Stoichiometry
Chapter 3
Atomic Mass

1961: Atomic Mass is based on
 12C

is assigned a mass of EXACTLY 12 AMU
Most Accurate Way to compare masses is to
use a mass spectrometer


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12C
Results in ratio of masses like 13C/12C
Apply ratio to mass of 12C to get atomic mass.
Atomic Mass on Periodic Table is a weighted
average of the isotopes


98.89% 12C, 1.11% 13C, negligible amounts of
(.9889)(12) + (.011)(13.0034) = 12.01amu
14C
The Mole

Mole: the number equal to the number of
carbon atoms in exactly 12 grams of pure



12C
Avogadro’s number: 6.022x1023 units = 1 mole
1 mole of an element weighs exactly the atomic
mass of the element in grams
1 mole of a molecule weighs exactly the atomic
mass of the molecule in grams
Calculating Number of Atoms

A silicon chip for a computer has a mass of
5.68 mg. How many atoms of silicon are in
the chip?
1. Convert milligrams to grams
2. Convert grams to moles
3. Convert moles to atoms
1gSi
3
5.68mgSi 
 5.68 10 gSi
1000mgSi
1molSi
5.68 10 gSi 
 2.02 104 molSi
28.09 gSi
3
6.022 1023 atomsSI
2.02 10 molSi 
 1.22 1020 atoms
1molSi
4
Percent Composition of Compounds

2 common ways to describe the composition of a
compound:

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Number of constituent atoms
Percent composition by mass
Example: % composition of ethanol (C2H5OH)

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2 mol C x 12.01 g/mol = 24.02g C
6 mol H x 1.008 g/mol = 6.048 g H
1 mol O x 16.00 g/mol – 16.00 g O
% C = 24.02g/(24.02g+6.048g+16.00g) x 100 = 52.14%
C
%H = 6.048g/46.07g x 100 = 13.13% H
%O = 16.00g/46.07g x 100 = 34.73% O
Determining Formula of Compounds

You have 0.1156g of a new compound
composed of C, H, and N

You decomposed or reacted it with water to get
CO2, and H2O, which was collected and weighed.
0.1638g CO2
 0.1676g H2O



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How can you calculate the formula of the
compound?
How much carbon was in the original compound?
How much nitrogen was in the original
compound?
Determining Formula of Compounds

Use % by mass of the product compounds:
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C: 1 mol x 12.01 g/mol = 12.01g C
O: 2 mole x16.00g/mol = 32.00g O
Molar mass CO2 = 44.01 g/mol
0.1638 g CO2 x 12.01gC/44.01gCO2 = 0.04470 g
C
0.04470g CO2/0.1156g compound x 100% =
38.67% C
0.1676 g H2O x 2.016 g H/18.02 g H2O =
0.001875 g H
0.01875 g H /0.1156g compound x 100% =
16.22% H
Rest (100 – 38.67 – 16.22) is N (45.11%)
Calculate Empirical Formula


From % by mass, calculate Empirical Formula
ASSUME YOU HAVE 100 g OF COMPOUND
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Find smallest WHOLE-NUMBER ratios of
elements:
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38.67 g C x 1 mol/12.01 g = 3.220 mol C
16.22 g H x 1 mol/1.008 g = 16.09 mol H
45.11 g N x 1 mol/14.01 g = 3.219 mol N
3.220 mol C/3.219 mol N = 1.000 = 1:1
16.09 mol O/3.220 mol C = 4.997 = 5:1
Formula would be: CH5N

Or any whole number multiple of those elements
Actual Molecular Formula

Must determine Actual molecular mass
experimentally
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Suppose compound is known to have mass of
62.12 g/mol
Actual mass/Empirical mass gives ratio
62.12g/mol / 31.06 G/mol = 2.0
 Actual formula is (CH5N)x2 = C2H10N2


Solve Questions 69-75 (one per group), page
119-120 in textbook.
Limiting Reactants
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Some chemical processes are carried out with
excess amounts of one or more reactants.
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Chemically we learn about this in equilibrium
The excess reactants are usually the cheaper
ones.
How do you handle questions where you have
quantities of reactants that are not
stoichiometrically balanced?

Concept of limiting reactant.
Limiting Reactants
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How to solve the problem:
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Solve the problem once for each reactant that
MIGHT be limiting.
Determine which solution has the lowest number
of product moles
That is your limiting reactant and maximum amount of
product.
 Any other reactant above the stoichiometric amount of
the limiting reactant is “excess”

Limiting Reactant

Solid lithium hydroxide is used in space
vehicles to remove exhaled carbon dioxide
from the living environment by forming solid
lithim carbonate and liquid water. What mass
of gaseous carbon dioxide can be absorbed
by 1.00kg of lithium hydroxide?

Step 1: Write the unbalanced equation:
LiOH(s) + CO2(g)  Li2CO3 (s) + H2O(l)
 Balance the equation:
 2LiOH(s) + CO2(g)  Li2CO3 (s) + H2O(l)

Limiting Reactant

2LiOH(s) + CO2(g)  Li2CO3 (s) + H2O(l)

Step 2: convert LiOH to moles
1000g LiOH 1molLiOH
1.00 kg LiOH 

 41.8molLiOH
1kgLiOH
23.95 gLiOH
Step 3: Determine amount of CO2 that reacts with a Given amount of LiOH
1molCO2
2molLiOH
Step 4: Calculate moles of CO2 that reacts with a given mass of LiOH
1molCO2
41.8molLiOH 
 20.9molCO2
2molLiOH
Limiting Reagent
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Step 5: Calculate mass of CO2 using molar mass
44.0 gCO2
2
20.9molCO2 
 9.20 10 gCO2
1molCO2
Limiting Reactant


Nitrogen gas can be prepared by passing
gaseous ammonia over solid copper(II) oxide
at high temperatures. The other products are
solid copper and water vapor. If a sample of
18.1g of NH3 is reacted with 90.4g of CuO,
which is the limiting reactant? How manuy
grams of N2 will be formed?
Balanced equation:

2NH3(g) + 3CuO(s)  N2(g) + 3 Cu(s) + 3H2O(g)
Limiting Reactant

Compute moles of NH3 and CuO
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Use Mole ratio of CuO and NH3 to determine limiting
reactant:
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1.06 mol NH3 x 3mol CuO/2mol NH3 = 1.59 mol CuO
Because it takes 1.59 mole of CuO to react with 1.06
mole of NH3, but we only have 1.14 mole of CuO,
CuO is the limiting reactant.
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18.1g NH3 x 1mol NH3/17.03gNH3 = 1.06mol NH3
90.4g CuO x 1 mol CuO/79.55g CuO = 1.14 mol CuO
We will run out of CuO before we run out of NH3
Verify:
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Balanced Equation: Mol CuO/Mol NH3 = 3/2 = 1.5
Amount Present: Mole CuO/Mol NH3 = 1.14/1.06 = 1.08
Limiting Reactant

Calculate amount of N2 formed based on
moles of CuO (limiting factor) present:
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1 mol N2/3 mol CuO x 1.14 mol CuO = 0.380 mol
N2
% Yield

Theoretical yield is the amount of product
formed based on actual amounts and
balanced equation stoichiometric calculations
Actual yield is the measured amount of
product formed
% yield is actual yield/theoretical yield * 100
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Assignment: Questions 97a, 98c, 99-103
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One for each group