Download Stoichiometry

Stoichiometry Study Guide
Chemistry Chapter 13
Dr, Hazlett
1
Stoichiometry
I. Review
A.
÷ Molar Mass (g/mol) 
Mass
(Grams)
 x Molar Mass
(g/mol)
x NA 
Moles
Number
Particles

÷ NA
NA = 6.022314 x 1023
B. Average Atomic Weight
1. (Isotope 1)(% Abundance) + (Isotope)(% Abundance) + …
2. If abundances are unknown
a. (Isotope 1)(x) + (Isotope 2)(1 – x) = Molar Mass Element
C. Molar Mass (M)
D. Percent Composition
1. M of a compound is all element’s molar masses summed
a. pay attention to the numbers indicated in subscripts!
2. Example (using rounded off molar masses):
H2SO4
2 H x 1.00 molar mass
1 S x 32 molar mass
4 O x 16 molar mass
Total
Mass Element
Mass Compound

=
=
=
=
2
32
64
98 g/mol
x 100 = % by mass in the molecule
H = 2 / 98 x 100
S = 32 / 98 x 100
O = 64 / 98 x 100
 3%
 33 %
 64 %
*Should add up to close to 100 %
II. Empirical and Molecular Formulas
A. Empirical Formula is the reduced form of the molecular formula
1. Ionic compounds are already in their empirical form!
2. Process:
a. Change percents given to masses (grams)
b. Convert masses to moles
c. Divide all mole answers by the smallest value
d. Multiply all new values by the same whole numbers until you get
close enough to round off to a whole number(s)
Stoichiometry Study Guide
Chemistry Chapter 13
Dr, Hazlett
2
e. Rewrite compound
3. Example (using rounded of molar masses):
Compound w/ 72.2% MG and 27.8% N by mass.
What is empirical formula? (NOTE: If given elements in grams, must
first find percentages by dividing their individual given amounts by the
total amount of compound)
1. Assume a 100 gram sample and each percentage is now grams
 72.2 g Mg and 27.8 g N
2. Convert each mass to mols
Mg = 72.2 g / Molar mass Mg 24.3 = 2.97 mols Mg
N = 27.8 g / 14.0 g/mol
= 1.99 mols N
3. Now, divide all the mol answers by the smallest value
Mg = 2.97 / 1.99
= 1.49
This is the mole ratio for this compound
N = 1.99 / 1.99
= 1
meaning you have 1.49 Mg for every 1 N.
4. Since you can not have 1.49 atoms of Mg,
you now multiply each and every value by a whole number until you get
as close as possible to a whole number within rounding limits.
Mg
=
1.49 x 2
= 3
N
=
1
x 2
= 2
5. These are your new subscripts for the empirical form of the compound:
Mg3N2
B. Molecular Formulas from Empirical Formulas
1. Empirical formula (EF) is the reduced simplest form, and the molecular
formula (MF) is how the molecule actually appears in nature
2. Calculation of molecular formula:
a. Step 1: Molar mass of compound
Mass of the EF
=
b. Step 2: Common Factor x EF subscripts
Scaling/Common Factor
=
MF
3. Example (using rounded amounts):
1.5 g hydrocarbon is combusted and produces 4.4 g CO2 and 2.7 g H2O.
The molecular weight is 78 u. What is MF?



CxHy 
1 C x 12
+
1H x1
78 u / 13 g/mol
= Common Factor of 6
6 x C1H1
=
C6H6
= 13 g/mol
Stoichiometry Study Guide
Chemistry Chapter 13
Dr, Hazlett
3
4. Example:
Compound w/ 43.7 g P and 56.3 g O. Molecular weight is 142.
Find the EF and MF.
a. Step 1: Finding the EF
Convert masses of the given elements to mols, and if units in %’s, use the
100 gram assumption method and convert the 5’s to grams.
42.7 g P / 30.97 g/mol P
=
1.41 mol P
56.3 g O / 16 g/mol O
=
3.52 mol O
b. Step 2: Divide all results by the lowest value
P
=
1.41 / 1.41
=
1
O
=
3.52 / 1/41
=
2.5
c. Step 3: Multiply each result by the same whole number until
either get a whole number or close enough to round off
P
=
1 x 2
=
2
O
=
2.5 x 2
=
5
d. Step 4: Write EF:
P2O5
e. Step 5: Find mass of EF
2 P x 31 =
62
5 O x 16 =
80
Total is 142
f. Step 6: Find Common Factor
Compound divided by EF mass, so 142 / 142
=
1
g. Step 7: Multiply EF subscripts by Common Factor to find MF
P2O5 x 1 = P2O5
(EF and MF happen to be same)
III. Reaction Stoichiometry Equations
A. Must start with a balanced equation w/ both reactants and products!!!!!
1. Coefficients appearing in balanced equation are the mole amounts
of that unit
2. Some notes:
a. The GIVEN is the unit that appears in the problem and you are
given some amount in mass or moles to work with
b. The UNKNOWN is the unit you are trying to find out how
much is needed and/or produced by the reaction
c. The fraction uses the # mols of the wanted/unknown unit
found in the balanced equation; and is divided by the # mols of the
given unit that is found in the balanced equation
d. You only solve for those units requested – and can ignore
those units not included/asked about in the problem
Stoichiometry Study Guide
Chemistry Chapter 13
Dr, Hazlett
4
B. Four Types:
1. The amount given is in moles and the wanted amount / unknown is
in moles
(MOLES  MOLES)
a. Mols Given
x Mols Unknown from Balanced Equation = Mols of
from Problem
Mols of Given taken from Equation
Unknown
2. Given Moles and want a Mass (g) answer
(MOLES  MASS)
a. Mols of Given x Mols Unknown from Balanced Equation x Molar Mass = Mass of
from Problem
Mols of Given taken from Equation
of Unknown
Unknown
3. Given an amount in grams (mass) and want to find moles
(MASS  MOLES)
a. Mass of Given
from Problem

Molar mass
of Given
x
Mols of Unknown from Equation
Mols of Given from Equation
= Mass of
Unknown
4. Given an amount of mass and want to find the unknown in mass (g)
(MASS  MASS)
a. Mass of Given
from Problem

Molar mass
of Given
x
Mols of Unknown from Equation
Mols of Given from Equation
x Molar mass
of Unknown
= Mass of Unknown
C. Examples (using rounded amounts):
1. Mole  Mole. You are given 18 mols of MgCO3 which decomposes
to make MgO and CO2. How many moles of carbon dioxide are
produced?
 Write the balanced equation: MgCO3  MgO + CO2
 (Coefficients in front of units represent # moles of that unit).
 Given is 18 mols of magnesium carbonate; the unknown is
The number of mols of carbon dioxide.
 Remember the # mols for fraction come from the equation!
Stoichiometry Study Guide
Chemistry Chapter 13
Dr, Hazlett
5
 Mols of Given
x
(18 mols MgCO3)
1 mol of unknown (CO2) = 18 mols
1 mol of given (MgCO3)
CO2

These values from the
equation’s coefficients
2. Mols  Mass. In the following reaction, how many grams of KCLO3 are
needed to make 4.5 moles of oxygen. Reaction: 2KClO3  2KCl + 3O2.
Given is the 4.5 mols of O and the unknown is the grams of potassium
chlorate.
 4.5 mols x 2 mols KClO3 (unknown from eq.) x 122.6 g/mol = 368 g
O
3 mols O (given from equation)
KClO3
KClO3

Molar Mass
Of Unknown
3. Mass  Moles. In the reaction of ammonia and oxygen, nitrous oxide and
water are produced. The balanced formula is: 4NH3 + 5O2  4NO + 6H2O
If given 824 g ammonia, how many mols of NO and H2O are produced?
Given from problem is the 824 g ammonia; the unknowns are the moles
of the two products.
And
824 g NH3
x
17.04 g/mol NH3
4 mols NO (unknown) = 48.4 mol NO
4 mols NH3 (given)

# mols from the equation’s coefficients
82.4 g NH3
x
17.04 g/mol NH3
6 mol H2O
4 mol NH3
=
72.6 mol H2O
4. Mass  Mass. In the reaction: Sn(s) + 2HF(g)  SnF2(s) + H2(g)
If there is 30 grams of HF present, how many grams of SnF2 will be produced?
Given is the 30 g HF and the unknown unit is SnF2.

30 g HF
x 1 mol SnF2 x 156.71 SnF2 = 117.5 g SnF2

20.01 g/mol HF
2 mol HF
Molar mass of the known unit
Stoichiometry Study Guide
Chemistry Chapter 13
Dr, Hazlett
6
IV. Limiting Reagents
A. This is the reactant used up first in a reaction, and after it is all gone, the
reaction ceases to continue. Remaining amounts are “excess”.
B. Example: 2 Na (s) + Cl2 (g)  2 NaCl (s). What will occur when 6.7 mol
of Na reacts w/ 3.2 mol Cl2? What is the limiting reactant; how many mols of
NaCl is produced; and how much reagent remains in excess?
1. Start w/ known amount and mole ratio to determine amounts of other
reactants:
6.7 mol Na x 1 mol Cl2 = 3.35 mol Cl2
2 mol Na
This is required amt.
2. Hence, 3.35 mol Cl2 needed, but only 3.2 is available,  Cl is the
limiting reactant/reagent and Na is in excess
Thus, if 3.2 mol Cl2 used:
3.2 Mol Cl2
x
2 mol NaCl
1 mol Cl2
=
6.4 mol NaCl
3.2 Mol Cl2
x
2 mol Na
1 mol Cl2
=
6.4 mol Na - amt. of excess used
 6.7 mol – 6.4 mol = .3 mol of Na in excess
V. Percent Yield
A. % Yield = Actual Yield
x
Theoretical Yield
100
B. Calculations:
1. CaCO3 (s)  CaO (s) + CO2 (g). What is the percent yield of the
reaction if 24.8 g CaCO3 is heated to make 13.1 g CaO?
Actual Yield
Theoretical Yield
=
=
13.1 g CaO
24.8 g CaCO3 x 1 mol CaO x 56.1 g CaO

100.1 g CaCO3
1 mol CaCO3


Amt. given 
Mol Unknown Molar Mass
Its molar mass
Mol Known
Unknown
= 13.9 CaO
 % yield = (13.1 g CaO  13.9 g CaO) x 100 = 94.2%