Chapter 3
Stoichiometry
Chemical stoichiometry: the study of the quantities of
the materials consumed and produced in chemical
reactions
3.1 Atomic Masses
The modern system for atomic masses is based on 12C as the
standard. In this system, C is assigned a mass of exactly 12
atomic mass units.
•
The unit of mass is the atomic mass unit (amu)
• one amu is defined as 1/12 the mass of an atom of
carbon with 6 protons and 6 neutrons in its nucleus
1 amu = 1.6605 x 10-24 g
6.022× 1023 amu = 1 g (exactly)
The most accurate method available today for comparing the
masses of atoms is the mass spectrometer.
Ratio of the mass of 12C and mass of 13C is 1.0836129, therefore
since the mass of 12C is assigned as exactly 12 amu, the mass of
a 13C atom is: (12 amu) X (1.0836129) = 13.003355 amu
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Why the atomic mass of carbon is listed in periodic tables not
equal to exactly 12?
Isotopes!!!!
12
C, 13C & 14C
Because natural carbon is a mixture of isotopes, the atomic mass
listed is the average value reflecting the natural abundance of the
three isotopes. It is calculated as follows:
12
C has a natural abundance of 98.89% and 13C an abundance of
1.11%. 14C is negligibly small at this level of precision.
98.89% of 12C (12 amu) + 1.11 % of 13C (13.0034) = 0.9889 x
12 amu + 0.0111 x 13.0034 = 12.01 amu
Average atomic mass = Σ (mass of isotope × relative abundance)
3.2 The Mole (mol)
•
•
a mole of any substance is that quantity which contains
as many atoms, molecules, or ions as are in exactly 12 g
of pure carbon-12
a mole, whether it is a mole of iron atoms, a mole of
methane molecules, or a mole of sodium ions, always
contains the same number of formula units
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•
•
•
•
the number of formula units in a mole is known as
Avogadro’s number
Avogadro’s number has been measured experimentally
its value is 6.02214199 x 1023 formula units (atoms, ions
or molecules) per mole
The mole therefore is the molecular weight of a
substance (atom, ion or molecule) in grams.
Formula Weight
• Formula weight: the sum of the atomic weights in
atomic mass units (amu) of all atoms in a compound’s
formula
Ionic Comp ou nds
Sod ium chlorid e (N aCl)
23.0 amu N a + 35.5 amu Cl = 58.5 amu
Nickel(II) ch loride h yd rate 58.7 amu N i + 2(35.5 amu Cl) +
12(1.0) amu H) + 6(12.0 amu O) = 237.7 amu
(N iCl 2• 6H 2O)
Molecu lar Comp ou nds
Water (H 2O)
Aspirin (C 9H 8 O 4)
2(1.0 amu H) + 16.0 amu O = 18.0 amu
9(12.0 amu C) + 8(1.0 amu H) +
4(16.0 amu O) = 180.0 = amu
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3.3 Molar mass: the formula weight of a
substance expressed in grams (the mass in grams
of one mole of a pure substance)
• Glucose, C6H12O6
• Molecular or formula weight: 180 amu
• molar mass: 180 g/mol
• one mole of glucose has a mass of 180 g
• Urea, (NH2)2CO
• Molecular or formula weight 60.0 amu
• molar mass: 60.0 g/mol
• one mole of urea has a mass of 60.0 g
• We can use molar mass to convert from grams to
moles, and from moles to grams
You are given one of these
and asked to find the other
Grams of A
Moles of A
Use molar mass (g/mol)
as the conversion factor
4
calculate the number of moles of water in 36.0 g water
36.0 g H 2O x
1 mol H2 O
18.0 g H 2O
= 2.00 mol H2 O
Grams to Moles
• Calculate the number of moles of sodium ions, Na+, in
5.63 g of sodium sulfate, Na2SO4
• first we find how many moles of sodium sulfate are
present in 5.63 g of sodium sulfate
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• the formula weight of Na2SO4 is
2(23.0) + 32.1 + 4(16.0) = 142.1 amu
• therefore, 1 mol of Na2SO4 = 142.1 g Na2SO4
5.63 g Na2 SO 4 x
1 mol Na2SO 4
142.1 g Na2 SO 4
= 0.0396 mol Na2SO 4
• the formula Na2SO4 tells us there are two moles of
Na+ ions per mole of Na2SO4
0.0396 mol Na2 SO 4
+
2
mol
Na
x
1 mol Na2SO 4
= 0.0792 mol Na+
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Grams to Molecules
• A tablet of aspirin, C9H8O4, contains 0.360 g of
aspirin. How many aspirin molecules is this?
• first we find how many mol of aspirin are in 0.360
g
0.360 g aspirin x
1 mol aspirin
180.0 g aspirin
= 0.00200 mol as pirin
each mole of aspirin contains 6.02 x 1023 molecules
• the number of molecules of aspirin in the tablet is
0.00200 mole x 6.02 x 1023 molecules = 1.20 x 1021 molecules
mole
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3.4 Percent Composition of Compounds
There are two common ways of describing the
composition of a compound: in terms of the numbers of
its constituent atoms and in terms of the mass percentages
of its elements.
Percent composition is the atomic weight for each
element divided by the formula weight of the compound
multiplied by 100: (textbook examples)
% Element 
Atoms of Element AW 
FW of Compound
 100
3.5 Determining the Formula of a Compound
The formula of a compound can be obtained by elemental
combustion analysis (CHN analysis):
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The result of such an analysis provides the mass of each
type of element in the compound, which can be used to
determine the mass percent of each element and from this
information the formula of a compound can be deduced.
(example on page 96)
After calculating percentages, work with a 100 g sample
of the compound.
38.67% Carbon by mass means 38.67 g of carbon.
16.22% Hydrogen by mass implies 16.22 g of hydrogen
45.11% Nitrogen by mass translates to 45.11 g of nitrogen
Now calculate the number of C atoms in 38.67 g of C,
etc…
C: 38.67 g C X 1 mol C/12.01 g/mol C = 3.220 mol of C
H: 16.22 g H X 1 mol H/1.008 g/mol H = 16.09 mol H
N: 45.11 g N X 1 mol of N/14.01 g/mol N = 3.219 molN
Thus 100 g of this compound contain 3.220 mol of C
atoms, 16.09 mol of H atoms and 3.219 mol of N atoms.
The smallest whole-number ratio of atoms in this
compound is obtained by dividing the three numbers
above by the smallest of the three.
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C: 3.220/3.219 = approx 1
H: 16.09/3.219 = approx 5
N: 3.219/3.219 = 1
Therefore the Empirical Formula (a formula providing the
smallest whole-number ratio of elements) for the
compound is: C1H5N1 or CH5N.
The compound could be CH5N or C2H10N2 etc…
To obtain the molecular formula we need to know the
molar mass of the compound. Suppose the molar mass of
our compound is 31.06 g/mole. How do we determine the
molecular formula?
The molecular formula is always a whole-number
multiple of the empirical formula. The empirical formula
mass of CH5N is given as 31.06 g/mol. Therefore what is
the molecular formula?
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3.6 Chemical equations
A chemical change involves a reorganization of the atoms
in one or more substances. Remember that mass (or atoms
in this case) is neither created nor destroyed.
The following chemical equation tells us that propane gas
and oxygen gas (reactants) react to form carbon dioxide
gas and water vapor (products)
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C3 H8 ( g) + O2 (g)
CO2 (g) + H2 O( g)
Propane
Carbon
dioxid e
Oxygen
Water
But while it tells us what the reactants and products are
and the physical state (s, l, g or aq) of each, it is
incomplete because it is not balanced.
A chemical equation for a reaction must provide all the
info about the nature of reactants and products, the
relative numbers of each and the physical state of each.
2Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g)
Stoichiometric coefficients: numbers in front of chemical
formulas; provide ratio (relative numbers) of reactants
and products in a chemical equation.
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3.7 Balancing Chemical Equations
To balance a chemical equation (more than one method),
begin with atoms that appear only in one compound on
the left and one on the right; in this case, begin with
carbon (C) which occurs in C3H8 and CO2
C3 H8 (g) + O2 (g)
3CO2( g) + H2 O(g)
now balance hydrogens, which occur in C3H8 and H2O
C3 H8 (g) + O2 (g)
3CO2( g) + 4H2 O(g)
if an atom occurs as a free element, as for example Mg or
O2, balance this element last; in this case O2
C3 H8 (g) + 5O2 ( g)
3CO2 (g) + 4H2 O( g)
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Practice problems: balance these equations
Ca( OH) 2 ( s) + HCl( g)
Calcium
hydroxide
CO2 ( g) + H2 O(l)
CaCl2 (s) + H2 O( l)
Calcium
chlorid e
photosynthes is
C6 H1 2 O6 (aq) + O2 (g)
Glucose
C4 H1 0 ( g) + O2 (g)
Butane
CO2 (g) + H2 O(g)
3.8 Stoichiometric Calculations:
Reactants & Products
Amounts
of
Stoichiometry: the study of mass relationships in
chemical reactions
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An overview of the types of calculations under study
You are given one of these
Grams of A
And asked to find one of these
Moles of A
From grams to moles,
use molar mass (g/mol)
as a conversion factor
C3 H8 (g) + 5O2 ( g)
Moles of B
Grams of B
From moles to moles, From moles to grams,
use the coefficients in use molar mass (g/mol)
the balanced equation as a conversion factor
as a conversion factor
3CO2 (g) + 4H2 O( g)
What mass of Oxygen will react with 96.1 g of Propane?
The above equation tells us that 1 mole of propane reacts
with 5 moles of oxygen to produce 3 moles of carbon
dioxide and 4 moles of water.
We must be able to convert between masses and moles of
substances. We ask ourselves how many moles of
propane are present in 96.1 g of propane.
The molar mass of propane is 44.1 g/mole:
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The number of moles of propane in 96.1 g is:
96.1 g propane X 1 mol propane/44.1 g propane = 2.18
mol propane
Now keep in mind that each mole of propane reacts with
five mole of oxygen (reaction stoichiometry). Therefore
construct a mole ratio:
5mol oxygen/1 mol propane
Multiplying the number of moles of propane available
with this conversion factor gives the number of moles of
oxygen required.
2.18 mol propane X 5 mol oxygen/1 mol propane = 10.9
mol oxygen
10.9 mol oxygen X 32.0 g oxygen/1 mol oxygen = 349 g
oxygen
Problem: how many grams of nitrogen, N2, are required to
produce 7.50 g of ammonia, NH3
N2 (g) + 3H2 (g)
2NH3 ( g)
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first find how many moles of NH3 are in 7.50 g of NH3
7.50 g N H 3 x
1 mol N H 3
17.0 g N H 3
= mol NH 3
next find how many moles of N2 are required to produce
this many moles of NH3
7.50 g N H 3 x
1 mol N H 3
17.0 g N H 3
x
1 mol N2
2 mol N H 3
= mol N 2
finally convert moles of N2 to grams of N2 and now do the
math
7.50 g NH 3 x
1 mol NH 3
17.0 g NH 3
x
1 mol N2
2 mol NH 3
x
28.0 g N 2
1 mol N2
= 6.18 g N 2
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How many grams each of CO2 and NH3 are produced
from 0.83 mol of urea?
( NH2 ) 2 CO(aq) + H2 O
Urea
urease
2NH3 (aq) + CO2 (g)
3.9 Calculations Involving a Limiting Reactant
(reagent).
In a chemical reaction, reagents are mixed together in
stoichiometric quantities (exact amounts) so that all
reagents are consumed at the same time and no unreacted
substance is left without reacting. Consider the formation
of water from its two constituent elements.
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Once hydrogen is consumed, the reaction stops, i.e.
production of water stops although there is still oxygen
left to react. In this particular case the amount of
hydrogen limits the production of water, which brings us
to the concept of limiting reagent (reactant).
Limiting reagent: the reagent that is used up first in a
chemical reaction. The reaction stops after the limiting
reagent is consumed although other reagent(s) are still
available.
Consider this reaction of N2 and O2
N2 (g) + O2 (g)
before reaction (moles) 5.0
after reaction (moles) 4.0
1.0
0
2 NO( g)
0
2.0
In this experiment, there is only enough O2 to react with
1.0 mole of N2
O2 is used up first; it is the limiting reagent. 4.0 Moles of
N2 remain unreacted.
(textbook gives the example of nitrogen and hydrogen
reacting to produce ammonia)
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Practice Problem
Suppose 12 g of carbon is mixed with 64 g of oxygen and
the following reaction takes place
C(s) + O2 ( g)
CO2 ( g)
Complete the following table. Which is the limiting
reagent?
C
before reaction (g)
12 g
+
O2
CO2
64 g
0
before reaction (mol)
after reaction (mol)
after reaction (g)
Theoretical Yields
The amount of product predicted from stoichiometry
taking into account
Limiting reagents is called the theoretical yield (the
amount of product formed
when the limiting reagent is completely consumed).
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The percent yield relates the actual yield (amount of
material recovered in the laboratory) to the theoretical
Actual yield
% Yield 
 100
Theoretica l yield
yield:
Practice problem:
Suppose we react 32.0 g of methanol with excess carbon
monoxide and get 58.7 g of acetic acid
CH3 OH + CO
before reaction (g) 32.0
before reaction (mol)
th eoretical yield (mol)
th eoretical yield (g)
actual yield (g)
percent yield (%)
excess
CH3 COOH
0
58.7
Text book examples on pages 117 & 119
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