FACULTY OF SCIENCE, TECHNOLOGY AND
MATHEMATICS (FSTeM)
2S MAT 21 (Mathematical Methods for the Sciences)
Activity Sheet #3. Basic Stoichiometry and the mole concept
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Introduction
Stoichiometry is the branch of chemistry that deals with the calculation of the amount of
reactants and products in a chemical reaction. Essential to the understanding of the
basics of stoichiometry is the concept of a mole. The Mole Concept was introduced by
was introduced by Amadeo Avogadro. In his experiment, he found that, “equal volume
of gases, at the same temperature and pressure, contain equal numbers of molecules.”
Initially, the number of molecules of gas in the volume is unknown to him. Eventually, he
was able to hypothesize and determine that the number is equal to:
NA = 6.02 x 1023
Which was eventually known to be Avogadro’s number.
Content Proper
The value, NA = 6.02 x 1023 is known as the measure of a mole. The mole refers
to the amount unit of the amount of substance which is one of the fundamental units in
the System International system. We can generally call the constituent elements of a
gas to be particles but may vary in terms depending on the composition of the material.
The amount of substance in a material may be expressed in terms of ion pairs for ionic
compounds, molecules for covalent compounds and atoms for metallic compounds.
Another fundamental concept to understanding stoichiometry is the molar mass
concept. The molar mass of a chemical compound is the mass of a sample of the
compound divided by the amount of substance in that compound expressed in grams
per mole. The molar mass of atoms of elements is given by the standard atomic weight
(or the relative atomic mass) of the element multiplied by the molar mass constant, M =
1 g/mol. You can consult the periodic table of elements to check the different atomic
masses to determine the molar mass of the 118 elements. Examples:
For Hydrogen, M(H) = 1.01(1 g/mol) = 1.01 g/mol
Chlorine, M(Cl) = 32.07(1 g/mol) = 32.07 g/mol
For a given compound, we determine the number of atoms and use it as
multiplier.
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FACULTY OF SCIENCE, TECHNOLOGY AND
MATHEMATICS (FSTeM)
2S MAT 21 (Mathematical Methods for the Sciences)
Examples:
For water: H2O,
M(H)(2) = 1.01(1 g/mol)(2) = 2.02 g/mol
M(O)(1) = 16.00(1 g/mol)(1) = 16.00 g/mol
Sum = 18.02 g/mol
For Sodium Chloride, NaCl:
M(Na)(1) = 22.99(1 g/mol)(1) = 22.99 g/mol
M(Cl)(1) = 35.45(1 g/mol)(1) = 35.45 g/mol
Sum = 58.44 g/mol
In solving stoichiometry problems, we consider the following:
First, you will need to balance a chemical reaction. Then you determine the ratio
of the quantities and solve:
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FACULTY OF SCIENCE, TECHNOLOGY AND
MATHEMATICS (FSTeM)
2S MAT 21 (Mathematical Methods for the Sciences)
Student Exercises
I. Task: Determine the atomic number, atomic mass, and molar mass of the
following elements:
Elements
Atomic Mass
(grams)
Atomic Number
Molar Mass (g/mol)
1) Lithium
2) Iron
3) Rutherfordium
4) Cesium
5) Carbon
6) Platinum
7) Barium
8) Gold
II. Task: Determine the molar mass and the percentage composition of the
following compounds made of two elements
A. Heteronuclear Molecules
Water
H2O
Sulfur Dioxide
SO2
Ammonia
NH3
Carbon Dioxide
CO2
Hydrogen Chloride
HCl
%
Total
Quantity
m
Element 2
ELEMENT 2
%
Total
Quantity
Chemical
Formula
m
Compound
Element 1
ELEMENT 1
Molar
Mass
(g/mol
)
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FACULTY OF SCIENCE, TECHNOLOGY AND
MATHEMATICS (FSTeM)
2S MAT 21 (Mathematical Methods for the Sciences)
B. Ionic Compounds
Sodium Chloride
NaCl
Lithium Fluoride
LiF
Sodium Iodide
NaI
Cesium Bromide
CsBr
Calcium Chloride
CaCl2
%
Total
Quantity
m
Element 2
ELEMENT 2
%
Total
Quantity
m
Compound
Chemical
Formula
Element 1
ELEMENT 1
Molar
Mass
(g/mol
)
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FACULTY OF SCIENCE, TECHNOLOGY AND
MATHEMATICS (FSTeM)
2S MAT 21 (Mathematical Methods for the Sciences)
Part 1: Mole ←→ Mass Conversions
Convert the following number of moles of chemical into its corresponding mass in
grams.
1. 0.436 moles of ammonium chloride
2. 2.360 moles of lead (II) oxide
3. 0.031 moles of aluminum iodide
4. 1.077 moles of magnesium phosphate
5. 0.50 moles of calcium nitrate
Convert the following masses into their corresponding number of moles.
6. 23.5 g of sodium chloride
7. 0.778 g of sodium cyanide
8. 0.250 g of water
9. 169.45 g of calcium acetate
10. 79.9 g of potassium permanganate
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FACULTY OF SCIENCE, TECHNOLOGY AND
MATHEMATICS (FSTeM)
2S MAT 21 (Mathematical Methods for the Sciences)
Part 2: Moles ←→ Number of Particles Conversions
Convert the following number of moles into their corresponding number of particles.
11. 0.0455 moles of hydrochloric acid
12. 1.2 moles of glucose (C6H12O6)
13. 0.32 moles of sodium bicarbonate
Part 3: Solve the following stoichiometry grams-grams problems:
1) The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of
water.
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
(a) How many moles of water formed?
(b) How many moles of butane burned?
(c) How many grams of butane burned?
(d) How much oxygen was used up in moles?
(e) How much oxygen was used up in grams?
2) Using the following equation:
2 NaOH + H2SO4  2 H2O + Na2SO4
How many grams of sodium sulfate will be formed if you start with 200 grams of sodium
hydroxide and you have an excess of sulfuric acid?
3) Using the following equation:
Pb(SO4)2 + 4 LiNO3  Pb(NO3)4 + 2 Li2SO4
How many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate,
assuming that you have an adequate amount of lead (IV) sulfate to do the reaction?
4) Calculate how many grams of iron can be made from 16.5 grams of Fe 2O3 by the
following equation:
Fe2O3 + 3 H2 -------------> 2 Fe + 3 H2O
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