Chemical Reactions and Reaction Stoichiometry

© 2015 Pearson Education, Inc.
Chapter 3
Chemical Reactions
and Reaction
Stoichiometry
Lecture Presentation

Zn2
+
Constant Charge Ions
Ag+
Though Zn and Ag are transition metal, they typically form
only one ion.
• Zn has an electron configuration of [Ar]4s23d10. Therefore,
to form an ion, it only loses the 4s electrons to form Zn2+.
The filled 3d orbital is very stable.
• Ag is an exception, just like the element Cu, which is
directly above it. The electron configuration of Cu is
[Ar]4s13d10. Therefore, to form an ion, it only loses the 4s
electron to form Ag+. The filled 3d orbital is very stable.

Stoichiometry
• The study of the mass relationships in
chemistry
• Based on the Law of Conservation of
Mass (Antoine Lavoisier, 1789)

Stoichiometry
Chemical Equations
Chemical equations are concise
representations of chemical reactions.

Stoichiometry
What Is in a Chemical Equation?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Reactants appear on the left
side of the equation.

Stoichiometry
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Products appear on the right
side of the equation.

Stoichiometry
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
The states of the reactants and products are written
in parentheses to the right of each compound.
(g) = gas; (l) = liquid; (s) = solid;
(aq) = in aqueous solution (in water)

Stoichiometry
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Coefficients are inserted to balance the equation
to follow the law of conservation of mass.

Stoichiometry
Why Do We Add Coefficients Instead of
Changing Subscripts to Balance?
• Hydrogen and oxygen can make water
OR hydrogen peroxide:
 2 H2(g) + O2(g) → 2 H2O(l)
 H2(g) + O2(g) → H2O2(l)

Stoichiometry
Steps To Balance Reactions
___ Na + ___Cl2 ___ NaCl
Na
Cl
Step 1. Count atoms
Step 2: Add coefficients to balance atoms that
are not equal
Step 3: Adjust other coefficients for unequal
elements

Stoichiometry
Additional Examples
___ H2O ___ H2 + ___ O2
H
O
Step 1. Count atoms
are not equal
elements

Stoichiometry
Special Scenario 1
Least Common Denominator Method
___ N2 + ___ H2 ___ NH3
N
H
Step 1. Count atoms
are not equal
elements

Stoichiometry
Special Scenario 2
Keep Intact Polyatomic Ions Together
___ CuNO3 + ___K2CO3 → ___ Cu2CO3 + ___ KNO3
Cu
NO3
K
CO3
Keep intact polyatomics together – don’t split them apart!
Step 1. Count atoms
are not equal
elements

Stoichiometry
Special Scenario 3
Combustion Reactions
___ C3H8 + ___ O2 → ___ CO2 + ___ H2O
C
H
O
Step 1. Count atoms
are not equal
elements
Combustion reactions: Save oxygen for last!! Also applies to other reactions
with diatomic gases.

Stoichiometry
Special Scenario 4
Balancing with a Decimal Coefficient
___ C8H18 + ___ O2 → ___ CO2 + ___ H2O
C
H
O
Step 1. Count atoms
are not equal
elements

Stoichiometry
Special Scenarios Covered
1.Least common denominator
2.Keeping intact polyatomic ions together
3.Balancing oxygen last in combustion
reactions
4.Balancing with a decimal coefficient

Stoichiometry
Balancing Equations

Stoichiometry
Three Types of Reactions
• Combination reactions
• Decomposition reactions
• Combustion reactions

Stoichiometry
Combination Reactions
• Examples:
– 2 Mg(s) + O2(g) 2 MgO(s)
– N2(g) + 3 H2(g) 2 NH3(g)
– C3H6(g) + Br2(l) C3H6Br2(l)
• In combination
reactions two or
more substances
react to form one
product.

Stoichiometry
• In a decomposition
reaction one
substance breaks
down into two or
more substances.
Decomposition Reactions
• Examples:
– CaCO3(s) CaO(s) + CO2(g)
– 2 KClO3(s) 2 KCl(s) + O2(g)
– 2 NaN3(s) 2 Na(s) + 3 N2(g)

Stoichiometry
Combustion Reactions
• Examples:
– CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
– C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
are generally rapid
reactions that produce
a flame.
most often involve
oxygen in the air as a
reactant.

Stoichiometry
Formula Weight (FW)
• A formula weight is the sum of the atomic
weights for the atoms in a chemical formula of
an ionic compound.
• This is the quantitative significance of a
formula.
• The formula weight of calcium chloride,
CaCl2, would be
Ca: 1(40.08 amu)
+ Cl: 2(35.45 amu)
110.98 amu
Enter into your calculator:
40.08 + (2 x 35.45) ENTER

Stoichiometry
Formula Weight (FW)
• The formula weight of calcium nitrate,
Ca(NO3)2, would be
Ca: 1(40.08 amu)
N: 2(14.01 amu)
+O: 6(16.00 amu)
164.10 amu
Enter into your calculator:
40.08 + (2 x 14.01) + (6 x 16) ENTER

Stoichiometry
Ionic Compounds and Formulas
• Remember, ionic compounds exist with
a three-dimensional order of ions.
There is no simple group of atoms to
call a molecule.
• As such, ionic compounds use empirical
formulas and formula weights (not
molecular weights, which are explained
on the next slide).

Stoichiometry
Molecular Weight (MW)
• A molecular weight is the sum of the atomic
weights of the atoms in a molecular
compound.
• For the molecule ethane, C2H6, the molecular
weight would be
C: 2(12.01 amu)
30.08 amu
+ H: 6( 1.01 amu)

Stoichiometry
Percent Composition
One can find the percentage of the
mass of a compound that comes from
each of the elements in the compound
by using this equation:
% Element =
(number of atoms)(atomic weight)
(FW of the compound)
× 100

Stoichiometry
Percent Composition
So the percentage of carbon in ethane
C2H6, is
%C =
(2)(12.01 amu)
(30.08 amu)
24.02 amu
30.08 amu
= × 100
= 79.85%

Stoichiometry
Avogadro’s Number
• In a lab, we cannot
work with individual
molecules. They are
too small.
• 6.02 × 1023 atoms
or molecules is an
amount that brings
us to lab size. It is
ONE MOLE.
• One mole of 12C has
a mass of 12.000 g.

Stoichiometry
Molar Mass
• A molar mass is the mass of 1 mol of a
substance (i.e., g/mol).
• The molar mass of an element is the
atomic weight for the element
from the periodic table.
If it is diatomic, it is twice that
atomic weight.
• The formula weight (in
amu’s) will be the same
number as the molar mass
(in g/mol).
• The molecular weight (in amu’s) will be
the same number as the molar mass
(in g/mol).

Stoichiometry
Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Molecules of a
molecular
compound OR
Of an ionic
compound

Stoichiometry
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.

Stoichiometry
Sample Problems
Calculate the mass in grams of 0.105 moles of C12H22O11.
Step 1. Calculate the molecular weight of C12H22O11.
C = 12(12.01 g/mol)
H = 22( 1.01 g/mol)
+O = 11(16.00 g/mol)
342.34 g/mol
Step 2. Use dimensional analysis to calculate the mass of
C12H22O11. Always start with what’s given in the problem.
0.105 mol x 342.34g = 35.9457 g C12H22O11
1 mol
= 35.9 g C12H22O11

Stoichiometry
Sample Problems
Calculate the moles of Zn(NO3)2 in 143.50 g of the substance.
Step 1. Calculate the formula weight of Zn(NO3)2
Zn = 1(65.38 g/mol)
N = 2(14.01 g/mol)
+O = 6(16.00 g/mol)
189.40 g/mol
Step 2. Use dimensional analysis to calculate the moles of
Zn(NO3)2. Always start with what’s given in the problem.
143.50 g x 1 mol = 0.757655 mol Zn(NO3)2
189.40 g
= 0.75766 mol Zn(NO3)2

Stoichiometry
Sample Problems
Calculate the number of molecules in 1.0 x 10-6 mol CH3CH2OH.
Use dimensional analysis and Avogadro’s number to calculate
the number of molecules in this molecular compound.
1.0 x 10-6 mol x 6.02 x 1023 molecules = 6.02 x 1017 molecules
1 mol
= 6.0 x 1017 molecules

Stoichiometry
Sample Problems
Calculate the number of N atoms in 0.410 mol NH3.
the number of N atoms in this molecular compound.
0.410 mol x 6.02 x 1023 molecules x 1 N atom = 2.4682 x 1023 N atoms
1 mol molecule
= 2.47 x 1023 N atoms

Stoichiometry
Sample Problems
How many moles of chloride ions are in 0.2550 g of AlCl3?
Step 1. Calculate the formula weight of AlCl3
Al: 1(26.98 g/mol)
Cl: 3(35.45 g/mol)
133.33 g/mol
the number of chloride ions in this ionic compound.
0.2550 g x 1 mol x 6.02 x 1023 formula units x 3 Cl- = 5.4540 x 1021 Cl-
133.33 g 1 mol formula unit
= 5.454 x 1021 Cl-

Stoichiometry
Sample Problems
What is the molar mass of cholesterol if 0.00105 mol has a mass
of 0.406 g?
You will likely encounter many types of problems in chemistry
next year where you will be asked to calculate the molar mass of
a substance, but it will not be immediately clear how to
accomplish that. If you can calculate the grams of the substance
and the moles of the substance, you can calculate the molar
mass because molar mass = g/mol.
0.406g = 386.666 g/mol = 387 g/mol
0.00105 mol

Stoichiometry
Homework
Chapter 3 Problems
1, 2, 10, 12, 13, 14, 16a, 16c-e, 22, 26,
30, 36, 38

Stoichiometry
Determining Empirical Formulas
One can determine the empirical formula
from the percent composition by following
these three steps.

Stoichiometry
Determining Empirical Formulas—
an Example
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.

Stoichiometry
1 mol
12.01 g
1 mol
14.01 g
1 mol
1.01 g
1 mol
16.00 g
Assuming 100.00 g of para-aminobenzoic acid,
C: 61.31 g × = 5.105 mol C
H: 5.14 g × = 5.09 mol H
N: 10.21 g × = 0.7288 mol N
O: 23.33 g × = 1.456 mol O
Steps 1 and 2: Convert percentages to grams in 100.00
grams and calculate the moles of each element

Stoichiometry
Step 3: Calculate the mole ratio by dividing by the smallest
number of moles:
5.105 mol
0.7288 mol
5.09 mol
0.7288 mol
0.7288 mol
0.7288 mol
1.458 mol
0.7288 mol
C: = 7.005 ≈ 7
H: = 6.984 ≈ 7
N: = 1.000
O: = 2.001 ≈ 2
These are the
subscripts for the
empirical formula:
C7H7NO2

Stoichiometry
A compound is found to contain 26.56% potassium, 35.41%
chromium, and the remainder oxygen. Find its empirical
formula.
Step 1. Calculate the percentage of oxygen

Stoichiometry
A compound is found to contain 26.56% potassium, 35.41% chromium,
and the remainder oxygen. Find its empirical formula.

Stoichiometry
Determining a Molecular Formula
• Remember, the number of atoms in a
molecular formula is a multiple of the
number of atoms in an empirical
formula.
• If we find the empirical formula and
know a molar mass (molecular weight)
for the compound, we can find the
molecular formula.

Stoichiometry
Determining a Molecular Formula
• The empirical formula of a compound
was found to be CH. It has a molar
mass of 78 g/mol. What is its molecular
formula?
• Solution: The mass of CH (12.01 +
1.01) is 13.02
Whole-number multiple = 78/13.02 = 6
The molecular formula is (CH)6 = C6H6.
The molecular formula must be written
with the subscript distributed to the
elements.

Stoichiometry
Combustion Analysis
• Compounds containing C, H, and O are routinely analyzed
through combustion in a chamber like the one shown in
Figure 3.14.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by the difference after C and H have been determined.

Stoichiometry
Combustion Analysis
Menthol is composed of C, H and O. A 0.1005-g sample of menthol is
combusted producing 0.2829 g of CO2 and 0.1159 g of H2O. What is its
empirical formula? If menthol has a molar mass of 156 g/mol, what is its
molecular formula?
Step 1. Calculate the amount of C in the sample
C: 1(12.01) g/mol)
O: 2(16.00) g/mol)
44.01 g/mol
0.2829 g CO2 x 1 mol CO2 x 1 mol C x 12.01 g C = 0.077201 g C
44.01 g 1 mol CO2 1 mol C

Stoichiometry
Combustion Analysis
combusted producing 0.2829 g of CO2 and 0.1159 g of H2O. What is
its empirical formula? If menthol has a molar mass of 156 g/mol, what
is its molecular formula?
Step 2. Calculate the amount of H in the sample
H: 2(1.01) g/mol)
O: 1(16.00) g/mol)
18.02 g/mol
0.1159 g H2O x 1 mol H2O x 2 mol H x 1.01g H = 0.012992 g H
18.02 g 1 mol H2O 1 mol H

Stoichiometry
Combustion Analysis
Step 3. Calculate the amount of O in the sample by subtracting the
mass of C and H from the mass of the original sample.
0.1005 g sample – (0.077201 g C + 0.012992 g H) = 0.010307 g O
Note that the total mass of the CO2 and the H2O (0.2829 g + 0.1159 g)
is greater than the mass of the starting sample. That is because,
during the combustion reaction, oxygen from the air is added to the
CO2 and the H2O.

Stoichiometry
Combustion Analysis
C: 0.077201 g × 1 mol = 6.428 x 10-3 mol C
12.01 g
H: 0.012992 g × 1 mol = 1.286 x 10-2 mol H
1.01 g
O: 0.010307 g O × 1 mol = 6.442 x 10-4 mol O
16.00 g
Calculate the moles of each element and divide by the
fewest number of moles

Stoichiometry
Combustion Analysis
C: 6.428 x 10-3 mol = 10 mol C
6.442 x 10-4 mol
H: 1.286 x 10-2 mol = 20 mol H
6.442 x 10-4 mol
O: 6.442 x 10-4 mol = 1 mol O
6.442 x 10-4 mol
Calculate the moles of each element and divide by the
fewest number of moles
Empirical formula:
C10H20O

Stoichiometry
Combustion Analysis
Calculate mass of empirical formula, C10H20O
C: 10(12.01 g/mol)
H: 20(1.01) g/mol)
O: 1(16.00) g/mol)
156.30 g/mol
156 g/mol = 1  the molecular and empirical formula are the same
156.03 g/mol

Stoichiometry
Quantitative Relationships
• The coefficients in the balanced equation show
 relative numbers of molecules of reactants and
products.
 relative numbers of moles of reactants and
products, which can be converted to mass.

Stoichiometry
Stoichiometric Calculations
We have already seen in this chapter how to
convert from grams to moles or moles to
grams. The NEW calculation is how to
compare two DIFFERENT materials, using
the MOLE RATIO from the balanced
equation!

Stoichiometry
Sodium silicate, Na2SiO3, reacts as follows:
Na2SiO3 (s) + HF (aq)  H2SiF6 (aq) + 2 NaF (aq) + 3 H2O (l)
How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

Stoichiometry
Sodium silicate, Na2SiO3, reacts as follows:
Na2SiO3 (s) + HF (aq)  H2SiF6 (aq) + 2 NaF (aq) + 3 H2O (l)
How many moles of Na2SiO3 can react with 0.800 g of HF?
H: 1(1.01 g/mol) F: 1(19.00 g/mol) HF: 20.01 g/mol

Stoichiometry
An Example of a Stoichiometric Calculation
How many grams of water can be produced from 1.00 g of glucose in the
following reaction?
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
Molar mass of glucose: 180.18 g/mol Molar mass of H2O: 18.02 g/mol

Stoichiometry
If you start with 10.5 grams of lithium hydroxide, how many grams of lithium
bromide will be produced?
LiOH + HBr  LiBr + H2O
MM LiOH = 23.95 g/mol, MM LiBr = 86.84 g/mol

Stoichiometry
If a reaction begins with 45 grams of ethylene (C2H4), how many grams of
carbon dioxide will be produced?
C2H4 + 3 O2  2 CO2 + 2 H2O
MM C2H4 = 28.06 g/mol, MM CO2 = 44.01 g/mol

Stoichiometry
Homework
Chapter 3 Problems
48, 52, 58, 62, 64, 67 a-b

Stoichiometry
Limiting Reactants
• The limiting reactant is the reactant present in the
smallest stoichiometric amount. In other words, it’s the
reactant you’ll run out of first.
2 H2 (g) + O2 (g)  2 H2O (g)
• In this case, the H2, only 5 O2 is needed to react with 10 H2

Stoichiometry
Limiting Reactants
In the example below, the O2 would be the
excess reagent.

Stoichiometry
Limiting Reactants
• The limiting reactant is used in all stoichiometry
calculations to determine amounts of products
and amounts of any other reactant(s) used in a
reaction.

Stoichiometry
Limiting Reactant
How many moles of NH3 can be formed from 3.0 moles of N2 and 6.0 mol of
H2?
N2 (g) + 3 H2 (g)  2 NH3 (g)
Step 1. Determine which reactant is limiting. Choose either reactant and
calculate how many moles of the other reactant are needed to react with it.

Stoichiometry
Limiting Reactant
How many moles of NH3 can be formed from 3.0 moles of N2 and 6.0 mol of
H2?
N2 (g) + 3 H2 (g)  2 NH3 (g)
Step 2. Use the quantity of the limiting reactant to calculate the answer.

Stoichiometry
Limiting Reactant
When 24 mol of methanol and 15 mol of oxygen combine in the combustion
reaction:
2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
How many moles of water are produced? How many moles of the excess
reactant remain after the reaction.
Step 1. Determine which reactant is limiting. Choose either reactant and
calculate how many moles of the other reactant are needed to react with it.

Stoichiometry
Limiting Reactant
reaction:
2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
Step 2. Calculate the amount of water produced using the limiting reactant
to start your calculation.

Stoichiometry
Limiting Reactant
reaction:
2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
Step 3.

Stoichiometry
Theoretical Yield
• The theoretical yield is the maximum
amount of product that can be made.
– In other words, it’s the amount of product
possible as calculated through the
stoichiometry problem. It is calculated
using the limiting reactant.
• This is different from the actual yield,
which is the amount one actually
produces and measures in a laboratory.

Stoichiometry
Percent Yield
Percent yield = × 100
actual yield
theoretical yield
One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield):

Chemical Reactions and Reaction Stoichiometry

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