Lesson

OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL 3
STEM 004: General Chemistry 1

Chemical Reaction and Stoichiometry

OUTCOMES OF LEARNING

1. Explain relative atomic mass and average atomic mass

2. Define a mole and illustrate Avogadro’s number with examples
3. Determine the molar mass of elements and compounds
4. Calculate the percent composition of a compound from its formula
5. Calculate the empirical formula from the percent composition of a compound
6. Write and balance chemical equations.
7. Interpret the meaning of a balanced chemical reaction in terms of the law of conservation of mass.
8. Describe evidences for the occurence of a chemical reaction.
9. Calculate the mass relationship of the reactant and product of a reaction

LESSON PRESENTATION

Atomic Mass

The mass of an atom depends on the number of protons, and neutrons it contains. Knowledge of an atom’s
mass is important in laboratory work. But atoms are extremely small particles—even the smallest speck of dust that our
unaided eyes can detect contains as many as 1X1016 atoms! Clearly, we cannot weigh a single atom, but it is possible
to determine the mass of one atom relative to another experimentally.
By international agreement, atomic mass (sometimes called atomic weight ) is the mass of the atom in atomic
mass units (amu). One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12
atom. Carbon-12 is the carbon isotope that has six protons and six neutrons. Setting the atomic mass of carbon-12 at
12 amu provides the standard for measuring the atomic mass of the other elements. For example, experiments have
shown that, on average, a hydrogen atom is only 8.400 percent as massive as the carbon-12 atom. Thus, if the mass of
one carbon-12 atom is exactly 12 amu, the atomic mass of hydrogen must be 0.084 X 12.00 amu or 1.008 amu.

Average Atomic Mass

When you look up the atomic mass of carbon in the periodic table you will
find that its value is not 12.00 amu but 12.01 amu. The reason for the difference is that
most naturally occurring elements (including carbon) have more than one isotope. This
means that when we measure the atomic mass of an element, we must generally settle

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Prepared by: Mr. Lawrence Niel A. Ignacio
 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
for the average mass of the naturally occurring mixture of isotopes. For example, the
natural abundances of carbon-12 and carbon-13 are 98.90% and 1.10% respectively.
The atomic mass of carbon-13 has been determined to be 13.00335 amu. Thus, the
average atomic mass of carbon can be calculated as follows:

𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶 = (𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 126𝐶 × abundance of 126𝐶)

+(𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 136𝐶 × abundance of 136𝐶)
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶 = (0.9890)(12.00 amu) + (0.0110)(13.00335 amu)
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶 = 11.868 amu + 0.1430 amu = 12.01 amu

Note that in calculations involving percentages, we need to convert percentages to fractions. For example, 98.90%
becomes 98.90/100, or 0.9890. Because there are many more carbon-12 atoms than carbon-13 atoms in naturally
occurring carbon, the average atomic mass is much closer to 12 amu than to 13 amu.

Avogadro’s Number

In the laboratory, students do not count the number of atoms, nor do they weigh a
few atoms of an element. A more tangible amount is used and which is amenable for
observation. The number of atoms or molecules is calculated from measured amount,
either by mass or volume.

The Avogadro’s number (NA) is a constant used to quantify the number of particles
of an element (atoms, ions) or compound (molecules, formula units). Named in honor
of the Italian scientist Amadeo Avogadro (1776-1856). The currently accepted value
is

Amadeo Avogadro NA= 6.0221415 × 1023

Generally, we round Avogadro’s number to 6.022 × 1023. Thus, just as one dozen oranges contain 12 oranges;
1 mole of hydrogen atoms contains 6.022 × 1023 H atoms.

Mole Concept

Atomic mass units provide a relative scale for the masses of the elements. But because atoms have such small
masses, no usable scale can be devised to weigh them in calibrated units of atomic mass units. In any real situation, we
deal with macroscopic samples containing enormous numbers of atoms. Therefore, it is convenient to have a special
unit to describe a very large number of atoms. The idea of a unit to denote a particular number of objects is not new.
For example, the pair (2 items), the dozen (12 items), and the gross (144 items) are all familiar units. Chemists measure
atoms and molecules in moles.

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
The mole (symbol: mol) is a term used to collectively refer to the quantity of particles (atoms ions, or molecules)
that is equal to the Avogadro’s number 6.022 × 1023. The International Union of Pure and Applied Chemistry (IUPAC)
defined the mole as:
“The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in
0.012 kilograms of carbon-12.”

Molecular Mass

If we know the atomic masses of the component atoms, we can calculate the mass of a molecule. The
molecular mass (sometimes called molecular weight ) is the sum of the atomic masses (in amu) in the molecule . For
example, the molecular mass of H2O is

2(atomic mass of H) + atomic mass of O

or
2(1.008 amu) + 16.00 amu = 18.02 amu

In general, we need to multiply the atomic mass of each element by the number of atoms of that element
present in the molecule and sum over all the elements.

Example

Calculate the molecular masses (in amu) of the following compounds: (a) sulfur dioxide (SO 2) and (b) caffeine
(C8H10N4O2 ).

Solution: To calculate molecular mass, we need to sum all the atomic masses in the molecule. For each element,
we multiply the atomic mass of the element by the number of atoms of that element in the molecule. We find
atomic masses in the periodic table.

(a) There are two O atoms and one S atom in SO2 , so that

molecular mass of SO2 = 32.07 amu + 2(16.00 amu)

= 64.07 amu

(b) There are eight C atoms, ten H atoms, four N atoms, and two O atoms in caffeine, so the molecular mass
of C8H10N4O2 is given by

8(12.01 amu) + 10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu) = 194.20 amu

From the molecular mass we can determine the molar mass of a molecule or compound. The molar
mass of a compound (in grams) is numerically equal to its molecular mass (in amu). For example, the molecular mass
of water is 18.02 amu, so its molar mass is 18.02 g. Note that 1 mole of water weighs 18.02 g and contains 6.022×10 23
H2O molecules, just as 1 mole of elemental carbon contains 6.022×1023 carbon atoms.

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Finally, note that for ionic compounds like NaCl and MgO that do not contain discrete molecular units, we use
the term formula mass instead. The formula unit of NaCl consists of one Na 1 ion and one Cl 2 ion. Thus, the formula
mass of NaCl is the mass of one formula unit:

formula mass of NaCl = 22.99 amu + 35.45 amu

= 58.44 amu

and its molar mass is 58.44 g.

Percent Composition of Compounds

The law of definite proportions states that any sample of a given compound always consists of the same
elements in the same proportion by mass. The mass of each element in the chemical formula is calculated and the
percent composition is taken from the formula mass.

The percent composition by mass is the percent by mass of each element in a compound. Percent composition
is obtained by dividing the mass of each element in 1 mole of the compound by the molar mass of the compound and
multiplying by 100 percent. Mathematically, the percent composition of an element in a compound is expressed as

𝑛 × 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑛 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = × 100
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑

where n is the number of moles of the element in 1 mole of the compound. For example, in 1 mole of sodium carbonate
(Na2CO3) there are 2 moles of Na atoms, 1 mole of C atom, and 2 moles of O atoms. The molar mass of Na2CO3= 105.96
g, Na = 22.99 g, C = 12.01 g, and O = 15.99 g. Therefore, the percent composition of Na2CO3 is calculated as follows:
2×22.99 𝑔
%Na = 105.96 𝑔 Na × 100 = 43.39%
2 CO3
1×12.01 𝑔
%C = 105.96 𝑔 Na × 100 = 11.33%
2 CO3
3×15.99 𝑔
%O = 105.96 𝑔 Na × 100 =45.27%
2 CO3
The sum of the percentages is 43.39% +11.33% + 45.27% = 99.99%. The small discrepancy from 100 percent is due
to the way we rounded off the molar masses of the elements. Another example of calculating the percent composition
by mass is shown below:

Example
Phosphoric acid (H3PO4 ) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in
carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in this
compound.
Solution: Assume that we have 1 mole of H3PO4. The percent by mass of each element (H, P, and O) is given
by the combined molar mass of the atoms of the element in 1 mole of H3PO4 divided by the molar mass of
H3PO4, then multiplied by 100 percent.

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Prepared by: Mr. Lawrence Niel A. Ignacio
 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the elements in H3PO4 is calculated as
follows:
3 × 1.008 𝑔
%H = × 100 = 3.09%
97.99 𝑔 𝐻3 𝑃𝑂4

1 × 30.97 𝑔
%P = × 100 = 31.61%
97.99 𝑔 𝐻3 𝑃𝑂4

4 × 15.99 𝑔
%O = × 100 = 65.27%
97.99 𝑔 𝐻3 𝑃𝑂4

To check let us get the sum of the percentages 3.09% +31.61% + 65.27 % = 99.97%. The small discrepancy
from 100 percent is due to the way we rounded off the molar masses of the elements.

Practice Exercise:
A. Tin (Sn) exists in Earth’s crust as SnO2 . Calculate the percent composition by mass of Sn and O in
SnO2.

B. Calculate the percent composition by mass of each of the elements in sulfuric acid (H2SO4).

The procedure used in the example can be reversed if necessary. Given the percent composition by mass of
a compound, we can determine the empirical formula of the compound by following the steps shown below.
Convert to Divide by the
grams and smallest Change to
Mass divide by Moles of each number of Mole ratios of integer Empirical
molar mass element moles elements subscript
Percent Formula

Because we are dealing with percentages and the sum of all the percentages is 100 percent, it is convenient to assume
that we started with 100 g of a compound, as the Example shows.

Example

Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92% carbon (C), 4.58% hydrogen (H), and 54.50%
oxygen (O) by mass. Determine its empirical formula.

Solution: Since we have 100% it is safe to assume that we have 100 g of ascorbic acid, then each percentage
can be converted directly to grams. In this sample, there will be 40.92 g of C, 4.58 g of H, and 54.50 g of O.
Because the subscripts in the formula represent a mole ratio, we need to convert the grams of each element
to moles. The conversion factor needed is the molar mass of each element. Let n represent the number of
moles of each element so that

1 𝑚𝑜𝑙 𝐶
𝑛𝐶 = 40.92 𝑔 𝐶 × = 3.407 𝑚𝑜𝑙 𝐶
12.01 𝑔 𝐶
1 𝑚𝑜𝑙 𝐻
𝑛𝐻 = 4.58 𝑔 𝐻 × = 4.544 𝑚𝑜𝑙 𝐻
1.008 𝑔 𝐻
1 𝑚𝑜𝑙 𝑂
𝑛𝑂 = 54.50 𝑔 𝑂 × = 3.408 𝑚𝑜𝑙 𝑂
15.99 𝑔 𝑂

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Thus, we arrive at the formula C3.407H4.544O3.408 , which gives the identity and the mole ratios of atoms present.
However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all
the subscripts by the smallest subscript (3.407).
3.407 4.544 3.408
C: = 1.00 H: = 1.33 O: ≈ 1.00
3.407 3.407 3.407

where the ≈ sign means “approximately equal to.” This gives CH1.33O as the formula for ascorbic acid. Next,
we need to convert 1.33, the subscript for H, into an integer. This can be done by a trial-and-error procedure:
1.33 × 1 = 1.33
1.33 × 2 = 2.66
1.33 × 3 = 3.99 ≈ 4

Because 1.33×3 gives us an integer (4), we multiply all the subscripts by 3:

C: 1 × 3 = 3
H: 1.33 × 3 = 3.99 ≈ 4
O: 1 × 3 = 3

This give us the empirical formula of ascorbic acid.

C3H4O3

Practice Exercise:
1. Determine the empirical formula of a compound having the following percent composition by mass: K:
24.75%; Mn: 34.77%; O: 40.51%.

2. What is the empirical formula of the compound with the following composition 40.1%C, 6.6%H,
53.3%O.

Chemical Reaction

Chemical reaction is a process in which a substance (or substances) is changed into one or more new
substances. To communicate with one another about chemical reactions, chemists have devised a standard way to
represent them using chemical equations. A chemical equation shows the identity of the reactants and products and
how much of each is consumed or produces in a chemical reaction.

Evidence of Chemical Reaction

A chemical reaction always involves formation of new substances different from the reactants. In others, there
is always a chemical change. The following are some evidences that a chemical reaction has occurred.
• Change in color
• Evolution of gas (bubbles are formed)
• Evolution or absorption of heat (warming, cooling, formation of sparks or flame)
• Change in composition (crystalline to powder form)
• Formation of a precipitate (an insoluble solid)
• Appearance or loss of odor

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Types of Chemical Reaction

There may be one or several reactants or products depending in the type of the chemical reaction. There are
five (5) basic types of chemical reactions.

Synthesis (or Combination) Reactions

Two or more reactants combine to form a single product:
𝐴 + 𝐵 → 𝐴𝐵
This can be a combination of a metal and nonmetal, forming an ionic compound.
2𝑀𝑔(𝑠) + 𝑂2 (𝑔) → 2 𝑀𝑔𝑂(𝑠)
It can also be a combination of nonmetals, producing covalent compound.
𝑁2 (𝑔) + 3 𝐻2 (𝑔) → 2 𝑁𝐻3 (𝑔)
It can even be an addition of two compounds such as the reaction between sulfur trioxide and water to form
sulfuric acid.
𝑆𝑂3 (𝑔) + 𝐻2 𝑂(𝑙) → 𝐻2 𝑆𝑂4 (𝑎𝑞)
Decomposition (or Analysis) Reactions
This is the reverse of synthesis reactions, where there is only one reactant dissociating into two or more
products. The products may be the constituent elements of the reactant or simpler compounds derived from the
reactant.
𝐴𝐵 → 𝐴 + 𝐵
An example of this type is the decomposition of an ionic compound into a metal and nonmetal.
2𝐶𝑢𝐶𝑙2 (𝑠) → 2𝐶𝑢(𝑠) + 𝐶𝑙2 (𝑔)
A covalent compound may also dissociate into its nonmetallic elements, for example the electrolysis of water.
2 𝐻2 𝑂(𝑙) → 2𝐻2 (𝑠) + 𝑂2 (𝑔)
A compound may also decompose to produce different new compounds.
𝐻2 𝐶𝑂3 (𝑎𝑞) → 𝐻2 𝑂(𝑙) + 𝐶𝑂2 (𝑔)
Single Replacement (or Substitution) Reactions
In a single replacement reaction, only one element is replaced from its compound during the reaction.
𝐴𝐵 + 𝑋 → 𝐴𝑋 + 𝐵
In the equation, X can either be a metal or a nonmetal. A more active free metal can replace a less active one
in a compound. The relative activities of several metals are organized in an activity series in such a way that it becomes
convenient to determine which between pair of metals is more active. From the activity series shown below, a more
reactive metal can replace a least reactive metal.

For example, magnesium can replace copper, but not sodium.

𝑀𝑔(𝑠) + 𝐶𝑢𝑆𝑂4 (𝑎𝑞) → 𝑀𝑔𝑆𝑂4 (𝑎𝑞) + 𝐶𝑢(𝑠)
𝑀𝑔(𝑠) + 𝑁𝑎𝐶𝑙 (𝑎𝑞) → 𝑛𝑜 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
A nonmetal X can also replace another nonmetal in a compound. For example, a halogen can replace another halogen
according the series.

Double Replacement (or Metathesis) Reactions

This type of reaction occurs when two ionic compounds exchange cations and anions with each other.
𝐴𝑋 + 𝐵𝑌 → 𝐴𝑌 + 𝐵𝑋
An example of which is the reaction between silver nitrate and sodium chloride to form an insoluble AgCl
precipitate.
𝐴𝑔𝑁𝑂3 (𝑎𝑞) + 𝑁𝑎𝐶𝑙(𝑎𝑞) → 𝑁𝑎𝑁𝑂3 (𝑎𝑞) + 𝐴𝑔𝐶𝑙(𝑠)

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Combustion
A hydrocarbon (compound containing carbon and hydrogen and sometimes oxygen) combines with oxygen
gas to produce carbon dioxide and water.
𝐶𝑥 𝐻𝑦 + 𝑂2 → 𝐶𝑂2 + 𝐻2 𝑂
Where x and y are the subscripts for carbon and hydrogen. For example, the reaction of propane and oxygen
gas to form carbon dioxide and water
𝐶3 𝐻8 + 𝑂2 → 𝐶𝑂2 + 𝐻2 𝑂

Writing Chemical Equations

A chemical equation uses chemical symbols to show what happens during a chemical reaction. In this section
we will learn how to write chemical equations and balance them.
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 → 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
For example, when magnesium is ignited, it produces a very bright glow as it reacts with oxygen in the air and
turns into ash after. This process can be presented in a chemical equation as follows:
Stoichiometric coefficients

2𝑀𝑔(𝑠) + 𝑂2 (𝑔) → 2 𝑀𝑔𝑂(𝑠)

Subscript

where the “plus” sign means “reacts with” and the arrow means “to yield.” Thus, this chemical equation is
read as: 2 moles of solid magnesium react with 1 mole of oxygen gas to produce 2 moles of 2 moles of solid magnesium
oxide. The reaction is assumed to proceed from left to right as the arrow indicates.
The numbers before the chemical formula or element symbol are called stoichiometric coefficients that
indicated the mole ratio among the reactants and products of reaction. If there is no coefficient written, it is understood
to be 1.
The shorthand (s), (l), or (g) may be written after a formula to indicate the physical state of compound. An (aq)
indicates aqueous, meaning the compound is dissolved in water.
The stoichiometric coefficients are not the same as the number subscripts usually found within chemical
formula. These subscripts only indicate how much of an atom is present for every one formula unit of a compound. Like
the stoichiometric coefficient, the subscript to understood to be 1 when there is no subscript written after an element
symbol.

Balancing Chemical Equations

The Law of conservation of mass states that matter can be neither created nor destroyed. Because matter is
made of atoms that are unchanged in a chemical reaction, it follows that mass must be conserved thus every chemical
reaction/equation must be balance.

Steps in Balancing Chemical Equation

1. Determine the number of atoms for each element.
2. Pick an element that is not equal on both sides of the equation.
3. Add a coefficient in front of the formula with that element and adjust your counts.
4. Continue adding coefficients to get the same number of atoms of each element on both sides

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Example

Mg + O2 → MgO
Step 1: Determine the number of atoms for each element on each side of the equation.
Mg + O2 → MgO
Reactants Products
Mg = 1 Mg=1
O= 2 O=1
Step 2: Pick an element that is not equal on both sides of the equation.
Mg + O2 → MgO Since the O atoms are
Reactants Products not equal, we’ll target
Mg = 1 Mg=1 those first.
O= 2 O=1
Step 3: Add a coefficient in front of the formula with that element and adjust your counts.
Mg + O2 → 2 MgO
Adding a 2 in front of MgO will
Reactants Products change the number of atoms
Mg = 1 Mg=1→ 2 on the product side of the
O= 2 O = 1→ 2 equation.

Step 4: Continue adding coefficients to get the same number of atoms of each element on both sides
2 Mg + O2 → 2 MgO
Reactants Products
Mg = 1→ 2 Mg=1→ 2
O= 2 O = 1→ 2

Now we need to increase the number of Mg atoms we have on the

reactants side. Adding a coefficient 2 in front of Mg will give us 2
atoms of Mg and balance the equation.

2 Mg + O2 → 2 MgO
Mg = 2 Mg= 2
O =2 O =2
Reactants = 80.58 g/mol Products = 80.58 g/mol

Practice Exercise:
Directions: Balance the following chemical equations.
1. Na + O2 → Na2O
2. N2 + O2 → N2O5
3. CH4 + O2 → CO2 + H2O
4. Li + AlCl3 → LiCl + Al
5. C7H16 + O2 → CO2 + H2O

Stoichiometry: Mass Relationship in Chemical Reaction

A balanced chemical equation shows the adherence of a chemical reaction to the law of conservation of mass,
which states that the mass of the reactants must be equal to the mass of the products in a chemical reaction. This
principle allows us to calculate the amount of a substance that a given amount of reactant can produce, or the amount
of reactant necessary for a certain quantity of product to be formed. Stoichiometry is a branch in chemistry that deals
with mass relationships between reactants and products based on a balanced chemical equation.

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
This mass relationship can be summarized in the following steps.
1. Write the balanced equation.
2. Convert the given amount of the
reactant (in grams or other units) to number of
moles.
3. Use the mole ratio from the balanced equation
to calculate the number of moles of product
formed.
4. Convert the moles of product to grams (or
other units) of product.

Example

A. If Iron pyrite, FeS2, is not removed from coal, oxygen from the air will combine with both the iron and the
sulfur as coal burns. If a furnace burns an amount of coal containing 100 g of FeS2, how much SO2 (an air
pollutant) is produced?

Solution: Follow the steps

Step 1: Balance the chemical equation
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2

Step 2: Convert the grams of FeS2 to moles

1 𝑚𝑜𝑙 FeS2
100g FeS2 × = 0.833 𝑚𝑜𝑙 FeS2
119.977 𝑔 FeS2

Step 3: From the mole ratio, we see that for every 4 mol of FeS2 there is 8 mol of SO2. Therefore, the number
of moles of SO2 formed is

8 𝑚𝑜𝑙 SO2
0.833 𝑚𝑜𝑙 FeS2 × = 1.666 𝑚𝑜𝑙 SO2
4 𝑚𝑜𝑙 FeS2

Step 4: Finally, the number of grams of SO2 formed is

64.064 𝑔 SO2
1.666 𝑚𝑜𝑙 SO2 × = 106.731 𝑔 SO2
1 𝑚𝑜𝑙 SO2

We can combine this conversion steps

𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝐹𝑒𝑆2 → 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝐹𝑒𝑆2 → 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑆𝑂2 → 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓𝑆𝑂2

into one equation:

1 𝑚𝑜𝑙 FeS2 8 𝑚𝑜𝑙 SO2 64.064 𝑔 SO2

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑆𝑂2 = 100g FeS2 × × ×
119.977 𝑔 FeS2 4 𝑚𝑜𝑙 FeS2 1 𝑚𝑜𝑙 SO2

= 106.731 𝑔 SO2

B. Determine how many grams of water are produced in the oxidation of 1.00 g of glucose C6H12O6.
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)

Solution:
1 𝑚𝑜𝑙 C6 H12 O6 6 𝑚𝑜𝑙 H2 O 18.015 𝑔 H2 O
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻2 O = 1.00 𝑔 C6 H12 O6 × × ×
180.156 𝑔 C6 H12 O6 1 𝑚𝑜𝑙 C6 H12 O6 1 𝑚𝑜𝑙 H2 O

= 0.599 𝑔 H2 O

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Practice Exercise:
A. Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas:
Si(s) + 2 Cl2 (g) → SiCl4(l)
In one reaction, 0.507 mole of SiCl4 is produced. How many grams of molecular chlorine were used in the
reaction?

B. Consider the combustion of butane (C4H10):

2 C4H10 (g) + 13 O2(g) → 8 CO2(g) + 10 H2O (l)
In a reaction, 5.0 moles of C4H10 are reacted with an excess of O2 How many moles of CO2 and how many
grams of H2O are formed.

Limiting and Excess Reagents

The goal of a chemical reaction is to produce the maximum quantity of a useful compound from the raw
materials, most of the times a large excess of one reactant is used to ensure that the more expensive reactant is
completely converted into desired product. Therefore, some reactant will be left over at the end of the reaction. The
reactant used up first in a reaction is called the limiting reagent, because the maximum amount of product formed
depends on how much of this reactant was used. When this reactant is used up, no more product can be formed.
Excess reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting
reagent. The concept of the limiting reagent is analogous to the relationship between student and books. If there are
14 students and only 9 books, then only 9 students can have a copy of the book. Five (5) students will be left without a
copy of the book. The number of books thus limits the number of students that can have a copy of the book, and there
is an excess of students.

For example, the synthesis of methanol (CH3OH) from CO and H2 at high temperature.
CO (g) + 2 H2(g) → CH3OH
Let’s say for our initial amount we have 4 moles of CO and 6 moles of H2. One way to determine which of the
two reactants is the limiting reactants is to calculate the moles of CH3OH produce based on CO and H2. From the
preceding definition, we see that only the limiting reagent will yield the smaller amount of the product. Starting with 4
moles of CO, we calculate the number of moles of CH3OH produced is
1 𝑚𝑜𝑙 CH3 OH
4 𝑚𝑜𝑙 CO × = 4 𝑚𝑜𝑙 CH3 OH
1 𝑚𝑜𝑙 CO

Next with 6 moles of H2 the number of moles CH3OH formed is

1 𝑚𝑜𝑙 CH3 OH
6 𝑚𝑜𝑙 H2 × = 3 𝑚𝑜𝑙 CH3 OH
2 𝑚𝑜𝑙 H2

Since H2 results in smaller amount of CH3OH it is the limiting reagent and CO is the excess reagent.
In stoichiometric calculations involving limiting reagents, the first step is to decide which reactant is the limiting
reagent. After the limiting reagent has been identified, problems involving mass relationship can be solved as shown in
the example below.
Example
Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide:
2NH3 (g) + CO2 (g) → (NH2)2CO (aq) + H2O (l)
In one process, 637.2 g of NH3 are treated with 1142 g of CO2 . (a) Which of the two reactants is the limiting
reagent? (b) Calculate the mass of (NH2)2CO formed. (c) How much excess reagent (in grams) is left at the end
of the reaction?

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Prepared by: Mr. Lawrence Niel A. Ignacio
 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Solution:
(a).
First to look for the limiting reagent, we need to calculate the number of moles of (NH2)2CO produced from
637.2 g of NH3 and 1142 g of CO2 using the following conversion.

grams of NH3 → moles of NH3 → moles of (NH2)2CO

1 𝑚𝑜𝑙 NH3 1 𝑚𝑜𝑙 (NH2 )2 CO

637.2 g NH3 × × = 18.707 𝑚𝑜𝑙 (NH2 )2 CO
17.031 g NH3 2 𝑚𝑜𝑙 NH3

grams of CO2 → moles of CO2 → moles of (NH2)2CO

1 𝑚𝑜𝑙 CO2 1 𝑚𝑜𝑙 (NH2 )2 CO

1442 g CO2 × × = 32.766 𝑚𝑜𝑙 (NH2 )2 CO
44.009 g CO2 1 𝑚𝑜𝑙 CO2

It follows therefore that NH3 must be the limiting reagent since it produces a smaller amount of (NH2)2CO.

(b)
To calculate the mass of (NH2)2CO formed we will be using 18.707 as the number of moles since NH3 is the
limiting reagent.

60.056 𝑔 (NH2 )2 CO
18.707 𝑚𝑜𝑙 (NH2 )2 CO × = 1123.468 𝑔 (NH2 )2 CO
1 𝑚𝑜𝑙 (NH2 )2 CO

(c)
We can determine the amount of CO2 that reacted with NH3 to produce 18.707 moles of (NH2)2CO. starting
with 18.707 moles of (NH2)2CO we can calculate the mass of CO2 reacted using the molar ratio using the
following steps.
moles of (NH2)2CO → moles of CO2 → grams of CO2

1 𝑚𝑜𝑙 CO2 44.009 𝑔 CO2

18.707 𝑚𝑜𝑙 (NH2 )2 CO × × = 823.276 𝑔 CO2
1 𝑚𝑜𝑙 (NH2 )2 CO 1 𝑚𝑜𝑙 CO2

The amount of CO2 excess is the difference between the initial amount (1442 g) and the amount reacted
(823.276 g).
Excess amount of CO2 = 1142 g – 823.276 g =318.724 g

Practice Exercise:
A. The reaction between aluminum and iron(III) oxide can generate temperatures approaching 3000⁰C and is
used in welding metals:
2 Al + Fe2O3 → Al2O3 + 2Fe

In one process, 124 g of Al are reacted with 601 g of Fe2O3. (a) Calculate the mass (in grams) of Al2O3 formed.
(b) How much of the excess reagent is left at the end of the reaction?

B. If steel wool (assume it is pure iron) is heated until it glows and is placed in a bottle containing pure oxygen,
the iron reacts spectacularly to produce iron(III) oxide.
Fe(s) + O2(g) → Fe2O3(s)
If 1.25 grams of iron is heated and placed in a bottle containing 0.0204 moles of oxygen gas, what mass of
iron(III) oxide is produced? What is the limiting reactant?

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Prepared by: Mr. Lawrence Niel A. Ignacio
 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
Percent Yield

To determine how efficient a given reaction is, chemists often figure the percent yield, which describes the
proportion of the actual yield to the theoretical yield. It is calculated as follows:
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 = × 100
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑

Example
The combustion of pyrite (FeS2) with oxygen gas produces hematite (Fe2O3) and sulfur dioxide
4 FeS2 (s) + 11 O2 (g) → 2 Fe2O3 (s) + 8 SO2 (g)
If 75.0 g of FeS2 is reacted with O2 (a) calculate the theoretical yield of Fe2O3 in grams. (b) Calculate the percent
yield if 45.7 g of FeS2 are obtained.
Solution:
(a)
grams of FeS2 → moles of FeS2 → moles of Fe2O3 → grams of Fe2O3

1𝑚𝑜𝑙 FeS2 2 𝑚𝑜𝑙 Fe2 O3 159.687 𝑔 Fe2 O3

75.0 𝑔 FeS2 × × × = 49.912 𝑔 Fe2 O3
119.977 𝑔 FeS2 4 𝑚𝑜𝑙 FeS2 1 𝑚𝑜𝑙 Fe2 O3

(b)

𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 = × 100
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑

45.7 𝑔
= × 100 = 91.56%
49.912 𝑔

Practice Exercise:
A. Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in
the production of aluminum metal. It is prepared by the reaction
CaF2 + H2SO4 → CaSO4 + 2 HF
In one process, 6.00 kg of CaF2 are treated with an excess of H2SO4 and yield 2.86 g of HF. Calculate the
percent yield of HF.

Network Links, Digital Sources and References

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https://courses.lumenlearning.com/suny-mcc-introductorychemistry/chapter/formula-mass-and-mole-
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https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistr
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o%20more%20than%20one%20category.

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 OLIVAREZ COLLEGE
DR. PABLO R. OLIVAREZ – SENIOR HIGH SCHOOL
STEM 004: General Chemistry 1
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https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)
/Chemical_Reactions/Stoichiometry_and_Balancing_Reactions#:~:text=Stoichiometry%20is%20a%20section
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Raymond Chang (2010) Chemistry 10th Edition

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