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Chapter 6Chapter 6Chemical Quantities Chemical Quantities

Powers of Ten Animation

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Section 6.1Section 6.1The Mole: A Measurement of The Mole: A Measurement of

MatterMatter OBJECTIVES:OBJECTIVES:

–Describe how Avogadro’s Describe how Avogadro’s number is related to a mole of number is related to a mole of any substance.any substance.

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Section 6.1Section 6.1The Mole: A Measurement of The Mole: A Measurement of

MatterMatter OBJECTIVES:OBJECTIVES:

–Calculate the mass of a mole of Calculate the mass of a mole of any substance.any substance.

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What is a Mole?What is a Mole?

You can measure You can measure massmass, , or or volumevolume,, or you can or you can count piecescount pieces.. We measure mass in We measure mass in gramsgrams.. We measure volume in We measure volume in litersliters.. We count pieces inWe count pieces in MOLESMOLES..

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Moles (abbreviated: mol)Moles (abbreviated: mol) Defined as the number of carbon Defined as the number of carbon

atoms in exactly 12 grams of atoms in exactly 12 grams of carbon-12.carbon-12.

1 mole is 6.02 x 101 mole is 6.02 x 102323 particles. particles. Treat it like a very large dozenTreat it like a very large dozen 6.02 x 106.02 x 102323 is called is called Avogadro’s Avogadro’s

number.number.

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Representative particlesRepresentative particles The smallest pieces of a substance.The smallest pieces of a substance.

–For a molecular compound: it is For a molecular compound: it is the molecule.the molecule.

–For an ionic compound: it is the For an ionic compound: it is the formula unit (ions).formula unit (ions).

–For an element: it is the atom.For an element: it is the atom.

»Remember the 7 diatomic Remember the 7 diatomic elements (made of molecules)elements (made of molecules)

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Types of questionsTypes of questions How many oxygen atoms in the How many oxygen atoms in the

following?following?–CaCOCaCO33

–AlAl22(SO(SO44))33

How many ions in the following?How many ions in the following?–CaClCaCl22

–NaOHNaOH–AlAl22(SO(SO44))33

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Types of questionsTypes of questions How many molecules of COHow many molecules of CO22 are are

there in 4.56 moles of COthere in 4.56 moles of CO22 ? ?

4.56 mol CO2

1 mol CO2

6.02x1023molecules CO2 2.75x10 24molecules CO2

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How many moles of water is 5.87 x How many moles of water is 5.87 x 10102222 molecules? molecules?

5.87x1022 molecules H2O 1 mol H2O

6.02x1023 molecules H2O

0.0975 mol H2O

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How many atoms of carbon How many atoms of carbon are there in 1.23 moles of are there in 1.23 moles of CC66HH1212OO66 ? ?

1.23 mol C6H12O6

1 mol C6H12O6

6.02x1023 molecules C6H12O6

1molecule C6H12O6

6 atoms C

= 4.44 x1024 atoms C

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How many moles is 7.78 x 10How many moles is 7.78 x 102424 formula units of MgClformula units of MgCl22

7.78x1024 fun MgCl2

6.02x1023 fun MgCl2

1 mol MgCl2 12.9 mol MgCl2

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Measuring MolesMeasuring Moles Remember relative atomic mass?Remember relative atomic mass? The amu was one twelfth the mass The amu was one twelfth the mass

of a carbon-12 atom.of a carbon-12 atom. Since the mole is the number of Since the mole is the number of

atoms in 12 grams of carbon-12,atoms in 12 grams of carbon-12, the decimal number on the periodic the decimal number on the periodic

table is also the mass of 1 mole of table is also the mass of 1 mole of those atoms in grams.those atoms in grams.

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Gram Atomic Mass (gam)Gram Atomic Mass (gam) Equals the mass of 1 mole of an Equals the mass of 1 mole of an

element in gramselement in grams 12.01 grams of C has the same 12.01 grams of C has the same

number of pieces as 1.008 grams of number of pieces as 1.008 grams of H and 55.85 grams of iron.H and 55.85 grams of iron.

We can write this as We can write this as 12.01 g C = 1 mole C 12.01 g C = 1 mole C

We can count things by weighing We can count things by weighing them.them.

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ExamplesExamples How much would 2.34 moles of How much would 2.34 moles of

carbon weigh?carbon weigh?

2.34 mol C

1 mol C

12.0 g C 28.1 g C

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How many moles of How many moles of magnesium is 24.31 g of Mg ?magnesium is 24.31 g of Mg ?

24.31 g Mg

24.3 g Mg

1 mol Mg 1.000 mol Mg

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How many atoms of lithium is How many atoms of lithium is 1.00 g of Li?1.00 g of Li?

1.00 g Li

6.9 g Li

1 mol Li

1 mol Li

6.02x10 23 atoms Li 8.72x10 22 atoms Li

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How much would 3.45 x 10How much would 3.45 x 102222 atoms atoms of U weigh?of U weigh?

3.34x10 22 atoms U

6.02x10 23 atoms U

1 mol U

1 mol U

238.0 g U 13.2 g U

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What about compounds?What about compounds? in 1 mole of Hin 1 mole of H22O molecules there are O molecules there are

two moles of H atoms and 1 mole of O two moles of H atoms and 1 mole of O atomsatoms

To find the mass of one mole of a To find the mass of one mole of a compound compound –determine the moles of the elements determine the moles of the elements

they havethey have–Find out how much they would weighFind out how much they would weigh–add them upadd them up

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What about compounds?What about compounds? What is the mass of one mole of CHWhat is the mass of one mole of CH44??

1 mole of C = 12.01 g1 mole of C = 12.01 g

4 mole of H x 1.01 g = 4.04g4 mole of H x 1.01 g = 4.04g

1 mole CH1 mole CH44 = 12.01 + 4.04 = 16.05g = 12.01 + 4.04 = 16.05g The The Gram Molecular Mass (gmm) Gram Molecular Mass (gmm) of of

CHCH44 is 16.05g is 16.05g

– this is the mass of one mole of a this is the mass of one mole of a molecular compound.molecular compound.

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Gram Formula Mass (gfm)Gram Formula Mass (gfm) The mass of one mole of an ionic The mass of one mole of an ionic

compound.compound. Calculated the same way as gmm.Calculated the same way as gmm. What is the GFM of FeWhat is the GFM of Fe22OO33??

2 moles of Fe x 55.85 g = 111.70 g2 moles of Fe x 55.85 g = 111.70 g

3 moles of O x 16.00 g = 48.00 g3 moles of O x 16.00 g = 48.00 g

The GFM = 111.70 g + 48.00 g = The GFM = 111.70 g + 48.00 g = 159.70 g159.70 g

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The molar mass (in g/mol) of any substance is always numerically equal to its formula weight (in amu).

Molar MassMolar MassMass of 1 mole of substanceMass of 1 mole of substance

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Section 6.2Section 6.2Mole-Mass and Mole-Volume Mole-Mass and Mole-Volume

RelationshipsRelationships OBJECTIVES:OBJECTIVES:

–Use the molar mass to convert Use the molar mass to convert between mass and moles of a between mass and moles of a substance.substance.

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Section 6.2Section 6.2Mole-Mass and Mole-Volume Mole-Mass and Mole-Volume

RelationshipsRelationships OBJECTIVES:OBJECTIVES:

–Use the mole to convert among Use the mole to convert among measurements of mass, measurements of mass, volume, and number of volume, and number of particles.particles.

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Molar MassMolar Mass Molar massMolar mass is the generic term for is the generic term for

the mass of one mole of any the mass of one mole of any substance (in grams) substance (in grams)

The same as: 1) gram molecular The same as: 1) gram molecular mass, 2) gram formula mass, and 3) mass, 2) gram formula mass, and 3) gram atomic mass- just a much gram atomic mass- just a much broader term.broader term.

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ExamplesExamples Calculate the molar mass of the Calculate the molar mass of the

following and tell what type it is:following and tell what type it is: NaNa22SS

NN22OO44

CC Ca(NOCa(NO33))22

CC66HH1212OO66

(NH(NH44))33POPO44

Na 2 atoms x 22.9 = 45.8S 1 atom x 32.1 = 32.1 77.9 g/ mol

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Molar MassMolar Mass The number of grams of 1 mole The number of grams of 1 mole

of atoms, ions, or molecules.of atoms, ions, or molecules. We can make conversion factors We can make conversion factors

from these.from these.

–To change grams of a To change grams of a compound to moles of a compound to moles of a compound.compound.

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For exampleFor example

How many moles is 5.69 g of NaOH?How many moles is 5.69 g of NaOH?

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For exampleFor example

How many moles is 5.69 g of NaOH?How many moles is 5.69 g of NaOH?

5 69. g

Na = O =H =

22.916.01.0

X 1 atomX 1 atomX 1 atom

= 22.9= 16.0= 1.0=39.9 g

39.9 g

1 mole= 0.142 mole of NaOH

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ExamplesExamples How many moles is 4.56 g of COHow many moles is 4.56 g of CO22??

4.56 g CO2

C 1 atom x 12.0 = 12.0O 2 atoms x 16.0 = 32 .0 = 44.0 g

44.0 g CO2

1 mol CO2 0.104 mol CO2

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How many grams is 9.87 moles of HHow many grams is 9.87 moles of H22O?O?

9.87 mol H2O

1 mol H2O

H 2 atom x 1.0 = 2.0O 1 atom x 16.0 = 16.0 = 18.0 g

18.0 g H2O 178 g H2O

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How many molecules is 6.8 g of CHHow many molecules is 6.8 g of CH44??

6.8 g CH4

16.0 g CH4

C 1 atom x 12.0 = 12.0H 4 atoms x 1.0 = 4.0 16.0 g

1 mol CH4

1 mol CH4

6.02x1023 molecules CH4

= 2.6 x1023 molecules CH4

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49 molecules of C49 molecules of C66HH1212OO66 weighs how weighs how

much?much?

49 molecules C6H12O6

6.02x1023 molecules C6H12O6

1 mol C6H12O6

1 mol C6H12O6

C 6 atoms x 12.0 = 72.0H 12 atoms x 1.0 = 12.0O 6 atoms x 16.0 = 96.0 180 g

180 g C6H12O6

= 1.5 x 10 – 20 g C6H12O6

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GasesGases Many of the chemicals we deal with Many of the chemicals we deal with

are gases.are gases.–They are difficult to They are difficult to weighweigh..

Need to know how many moles of gas Need to know how many moles of gas we have.we have.

Two things effect the volume of a gasTwo things effect the volume of a gas–Temperature and pressureTemperature and pressure

We need to compare them at the same We need to compare them at the same temperature and pressure.temperature and pressure.

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Standard Temperature and Standard Temperature and PressurePressure

0ºC and 1 atm pressure0ºC and 1 atm pressure abbreviated abbreviated STPSTP At STP 1 mole of gas occupies At STP 1 mole of gas occupies

22.4 L22.4 L Called the Called the molar volumemolar volume 1 mole = 22.4 L of any gas at STP1 mole = 22.4 L of any gas at STP

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ExamplesExamples What is the volume of 4.59 mole What is the volume of 4.59 mole

of COof CO22 gas at STP? gas at STP?

4.59 mol CO2

1 mol CO2

22.4 L CO2 103 L CO2

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How many moles is 5.67 L of OHow many moles is 5.67 L of O2 2 at STP?at STP?

5.67 L O2

22.4 L O2

1 mol O2 0.253 mol O2

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What is the volume of 8.8 g of CHWhat is the volume of 8.8 g of CH44 gas gas

at STP?at STP?

8.8 g CH4

16.0 g CH4

C 1 atom x 12.0 = 12.0H 4 atoms x 1.0 = 4.0 16.0 g

1 mol CH4

1 mol CH4

22.4 L CH4 12 L CH4

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Density of a gasDensity of a gas D = m / VD = m / V–for a gas the units will be g / Lfor a gas the units will be g / L

We can determine the density of any We can determine the density of any gas at STP if we know its formula.gas at STP if we know its formula.

To find the density we need the mass To find the density we need the mass and the volume.and the volume.

If you assume you have 1 mole, then If you assume you have 1 mole, then the mass is the molar mass (from PT)the mass is the molar mass (from PT)

At STP the volume is 22.4 L.At STP the volume is 22.4 L.

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ExamplesExamples Find the density of COFind the density of CO2 2 at STP.at STP.

C 1 atom x 12.0 = 12.0O 2 atoms x 16.0 = 32.0 = 40.0 g

40.0 g CO2

1 mol CO2 22.4 L CO2

1 mol CO2 1.79 g CO2

L CO2

= 1.79 g / L CO2

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Find the density of CH4 at STP.

16.0 g CH4

C 1 atom x 12.0 = 12.0H 4 atoms x 1.0 = 4.0 16.0 g

1 mol CH4

1 mol CH4

22.4 L CH4

0.714 g CH4

L CH4

= 0.714 g / L CH4

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The other wayThe other way Given the density, we can find the molar Given the density, we can find the molar

mass of the gas.mass of the gas. Again, pretend you have 1 mole at STP, Again, pretend you have 1 mole at STP,

so V = 22.4 L.so V = 22.4 L. m = D x Vm = D x V m is the mass of 1 mole, since you have m is the mass of 1 mole, since you have

22.4 L of the stuff.22.4 L of the stuff. What is the molar mass of a gas with a What is the molar mass of a gas with a

density of 1.964 g/L?density of 1.964 g/L? 2.86 g/L?2.86 g/L?

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SummarySummary These four items are all equal:These four items are all equal:

a) 1 molea) 1 mole

b) molar mass (in grams)b) molar mass (in grams)

c) 6.02 x 10c) 6.02 x 102323 representative representative particlesparticles

d) 22.4 L at STPd) 22.4 L at STP

Thus, we can make conversion Thus, we can make conversion factors from them.factors from them.

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Section 6.3Section 6.3Percent Composition and Percent Composition and

Chemical FormulasChemical Formulas OBJECTIVES:OBJECTIVES:

–Calculate the percent Calculate the percent composition of a substance composition of a substance from its chemical formula or from its chemical formula or experimental data.experimental data.

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Section 6.3Section 6.3Percent Composition and Percent Composition and

Chemical FormulasChemical Formulas OBJECTIVES:OBJECTIVES:

–Derive the empirical formula Derive the empirical formula and the molecular formula of a and the molecular formula of a compound from experimental compound from experimental data.data.

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Calculating Percent Composition of Calculating Percent Composition of a Compounda Compound

Like all percent problems:Like all percent problems:

Part Part wholewhole

Find the mass of each Find the mass of each component,component,

then divide by the total mass.then divide by the total mass.

x 100 %

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Example:

What is the percent composition of H2O

H 2atoms x 1.01 = 2.02 g/molO 1atom x 16.00 = 16.00 g/mol

18.02 g/mol

% H = (2.02/18.02) x 100 = 11.2%

% O = (16.00/18.02) x 100 = 88.8 %

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ExampleExample Calculate the percent Calculate the percent

composition of a compound that composition of a compound that is 29.0 g of Ag with 4.30 g of S.is 29.0 g of Ag with 4.30 g of S.

29.0 g Ag + 4.30 g S = 33.3 g Total

% S = (4.30g / 33.3g) x 100 = 12.9%

% Ag = (29.0g / 33.3g) x 100 = 87.1%

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Getting it from the formulaGetting it from the formula If we know the formula, assume If we know the formula, assume

you have 1 mole.you have 1 mole. NaCl = I moleNaCl = I mole Then you know the mass of the Then you know the mass of the

pieces and the whole.pieces and the whole.

Na 1 x 22.99 = 22.99 g/molCl 1 x 35.45 = 35.45 g/mol

58.44 g/mol

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ExamplesExamples Calculate the percent composition Calculate the percent composition

of Cof C22HH44??

C 2 atoms x 12.0 = 24.0H 4 atoms x 1.0 = 4.0 28.0 g

%C = 24.0/28.0 x 100 = 85.7 % C

%H = 4.0 / 28.0 x 100 = 14.3 % H

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How about Aluminum carbonate?How about Aluminum carbonate?

–Sample Problem 6-13, p.132Sample Problem 6-13, p.132

Al 3+ and CO3 2- is written Al2(CO3)3

Al 2 atoms x 27.0 = 54.0C 3 atoms x 12.0 = 36.0O 9 atoms x 16.0 = 144.0 = 234.0 g

% Al = 54.0 / 234.0 x 100 = 23.08 % Al

% C = 36.0 / 234.0 x 100 = 15.38 % C

% O = 144.0 / 234.0 x 100 = 61.5 % O

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The Empirical FormulaThe Empirical Formula The lowest The lowest whole number ratiowhole number ratio of of

elements in a compound.elements in a compound. The molecular formula = the The molecular formula = the actual actual

ratio of elements in a compound.ratio of elements in a compound. The two The two cancan be the same. be the same. CHCH22 is an empirical formula is an empirical formula CC22HH44 is a molecular formula is a molecular formula CC33HH66 is a molecular formula is a molecular formula HH22O is both empirical & molecularO is both empirical & molecular

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Calculating EmpiricalCalculating Empirical Just find the lowest whole number ratioJust find the lowest whole number ratio CC66HH1212OO66 CHCH44NN It is not just the ratio of atoms, it is also It is not just the ratio of atoms, it is also

the ratio of moles of atoms.the ratio of moles of atoms. In 1 mole of COIn 1 mole of CO22 there is 1 mole of there is 1 mole of

carbon and 2 moles of oxygen.carbon and 2 moles of oxygen. In one molecule of COIn one molecule of CO22 there is 1 atom there is 1 atom

of C and 2 atoms of O.of C and 2 atoms of O.

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Calculating EmpiricalCalculating Empirical We can get a ratio from the We can get a ratio from the

percent composition.percent composition. Assume you have a 100 g.Assume you have a 100 g. The percentages become grams.The percentages become grams. Convert grams to moles. Convert grams to moles. Find lowest whole number ratio Find lowest whole number ratio

by dividing by the smallest.by dividing by the smallest.

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ExampleExample Calculate the empirical formula of a Calculate the empirical formula of a

compound composed of 38.67 % C, 16.22 % compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.H, and 45.11 %N.

Assume 100 g soAssume 100 g so 38.67 g C x 1mol C = 3.220 mole C 38.67 g C x 1mol C = 3.220 mole C

12.01 gC 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 16.22 g H x 1mol H = 16.09 mole H

1.01 gH1.01 gH 45.11 g N x 1mol N = 3.219 mole N 45.11 g N x 1mol N = 3.219 mole N

14.01 gN14.01 gN

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ExampleExample The ratio is 3.220 mol C = 1 mol CThe ratio is 3.220 mol C = 1 mol C

3.219 molN 1 3.219 molN 1 mol Nmol N

The ratio is 16.09 mol H = 5 mol HThe ratio is 16.09 mol H = 5 mol H 3.219 molN 1 3.219 molN 1

mol Nmol N

= C= C11HH55NN11

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A compound is 43.64 % P and A compound is 43.64 % P and 56.36 % O. What is the 56.36 % O. What is the empirical formula?empirical formula?

Empirical formulaAnd molar mass

P2O5

43.64g P

56.36g O

30.96g

1 mol 1.41 mol

16.00g

1mol 3.5 mol

1.41

1.41

1

2.5

203.88

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Empirical to molecularEmpirical to molecular Since the empirical formula is the Since the empirical formula is the

lowest ratio, the actual molecule lowest ratio, the actual molecule would weigh more.would weigh more.

By a whole number multiple.By a whole number multiple. Divide the actual molar mass by the Divide the actual molar mass by the

empirical formula mass.empirical formula mass.

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ExampleExample

Empirical formula and molar mass of LauriumLaurium is Li165C32O4 (1586.82g)

LauriumLaurium has a molar mass of 4760.46 g. has a molar mass of 4760.46 g. LauriumLaurium is known to be composed of is known to be composed of

71.65 % Li, 24.27% C and 4.07% O.. What is 71.65 % Li, 24.27% C and 4.07% O.. What is LauriumLaurium’s molecular formula?’s molecular formula?

71.65g Li71.65g Li

24.27g C24.27g C

4.07g O4.07g O

6.94g

12.01g

16.00g

1mol

1mol

1mol

10.32 mol

2.02 mol

0.25 mol

0.25 mol

0.25 mol

0.25 mol

1

8

41.25 X4 =165

X4 =32

X4 =4

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Molar mass of the molecular formula for Laurium is 4760.464760.46 gThe empirical formula of Laurium is Li165C32O4

Molar mass of the Laurium empirical formula is 1586.82 g g

Molar mass of molecular formula(given)

Molar mass of empirical formula(calculated)

Empirical formula Molecular Formula

=

4760.46 g4760.46 g

1586.82g= 3

Li165C32O4 Li495C96O12x3