Chapter 3 Stoichiometry
Mass
1. Atomic mass
A. Atomic masses are representing the mass
of an atom of one element compared to
the mass of another. We use carbon-12 as
the standard and assign it a mass of exactly
12 amu.
B. Exercise:
1. Using the atomic masses in the periodic
table, find an element that is four times
heavier than carbon.
Solution: Since carbon has as an atomic mass
of 12 amu, the element must have an
atomic mass of about 48 amu, Ti atomic
mass is 47.90 amu.
2. Germanium (Ge) is about six times heavier
than carbon. What is its approximate atomic
mass? _______________
3. If helium (He) was the standard and
assigned an atomic mass of 10 amu, what
would the atomic mass of carbon be?
______________
C. Experimental determination of relative
atomic mass
1. Instrument used: _________________
2. Describe briefly how the instrument
determines the relative atomic mass.
D. Atomic mass and isotopic abundance
1. Mathematical relationship:
(abundance)(amu) + (abundance)(amu) =
average atomic mass
2. Exercise:
a. Antimony occurs in nature as a mixture
of two isotopes: The isotope 121 Sb has a
51
mass of 120.9 amu and an abundance of
57.25%. Antimony has an average atomic
mass of 121.75 amu. What is the
abundance of the other isotope? What is
the atomic mass?
b. Practice problems: Silicon consists of
three stable isotopes whose atomic
masses and abundances are as follows:
27.98 amu
92.2 %
28.98 amu
4.7 %
29.97 amu
3.1 %
Calculate the mass of Si.
2. Mass of a single atom
a. Avogadro’s number
What does it represent? ____________
________________________________
b. Sample problems:
1. Calculate the mass of a carbon atom.
Solution: Use the conversion factor:
6.02 x 1023 C atoms in 12.01 g C.
Can be written as 6.02 x 1023atoms
12.01 g
or
12.01 g C
6.02 x1023 C atoms
Depending on which unit we want to
cancel. In this exercise we want to
eliminate the unit C atom and obtain
the unit grams of C. Thus, we have
1 C atoms x 12.01 g C
= 2.00 x 10-23 g
6.02 x 1023 C atoms
2. Practice Problem: How many atoms are
there in 2.483 g of C?
3. Formula masses
a. The sum of the atomic masses of all
atoms represented in the formula of
any
compound or polyatomic ion
b. Sample problem:
1. What is the formula mass of chloric acid?
Solution: See packet
2. Practice problem: What is the formula
mass for sodium nitrite?
4. Relationship of Avogadro’s number to
formula masses.
A formula mass in grams has 6.02 x 1023
formula units.
Mole
1. The amount of a substance that contains the
same number of particles as there are atoms
in exactly 12 g of carbon-12.
2. Molar masses
a. The mass in grams of one mole of a pure
substance
b. Sample problem: See packet
c. Practice problems:
1. Calculate the molar mass of : (NH4)2SO4
N = 2(14.0) H = 8 (1.0) S = 32.1 O= 4(16.0)
28.0 +8.0 + 32.1 + 64.0 = 132.1 g/mole
2. Calculate the molar mass of: C4H7NO4
C = 4(12.0) = 48.0
H= 7(1.0) = 7.0
N = 14.0 = 14.0
O = 4(16.0) = 64.0
133.07 g/mole
3. Mole-gram conversion exercises:
a. How many moles are there in 1.00 g of
Al2O3? See packet
b.Practice problems:
1. Find the number of grams in 3874 mole
of NH4Cl.
3.874 mole NH4Cl 53.0 g NH4Cl =
1 mole NH4Cl
205 g
NH4Cl
2. Find the number of moles in 13.70 g of
Ba(OH)2.
• Mass Relations in Chemical Formulas
1. What a chemical reaction can tell you:
a. The atom ratios of the different elements
in the compound
Ex. NH4Cl: 1 atom N 4 atoms H
1 atom Cl
b. The mole ratios of the different elements
in the compound
Ex. NH4Cl: 1 mole N 4 mole H 1 mole Cl
c. The mass ratios of the different elements
in the compound
Ex. NH4Cl: 14.01 g N, 4(1.01) g H, 35.45 g Cl
2. Percent composition from formulas:
a. Formula: %element = total mass element x100
total mass compound
b. See packet
c. Practice problems:
1. Calculate the mass percent of each
element in cholesterol. Its simplest
formula is C27H46O.
C = 27(12.0) = 324.0 g C
H = 46(1.0) = 46.0 g H
O = 1(16.0) = 16.0 gO
386.0 g
% mass of C = 324.0 g x 100 = 83.9
386.0 g
% mass of H = 46.0 g x 100 = 11.9
386.0 g
% mass of O = 16.0 g x 100 = 4.15
386.0 g
2. Teflon has the simplest formula CF. How
many grams of fluorine are required to
make 100.0 g Teflon?
3. Empirical formula from chemical analysis:
a. Steps in determining the simplest formula.
1. Determine the mass of each element.
2. Convert each mass to moles.
3. Pick out the smallest number of moles.
4. Divide the number of moles of each
element by the smallest number of moles.
5. The answers are usually integers and
become subscripts.
b. Sample problem: See packet
c. Practice problems:
1. A compound is made up of 11.80 g of
sulfur, 16.92 g of sodium, and 23.55 g
oxygen. What is the empirical formula?
2. A compound is made up of C, Cl and O
atoms. It has 12.13% of C and 70.91% of
Cl by mass. What is the empirical
formula?
4. Empirical formula from experimental
analysis.
See packet
b. Practice problems: An unknown
compound contains only carbon,
hydrogen, and oxygen. If 1.500 g of the
compound is burned with an excess of
oxygen, 1.433 g CO2 and 0.582 g H2O are
produced. What is the empirical formula
of the compound?
2. A compound is made up of C, Cl, and O
atoms. It has 12.13% of C and 70.91% Cl by
mass. What is the empirical formula?
4. Empirical formula from experimental
analysis
The result of an experimental analysis to
determine the simplest formula of a
compound can be given in terms of masses of
compounds. You have to determine the mass
of the element desired from the given mass of
compound obtained in the analysis. All the
other steps to empirical formulas are the
same.
a. Sample problem: See packet
b. Practice problem: An unknown compound
contains only carbon, hydrogen, and oxygen.
If 1.500 g of the compound is burned with an
excess of oxygen, 1.433 g CO2 and 0.582 H2O
are produced. What is the empirical formula
of the compound?
5. Molecular Formula and empirical formula
To determine the molecular formula from
the simplest formula, one has to know the
molar mass of the compound. Since the
molecular formula is a whole number
multiple of the simplest formula, it follows
that
Multiple = molar mass
empirical mass
Here the empirical formula mass is the sum of
the atomic masses of the elements that make
up the empirical formula. The subscripts in
the simplest formula are multiplied by the
multiple to get the subscripts for the
molecular formula.
Ex. See packet
Practice problem: Determine the molecular
formula of ethylene glycol. It has an atom
ratio of 1 C : 1 O : 3 H. Its molar mass is
62.06 g.
 Reactions
1. Writing and balancing equations
A. Steps in writing a balanced equation
1. Determine the correct formulas for all
the reactants and products. Include
states of matter.
2. Balance the elements one at a time by
using coefficients.
3. Make sure all the coefficients are in the
lowest possible ratio.
B. Thing to remember:
1. Use only coefficients, do not change the
subscripts
2. Leave H and O go to last
3. Remember the diatomic molecules ( H O
F Br I N Cl)
4. All metals are solids at room temperature
except Hg (l).
5. All ionic compounds (+ -) are solids.
C. Sample Problem
D. Practice problems:
1. Write a balanced equation for the
reaction of chromium with sulfur.
Cr (s) +
S (s)  Cr2S3
2. Write a balanced equation for the
reaction of bismuth with oxygen
2. Mass Relationship with equations
a. Mole-mole relationship
1. The coefficients indicate how many
moles of one reactant are required to
combine with another reactant. They
also show how many moles of product
are obtained.
2. These coefficients can be used as
conversion factors for the particular
equation you are dealing with.
3. Ex. See packet
3. Gram-mole relationships
Ex. See packet
4. Gram-gram packet
Ex. See packet
5. Practice problems:
Limiting reactant and theoretical yield
1. Definitions:
a. limiting reactant: the reactant that is
used
up first in the equation
b. Theoretical yield: the amount of product
that would be formed if the reaction
went to completion
2. Steps to be followed in determining the
limiting reactant of a reaction.
a. Convert the amounts to moles or g
b. Compare what you calculated to what
was given
c. If the calculated > given = limiting
If the calculated < given = excess
3. Sample problems: See packets
4. Practice problems: 3.837 g of NH3 reacts
with 4.129 g of H2S
a. Write and balance the equation.
b. What is the limiting reactant?
c. What is the theoretical yield of (NH4)2S?
d. What is the reactant in excess and how
much is left over?
• Percent Yield
1. % yield = actual mass of product
x 100
expected mass of product
2. Actual yield: the quantity of a product that
is obtained from a chemical reaction
3. See packet
4. Practice problems:
When 1.417 g of carbon reacts with
hydrogen, 1.67 g of ethane gas (C2H6) are
produced.
a. Write a balanced equation for the reaction.
b. What is the theoretical yield?
c. What is the percent yield?