1

2

Section 3.1 Atomic Mass Atoms are so small, it is difficult to

discuss how much they weigh in grams. Use atomic mass units. An atomic mass unit (amu) is one

twelfth the mass of a carbon-12 atom. This gives us a basis for comparison. The decimal numbers on the table are

atomic masses in amu.

3

Figure 3.1Schematic Diagram of a Mass

Spectrometer

4

Figure 3.2Neon Gas

5

They are not whole numbers Because they are based on averages

of atoms and of isotopes. can figure out the average atomic

mass from the mass of the isotopes and their relative abundance.

add up the weighted percents as decimals times the masses of the isotopes.

%(m1) + %(m2) + . . . = Avg. mass

6

Examples of average mass There are two isotopes of carbon There are two isotopes of carbon

1212C with a mass of 12.00000 C with a mass of 12.00000

amu (98.892%), amu (98.892%), 1313C with a mass of C with a mass of 13.00335 amu (1.108%). What is the 13.00335 amu (1.108%). What is the average mass? . . .average mass? . . .

(12.00000)(.98892) + (13.00335)(.01108) = (12.00000)(.98892) + (13.00335)(.01108) = 12.01amu12.01amu

7

Examples of average atomic massExamples of average atomic mass pp

There are two isotopes of nitrogen , one There are two isotopes of nitrogen , one with an atomic mass of 14.0031 amu and with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is one with a mass of 15.0001 amu. What is the percent abundance of each? . . .the percent abundance of each? . . .

HintHint: the abundances have to had up to : the abundances have to had up to 100% (or 1.00 in decimal form). So, set 100% (or 1.00 in decimal form). So, set one of them equal to “x” and the other one of them equal to “x” and the other equal to 1-x . . .equal to 1-x . . .

Where would you get the value of the Where would you get the value of the average atomic mass? . . .average atomic mass? . . .

From the periodic table = 14.007From the periodic table = 14.007 Now, set up the problem . . .Now, set up the problem . . .

8

Examples pp

Two isotopes of N, one of 14.0031 amu Two isotopes of N, one of 14.0031 amu & one of 15.0001 amu. What’s % & one of 15.0001 amu. What’s % abundance? . . .abundance? . . .

14.0031(x) + 15.0001(1-x) = ?14.0031(x) + 15.0001(1-x) = ? You get . . .You get . . . 99.61%99.61% of 14.0031 and of 14.0031 and 0.39%0.39% of of

15.0001 . . .15.0001 . . .

9

Finding An Isotopic Mass pp

A sample of boron consists of A sample of boron consists of 1010B (mass B (mass 10.0 amu) and 10.0 amu) and 1111B (mass 11.0 amu). If the B (mass 11.0 amu). If the average atomic mass of B is 10.8 amu, average atomic mass of B is 10.8 amu, what is the % abundance of each boron what is the % abundance of each boron isotope? . . .isotope? . . .

10

Assign X and Y values:Assign X and Y values: pp

X = % X = % 1010B Y = % B Y = % 1111B B Determine Y in terms of XDetermine Y in terms of X

X + Y = 100X + Y = 100Y = 100 - XY = 100 - X

Solve for X:Solve for X:X (10.0)X (10.0) + + (100 - X )(11.0) (100 - X )(11.0) = 10.8 = 10.8

100 100100 100

Multiply through by 100Multiply through by 10010.0 X + 1100 - 11.0X = 108010.0 X + 1100 - 11.0X = 1080

11

Collect X termsCollect X terms pp

10.0 X - 11.0 X = 1080 - 110010.0 X - 11.0 X = 1080 - 1100

- 1.0 X = -20 - 1.0 X = -20

X = X = -20-20 = 20 % = 20 % 1010BB

- 1.0- 1.0

Y = 100 - XY = 100 - X

% % 1111B = 100 - 20% = 80% B = 100 - 20% = 80% 1111BB

12

Learning Check

Copper has two isotopes 63Cu (62.9 amu) and 65Cu (64.9 amu). What is the % abundance of each isotope? (Hint: Check periodic table for atomic mass)

1) 30% 2) 70% 3) 100%

13

Solution AT8

2) 70% . . .

Solution62.9X + 64.90(100-x) = 63.50

100 100

62.9x + 64.90(100-x) = 6350

-2.0 X = -140

X = 70%

14

3.2 The Mole The mole is a number. A very large number, but still, just

a number. 6.022 x 1023 of anything is a mole A large dozen. The number of atoms in exactly 12

grams of carbon-12.

15

The Mole Makes the numbers on the periodic

table the mass of the average atom.

16

More Stoichiometry

17

3.3 Molar mass Mass of 1 mole of a substance. Often called molecular weight. To determine the molar mass of an

element, look on the periodic table. To determine the molar mass of a

compound, add up the molar masses of the elements that make it up.

18

Find the molar mass of CHCH44

MgMg33PP22

Ca(NOCa(NO33))33

AlAl22(Cr(Cr22OO77))33

CaSOCaSO44 · ·

2H2H22OO

16.04316.043

134.862134.862

226.095226.095

701.926701.926

172.12172.12

19

20

Calculating atoms, moles and mass

A silicon chip has a mass of 5.68 mg. A silicon chip has a mass of 5.68 mg. How many atoms of Si are in the chip? How many atoms of Si are in the chip? (Use spider) . . .(Use spider) . . .

(5.68 mg)(1 g/1000 mg)(1 mol /28.086 g)(6.02 x 10(5.68 mg)(1 g/1000 mg)(1 mol /28.086 g)(6.02 x 102323 atoms atoms /1mol)/1mol)

= 1.22 x 10= 1.22 x 102020 atoms of Silicon. atoms of Silicon.

21

Calculating atoms, moles and mass

Cobalt is added to steel to improve Cobalt is added to steel to improve corrosion resistance. Calculate the corrosion resistance. Calculate the moles and the mass of Co in a sample moles and the mass of Co in a sample of Co containing 5.00 x 10of Co containing 5.00 x 102020 atoms of atoms of Co. . .Co. . .

Answers? . . .Answers? . . . 8.30 x 108.30 x 10-4-4 molesmoles Co; 4.89 x 10 Co; 4.89 x 10-2-2 gg Co Co

22

3.4 Percentage Composition Like all percents Part x 100%

whole Strategy . . . Find the mass of each component, Divide by the total mass. Multiply by 100%

23

Example pp

Calculate the percent composition of a Calculate the percent composition of a compound that is composed only of 29.0 g compound that is composed only of 29.0 g of Ag and 4.30 g of S.of Ag and 4.30 g of S.

The answer is . . . The answer is . . . 87.1% Ag 87.1% Ag from [29.0 ÷ (29.0 + 4.30)] x 100%.from [29.0 ÷ (29.0 + 4.30)] x 100%.

What is the % of S? . . . What is the % of S? . . . 12.9% (must add up to 100%).12.9% (must add up to 100%).

24

Getting % Composition from the formula

If we know the formula, then we assume we have 1 mole.

From that we can then figure out the pieces and the whole.

25

Example pp

Calculate the percent composition of C2H4?

Assume one mole. How many moles of carbon in that one mole? How many grams of carbon is that? How many moles of hydrogen? How many grams of hydrogen is that? What is total grams of C2H4? What is % comp. of carbon? Of hydrogen? . . . 85.60% carbon, 14.40% hydrogen.

26

Example Calculate the percent of aluminum in

Aluminum sulfate. . . Figure out the formula, which is . . . Al2(SO4)3 (M = 342 g/ mol) Then do the procedure to get an answer

of . . . 15.77% -- Did you get half of this? If so, you forgot there were two

aluminums in the formula.

27

3.4 Percent Composition Review pp

Percent of each element a compound is composed of.

Find the mass of each element, divide by the total mass, multiply by 100.

Easiest if use one mole of the compound.

Find the percent composition of CH4 . . . 12/16 x 100 = 75% of carbon, so H is

100% - 75% = 25%

28

3.4 Percent Composition Review

Try on your ownTry on your own Find the percent composition of Find the percent composition of

AlAl22(Cr(Cr22OO77))33 Al 7.69%; Cr 44.45%; O 47.87%Al 7.69%; Cr 44.45%; O 47.87% % Composition of CaSO% Composition of CaSO44 · 2H · 2H22OO Ca 23.28%; S 18.62%; H 2.34%; O Ca 23.28%; S 18.62%; H 2.34%; O

55.76%55.76%

29

The Empirical Formula The lowest The lowest whole numberwhole number ratio of elements in ratio of elements in

a compound.a compound. The molecular formula is the The molecular formula is the actualactual ratio of ratio of

elements in a compound.elements in a compound. The two can be the same (but may not be). The two can be the same (but may not be). CHCH22 is an is an empiricalempirical formula formula CC22HH44 is a is a molecularmolecular formula formula CC33HH66 is a is a molecularmolecular formula formula HH22O is O is bothboth..

30

Calculating EmpiricalJust find the lowest whole number ratioC6H12O6 is ?

CH3N is? It is not just the ratio of atoms, it is also the

ratio of moles of atoms.

In 1 mole of CO2 there is 1 mole of carbon

and 2 moles of oxygen. In one molecule of CO2 there is 1 atom of

C and 2 atoms of O.

31

Calculating Empirical pp

We can get the ratio from percent We can get the ratio from percent composition.composition.

1st, assume you have 100 g.1st, assume you have 100 g. 2nd, the percentages become grams.2nd, the percentages become grams. 3rd, turn grams to moles (spider). 3rd, turn grams to moles (spider). 4th, find lowest whole number ratio by 4th, find lowest whole number ratio by

dividing each “subscript” by the smallest.dividing each “subscript” by the smallest. 5th, clear fractions by multiplying with the 5th, clear fractions by multiplying with the

correct factor.correct factor.

32

Example pp

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

1.01 g H 1.01 g H 45.11 g N x 1mol N = 3.219 mole N 45.11 g N x 1mol N = 3.219 mole N

14.01 g N 14.01 g N

33

Example pp

The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N

The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1

mol N

C1H5N1

CH5N

34

Short cut way pp

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

PP(43.64 g/30.97 g/mol)(43.64 g/30.97 g/mol) O O(56.36g/16.00 g/mol)(56.36g/16.00 g/mol)

PP1.411.41 O O3.523.52

PP1.41/1.411.41/1.41 O O3.52/1.413.52/1.41

PP11OO2.502.50

Multiply by 2 to “clear the fraction.”Multiply by 2 to “clear the fraction.” We get PWe get P22OO55

35

Clearing the Fractions pp

If after dividing each ratio by the smallest you get a fraction and . . .

The last two decimals are approximately .33, .67, .25, .50, .75 or .20, then multiply all the ratios as follows . . .

36

Clearing the Fractions pp

.33 = 1/3 multiply by 3 .67 = 2/3 3/2 .25 = 1/4 4 .50 = 1/2 5 .75 = 3/4 4/3 .20 = 1/5 5

37

Clearing the Fractions pp

Example, get to X4Y1.67Z2

Multiply each ratio by what? . . . 3/2 (the reciprocal of 2/3, which is .67) You get what? . . . X6Y2.5Z3 so now you have to multiply by

what? . . . Multiply by 2 to get what? . . . X12Y5Z6

38

Working backwards From percent composition, you can

determine the empirical formula. Empirical Formula the lowest ratio of

atoms in a molecule. Based on mole ratios. A sample is 59.53% C, 5.38%H,

10.68%N, and 24.40%O. What is its empirical formula? Ans. . .

C13H14N2O4

39

Pure O2 in CO2 is absorbed

H2O is absorbed

Sample is burned completely to form CO2 and H2O

40

Figure 3.5Schematic Diagram of the

Combustion Device Used to Analyze Substances for Carbon and

Hydrogen

41

Empirical Example A 0.2000 gram sample of a A 0.2000 gram sample of a

compound (vitamin C) composed of compound (vitamin C) composed of only C, H, and O is burned only C, H, and O is burned completely with excess Ocompletely with excess O22 . 0.2998 . 0.2998

g of COg of CO22 and 0.0819 g of H and 0.0819 g of H22O are O are

produced. What is the empirical produced. What is the empirical formula? Answer . . . formula? Answer . . .

CC33HH44OO33

42

3.5 Determinine the Formula of a Compound: Empirical To Molecular

Formulas Empirical is lowest ratio.Empirical is lowest ratio. Molecular is actual molecule.Molecular is actual molecule. Need Molar mass also to solve this.Need Molar mass also to solve this. Ratio of molar to empirical Ratio of molar to empirical massmass will will

tell you the molecular tell you the molecular formulaformula.. Must be a whole number because...Must be a whole number because...

43

Example A compound is made of only sulfur A compound is made of only sulfur

and oxygen. It is 69.6% S by mass. and oxygen. It is 69.6% S by mass. Its molar mass is 192 g/mol. What Its molar mass is 192 g/mol. What is its formula? Answer . . . is its formula? Answer . . .

SS44OO44

44

3.6 Chemical Equations Are sentences.Are sentences. DescribeDescribe what happens in a what happens in a

chemical reaction.chemical reaction. Reactants Reactants Products Products Equations should be Equations should be balanced.balanced. Have the same number of each Have the same number of each kindkind

of atoms on both sides because . . .of atoms on both sides because . . .

45

3.7 Balancing equations pp

CH4 + O2 CO2 + H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2

46

Balancing equationsBalancing equations pp

CH4 + O2 CO2 + 2H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

47

Balancing equationsBalancing equations pp

CH4 + O2 CO2 + 2H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

4

48

Balancing equationsBalancing equations pp

CH4 + 2O2 CO2 + 2H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

44

49

Another Example

H2 + H2OO2

Make a table to keep track of where you are at

50

Example pp

H2 + H2OO2

Need twice as much O in the product

R PH

O

2

2

2

1

51

Example pp

H2 + H2OO2

Changes the O

R PH

O

2

2

2

1

2

52

Example pp

H2 + H2OO2

Also changes the H

R PH

O

2

2

2

1

2

2

53

Example pp

H2 + H2OO2

Need twice as much H in the reactant

R PH

O

2

2

2

1

2

2

4

54

Example pp

H2 + H2OO2

Recount

R PH

O

2

2

2

1

2

2

4

2

55

Example pp

H2 + H2OO2

The equation is balanced, has the same number of each kind of atom on both sides

R PH

O

2

2

2

1

2

2

4

2

4

56

Example

H2 + H2OO2

This is the answer

R PH

O

2

2

2

1

2

2

4

2

4

Not this

57

Abbreviations (s) (g) (aq) heat

catalyst

58

Practice Ca(OH)Ca(OH)22 + H + H33POPO44 H H22O + CaO + Ca33(PO(PO44))22

Hint: Start with most complicated molecule Hint: Start with most complicated molecule first. Answer . . . first. Answer . . .

3Ca(OH)3Ca(OH)22 +2 H +2 H33POPO44 6H 6H22O + CaO + Ca33(PO(PO44))22

Cr + SCr + S88 Cr Cr22SS33 . . . . . . 16Cr + 3S16Cr + 3S88 8Cr 8Cr22SS33

KClOKClO33((ss) ) KCl + OKCl + O22((gg))

2KClO2KClO33((ss) ) KCl + 3OKCl + 3O22((gg))

59

Practice Solid iron(III) sulfide reacts with gaseous Solid iron(III) sulfide reacts with gaseous

hydrogen chloride to form solid iron(III) hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.chloride and hydrogen sulfide gas.

FeFe22SS33 + 6HCl + 6HCl 2FeCl2FeCl33 + 3H + 3H22SS FeFe22OO33((ss) + Al() + Al(ss) ) Fe( Fe(ss) + Al) + Al22OO33((ss))

FeFe22OO33((ss) + 2Al() + 2Al(ss) ) 2Fe( 2Fe(ss) + Al) + Al22OO33((ss))

60

Meaning A balanced equation can be used A balanced equation can be used

to describe a reaction in molecules to describe a reaction in molecules and atoms.and atoms.

NotNot grams. grams. Chemical reactions happen by Chemical reactions happen by

molecules at a time . . . molecules at a time . . . or dozens of molecules at a time . . or dozens of molecules at a time . .

. . or moles of molecules (but or moles of molecules (but notnot by by

grams). grams).

61

3.8 Stoichiometry Given an amount of either starting Given an amount of either starting

material material oror product, can determine product, can determine the other quantities.the other quantities.

use conversion factors fromuse conversion factors from– molar mass (g - mole)molar mass (g - mole)– balanced equation (mole - mole)balanced equation (mole - mole)

keep track (use “spider equation”).keep track (use “spider equation”).

62

For example... pp

If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? . . .

Fe + CuSO4 Fe2(SO4)3 + Cu

(unbalanced) 2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu 10.1 g Fe

55.85 g Fe

1 mol Fe= 0.181 mol Fe

63

2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu pp

0.181 mol Fe2 mol Fe

3 mol Cu= 0.272 mol Cu

0.272 mol Cu1 mol Cu63.55 g Cu

= 17.3 g Cu

64

Could have done it via “spider” pp

10.1 g Fe55.85 g Fe1 mol Fe

2 mol Fe3 mol Cu

1 mol Cu63.55 g Cu

= 17.3 g Cu

65

pp

66

Examples pp

One way of producing OOne way of producing O22((gg) involves the ) involves the decomposition of potassium chlorate decomposition of potassium chlorate into potassium chloride and oxygen gas. into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is A 25.5 g sample of Potassium chlorate is decomposed. How many moles of Odecomposed. How many moles of O22(g) (g) are produced? Ans. are produced? Ans.

0.314 moles O20.314 moles O2 How many grams of potassium chloride?How many grams of potassium chloride? 15.51g KCl15.51g KCl How many grams of oxygen?How many grams of oxygen? 9.98g O9.98g O22

67

Examples pp

A piece of aluminum foil 5.11 in x 3.23 in x A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl (aq). 0.0381 in is dissolved in excess HCl (aq). How many grams of HHow many grams of H22((gg) are produced? . . ) are produced? . . ..

Al: find volume, convert to mL, find mass Al: find volume, convert to mL, find mass using density 2.702 g/ml to get . . .using density 2.702 g/ml to get . . .

27.84 g. Now write the equation . . .27.84 g. Now write the equation . . . 2Al + 6HCl 2Al + 6HCl 2AlCl 2AlCl33 + 3H + 3H22

Since the g of Al just happens to be the Since the g of Al just happens to be the same as its molar mass and the mole ratio same as its molar mass and the mole ratio would be 3/2, so get 3/2 moles Hwould be 3/2, so get 3/2 moles H22 3 g H 3 g H22

68

Examples pp

How many grams of Al are needed to How many grams of Al are needed to produce 15 grams of iron from the produce 15 grams of iron from the following reaction? following reaction? FeFe22OO33((ss) + Al() + Al(ss) ) Fe( Fe(ss) + Al) + Al22OO33((ss) )

Ans . . .Ans . . . 7.2 g Al7.2 g Al

69

Examples if time pp

K2PtCl4(aq) + NH3(aq)

Pt(NH3)2Cl2 (s)+

KCl(aq) what mass of Pt(NH3)2Cl2 can be

produced from 65 g of K2PtCl4 ?

How much KCl will be produced? How much from 65 grams of NH3?

70

3.9 Limiting Reagent pp

Reactant that determines the amount of product formed.

See p. 116 # 3 (assigned as HW) The one you run out of first. http://www.chemcollective.org/tutorials.ph

p

Makes the least product. Book shows you a ratio method. It works. So does mine

71

Limiting reagent pp

To determine the limiting reagent To determine the limiting reagent requires that you do two requires that you do two stoichiometry problems.stoichiometry problems.

Figure out how much product each Figure out how much product each reactant makes.reactant makes.

The one that makes the The one that makes the leastleast is the is the limiting reagent.limiting reagent.

72

If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? (steps to follow) pp

2Cu + S Cu2S

10.6 g Cu 63.55g Cu 1 mol Cu

2 mol Cu 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.07g S 1 mol S

1 mol S 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

= 13.3 g Cu2S

Cu is Limiting Reagent

73

Your turn If 10.1 g of magnesium and 2.87 L of HCl gas If 10.1 g of magnesium and 2.87 L of HCl gas

are reacted, how many L of gas will be produced are reacted, how many L of gas will be produced at STP? Grams of solid? Excess remaining? . . .at STP? Grams of solid? Excess remaining? . . .

(Write balanced equation, then do 2 stoichs to (Write balanced equation, then do 2 stoichs to find LR) . . .find LR) . . .

Mg + 2HCl Mg + 2HCl MgCl MgCl22 + H + H22 LR is . . . LR is . . . (10.1 g (10.1 g MgMg)(1 mol/24.3 g)(1/1)(22.4 L/mol) = )(1 mol/24.3 g)(1/1)(22.4 L/mol) =

9.3 L H9.3 L H22

(2.87 L (2.87 L HClHCl)(1 mol/22.4 L)(1/2)(22.4 L/mol) = )(1 mol/22.4 L)(1/2)(22.4 L/mol) = 1.43 L H1.43 L H22

HClHCl is the LRis the LR, so use , so use HClHCl for all calculations.for all calculations.

74

Your turn If 10.1 g of magnesium and 2.87 L of HCl If 10.1 g of magnesium and 2.87 L of HCl

gas are reacted, gas are reacted, how many liters of gas how many liters of gas will be produced? Grams of solid?will be produced? Grams of solid? Excess Excess remaining?remaining?

Mg + 2HCl Mg + 2HCl MgCl MgCl22 + H + H22. (. (HClHCl is the LRis the LR)) How many How many liters of gas produced?liters of gas produced? . . . . . . (2.87 L)(1 mol/22.4 l)(1/2)(22.4 L/mol) = (2.87 L)(1 mol/22.4 l)(1/2)(22.4 L/mol) =

1.43 Liters of H1.43 Liters of H2 2 is produced.is produced. How many How many grams of solid produced?grams of solid produced? . . . . . . (2.87 L)(1 mol/22.4L)(1/2)(95.2 g/mol) = (2.87 L)(1 mol/22.4L)(1/2)(95.2 g/mol) =

6.10 g MgCl6.10 g MgCl22 is produced is produced..

75

Your turn If 10.1 g of magnesium and 2.87 L of HCl If 10.1 g of magnesium and 2.87 L of HCl

gas are reacted, how many liters of gas will gas are reacted, how many liters of gas will be produced? Grams of solid? XS remaining? be produced? Grams of solid? XS remaining? . . .. . .

Mg + 2HCl Mg + 2HCl MgCl MgCl22 + H + H22. . Mg is xs, so calc how much reacted with Mg is xs, so calc how much reacted with

HCl.HCl. (2.87 L HCl)(1 mol/22.4L)(1mol B/2 mol A)(2.87 L HCl)(1 mol/22.4L)(1mol B/2 mol A)

(24.3 g) = 1.57 g Mg (24.3 g) = 1.57 g Mg consumedconsumed. So, xs . So, xs = . . .?= . . .?

10.1 g - 1.57 g = 8.53 g Mg left over (but 10.1 g - 1.57 g = 8.53 g Mg left over (but SF)SF)

8.5 g Mg remaining8.5 g Mg remaining

76

Example - Class does Ammonia is produced by the following Ammonia is produced by the following

reactionreaction : N: N22 + H + H22 NH NH3 3

What mass of NHWhat mass of NH33 can be produced from can be produced from a mixture of 100. g Na mixture of 100. g N22 and 500. g H and 500. g H22? ?

Ans. . . .Ans. . . . 121g NH121g NH33 (Did you balance the reaction?) (Did you balance the reaction?) How much unreacted material remains?How much unreacted material remains? 21.4 g H21.4 g H22 reacted so 478 g H reacted so 478 g H22 remain. remain.

77

Excess Reagent The reactant you don’t run out of.The reactant you don’t run out of. The amount of stuff you make is the The amount of stuff you make is the

yield.yield. The The theoretical yield theoretical yield is the amount is the amount

you would make if everything went you would make if everything went perfect.perfect.

The The actual yield actual yield is what you make is what you make in the lab.in the lab.

78

Percent Yield

% yield = Actual x 100% Theoretical

% yield = what you got x 100% what you could have got

79

Example pp

6.78 g of copper is produced when 3.92 g of Al 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.are reacted with excess copper (II) sulfate.

2Al + 3 CuSO2Al + 3 CuSO44 Al Al22(SO(SO44))33 + 3Cu + 3Cu

What is the actual yield? (read question) What is the actual yield? (read question) What is the theoretical yield? (do stoich) . . .What is the theoretical yield? (do stoich) . . . (3.92g A)(1mol/27g A)(3/2)(63.5 g B/mol) = 13.84 g Cu(3.92g A)(1mol/27g A)(3/2)(63.5 g B/mol) = 13.84 g Cu

What is the percent yield? (use formula)What is the percent yield? (use formula) AY = 6.78 g; TY = 13.84 g (Al is LR)AY = 6.78 g; TY = 13.84 g (Al is LR) % yield = (6.78/13.84)(100) = 49.0%.% yield = (6.78/13.84)(100) = 49.0%.

80

Another Example 2ZnS + 3O2ZnS + 3O22 2ZnO + 2SO 2ZnO + 2SO22 If If

the typical yield is 86.78%, how much SOthe typical yield is 86.78%, how much SO22

should actually be expected if 4897 g of ZnS should actually be expected if 4897 g of ZnS are used with an excess of Oare used with an excess of O22??

First, calculate theoretical yield (stoich) . . .First, calculate theoretical yield (stoich) . . . ((4897)(1/97.46)(2/2)(64.07/1) = 3219 g SO4897)(1/97.46)(2/2)(64.07/1) = 3219 g SO22

Solve for actual yieldSolve for actual yield Since %Y = AY/TY, AY = %Y x TYSince %Y = AY/TY, AY = %Y x TY So, 3219 g SOSo, 3219 g SO2 2 x 86.78% = . . .x 86.78% = . . . 2794 g2794 g

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Examples Aluminum burns in bromine producing Aluminum burns in bromine producing

aluminum bromide. In a laboratory 6.0 g of aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. 50.3 g of aluminum bromide are produced.

What are the 3 types of yield & their What are the 3 types of yield & their values?values?

TheoreticalTheoretical = 59.4 g = 59.4 g ActualActual = 50.3 g = 50.3 g PercentPercent = 84.7% = 84.7%

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Examples Years of experience have proved that the Years of experience have proved that the

percent yield for the following reaction is 74.3%percent yield for the following reaction is 74.3%Hg + BrHg + Br22 HgBr HgBr22

If 10.0 g of Hg and 9.00 g of BrIf 10.0 g of Hg and 9.00 g of Br22 are reacted, how are reacted, how

much much actualactual HgBr HgBr22 will be produced? Ans. . . . will be produced? Ans. . . .

10g Hg = .05 mole; 9g Br2 = .0562 moles; so Hg 10g Hg = .05 mole; 9g Br2 = .0562 moles; so Hg = LR TY = 18.97 g; actual = .743 x TY = 14.10 g= LR TY = 18.97 g; actual = .743 x TY = 14.10 g

If the reaction did go to completion, how much If the reaction did go to completion, how much excess reagent would be left?excess reagent would be left?

Xs left = .0062 mol Br2 x 160 g/mol Br2 = .992 gXs left = .0062 mol Br2 x 160 g/mol Br2 = .992 g

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Example (lab 2) Commercial brass is an alloy of Cu and Zn. It Commercial brass is an alloy of Cu and Zn. It

reacts with HCl by the following reaction reacts with HCl by the following reaction Zn(s) + 2HCl(aq) Zn(s) + 2HCl(aq) ZnCl ZnCl22 (aq) + H (aq) + H22(g)(g)

Cu does not react. When 0.5065 g of brass is Cu does not react. When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnClreacted with excess HCl, 0.0985 g of ZnCl22 are eventually isolated. are eventually isolated. What is the What is the composition of the brass?composition of the brass? . . . . . .

Find g of Zn using stoich, then do % comp for Find g of Zn using stoich, then do % comp for Zn, then 100% - %Zn to get % for CuZn, then 100% - %Zn to get % for Cu

Ans. 9.32% Zn (from 0.047g Zn); 90.68% CuAns. 9.32% Zn (from 0.047g Zn); 90.68% Cu You will do something similar to this (using Ag You will do something similar to this (using Ag

and Cu) with Lab #1.and Cu) with Lab #1.