Chapter 1 Fundamental Concepts of Chemistry
1.1 Atoms, Molecules, and Compounds
1.2 Measurements in Chemistry
1.3 Chemical Problem Solving
1.4 Counting Atoms: The Mole
1.5 Amounts of Compounds
1.6 Aqueous Solutions
1.7 Writing Chemical Equations
1.8 The Stoichiometry of Chemical Reactions
1.9 Yields of Chemical Reactions
1.10 The Limiting Reagent
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.1 Atoms, Molecules and Compounds
Learning Objectives:
recognize elemental symbols and names of the
elements
name compounds from molecular pictures
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.1 Atoms, Molecules and Compounds
 What is matter made of?

Democritus had the answer: All matter was made of very
small particles – much too small to see and he called the small
particles atoms.

From the Greek word atomos meaning “uncuttable”
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Macro vs. Micro
 Macroscopic – able to be seen by the human eye.
 Microscopic – requires the use of a magnifying device

Allows us to see the underlying structure of a chemical
substance
Can we go any smaller?
Atomic View – reveals the building blocks of matter
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Classifications of Matter
Matter can be divided into two main classes
1.
Atoms (and elements)
2.
Molecules
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How Small are Atoms?
A sample of carbon the size of a period at the end of a
printed sentence contains more atoms than the
number of stars in the Milky Way.
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Atomic Colour Coding
Fig 1-4 Pg 7
Color-coded scale models of 10 types of atoms that appear
frequently in this book.
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Elements
 A substance that contains only one type of atom is
called a chemical element.
 Each has
A unique name
 A unique number (Z)

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Molecules
 A molecule is a
combination of two or
more atoms held
together in a specific
shape by attractive
forces.
 The simplest is a
molecule containing two
of the same atoms.
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Chemical Formulas
 Compound – a substance that contains more
than one element.
For example, water is composed of hydrogen
and oxygen
The chemical formula describes the atomic
composition of the elements.
Water’s chemical formula is H2O – 2 hydrogen
atoms and 1 oxygen atom
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Some Simple Molecules
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Exercises
What are the chemical formulas of the following compounds?
1.1.1 What are the elemental symbols for cerium, cesium,
copper, calcium and carbon?
1.1.2 What are the names of the elements represented by
the symbols Zr, Ni, Sn, W, Se, Be and Au?
1.1.3 Molecular pictures of some molecules are shown here.
What are their chemical formulas?
1
2
3
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4
5
1.2 Measurements in Chemistry
Learning Objectives:
recognize the SI units commonly used in chemistry, and
perform some common unit conversions
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1.2 Measurements in Chemistry
 Size – defined by length, area and volume
 Mass (m) – the certain quantity of matter
 Time (t) – to determine how long it takes for a chemical
transformation to take place
 Temperature (T) – determines the direction of heat
flow
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Frequently Used Symbols for Magnitudes
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Units
 Système International (SI) expresses each fundamental
physical quantity in decimally related units
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Unit Conversions
 We frequently need to convert a measurement
from one unit to another.
 When multiplying numbers we also multiply units
and when dividing numbers we also divide units
 Never forget the units!
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Temperature
Two scales are commonly used:
1. Centigrade or Celsius (°C)
The boiling point of water is set at 100 °C
and the freezing point at 0 °C.
2. Kelvin (K) (also called Absolute scale)
Has the same divisions as the Celsius scale,
but a different zero point -273.15 °C = 0 K
0 K is referred to as absolute zero…WHY?
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Temperature
T(K) = T(°C) + 273.15
T (°C) = T (K) - 273.15
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Precision and Accuracy
The patterns of bullets striking a target illustrate the notions
of precision and accuracy.
(a) A precise and accurate pattern
(b) A precise but inaccurate pattern
(c) An imprecise and inaccurate pattern
(d) An imprecise pattern with one accurate shot
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Significant Figures
 The number of digits expressed in a numerical value is
called the number of significant figures.
 How to determine “sig figs”
Read the number from left to right and count all the digits
starting with the first non-zero digit
 Place a decimal point after the value when its trailing zeros are
significant

110 has 2 sig figs
 110. has 3 sig figs

Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Precision in Calculations
1.
2.
3.
When adding or subtracting, the number of decimal
places in the result is the number of decimal places in
the number with the fewest places.
When multiplying or dividing, the number of
significant figures in the result is the same as in the
quantity with the fewest significant figures.
Postpone adjusting result to the correct number of
significant figures until a calculation is complete
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Section Exercises
What are the chemical formulas of the following compounds?
 1.2.1 Convert the following measurements to scientific
notation and express in base SI units: 0.000 463 L, 17 935
km, and 260 000 h (precise to three significant figures).
 1.2.2 One light-year is the distance light travels in exactly
one year. The speed of light is 1.08 x 109 km/h. Express the
speed of light and the length of one light-year in SI units.
 1.2.3 Convert each of these measurements to SI units: 155
pounds (mass of a typical person, 1 kg = 2.2 pounds), 110.0
yards (full length of a Canadian football field, 1 yard =
91.44 cm), 39 °C (body temperature of someone with a
slight fever), and 365.2422 days (length of one year).
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1.3 Chemical Problem Solving
Learning Objective:
Analyze and solve problems in a consistent, organized fashion
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.3 Chemical Problem Solving
Step 1 Determine what is asked for. What type of problem is it? What are you
asked to find?
Step 2 Visualize the problem. Draw pictures that illustrate what takes place.
If chemical changes occur, draw molecular pictures.
Step 3 Organize the data. What data are available? How are the data related to
what is asked for?
Step 4 Identify a process to solve the problem. What concepts are required?
What equations apply?
Step 5 Manipulate the equations. If necessary, do calculations in steps. Solve for
what is asked for.
Step 6 Substitute and calculate. Keep track of units. Use the correct number of
significant figures.
Step 7 Does the result make sense? Are the units consistent? Is the result
sensible?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 1
Electroplating is a process by which a metal such as
copper is coated with another metal such as silver
or chromium. The transfer of metal atoms is
driven by an electrical current. In an electroplating
process, a spoon is coated with silver from a silver
rod. In the process, 1.0 x 1021 atoms are
transferred from the rod to the spoon, and the
rod loses 0.179 g of its mass. Use this information
and the density of silver (10.50 g/cm3) to estimate
the volume occupied by one silver atom.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.4 Counting Atoms: The Mole
Learning Objective:
Solve mass-number-molecular weight type problems
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1.4 Counting Atoms: The Mole
Mole – the amount of substance that contains as many
elementary entities (atoms, molecules, or other
particles) as there are atoms in exactly 12 g of 12C
1 mol = 12 g of 12C
But how many atoms are there in 12 g of 12C?
Using a mass spectrometer, it was determined that
1 atom of 12C = 1.992646 x 10-23 g
So, let’s do the math…
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
The Mole (aka Avogadro’s Number)
1 mole  6.022  10
23
 Avogadro’s number is a conversion unit: 1 mole =
6.022 x 1023 “items”
 “items” = atoms, molecules, atoms, etc…
 The symbol is NA = 6.022 x 1023 “items”/mol
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Moles and Isotopic Mass
 Most elements are found on the earth as a mixture
of isotopes in a constant ratio.
 A single atom of each isotope has a unique mass.
 However, is it useful to discuss a single atom?
 So instead, we think of a mole of atoms. How much
does a mole of atoms of an isotope weigh?
 Let’s look at 13C. Its atomic mass is 2.15 x 10-23 g;
what is the mass of one mole of 13C?
 Is this the mass for carbon on the periodic table?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Molar Mass
 We must take into account all the isotopes of each
element and their percent abundance. (What is this?)
 Molar mass of an element is a weighed average of the
masses of all the isotopes of one element. Once
calculated, it is the mass of one mole of atoms of the
elements.
Elemental molar mass =
 (Fractional abundance)(Isotopic molar mass)
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Mass-Mole-Atom Conversions
 Molar mass can be thought of as a conversion factor.
m
n
M
 For instance M (Ag) = 107.86 g/mol
107.86 g Ag = 1 mol of Ag
 If a bracelet contains 168 g of Ag, how many moles of
silver is this?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
The flowchart shows the conversion processes among
the number of moles, the mass, and the number of
atoms for a sample of an element.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.5 Amounts of Compounds
Learning Objective:
Perform mole-mass-number conversions
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1.5 Amounts of Compounds
 The molar mass of compound is found by adding
together the molar masses of all of its elements,
taking into account the number of moles of each
element present.
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Example 1-3 Converting Line Structures
Construct the formulas and determine the chemical
formulas from the following line drawings:
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Example 1-4 Molar Mass of a Molecule
The line drawing and chemical formula of Sevin, a
common insecticide, appears below. Determine the
molar mass of Sevin.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1-5
Metal salts often exist as hydrates. One example is
iron(II) nitrate hexahydrate. Determine the molar mass
of this compound.
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Mass-Mole-Number
Conversions for Compounds
Just as one can convert between grams and moles using
atomic molar mass, one can convert using the molar
mass of compounds
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 6
Ammonium nitrate (NH4NO3) is used as a fertilizer
because it is a good source of nitrogen atoms. It is such
a good source, in fact, that it ranks among the top 15
industrial chemicals produced yearly in the world. How
many moles of nitrogen atoms are present in a 454 g
bag of NH4NO3 fertilizer? How many atoms is this?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.6 Aqueous Solutions
Learning Objective:
Calculate concentrations of solutions and diluted
solutions
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1.6 Aqueous Solutions
 Solution – a homogeneous mixture of two or
more substances, in which one is called the
solvent and the other is the solute
Solvent – main medium, what is there in the largest
quantity
 Solute – dissolved in the solvent

Sugar and Water
Sugar (solute) and Water (solvent)
NaCl and Water
NaCl (solute) and Water (solvent)
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Concentration
Concentration – the amount of a solute dissolved in a
given quantity of solvent
A concentrated solution has a large amount of solute in
it.
A dilute solution has a small amount of solute in it.
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Molarity
The concentration unit which is most useful is molarity.
Molarity – the number of moles of solute dissolved in 1 L
of solution
quantity of A (mol)
n
Molarity of A (c) 

Volume of Solution (L) V
Just like molar mass, molarity can be used as a
conversion unit.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 7
 What mass of nickel(II) chloride hexahydrate is required
to prepare 250. mL of aqueous solution with a
concentration of 0.255M?
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Species in Solution
What chemical species are in an aqueous solution?
Water will always be present in excess.
There will also be solute species that are either molecules
or ions.
When ionic compounds dissolve in water, they dissociate
into their ions.
Remember: polyatomic ions remain intact when a salt
dissolves in water.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 8
Find the molarities of the ionic species present in 250. mL
of an aqueous solution containing 1.75 g of ammonium
sulfate (NH4)2SO4.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Dilutions
What if you have a concentrated solution and you
would like to dilute it to another concentration?
Concentrationi × Volume i = Concentration f × Volume f
ci Vi =c f Vf
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 – 9 Dilution of a Solution
Aqueous hydrochloric acid, HCl, is usually sold as a 12.0
M solution, commonly referred to as concentrated HCl.
A chemist needs to prepare 2.00 L of 1.50 M HCl for a
number of different applications. What volume of
concentrated HCl solution should the chemist use in
the dilution?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 – 10 Preparing Solutions
An agricultural chemist wished to study the effect of
varying fertilizer applications on the growth of tomato
plants. The chemist prepared a stock aqueous solution
of urea, (NH4)2CO, by dissolving 1.75 g of this
compound in water to make 1.00 L of solution. Then,
she prepared a series of more dilute solutions to apply
to her tomato plants. One of these solutions contained
5.00 mL of stock solution diluted to give a final volume
of 25.00 mL. What was the concentration of urea in this
diluted solution?
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1.7 Writing Chemical Equations
Learning Objective:
Balance chemical reactions
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1.7 Writing Chemical Equations
Molecular nitrogen reacts with molecular hydrogen to
yield ammonia.
Molecular nitrogen + molecular hydrogen → ammonia
N2 + H2 → NH3
Is this a correct representation of what happens?
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Using the molecular formulas, we are able to
express what occurs in the chemical reaction
N2 + H2 → NH3
We do have one problem; it does not give amounts
correctly (it is not balanced).
In chemical reactions, atoms cannot be created or
destroyed.
N2 + 3 H2 → 2 NH3
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Stoichiometry…
 The relationship between quantities in chemical
reactants and products in a balanced chemical
equation.
 The coefficients in a balanced equation are called the
stoichiometric coefficients.
 The reactants are consumed in a reaction, while the
products are produced.
 Identify the reactants, products and stoichiometric
coefficients below.
2 Mg + O 2  2 MgO
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Balancing Equations
To balance, we place counting numbers in front of the
formulas until the equation is balanced.
2. The numbers are called stoichiometric coefficients and
they represent numbers of atoms, molecules or moles of
reactants and products.
3. In a balanced equation, there must be the same number
of atoms of each element on both sides.
4. And, the charge must be conserved (What does this
mean?)
1.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 11
Ammonium nitrate (NH4NO3) explodes when it is heated
above 300 °C. The products are three gases: molecular
nitrogen, molecular oxygen and steam (water vapor).
Write a balanced equation for the explosion of
ammonium nitrate.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Balancing Combustion Equations
C 5H12 +
O2 
CO 2 +
H2 O
Combustion equations involve hydrocarbons
burning in oxygen to form carbon dioxide and
water
• Step 1: Balance the carbon atoms
• Step 2: Balance the hydrogen atoms
• Step 3: Balance the oxygen atoms
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.8 Stoichiometry of Chemical Reactions
Learning Objective:
Calculate the amount of a product from the amounts of the
reactants and a balanced chemical reaction
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1.8 Stoichiometry of Chemical Reactions
Once an equation is balanced, you can use it to
relate reactants and products.
The key is the stoichiometric
coefficients!
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Making a Pancake
1
1
1 egg + cup milk + cup flour + 1 tbsp sugar  1 pancake
2
2
If I have 2 eggs and an unlimited supply of the rest
of the ingredients how many pancakes can I
make?
From practical knowledge, you would say 2
pancakes.
In chemistry, stoichiometric calculations are based
on the mole.
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N2 + 3 H2 → 2 NH3
How many grams of hydrogen are need to produce 68 g
of ammonia?
Step 1: Write what you know
Step 2: Ask yourself, how can I relate this to what I need?
Once your problem is in units of moles, you should think to
yourself, “I should be using the balanced equation!”
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
N2 + 3 H2 → 2 NH3
First you must convert grams to moles
1 mol NH3
68 g NH3 ×
17.0 g NH3
Once you have moles, you can use the balanced equation
1 mol NH3 3 mol H2
68 g NH3 ×
×
17.0 g NH3 2 mol NH3
Finally, convert moles back to grams
1 mol NH3 3 mol H2 2.02 g H2
68 g NH3 ×
×
×
= 12 g H2
17.0 g NH3 2 mol NH3 molH2
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 12
Geranyl formate is used as a synthetic rose essence in cosmetics.
The compound is prepared from formic acid and geraniol:
HCO2H + C10H18O → C11H18O2 + H2O
A perfumery needs some geranyl formate for a batch of perfume.
How many grams of geranyl formate can a chemistry make from
375 g of geraniol?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 13
Poisonous hydrogen cyanide (HCN) is an important
industrial chemical. It is produced from methane (CH4),
ammonia, and molecular oxygen. The reaction also
produces water. An industrial manufacturer wants to
convert 175 kg of methane into HCN. How much
molecular oxygen will be required for this synthesis?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Reality Check…
 When you mix pancakes, do you always make the
amount that the box says with the correct ingredients?
 Or, when baking cookies…it says you can make 3 dozen,
can you really?
 The same follows for chemical reactions. Almost
always, chemical reactions produce smaller amounts of
products than predicted by stoichiometric analysis.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
WHY?
1.
2.
3.
Many reactions stop before reaching completion.
Competing reactions often consume starting materials.
When the product of a reaction is purified and isolated,
some of it is inevitably lost during the collection process.
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1.9 Yields of Chemical Reactions
Learning Objective:
Calculate reaction yields
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1.9 Yields of Chemical Reactions
 Theoretical amount – the maximum quantity of a
product that can be obtained from a chemical
reaction (in a perfect world)
 Actual amount – the quantity of a product that is
actually obtained from a chemical reaction in the
laboratory
actual amount
Percent Yield 
 100
theoretical amount
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 14
According to Example 1-12, it is possible to make 443 g
of geranyl formate from 375 g of geraniol. A chemist
making geranyl formate for the preparation of a
perfume uses 375 g of starting material and collects
417 g of purified product. What is the percent yield
of this synthesis?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 15
The industrial production of hydrogen cyanide is
described in Example 1-13. If the yield of this synthesis
is 97.5%, how many kilograms of methane should be
used to produce 1.50 x 105 kg of HCN?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
1.10 The Limiting Reactant
Learning Objective:
Solve limiting-reagent-type problems
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1.10 The Limiting Reactant
1
1
1 egg + cup milk + cup flour + 1 tbsp sugar  1 pancake
2
2
Let’s return to the pancake example.
If I have 12 eggs, 4 L of milk, 1 cup of flour and 1 kg
of sugar, what is my limiting reactant?
Limiting reactant – determines or limits the
amount of product formed.
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Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
If we start with six molecules of H2 and four
molecules of N2, the six molecules of H2 will
combine with two molecules of N2 to make four
molecules of NH3
 Two molecules of N2 will be left over.
 The H2 is the limiting reactant
 Once again, it is best to work in units of moles.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
N2 + 3 H2 → 2 NH3
If we start with 84.0 g N2 and 24.2 g of H2, what mass of ammonia can be
prepared?
1. Convert each mass of reactant to moles using molar mass.
84 g N2 ×
1 mol N2
=3.00 mol N2
28.01 g N2
24.2 g H2 ×
1 mol H2
=12.00 mol H2
2.016 g H2
2. Identify the limiting reactant by dividing each amount in moles by the
stoichiometric coefficient for that reactant.
12.0 mol H2
=4
3 mol H2
3.00 mol N2
=3
1 mol N2
3. The reactant with the smaller reaction will be the limiting reactant. N2
in this case. Use the moles of N2 to calculate the mass of the NH3
produced.
3.00 mol N2 ×
2 mol NH3 17.03 g NH3
×
= 102. g NH3
1 mol N2
1 mol NH3
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Table of Amounts
 Organizes data, and summarizes the calculations of a
stoichiometry problem
Reaction
N2
Starting amount (mol) 3.00
Change in amount (mol) -3.00
Final amount (mol)
0
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
+
3 H2
12.0
-9.00
3.0
→
2 NH3
0
+6.00
+6.00
Example 1 - 16
Nitric acid, a leading industrial chemical, is used in the
production of fertilizers and explosives. One step in the
industrial production of nitric acid is the reaction of
ammonia with molecular oxygen to form nitrogen
oxide:
4 NH3 + 5 O2 → 4 NO + 6 H2O
In a study of this reaction, a chemist mixed 125 g of
ammonia with 256 g of oxygen and allowed them to
react to completion. What masses of NO and H2O were
produced, and what mass of which reactant was left
over?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 1 - 17
The synthesis of aspirin appears in Section Exercise
1.9.3:
Suppose a chemist started with 152 g of salicylic
acid and 86.8 g of acetic anhydride and produced
133 g of aspirin. What is the yield of this
reaction?
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 1 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.