Moles and Stoichiometry
AP chemistry
CCHS, Mr. Bartelt’s class
Mass spectroscopy
 Information found on pg 82
What’s amu
 Amu stands for atomic mass unit.
 1 amu is defined as 1/12th of the mass of the carbon—12
nucleus.
 If you recall from nuclear chemistry, as the size of the nucleus
increases some of the mass is converted into energy to keep
the nucleus from being blown apart by the repulsive force
between the protons.
 This causes the mass number (protons and neutrons) to not
correspond perfectly with amus. Don’t worry too much
about this. Amus will be given when they are needed.
Calculating atomic masses
 Look on pg83-85
 Chlorine has two natural isotopes:
 Chlorine—35 (75.77% in nature)
 Chlorine—37 (24.23% in nature)
Determining the molar mass is simple. It’s a mater of
performing a weighted average. You do it ever semester when
you calculate your final grade. Look
 Semester average 86.4 (80% of final grade)
 Final exam 96.0 (20% of final grade)
So you plug: 86.4*.8+96.0*.2 into your calculator and find your
final average to be 88.32. Sorry, you got a “B”
Calculating atomic masses
 So think of chlorine just like your grade
 Chlorine has two natural isotopes:
 Chlorine—35 (75.77% in nature)
 Chlorine—37 (24.23% in nature)
So you plug: 35*.7577+.37*.2423 into your calculator and find
the atomic weight to be 35.48. Which is close to the value on
the periodic table 35.453
NOTE: If you use mass number’s you be a little off because amu
and mass number are not the same thing. Amus for the
different isotopes will be given if you need them
The mole
 Pages 90-93
 The molar mass that you look up on the periodic table is the
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mass in grams that one mole will weigh.
A mole is defined as 6.022*1023 (it has no units)
A mole is like a dozen (12) in that it’s a counting unit. A
mole however is significantly larger.
If you gave away a million dollars every second for the last
4.5 billion years (since the earth formed) you would have
only given a way ¼ of a mole of dollars.
A mole is HUGE
Conversions
 How many moles are in 1.204*1023 molecules of O2?
1 mol
1.204 *10 molecules O 2 *
 .2000 mol O 2
23
6.022 *10 molecules
23
 How much would that weigh in grams?
 First calculate the molar mass of O2
Molar mass of O  16.00 g/mol
2 *16.00 g/mol  32.00 g/mol
 Next find the mass
.2000 mol O 2 *
32.00 g
 6.400 g O 2
1 mol
Sample problems (work in groups)
 Find the mass of 1.5*1019 atoms of argon.
 How many moles of sugar (C12H22O11) are in 200.0 grams
of sugar?
 How many alcohol (C2H5OH) molecules are in 14.5 pounds?
 What weighs more 400. atoms of krypton or 900. molecules
of N2?
% composition
 This is easy (93-95)
 I’ll show you. Here’s how you find the % composition of
cocaine (C17H21NO4)
 First, find the mass contribution of all the elements and the
total mass
Carbon 12.01 g/mol * 17 
204.2 g/mol
Nitrogen 14.01 g/mol * 1 
Oxygen 16.00 g/mol * 4 
14.01 g/mol
64.00 g/mol
Hydrogen 1.008 g/mol * 21  21.17 g/mol
Total  303.4 g / mol
% composition continued
 Second, find the percent by dividing each part by the whole
204.2 g/mol
Carbon
 0.673 *100  67.3%
303.4g/mol
14.01 g/mol
Nitrogen
 0.0462 *100  4.62%
303.4g/mol
64.00 g/mol
Oxygen
 0.211*100  21.1%
303.4g/mol
21.17 g/mol
Hydrogen
 0.0698 *100  6.98%
303.4g/mol
 That’s it
Practice
 Find the % composition of octane (C8H18)
 Find the % composition of MDMA (C11H15NO2)
 Find the % composition of prozac (C17H18F3NO)
Empirical formula from % composition
 Pg 96-102
 It is possible to determine the empirical formula from %
composition data.
 A compound is found to be 20.00% C, 53.30% O, 23.34%
N, 3.36% H by mass. What is the empirical formula?
 This seems like a difficult problem but it’s easy once you
learn the technique for solving it.
 First assume that you have a 100.0 gram sample. This causes
the % signs to become mass in grams
Empirical formula from % composition
20.00g C, 53.30g O, 23.34g N, 3.36g H
Next, convert grams into moles
1 mol
20.00 g C *
 1.666 mol C
12.01g
1 mol
53.30 g O *
 3.331 mol O
16.00g
1 mol
23.34 g N *
 1.666 mol N
14.01g
1 mol
3.36 g H *
 3.331 mol H
1.008g
Empirical formula from % composition
 Now you have a mole ratio. The easiest way to simplify this
ratio is to divide each mole amount by the lowest value.
1.666 mol C  1.666  1
3.331 mol O  1.666  2
1.666 mol N  1.666  1
3.331 mol H  1.666  2
 This means that the empirical formula is:
CO2NH2
 The numbers in the simplified ratios are the subscripts in the
empirical formula
Sample problems
1. 63.50% Ag, 28.25% N, 8.25% N
2. 12.63% Li, 58.21% O, 29.17% S
3. 27.93% Fe, 48.01% O, 24.06% S
4. 29.28% C, 52.01% O, 17.08 N, 1.64% H
The last one’s a little tricky. Ask if you need help
Molecular formula
 Pg 100-101
 As you saw in the bellwork, many different molecules can
have the same empirical formula. In order to determine the
molecular formula you need more than just the %
composition. You also need the molar mass.
 If you found that the empirical formula of a molecule was
C2O2NH2. That could be the molecular formula, but so
could C4O4N2H4 or C6O6N3H6.
Molecular formula
 If you know that the molar mass of the molecule with
empirical formula C2O2NH2 is 144.1 g/mol then you can
determine the molecular formula.
 First find the molar mass of the empirical formula
Carbon 12.01 g/mol * 2 
Nitrogen 14.01 g/mol * 1 
Oxygen 16.00 g/mol * 2 
24.02 g/mol
14.01 g/mol
32.00 g/mol
Hydrogen 1.008 g/mol * 2  1.016 g/mol
Total  72.05 g / mol
Molecular formula
 Now divide144.1 g/mol by the molar mass of the empirical
formula, 72.05 g/mol.
144.1g / mol
2
72.05 g / mol
 Now distribute that number across the empirical formula and
you have the molecular formula
2C2O2 NH2   C4O4 N2 H4
 And you’re done
Sample problems
 You find a sample to be 44.45% C, 19.74% O,
34.57% N, 1.24% H by mass. You know that
the molar mass of the molecule is 243.2 g/mol.
 Find the empirical and molecular formulas
Balancing equations
 Pg 102-108
 Practice makes perfect (balance these)
1.
2.
3.
4.
5.
6.
7.
Zn + HCl ---> ZnCl2 + H2
C10H16 + Cl2 ---> C + HCl
C7H6O2 + O2 ---> CO2 + H2O
KClO3 ---> KClO4 + KCl
Al(OH)3 + H2SO4 ---> Al2(SO4)3 + H2O
Al2(SO4)3 + Ca(OH)2 ---> Al(OH)3 + CaSO4
NaOH + Cl2 ---> NaCl + NaClO + H2O
And much, much more
Stoichiometry
 Most stoichiometry problems will give you grams of one
compound and ask for grams of another compound.
 Example
Given the equation below (unbalanced), how many grams of
sulfur hexaflouride will be produced if 85.23 grams of sulfur
are reacted?
S8 + F2  SF6
A stoichiometry problem has 4 basic steps
Stoichiometry (4 steps)
1) Balance the equation (sometimes this is already done)
S8 + F2  SF6 becomes S8 + 24F2  8SF6
2) Convert grams to moles (given: 85.23 grams of sulfur)
Molar mass of S8  32.07 * 8  256.56 g / mol
1 mol
85.23g Sulfur *
 0.3322mol S8
256.56
3) Perform mole ratio
S8 + 24F2  8SF6
0.3322
X
1
8
  x  2.658mol SF6
0.3322 x
Stoichiometry (Last step)
4) Convert moles to grams
1
8
  x  2.658mol SF6
0.3322 x
Molar Mass of SF6  32.07  6(19.00)  146.07 g / mol
146.07 g
2.658mol *
 377.2 g SF6
1mol
What you just found is the theoretical yield. That’s how
much product would be produced in a “perfect world”. We
don’t live in perfect world so we often times look at the %
yield.
Stoichiometry % yield
 Example
Given the equation below (unbalanced), 155.23 grams of
tetraphosphorous decoxide yields only 102.25 grams of
phosporic acid, what’s the % yield?
P4O10 + H2O  H3PO4
1) Balance
P4O10 + 6H2O  4H3PO4
2) Grams to moles
Molar mass of P4 O10  30.97 * 4  16.00 *10  283.88 g / mol
1 mol
155.23 g P4 O10 *
 0.547 mol P4 O10
283.88 g
Next
3) Perform mole ratio
P4O10 + 6H2O  4H3PO4
1
4
  x  2.1872mol H 3 PO 4
0.54682 x
4) Convert moles to grams
Molar Mass of H 3 PO 4  3(1.008)  30.97  4(16.00)  97.99 g / mol
97.99 g
2.1872mol *
 214.34 g H 3 PO 4 Theoretica l Yield 
1mol
5) Determine percent yield
Actual yield
102.25( given)
Percent yield 
*100 
*100  47.70%
Theoretica l yield
214.34 (found)
Stoichiometry (LR)
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Finding the limiting reagent is not very hard.
Example: Given the equation below, if you have 100.0
grams of aluminum and 75.00 grams of oxygen, what will
the limiting reagent be?
Al +O2  Al2O3
Balance
4Al +3O2  2Al2O3
First convert both from grams to moles
1 mol
100.0 g Al *
 3.708 mol Al
26.98g
1 mol
75.00 g O2 *
 2.344 mol O 2
32.00g
Stoichiometry (LR)
4Al +3O2  2Al2O3
1 mol
 3.708 mol Al
26.98g
1 mol
75.00 g O 2 *
 2.344 mol O 2
32.00g
100.0 g Al *
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Next, divide both by their coefficient
3.708 mol Al
 0.927
4
2.344 mol O 2
 0.781
3
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Oxygen is the limiting reagent.
Solve for the theoretical yield using oxygen. Forget about
Al! The oxygen will run out and leave a little Al behind.