2 5

2 mol V

1 mol V O = 0.577 mol V

(b) The mass of vanadium is calculated from the moles using the molar mass of vanadium (50.94 g/mol):


Mass V = 0.577 mol V50.94 g V

1 mol V = 29.4 g V

(c) The number of atoms of vanadium is calculated from the number of moles of vanadium, using

Avogadro’s number:

Map: AN Moles Number of atoms

Atoms V = 0.577 mol V236.022 10 atoms V

1 mol V

= 3.48 10

23 atoms V

4.117 The average mass of one argon atom, in amu, is the value given on the periodic table, 39.95 amu. This is

also the mass of one mole of average argon atoms (i.e. 39.95 g/mol). To calculate the average mass of a

single Ar atom we look for units of grams per atom. We can derive this result using Avogadro’s number to

convert g/mol to g/atom:

Average Ar atom mass (g) = 39.95 g Ar

1 mol Ar

1 mol Ar

236.022 10 atom = 6.634 10

23 g Ar/atom

4.118 To calculate the number of molecules in the sample we must first determine the number of moles of

C4H9OH that are present, using the molar mass:

Map: ANMass in grams Moles Number of molecules

MM

C4H9OH




Mass of 1 mol of C4H9OH 74.12 g

Molecules C4H9OH = 15.43 g1 mol

74.12 g


1 mol

= 1.254 10

23 molecules

4.119 (a) To calculate the number of moles of argon, use the molar mass:

Moles Ar = 36.1 g1 mol Ar

39.95 g = 0.904 mol Ar

Calculate the number of atoms from the number of moles, using Avogadro’s number:

Atoms Ar = 0.904 mol Ar236.022 10 atom

1 mol Ar

= 5.44 10

23 atoms

441

(b) Use the conversion 1 carat = 0.200 g to calculate the mass of carbon in the Hope diamond. Then

convert this mass to moles and to number of atoms, using the following conversion map:

Map: A0.200 g = 1 carat NMass in carats Mass in grams Moles Number of atoms

MM

Moles C = 44.5 carat0.200 g C

1 carat

1 mol C

12.01 g C = 0.741 mol C

Atoms C = 0.741mol C236.022 10 atom

1 mol C

= 4.46 10

23 atoms C

(c) We can use the density of mercury to convert from volume to mass. Then we can convert from mass

to numbers of moles and atoms using the conversion pathway:

Map: ANMass in grams Moles Number of atoms

MM

Mass Hg = 2.50 mL13.6 g Hg

mL = 34.0 g

Moles Hg = 34.0 g Hg1 mol Hg

200.6 g Hg = 0.169 mol Hg

Atoms Hg = 0.169 mol Hg236.022 10 atom Hg

1 mol Hg

= 1.02 10

23 atoms

4.120 The mass of one mole of iodine atoms is 126.9 g. The mass of one mole of bromine atoms is 79.90 g. If

you have 50.0 g of iodine atoms, the same number of bromine atoms has a mass of 179 g. This problem

can also be solved using mole conversions:

Map: I 1 mol I 1 mol Br Br

Mass I Moles I Moles Br Mass BrMM MM

The 1 mol I = 1 mol Br conversion satisfies the requirement that there are equal numbers of atoms (or

moles of atoms) in the samples:

Mass Br = 50.0 g I1 mol I

126.9 g I

1 mol Br

1 mol I

79.90 g Br

1 mol Br = 31.5 g Br

4.121 To calculate moles, first convert the mass to grams and then use the molar mass of calcium carbonate

(100.09 g/mol) to determine the number of moles

Map: 3

3 3 3

1 mg 10 g 1 mol 100.09 gMass mg CaCO Mass g CaCO Moles CaCO

Moles CaCO3 = 750.0 mg

3

3

10 g CaCO

1 mg

1 mol

100.09 g = 7.493 10

3 mol CaCO3

4.122 The percent composition of our sample is 92.3% C and 7.7% H. A 100-gram sample contains 92.3 g C and

7.7 g H. We convert these masses to the equivalent numbers of moles:

442

Moles C = 92.3g C1 mol C

12.01 g C = 7.69 mol C


1.008 g H = 7.6 mol H

Since the mole ratios are essentially the same, we conclude that the empirical formula must be CH. The

empirical formula mass of CH is 13.02 g/mol. From the molar mass of the compound, 78.1 g/mol, we

calculate the ratio of the molar mass of the compound to the empirical formula mass:

Molar mass ratio = 78.1 g / mol

13.02 g / mol = 6.00

Multiplying the subscripts of the empirical formula by 6, we determine that the molecular formula for the

compound is C6H6.

4.123 The percent composition of tear gas is 40.25% C, 6.19% H, 8.94% O, and 44.62% Br. A 100-gram sample

contains 40.25 g C, 6.19 g H, 8.94 g O, and 44.62 g Br. To determine the empirical formula, we convert

the masses of each element to the equivalent number of moles and determine the relative number of moles

of each element in the substance.


12.01 g C = 3.351 mol C


1.008 g H = 6.14 mol H


16.00 g O = 0.559 mol O

Moles Br = 44.62 g Br1 mol Br

79.90 g Br = 0.5584 mol Br


Br).

Moles C 3.351 mol C 6.001 mol C

Moles Br 0.5584 mol Br 1 mol Br

Moles H 6.14 mol H 11.0 mol H


Moles I 0.559 mol O 1.00 mol O


The empirical formula has 6 moles of carbon and 11 moles of hydrogen for each mole of O and Br. The

empirical formula is C6H11OBr.

4.124 The molar mass of CO2 is 44.01 g/mol. The mass of dry ice needed is:

443

Mass CO2 (g) = 23.54 mol CO 2

2

44.01 g CO

1 mol CO = 156 g CO2

4.125 To calculate the number of oxygen atoms, first we calculate the number of moles of H3PO4. Then we use

the mole ratio and Avogadro’s number to calculate the number of oxygen atoms. The molar mass of H3PO4

is 97.99 g/mol.

Map: ANmole ratioMass in grams Moles Moles O Number of O atoms

MM

Oxygen Atoms = 3 410.00 g H PO3 41 mol H PO

3 497.99 g H PO

4 mol O

3 41 mol H PO

236.022 10 atoms O

1 mol O

= 2.458 1023

atoms O

4.126 Percent composition (or mass percent) Na2CO3




Mass of 1 mol of Na2CO3 105.99 g

% Na = 45.98 g

100%105.99 g

= 43.38% Na

% C = 12.01 g

100%105.99 g

= 11.33% C

% O = 48.00 g

100%105.99 g

= 45.29% O

4.127 The molar mass of CO2 is 44.01 g/mol.

Moles CO2 = 2960 g CO21mol CO

244.01 g CO

232

2

6.022 10 molecules CO

1mol CO

= 1.3 × 10

25 molecules CO2

4.128 To calculate moles, first convert the mass to grams and then use the molar mass of ibuprofen, C13H18O2

(206.27 g/mol) to determine the number of moles

Map: 3

13 18 2 13 18 2 13 18 2

1 mg 10 g 1 mol 206.27 gMass mg C H O Mass g C H O Moles C H O

Moles C13H18O2 = 200.0 mg

-3

13 18 2

10 g C H O ×

1 mg

1 mol×

206.27 g = 9.696 10

4 mol C13H18O2

4.129 (a) The percent composition of vanillin is 63.15% C, 5.30% H, and 31.55% O. A 100-gram sample will

contain 63.15 g C, 5.30 g H, and 31.55 g O. To determine the empirical formula, we convert the mass

of each element to the equivalent number of moles, and determine the relative number of moles of each

element in the substance.


12.01 g C = 5.258 mol C

444


1.008 g H = 5.258 mol H


16.00 g O = 1.972 mol O

Next divide the number of moles of each element by the smallest of the molar amounts (i.e., 1.972 mol

O).

Moles C 5.258 mol C 2.667 mol C

Moles O 1.972 mol O 1 mol O

Moles H 5.258 mol H 2.667 mol H




The fractional portion of the ratios (0.667) indicates two thirds (i.e., 2/3) and can be converted to

whole numbers by multiplying by 3. The empirical formula has 8 moles of carbon, 8 moles of

hydrogen for every 3 moles of oxygen. The empirical formula for vanillin is C8H8O3. The molar

mass of the empirical formula is:




Mass of 1 mol of Na2CO3 152.14 g

(b) Since the molar mass of the empirical formula and the vanillin molecule are the same, the molecular

formula is also C8H8O3.

4.130 Percent composition of monosodium glutamate, NaC5H8NO4






Mass of 1 mol of NaNO3 169.11 g

% Na = 22.99 g

100%169.11 g

= 13.59% Na

% C = 60.05 g

100%169.11 g

= 35.51% C

% H = 8.064 g

100%169.11 g

= 4.768% H

445

% N = 14.01 g

100%169.11 g

= 8.284% N

% O = 64.00 g

100%169.11 g

= 37.85% O

4.131 To calculate the empirical formula, we need to know the mass of each element in the compound. Since

5.00 g of aluminum produced 9.45 g of aluminum oxide, the principle of conservation of mass tells us that

4.45 g of oxygen has been incorporated into the aluminum oxide (i.e. 9.45 g 5.00 g = 4.45 g). Starting

with those masses, we convert them to moles, and then determine the mole ratio of the elements:

Moles Al = 5.00 g Al1 mol Al

26.98 g Al = 0.185 mol Al


16.00 g O = 0.278 mol O

The mole ratio of aluminum to oxygen (the substance present in smaller number of moles) is:


moles Al 0.185 mol Al 1 mol Al

The fractional portion of the number indicates one half (1/2). Multiplying both parts of the ratio by 2

will produce a whole number. The empirical formula has 2 moles of aluminum for every 3 moles of

oxygen. The empirical formula is Al2O3.

4.132 To calculate the empirical formula, we need to know the mass of each element in the compound. Since

4.32 g of copper produced 5.41 g of chalcocite, we know that all 1.09 grams of sulfur reacted (i.e. 4.32 g +

1.09 g = 5.41 g). We convert the masses of copper and sulfur in chalcocite to the equivalent numbers of

moles, and determine the mole ratio of the elements:

Moles Cu = 4.32 g Cu1 mol Cu

63.55 g Cu = 0.0680 mol Cu

Moles S = 1.09 g S1 mol S

32.06 g S = 0.0340 mol S

The mole ratio of copper to sulfur (the substance present in smaller number of moles) is:

moles Cu 0.0680 mol Cu 2 mol Cu

moles S 0.0340 mol S 1 mol S

The empirical formula has 2 moles of copper for every 1 mole of sulfur.

The empirical formula is Cu2S.

4.133 First we must calculate the molar mass of calcium nitrate so that we can use it to convert between moles

and grams:

3 2Ca(NO ) 40.08 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol) = 164.10 g/molMM

446

Then we multiply molar mass by moles to get grams:

3 2

3 2Ca(NO )

3 2

164.10 g Ca(NO )Mass 0.742 mol = 122 g

1 mol Ca(NO )

4.134 (a) CH4 has a greater mass percent carbon because four hydrogen atoms have a lower mass than two

oxygen atoms attached to carbon in CO2.

(b) This comparison is easier if we compare empirical formulas (CH4 and CH3). We can do this because

empirical formulas have the same mass percent composition as their molecular formulas. The

empirical formula CH3 has the greater mass percent carbon because it has fewer hydrogen atoms and

the same number of carbon atoms, so the compound C2H6 must have a greater mass percent carbon

than CH4.

(c) These two compounds have the same ratio of carbon to hydrogen so they have the same empirical

formula, CH2. Therefore they have the same mass percent carbon.

4.135 Both compounds have the same ratio of calcium to anion, so we can compare the molar masses of the

anions to determine relative mass percent calcium.

3–4PO

30.97 g/mol + 4(16.00 g/mol) = 94.97 g/molMM

3–6 5 7C H O

6(12.01 g/mol) + 5(1.008 g/mol) + 7(16.00 g/mol) = 189.1 g/molMM

The phosphate ion contributes a lower mass to its calcium compound compared to the citrate ion, so

Ca3(PO4)2 has the greater mass percent calcium.

4.136 The number of Ca3(C6H5O7)2 formula units is related to moles by Avogadro’s number:

'

Moles Number of Formula UnitsAvogadro s Number

23

3 6 5 7 2

3 6 5 7 2

243 6 5 7 2

Number of formula units

6.022 10 formula units5.00 mol Ca (C H O )

1 mol Ca (C H O )

= 3.01 10 Ca (C H O ) formula units

4.137 This is similar to the question: Which box contains more balls, a 10-pound box of baseballs or a 10-pound

box of ping-pong balls? Many more of the ping-pong balls are needed to make a total mass of 10 pounds

than the heavier baseballs. Similarly more of the lighter-weight F2 molecules are needed to make a 1.0

gram sample than the heavier-weight SF2 and CF4 molecules. We can also reason this out mathematically:

Moles are proportional to number of molecules, so if we determine which sample has the greater number of

moles, which is also the sample with the greater number of molecules. To calculate moles of each

substance, we divide 1.0 g by the molar mass of each substance. The larger the molar mass the smaller the

calculated number of moles. The molar mass of F2 is the smallest, so a 1.0 gram sample of F2 has the

greatest number of moles and molecules.

4.138 (a) The samples are each 1.0 mole so each contains the same number of molecules: 6.0221023

molecules.

(b) While each sample contains the same number of molecules, the substance with the greatest number of

atoms per molecule will have the greatest number of atoms in the sample. A CF4 molecule has the

most atoms (5) compared to SF2 (3 atoms) and F2 (2 atoms) so a 1.0 mol sample of CF4 will have the

greatest number of atoms.

4.139 (a) 98% of 20.0 g is: 0.98 20.0 g = 19.6 g

447

(b) To determine the mass of Na in 19.6 g NaCl, we must first determine the mass percent Na in the

sample:

Mass of 1 mol Na 22.99 g Na% Na = 100 100 39.34% Na

Mass of 1 mol NaCl 58.44 g NaCl

Mass Na in 19.6 g NaCl = 0.3934 19.6 g = 7.71 g Na

(c) Since sodium and chlorine are the only elements in the compound, we can calculate the mass of

chlorine by subtracting the mass of sodium from the total mass of the compound:

Mass Cl in 19.6 g NaCl = 19.6 g NaCl – 7.71 g Na = 11.9 g Cl

4.140 (a) Molarity is a ratio of moles to liters:

L

mol = M

V.

Rearranging to solve for moles we get: mol = M VL

Before we can substitute volume into this equation we must convert volume units from mL to L. The two

step process is: 1 L = 1000 mL

mL L moles

MolarityV V

Volume in liters = 1 L

10.0 mL = 0.0100 L1000 mL

Moles citric acid = M VL = 0.30 mol

1 L × 0.0100 L = 0.0030 mol H3C6H5O7

(b) We can convert from moles of H3C6H5O7 to grams using molar mass. First we must calculate the

molar mass of H3C6H5O7:

3 6 5 7H C H O 8(1.008 g/mol) + 6(12.01 g/mol) + 7(16.00 g/mol) = 192.12 g/molMM

3 6 5 7H C H O 3 6 5 7Mass 0.00300 mol H C H O 3 6 5 7

3 6 5 7

192.1 g H C H O

1 mol H C H O

3 6 5 7 = 0.576 g H C H O

4.141 In each case we will calculate the number of formula units using Avogadro’s number, and then we will

multiply by the number of Cl ions per formula unit to determine the number of Cl

ions in each quantity.

(a) Number of Cl ions =

31.0 mol AlCl

2336.022 10 AlCl formula units

31 mol AlCl 3

3 Cl ions

1 AlCl formula unit

241.8 10 Cl ions

(b) Number of Cl ions =

23233

33

6.022 10 AlCl formula units 3 Cl ions0.25 mol AlCl 4.5 10 Cl ions

1 mol 1 AlCl formula unit

(c) Number of Cl ions =

20.25 mol MgCl23

26.022 10 MgCl formula units

1 mol MgCl

23

22

2 Cl ions3.0 10 Cl ions

1 MgCl formula unit

4.142 The definition of molarity (L

mol = M

V) tells us that 1.0 L of a 2.5 M Mg(NO3)2 solution contains 2.5 mol of

Mg(NO3)2. We also know that each formula unit of Mg(NO3)2 consists of 1 Mg2+

ion and 2 NO3 ions, so

448

when Mg(NO3)2 dissolves, the concentration of NO3 is twice that of Mg

2+ ion. We have 2.5 mol of

Mg(NO3)2, so it dissolves to form 2.5 mol Mg2+

ion and 5.0 mol of NO3 ion.

CONCEPT REVIEW

4.143 Answer: D; the percent composition will be the same for compounds that have the same ratio of the carbon

to hydrogen atoms. In this case 2/4 = 0.5 for C2H4 and 3/6 = 0.5 for C3H6.

In each case, the compound with the lower carbon/ hydrogen ratio would have the lower percent carbon.

A. The carbon/ hydrogen ratios are 3/4 = 0.75 for C3H4 and 3/6 = 0.5 for C3H6. C3H6 has the lower

percent carbon by mass.

B. The carbon/ hydrogen ratios are 2/4 = 0.5 for C2H4 and 3/4 = 0.75 for C3H4. C2H4 has the lower


C. The carbon/ hydrogen ratios are 2/4 = 0.5 for C2H4 and 4/2 = 2 for C4H2. C2H4 has the lower percent

carbon by mass.

E. The carbon/ hydrogen ratios are 4/8 = 0.5 for C4H8 and 3/8 = 0.375 for C3H8. C3H8 has the lower


4.144 Answer: A and E; the molar mass of C3H8 is 44.0 g/mol, so this represents 1 mol; 1 mol of propane

contains 3 mol of carbon, or 36.0 g.

B. 44.0 g of C3H8 would contain 8 mol or 8 .0 g of hydrogen: the amount of propane that contains 8.0 g of

hydrogen

C. The 44.0 g of propane represents 1 mol of propane, not 44.0 mol: the amount of propane that contains

6.02 × 1023

molecules of propane

D. The 44.0 g of propane represents 1 mol of propane, not 8 mol: the amount of propane that contains

6.02 × 1023

molecules of propane

4.145 Answer: D; the number of molecules is:

4molecules As = 1 g1 mol

299.68 g


1 mol

21

4 = 2 10 molecules As

The number of molecules of the other substances can be obtained by a similar equation, but using the

appropriate molar mass of the substance. Since the other molar masses are smaller, the number of

molecules would be larger. It is only necessary to determine the molar mass of each substance to compare

the number of molecules in the same mass of each. The molar masses are 123.88 g/mol P4, 70.90 g/mol

Cl2, 28.02 g/mol N2, 299.68 g/mol As4, and 256.48 g/mol S8. The larger the molar mass, the smaller the

number of molecules in 1 g, so the order will be As4 < S8 < P4 < Cl2 < N2.

4.146 Answer: C; the number of atoms will equal:

2atoms of N = 1 gN21 mol N

228.02 g N

2326.022 10 molecules N

21 mol N 2

2 atoms N

1 molecule N 23 = 4.3 10 N atoms

The number of atoms present in the other substances can be calculated in a similar way. However, the

substance with the smallest product of 1/molar mass and number of atoms per molecule will contain the

largest number of atoms. The products of 1/molar mass and number of atoms/molecule are:

A. P4: 4/123.88 g/mol = 0.032

B. Cl2: 2/70.90 g/mol = 0.028

C. N2: 2/28.02 g/mol = 0.071

D. As4: 4/299.68 g/mol = 0.013

E. S8: 8/256.48 g/mol = 0.031

The order of increasing number of atoms in 1 g is As4 < Cl2 < S8 < P4 < N2.

449

4.147 Answer: C; The number of moles of oxygen atoms in this sample of O2 is:

2mol of O atoms = 0.60 mol O2

2 mol O atoms

1 mol O = 1.2 mol O atoms

A. 2mol of O atoms = 1.0 mol H O2

1 mol O atoms

1 mol H O = 1.0 mol O atoms

B. 2 4mol of O atoms = 0.20 mol N O2 4

4 mol O atoms

1 mol N O = 0.80 mol O atoms

D. 4 10mol of O atoms = 0.11 mol As O4 10

10 mol O atoms

1 mol As O = 1.1 mol O atoms

E. 3mol of O atoms = 0.30 mol O3

3 mol O atoms

1 mol O = 0.90 mol O atoms

The order of increasing number of oxygen atoms is B < E < A < D < C.

4.148 Answer: E; none of the subscripts can be simplified to smaller numbers.

A. Both subscripts can be divided by 2, giving an empirical formula of NF.

B. Both subscripts can be divided by 2, giving an empirical formula of NF2.

C. Both subscripts can be divided by 2, giving an empirical formula of HC.

D. All three subscripts can be divided by 2, giving an empirical formula of HNO.

4.149 Answer: C; the molecular formula is either the same as the empirical formula (a multiple of 1) or a whole

number multiple of the empirical formula.

A. H2C2O4 and HCO2 are of equal complexities.

B. H2O is both the molecular and the empirical formula of water.

D. Not all empirical formulas are the same as the molecular formula: CH and C2H2.

E. The molecular formula is a multiple, not a fraction of the empirical formula. The empirical formula

H2O cannot be divided to give smaller whole number subscripts.

4.150 Answer: C; 0.10 mol NaCl

molarity = = 0.20 NaCl0.5000 L

M

A. Dissolving 0.040 mol NaCl in water to make 0.20 L of solution

mol NaCl = 2.00 L0.20 mol NaCl

1 L = 0.040 mol NaCl

B. Dissolving 0.32 mol NaCl in water to make 1.6 L of solution


1 L = 0.32 mol NaCl

D. Dissolving 0.020 mol NaCl in water to make 100.0 mL of solution


1 L = 0.020 mol NaCl

E. Dissolving 0.0020 mol NaCl in water to make 10.0 mL of solution


1 L = 0.0020 mol NaCl

4.151 Answer: A; the solution contains:

450

3 4 2mol Ca (PO ) = 2.00 L 3 4 20.100 mol Ca (PO )

1 L 3 4 2 = 0.200 mol Ca (PO )

B. The solution contains 1.60 mol of oxygen atoms.

mol O atoms = 2.00 L3 4 20.100 mol Ca (PO )

1 L 3 4 2

8 mol O atoms

1 mol Ca (PO ) = 1.60 mol O atoms

C. 1.00 L of this solution contains 0.300 mol Ca2+

ions.

2+mol Ca ions = 1.00 L3 4 20.100 mol Ca (PO )

1 L

2+

3 4 2

3 mol Ca ions

1 mol Ca (PO ) 2+ = 0.300 mol Ca ions

D. There 6.02 × 1022

P atoms in 500.0 mL of this solution.

P atoms = 0.5000 L3 4 20.100 mol Ca (PO )

1 L

2 mol P atoms

3 4 21 mol Ca (PO )

236.022 10 P atoms

1 mol P atoms

22 = 6.02 10 P atoms

E. This solution contains 0.600 mol of Ca2+

.

2+mol Ca ions = 2.00 L3 4 20.100 mol Ca (PO )

1 L

2+

3 4 2

3 mol Ca ions

1 mol Ca (PO ) 2+ = 0.600 mol Ca ions

4.152 Answer: D; this solution contains

mol Cl = 0.300 L 0.200 mol NaCl

1 L

1 mol Cl

1 mol NaCl

= 0.600 mol Cl

A. mol Cl = 0.100 L 0.200 mol NaCl

1 L

1 mol Cl

1 mol NaCl

= 0.0200 mol Cl

B. mol Cl = 0.200 L 20.100 mol CaCl

1 L 2

2 mol Cl

1 mol CaCl

= 0.0400 mol Cl

C. mol Cl = 0.150 L 30.100 mol FeCl

1 L 3

3 mol Cl

1 mol FeCl

= 0.0450 mol Cl

E. mol Cl = 0.100 L 20.050 mol CaCl

1 L 2

2 mol Cl

1 mol CaCl

= 0.010 mol Cl

The order of decreasing moles chloride ion is D > C > B > A > E.

Documents

Chapter 4 – Chemical Composition - University of …chemweb/chem1000/docs...4.3 The percent composition of calcium in calcium carbonate is set up as: % Ca = 33 grams Ca 9.88g Ca