C H E M I S T R Y
Chapter 3
Mass Relationships in Chemical
Reactions
Chemical Reaction
In a chemical reaction
• a chemical change
produces one or more
new substances.
• there is a change in the
composition of one or
more substances.
Chemical Reaction
In a chemical reaction
• old bonds are broken and new bonds are formed.
• atoms in the reactants are rearranged to form one or
more different substances.
Chemical Reaction
In a chemical reaction
• the reactants are Fe and O2.
• the new product Fe2O3 is
called rust.
Chemical Equations
A chemical equation gives
• the formulas of the reactants on the left of the arrow.
• the formulas of the products on the right of the arrow.
Reactants
O2 (g)
Product
CO2 (g)
C(s)
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Symbols Used in Equations
Symbols in chemical
equations show
TABLE
• the states of the
reactants.
• the states of the
products.
• the reaction
conditions.
6
Chemical Equations are Balanced
In a balanced
chemical reaction
• no atoms are lost or
gained.
• the number of reacting
atoms is equal to the
number of product
atoms.
7
Balancing Chemical Equations
A balanced chemical equation shows that the law of conservation of
mass is adhered to.
In a balanced chemical equation, the numbers and kinds of atoms on
both sides of the reaction arrow are identical.
2Na(s) + Cl2(g)
2NaCl(s)
left side:
right side:
2 Na
2 Cl
2 Na
2 Cl
Balancing Chemical Equations
1.
Write the unbalanced equation using the correct chemical formula for
each reactant and product.
H2(g) + O2(g)
2.
H2O(l)
Find suitable coefficients—the numbers placed before formulas to
indicate how many formula units of each substance are required to
balance the equation.
2H2(g) + O2(g)
2H2O(l)
Balancing Chemical Equations
3.
Reduce the coefficients to their smallest whole-number values, if
necessary, by dividing them all by a common denominator.
4H2(g) + 2O2(g)
4H2O(l)
divide all by 2
2H2(g) + O2(g)
2H2O(l)
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Balancing Chemical Equations
4.
Check your answer by making sure that the numbers and kinds of atoms
are the same on both sides of the equation.
2H2(g) + O2(g)
2H2O(l)
left side:
right side:
4H
2O
4H
2O
Copyright © 2008 Pearson Prentice
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Balancing Chemical Equations
Do not change subscripts when you balance a chemical equation. You are only
allowed to change the coefficients.
H2(g) + O2(g)
2H2(g) + O2(g)
2H2O(l)
Balanced properly
H2O(l)
unbalanced
H2(g) + O2(g)
H2O2(l)
Chemical equation changed!
Examples
Balance the coefficients from reactants to products.
A.
__N2(g) + __H2(g)
A.
B. __Co2O3(s) + __ C(s)
__ NH3(g)
__Co(s) + __CO2(g)
Write a balanced equation for the reaction between carbon dioxide and
potassium hydroxide to form potassium carbonate and water.
13
Chemical Symbols on Different Levels
2H2(g) + O2(g)
microscopic:
macroscopic:
2H2O(l)
2 molecules of hydrogen gas react with 1 molecule of
oxygen gas to yield 2 molecules of liquid water.
0.56 kg of hydrogen gas react with 4.44 kg of oxygen gas
to yield 5.00 kg of liquid water.
How can we relate the two to each other?
Avogadro’s Number and the Mole
Molecular Mass: Sum of atomic masses of all atoms in a molecule.
Formula Mass: Sum of atomic masses of all atoms in a formula unit of any
compound, molecular or ionic.
HCl:
C 2H 4:
1.0 amu + 35.5 amu = 36.5 amu
2(12.0 amu) + 4(1.0 amu) = 28.0 amu
Molar Mass of Na2SO4
Calculate the molar mass of Na2SO4.
Element Number
of Moles
Na
Atomic Mass
Total Mass
in Na2SO4
S
O
16
Stoichiometry: Chemical Arithmetic
Stoichiometry: The relative proportions in which elements form compounds or
in which substances react.
aA + bB
Grams of
A
Molar Mass of A
Moles of
A
cC + dD
Moles of
B
Mole Ratio
Between A and B
(Coefficients)
Grams of
B
Molar Mass of B
Stoichiometry: Chemical Arithmetic
Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach,
are prepared by reaction of sodium hydroxide with chlorine gas:
2NaOH(aq) + Cl2(g)
NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react with 25.0 g Cl2?
Grams of
Cl2
Molar Mass Cl2
Moles of
Cl2
Mole Ratio
Moles of
NaOH
Grams of
NaOH
Molar Mass
NaOH
Stoichiometry: Chemical Arithmetic
The commercial production of iron from iron ore involves the
reaction of Fe2O3 with CO to yield iron metal plus carbon dioxide:
Fe2O3(s) + 3CO(g)
2Fe(s) + 3CO2(g)
Predict how many grams of CO will react with 0.500 moles Fe2O3?

Yields of Chemical Reactions
Actual Yield: The amount actually formed in a reaction.
Theoretical Yield: The amount predicted by calculations from the
limiting reactant.
Percent Yield =
actual yield
theoretical yield
x 100%
Reactions with Limiting Amounts of
Reactants
Limiting Reactant: The reactant that is present in limiting
amount. The extent to which a chemical reaction takes place
depends on the limiting reactant.
Excess Reactant: Any of the other reactants still present
after determination of the limiting reactant.
Reactions with Limiting Amounts of
Reactants
At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is
an automobile antifreeze and a starting material in the preparation of polyester polymers:
C2H4O(aq)
+
H2O(l)
C2H6O2(l)
Examples

How many sandwiches can be made from 10 slices of bread and 8
slices of cheese? Which one is the limiting reactant?
The balanced chemical equation is
2Bd + Ch
Bd2Ch
Limiting reactant and theoretical
yield
Consider the reaction A + B  AB
Step1: Grams A  moles A  moles AB
Grams B  moles B  moles AB
Step 2: Determine which reactant (A or B) produced the least amount of
product (AB) and label it as limiting reactant
Step 3: Calculate the theoretical yield of AB (or least moles of AB 
grams AB)
Excess reactant

Whichever reactant (A or B) produced the most moles of product
(AB) is the excess reactant
most moles of product - least moles of product = moles of product
in excess

moles of product in excess --> moles excess reactant --> grams
of reactant = left over from reaction
Example

Ammonia, NH3, can be synthesized by the following reaction:
2 NO(g) + 5 H2(g)  2 NH3(g) + 2 H2O(g)
Starting with 86.3 g NO and 25.6 g H2
a.
Determine the limiting reactant
b.
Find the theoretical yield of NH3 in grams
Reactions with Limiting Amounts of
Reactants
Lithium oxide is used aboard the space shuttle to remove water from the air
supply according to the equation:
Li2O(s) + H2O(g)
2LiOH(s)
If 80.0 g of water are to be removed and 65.0 g of Li2O are available, which
reactant is limiting? How many grams of LiOH are produced? How many grams
of excess reactant remain?
Percent Composition and Empirical
Formulas
Percent Composition: Expressed by identifying the elements present and
giving the mass percent of each.
Empirical Formula: It tells only the ratios of the atoms in a compound.
multiple =
molecular mass
empirical formula mass
Steps in determine the Empirical formula

Step 1: Obtain the mass of each element (in grams)
E.G 100% = 100g therefore mass percent is the same numerical value in grams



Step 2: Determine the numbers moles of each atom present
Step 3: Divide the smallest moles by numbers of each atoms to obtain
the closet integer as possible.
Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then
multiply with a factor to get the nearest integer as possible.
E.g 1.5 x 2 = 3.0 atoms
1.33 x 3 = 3. 99 = 4 atoms
Example

A compound containing nitrogen and oxygen is decomposed in the
laboratory and produces 24.5 g nitrogen and 70.0 g oxygen. Calculate
the empirical formula of the compound.
Example

An unknown sample gives the following mass percent: 17.5% Na,
39.7% Cr and 42.8% O. What is the empirical formula?
Examples

Ibuprofen, an aspirin substitute, has the following mass percent
composition: C 75.69%, H 8.80% an O 15.51%. What is the
empirical formula of ibuprofen?
Molecular formula

Molecular Formula: It tells the actual numbers of atoms in a
compound. It can be either the empirical formula or a multiple of it.
multiple =
molecular mass
empirical formula mass
Example

Butanedione – a main component in the smell and taste of butter and
cheese – contains the elements carbon, hydrogen, and oxygen. The
empirical formula of butanedione is C2H3O and its molar mass is
86.09 g/mol. Find its molecular formula
Percent Composition and Empirical
Formulas
A colorless liquid has a composition of 84.1 % carbon and 15.9
% hydrogen by mass. Determine the empirical formula. Also,
assuming the molar mass of this compound is 114.2 g/mol,
determine the molecular formula of this compound.
Determining Empirical Formulas:
Elemental Analysis
Combustion Analysis: A compound of unknown composition (containing a
combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to
produce the volatile combustion products CO2 and H2O, which are separated and
weighed by an automated instrument called a gas chromatograph.
hydrocarbon + O2(g)
carbon
hydrogen
xCO2(g) + yH2O(g)
Determining Empirical Formulas:
Elemental Analysis
Step 1: g CO2  moles CO2  moles C  g C
 Step 2: g H2O  moles H2O  moles H  g H
 Step 3:
g O (or any missing atom) = grams sample – (g C + g H)
 Step 4: determine the moles of O or the missing atom
 Step 5: Follow all steps from determining the empirical formula

Example

A compound contains only carbon, hydrogen, and oxygen.
Combustion of 10.68 mg of compound yields 16.01 mg CO2 and 4.37
mg H2O. The molar mass of the compound is 176.1 g/mol.
Examples

Paradichlorobenzene, a common ingredient in solid air fresheners,
contains only C, H and Cl and has a molar mass of about 147g/mol.
Given that combustion of 1.68 g of this compound produces 3.02 g
of CO2 and 0.412 g H2O, determine its empirical and molecular
formulas