Chapter 3
Stoichiometry: Mass, Formulas, and Reactions
Molecular & Formula Masses
e.g., Calculate the molecular mass of sulfur hexafluoride.
Formula: SF6
Molecular Mass = [1 S(amu)] + [6 F(amu)]
= [1(32.07 amu)] + [6(19.00 amu)] = 146.07 amu
e.g., Calculate the formula mass of calcium chloride.
Formula: CaCl2
Molecular Mass = [1 Ca(amu)] + [2 Cl(amu)]
= [1(40.08 amu)] + [2(35.45 amu)] = 110.98 amu
[Ionic compounds don’t consist of molecules, so it is strictly incorrect to talk about a molecular
mass. In this case, we use “formula mass” to refer to the mass of one formula unit in the
extended array of ions.]
The Mole
We can’t see the microscopic world where reactions occur. We can’t count the billions
of billions of particles at that level. We use a counting quantity to represent these large
quantities of microscopic particles: the mole.
 mole: the amount of substance containing the same number of entities as there are atoms in
exactly 12 g of carbon-12. Experimentally:
1 mole contains 6.0221418 1023 entities
6.0221418
1023, aka Avogadro’s Number (NA)
e.g., 1 mole of H2O contains 6.022 1023 H2O molecules
e.g., 1 mole of KNO3 contains 6.022 1023 KNO3 formula units
e.g., 1 mole of Hg contains 6.022 1023 Hg atoms
You can’t possibly individually count atoms/molecules/ions. But we can use mass to “count”
atoms/ions/molecules, thanks to how the mole is defined!
mass of one
mass of one mole
Substance
atom/molecule
of atoms/molecules
Cu
63.55 amu
63.55 g
O2
32.00 amu
32.00 g
SCl2
102.97 amu
102.97 g
Mass Percent from Chemical Formula
 atoms X in formula 
 1 molecule compound   atomic mass of X 

Mass % X = 
100%
molecular mass of compound
mass of element
(Compare to percent by mass =
100%)
total mass of substance
 2 atoms H   1.008 amu 



1 molecule H 2O   1 atom H 

e.g., Mass % H in H2O =
100% = 11.19% H by mass
 18.02 amu 


 1 molecule H 2O 
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Chemical Equations
Chemical equations express using chemical formulas the identities and quantities of substances involved
in physical or chemical changes. Chemical equations are based on the Law of Conservation of Mass.
Steps to writing a balanced chemical equation
1. Write skeleton equation
2. Balance atoms using “stoichiometric coefficients”
 can’t change chemical formulas
 can’t add/remove substances from equation
 hint: start with most complex cpd.; save elements for last
Groups of atoms (like
polyatomic ions) that
travel together
can/should be
balanced together.
3. Adjust stoichiometric coefficients (if necessary)
4. Check for balancing atoms
5. Specify states of matter
Molar Mass M or
(script “M”): the mass (usually in grams) of one mole of entities
Calculate molar mass the same way as molecular/formula mass, but with units of g∙mol –1
CH4
M 1(12.01) + 4(1.008) = 16.04 g∙mol–1
NaCl
e.g.,
M 1(22.99) + 1(35.45) = 58.44 g∙mol–1
M = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g∙mol –1
C6H12O6
Consider 50.0 g of glucose (C6H12O6). How many moles of glucose is this?
How many molecules of glucose does this sample contain?
How many hydrogen atoms are there in this sample of glucose?
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Empirical Formula: expresses lowest whole-number ratio of atoms in compound. May or may not be
the same as the molecular formula!
e.g., C6H12O6 has empirical formula C 6 H 12 O 6 = C1H2O1 or CH2O
6
6
6
¡Many other compounds have this same empirical formula!
Two common ways to determine empirical formula……


elemental analysis
combustion analysis
Once the empirical formula is known, the molar mass can be obtained from a second experiment. The
molar mass of the compound will be a whole-number multiple of the empirical formula mass.
molar mass
whole-number multiple =
empirical formula mass
–1
e.g., Benzene has M = 78 g∙mol . Its empirical formula is CH. What is its molecular formula?
78
whole-number multiple =
=6
13
 Molecular formula of benzene: C1×6H1×6= C6H6
Note also that % by mass can be calculated from the molecular formula and molar mass, using the
same equation as above, substituting moles for molecules/atoms and molar mass for molecular mass.
Calculating Amounts of Reactant and Product
A balanced chemical equation contains lots of information relating molar and mass quantities of
substances in the reaction.
You MUST have a correctly balanced equation to make stoichiometric calculations!!!
General Procedure:
GRAMS of A
NO DIRECT CALCULATION!
M of A
MOLES of A
GRAMS of B
M of B
mole ratio
moles B / moles A
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MOLES of B
Example:
What mass of ammonia (NH3) is produced from the reaction of 1.00 kg H2(g) with excess N2(g)?
Game Plan:
Write balanced equation: 3H2(g) + N2(g) → 2NH3(g)
GRAMS of H2
GRAMS of NH3
M of H2
MOLES of H2
M of NH3
mole ratio
 2 mol NH 3 


 3 mol H 2 
MOLES of NH3
 1000 g   1 mol H 2  2 mol NH3   17.03 g NH3 
3
Mass NH3(g) = 1.00 kg H2 
 = 5.63 × 10 g NH3



 1 kg  2.016 g H 2  3 mol H 2  1 mol NH 3 
(= 5.63 kg NH3)
What if a quantity of N2 were specified in the previous example? We would have to worry about
whether or not there were enough N2 to react with all the H2. One reactant or the other will run out
first and limit the amount of product that can be produced….
Limiting Reactant Problems
 When starting amounts of two or more reactants are known, we must consider that one or the other
will run out before the other. To determine which reactant will run out first, thus limiting the
amount of product that can be produced, we must calculate the amount of product that can be
produced based on each reactant.
Example: What mass of NH3 is produced from the reaction of 1.00 kg H2(g) with 1.00 kg N2(g)?
Now masses of both reactants are known. Must calculate based on each.
From H2: done above; 5.63 kg NH3 could be produced
From N2:
Compare the two answers. The smaller quantity of product is the largest amount that can be
produced. The reactant that would produce this amount of product is called the limiting
reactant. Other reactants are called excess reactants.
 theoretical yield: maximum amount of product that can be formed when the limiting reactant is
completely consumed.
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In most chemical reactions, you recover less product than is theoretically possible to form.
 percent yield =
actual yield
× 100%
theoretical yield
e.g., If 7.81 × 102 g of ammonia were recovered in the limiting reactant example above, what
would be the percent yield?
NOTE: UNITS of actual and theoretical yields MUST be the SAME in % Yield calculation
Percent Yield for Multi-Step Reactions
e.g.,
C2H2 + H2 → C2H4
yield = 80.0%
C2H4 + Cl2 → C2H4Cl2
yield = 70.0%
C2H2 + H2 + Cl2 → C2H4Cl2 yield = ????%
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