Atomic Mass and the Mole

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 3.1
Schematic Diagram of a Mass Spectrometer

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 carbonThere are two isotopes of carbon
1212
C with a mass of 12.00000 amu (98.892%),C with a mass of 12.00000 amu (98.892%),
1313
C with a mass of 13.00335 amu (1.108%).C with a mass of 13.00335 amu (1.108%).
What is the average mass? . . .What is the average mass? . . .
(12.00000)(.98892) + (13.00335)(.01108) = 12.01amu(12.00000)(.98892) + (13.00335)(.01108) = 12.01amu

7
Examples of average atomic massExamples of average atomic mass pp
There are two isotopes of nitrogen , one withThere are two isotopes of nitrogen , one with
an atomic mass of 14.0031 amu and one with aan atomic mass of 14.0031 amu and one with a
mass of 15.0001 amu. What is the percentmass of 15.0001 amu. What is the percent
abundance of each? . . .abundance of each? . . .
HintHint: the abundances have to had up to 100%: the abundances have to had up to 100%
(or 1.00 in decimal form). So, set one of them(or 1.00 in decimal form). So, set one of them
equal to “x” and the other equal to 1-x . . .equal to “x” and the other equal to 1-x . . .
Where would you get the value of the averageWhere would you get the value of the average
atomic mass? . . .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 & oneTwo isotopes of N, one of 14.0031 amu & one
of 15.0001 amu. What’s % abundance? . . .of 15.0001 amu. What’s % 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 andof 14.0031 and 0.39%0.39% of 15.0001 . . .of 15.0001 . . .

9
Finding An Isotopic Mass pp
A sample of boron consists ofA sample of boron consists of 1010
B (massB (mass
10.0 amu) and10.0 amu) and 1111
B (mass 11.0 amu). If theB (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 boronwhat is the % abundance of each boron
isotope? . . .isotope? . . .

10
Assign X and Y values:Assign X and Y values: pp
X = %X = % 1010
B Y = %B Y = % 1111
BB
Determine Y in terms of XDetermine Y in terms of X
X + Y = 100X + Y = 100
Y = 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 100
10.0 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 % 1010
BB
- 1.0- 1.0
Y = 100 - XY = 100 - X
%% 1111
B = 100 - 20% = 80%B = 100 - 20% = 80% 1111
BB

12
Learning Check
Copper has two isotopes 63
Cu (62.9
amu) and 65
Cu (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% . . .
Solution
62.9X + 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.

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 · 2H· 2H22OO
16.04316.043
134.862134.862
226.095226.095
701.926701.926
172.12172.12

20
Calculating atoms, moles and mass
A silicon chip has a mass of 5.68 mg. HowA silicon chip has a mass of 5.68 mg. How
many atoms of Si are in the chip? (Usemany atoms of Si are in the chip? (Use
spider) . . .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 /1mol)atoms /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 improveCobalt is added to steel to improve
corrosion resistance. Calculate the molescorrosion resistance. Calculate the moles
and the mass of Co in a sample of Coand the mass of Co in a sample of Co
containing 5.00 x 10containing 5.00 x 102020
atoms of Co. . .atoms of Co. . .
Answers? . . .Answers? . . .
8.30 x 108.30 x 10-4-4
molesmoles Co; 4.89 x 10Co; 4.89 x 10-2-2
gg CoCo

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 aCalculate the percent composition of a
compound that is composed only of 29.0 gcompound 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% Ag87.1% 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 AlFind the percent composition of Al22(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 55.76%Ca 23.28%; S 18.62%; H 2.34%; O 55.76%

29
The Empirical Formula
The lowestThe lowest whole numberwhole number ratio of elements inratio of elements in
a compound.a compound.
The molecular formula is theThe molecular formula is the actualactual ratio ofratio 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 anis an empiricalempirical formulaformula
CC22HH44 is ais a molecularmolecular formulaformula
CC33HH66 is ais a molecularmolecular formulaformula
HH22O isO is bothboth..

30
Calculating Empirical
Just find the lowest whole number ratio
C6H12O6 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 percentWe 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 by4th, find lowest whole number ratio by
dividing each “subscript” by the smallest.dividing each “subscript” by the smallest.
5th, clear fractions by multiplying with the5th, clear fractions by multiplying with the
correct factor.correct factor.

32
Example pp
Calculate the empirical formula of aCalculate 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 C38.67 g C x 1mol C = 3.220 mole C
12.01 g C12.01 g C
16.22 g H x 1mol H = 16.09 mole H16.22 g H x 1mol H = 16.09 mole H
1.01 g H1.01 g H
45.11 g N x 1mol N = 3.219 mole N45.11 g N x 1mol N = 3.219 mole N
14.01 g N14.01 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) OO(56.36g/16.00 g/mol)(56.36g/16.00 g/mol)
PP1.411.41 OO3.523.52
PP1.41/1.411.41/1.41 OO3.52/1.413.52/1.41
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
.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
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 3.5
Schematic Diagram of the Combustion
Device Used to Analyze Substances for
Carbon and Hydrogen

41
Empirical Example
A 0.2000 gram sample of a compoundA 0.2000 gram sample of a compound
(vitamin C) composed of only C, H, and(vitamin C) composed of only C, H, and
O is burned completely with excess OO is burned completely with excess O22 ..
0.2998 g of CO0.2998 g of CO22 and 0.0819 g of Hand 0.0819 g of H22O areO are
produced. What is the empiricalproduced. 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 empiricalRatio of molar to empirical massmass willwill
tell you the moleculartell 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 andA compound is made of only sulfur and
oxygen. It is 69.6% S by mass. Its molaroxygen. It is 69.6% S by mass. Its molar
mass is 192 g/mol. What is its formula?mass is 192 g/mol. What is its formula?
Answer . . .Answer . . .
SS44OO44

44
3.6 Chemical Equations
Are sentences.Are sentences.
DescribeDescribe what happens in a chemicalwhat happens in a chemical
reaction.reaction.
ReactantsReactants →→ ProductsProducts
Equations should beEquations should be balanced.balanced.
Have the same number of eachHave the same number of each kindkind ofof
atoms on both sides because . . .atoms on both sides because . . .

45
3.7 Balancing equations pp
CH4 + O2 → CO2 + H2O
ReactantsReactants ProductsProducts
C1 1
O2 3
H4 2

46
Balancing equationsBalancing equations pp
CH4 + O2 → CO2 + 2H2O
C1 1
O2 3
H4 2 4

47
CH4 + O2 → CO2 + 2H2O
C1 1
O2 3
H4 2 4
4

48
CH4 + 2O2 → CO2 + 2H2O
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 P
H
O
2
2
2
1

51
Example pp
H2 + H2OO2
→
Changes the O
R P
H
O
2
2
2
1
2

52
Example pp
H2 + H2OO2
→
Also changes the H
R P
H
O
2
2
2
1
2
2

53
Example pp
H2 + H2OO2
→
Need twice as much H in the reactant
R P
H
O
2
2
2
1
2
2
4

54
Example pp
H2 + H2OO2
→
Recount
R P
H
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 P
H
O
2
2
2
1
2
2
4
2
4

56
Example
H2 + H2OO2
→
This is the answer
R P
H
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 →→ HH22O + CaO + Ca33(PO(PO44))22 Hint:Hint:
Start with most complicated molecule first.Start with most complicated molecule first.
Answer . . .Answer . . .
3Ca(OH)3Ca(OH)22 +2 H+2 H33POPO44 →→ 6H6H22O + CaO + Ca33(PO(PO44))22
Cr + SCr + S88 →→ CrCr22SS33 . . .. . .
16Cr + 3S16Cr + 3S88 →→ 8Cr8Cr22SS33
KClOKClO33((ss)) →→ KCl + OKCl + O22((gg))
2KClO2KClO33((ss)) → 2→ 2KCl + 3OKCl + 3O22((gg))

59
Practice
Solid iron(III) sulfide reacts with gaseousSolid 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 toA balanced equation can be used to
describe a reaction in molecules anddescribe a reaction in molecules and
atoms.atoms.
NotNot grams.grams.
Chemical reactions happen byChemical 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 (butor moles of molecules (but notnot byby
grams).grams).

61
3.8 Stoichiometry
Given an amount of either startingGiven an amount of either starting
materialmaterial oror product, can determine theproduct, can determine the
other quantities.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 Fe
2 mol Fe
3 mol Cu
= 0.272 mol Cu
0.272 mol Cu
1 mol Cu
63.55 g Cu
= 17.3 g Cu

64
Could have done it via “spider” pp
10.1 g Fe
55.85 g Fe
1 mol Fe
2 mol Fe
3 mol Cu
1 mol Cu
63.55 g Cu
= 17.3 g Cu

66
Examples pp
One way of producing OOne way of producing O22((gg) involves the) involves the
decomposition of potassium chlorate intodecomposition of potassium chlorate into
potassium chloride and oxygen gas. A 25.5potassium chloride and oxygen gas. A 25.5
g sample of Potassium chlorate isg 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 xA piece of aluminum foil 5.11 in x 3.23 in x
0.0381 in is dissolved in excess HCl (aq). How0.0381 in is dissolved in excess HCl (aq). How
many grams of Hmany grams of H22((gg) are produced? . . .) are produced? . . .
Al: find volume, convert to mL, find massAl: 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 + 6HCl2Al + 6HCl →→ 2AlCl2AlCl33 + 3H+ 3H22
Since the g of Al just happens to be the sameSince the g of Al just happens to be the same
as its molar mass and the mole ratio would beas its molar mass and the mole ratio would be
3/2, so get 3/2 moles H3/2, so get 3/2 moles H22 →→ 3 g H3 g H22

68
Examples pp
How many grams of Al are needed toHow many grams of Al are needed to
produce 15 grams of iron from theproduce 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)2Cl2can 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.php
Makes the least product.
Book shows you a ratio method.
It works.
So does mine

71
Limiting reagent pp
To determine the limiting reagentTo determine the limiting reagent
requires that you do two stoichiometryrequires that you do two stoichiometry
problems.problems.
Figure out how much product eachFigure out how much product each
reactant makes.reactant makes.
The one that makes theThe one that makes the leastleast is theis 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 areIf 10.1 g of magnesium and 2.87 L of HCl gas are
reacted, how many L of gas will be produced atreacted, how many L of gas will be produced at
STP? Grams of solid? Excess remaining? . . .STP? Grams of solid? Excess remaining? . . .
(Write balanced equation, then do 2 stoichs to find(Write balanced equation, then do 2 stoichs to find
LR) . . .LR) . . .
Mg + 2HClMg + 2HCl →→ MgClMgCl22 + 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
HClHClis the LRis the LR, so use, so use HClHClfor all calculations.for all calculations.

74
Your turn
If 10.1 g of magnesium and 2.87 L of HCl gasIf 10.1 g of magnesium and 2.87 L of HCl gas
are reacted,are reacted, how many liters of gas will behow many liters of gas will be
produced? Grams of solid?produced? Grams of solid? Excess remaining?Excess remaining?
Mg + 2HClMg + 2HCl →→ MgClMgCl22 + H+ H22. (. (HClHCl is the LRis the LR))
How manyHow 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) = 1.431.43
Liters of HLiters of H22 is produced.is produced.
How manyHow 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 g6.10 g
MgClMgCl22 is producedis produced..

75
Your turn
If 10.1 g of magnesium and 2.87 L of HCl gas areIf 10.1 g of magnesium and 2.87 L of HCl gas are
reacted, how many liters of gas will bereacted, how many liters of gas will be
produced? Grams of solid? XS remaining? . . .produced? Grams of solid? XS remaining? . . .
Mg + 2HClMg + 2HCl →→ MgClMgCl22 + H+ H22..
Mg is xs, so calc how much reacted with HCl.Mg is xs, so calc how much reacted with HCl.
(2.87 L HCl)(1 mol/22.4L)(1mol B/2 mol A)(24.3(2.87 L HCl)(1 mol/22.4L)(1mol B/2 mol A)(24.3
g) = 1.57 g Mgg) = 1.57 g Mg consumedconsumed. So, xs = . . .?. So, xs = . . .?
10.1 g - 1.57 g = 8.53 g Mg left over (but SF)10.1 g - 1.57 g = 8.53 g Mg left over (but SF)
8.5 g Mg remaining8.5 g Mg remaining

76
Example - Class does
Ammonia is produced by the followingAmmonia is produced by the following
reactionreaction: N: N22 + H+ H22 →→ NHNH33
What mass of NHWhat mass of NH33 can be produced from acan be produced from a
mixture of 100. g Nmixture of 100. g N22 and 500. g Hand 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 Hreacted 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 theThe amount of stuff you make is the
yield.yield.
TheThe theoretical yieldtheoretical yield is the amount youis the amount you
would make if everything went perfect.would make if everything went perfect.
TheThe actual yieldactual yield is what you make inis what you make in
the lab.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 Al6.78 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 →→ AlAl22(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 + 2SO2ZnO + 2SO22 IfIf
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 ZnSshould 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 SO22 x 86.78% = . . .x 86.78% = . . .
2794 g2794 g

81
Examples
Aluminum burns in bromine producingAluminum burns in bromine producing
aluminum bromide. In a laboratory 6.0 g ofaluminum bromide. In a laboratory 6.0 g of
aluminum reacts with excess bromine. 50.3aluminum reacts with excess bromine. 50.3
g of aluminum bromide are produced.g of aluminum bromide are produced.
What are the 3 types of yield & their values?What are the 3 types of yield & their values?
TheoreticalTheoretical = 59.4 g= 59.4 g
ActualActual = 50.3 g= 50.3 g
PercentPercent = 84.7%= 84.7%

82
Examples
Years of experience have proved that the percentYears of experience have proved that the percent
yield for the following reaction is 74.3%yield for the following reaction is 74.3%
Hg + BrHg + Br22 →→ HgBrHgBr22
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, howare reacted, how
muchmuch actualactual HgBrHgBr22 will be produced? Ans. . . .will be produced? Ans. . . .
10g Hg = .05 mole; 9g Br2 = .0562 moles; so Hg = LR10g Hg = .05 mole; 9g Br2 = .0562 moles; so Hg = LR
TY = 18.97 g; actual = .743 x TY = 14.10 gTY = 18.97 g; actual = .743 x TY = 14.10 g
If the reaction did go to completion, how muchIf 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

83
Example (lab 2)
Commercial brass is an alloy of Cu and Zn. ItCommercial brass is an alloy of Cu and Zn. It
reacts with HCl by the following reactionreacts with HCl by the following reaction
Zn(s) + 2HCl(aq)Zn(s) + 2HCl(aq) →→ ZnClZnCl22 (aq) + H(aq) + H22(g)(g)
Cu does not react. When 0.5065 g of brass isCu 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 areare
eventually isolated.eventually isolated. What is the composition ofWhat is the composition of
the brass?the brass? . . .. . .
Find g of Zn using stoich, then do % comp for Zn,Find g of Zn using stoich, then do % comp for Zn,
then 100% - %Zn to get % for Cuthen 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 AgYou will do something similar to this (using Ag
and Cu) with Lab #1.and Cu) with Lab #1.

Atomic Mass and the Mole

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