TOPIC 1
STOICHIOMETRIC
RELATIONSHIPS
1.3
REACTING MASSES AND
VOLUMES
ESSENTIAL IDEA
Mole ratios in chemical equations
can be used to calculate reacting
ratios by mass and gas volume.
NATURE OF SCIENCE (1.8)
Making careful observations and obtaining
evidence for scientific theories – Avogadro’s
initial hypothesis.
STOICHIOMETRY
 For
any stoichiometry problem, you must
know what unit you are given and what
unit you are looking for.
 Step one is to ALWAYS convert to moles
unless you are already in moles.
 Step two is to multiply by the mole ratio
with the unknown on top.
 Step three is to convert to the unit you are
looking for using the mole conversion
chart.
Stoich Problems – Mole to Mole
 Given
moles – looking for moles
2H2 + O2
2H2O
How many moles of H2O are produced from 6.0 moles of oxygen?
Step 1: Label your equation
Step 2: Write down the given and multiply by the mole ratio with the
unknown on top.
MOLE TO MOLE EXAMPLES
1. N2 + 3H2
2 NH3
How many moles of N2 are needed to make 12.2 moles of NH3?
2. 2C2H2 + 5O2
4CO2 + 2H2O
How many moles of CO2 are produced from .80 moles of O2?
3. 2H2S + 3O2
2SO2 + 2H2O
How many moles of H2S must react with .68 moles O2?
PRACTICE - MOLE TO MOLE

2C2H2 + 5O2
4CO2 + 2H2O

1. How many moles of CO2 are produced when
14.3 moles of O2 are burned?

2. How many moles of water are produced when
.8432 moles of CO2 are produced?

3. How many moles of oxygen are needed to
react with 11.44 moles of C2H2?
Stoich Problems – Mole to Mass
 Given
moles – looking for grams
2H2 + O2
2H2O
How many grams of water are formed from 6.0 moles of H2?
Step 1: Label the equation correctly.
Step 2: Write down the given then multiply by the mole ratio –
unknown on top.
Step 3: Multiply by molar mass of unknown to convert from
moles to grams.
MOLE TO MASS EXAMPLES
2H2 + O2
 1.
2H2O
How many grams of water are
produced when 4.84 mol hydrogen
reacts?
 2. How many grams of oxygen are
need to react with 54.2 mol of
hydrogen?
 3. How many grams of hydrogen are
needed to produce 1.634 mol of
water?
PRACTICE – MOLE TO MASS
Ca3(PO4)2 + 3SiO2 + 5C 3CaSiO3 + 5CO + 2P
 1. How many grams of SiO2 are needed to
produce 10.0 mol CO?

 2.
How many grams of carbon are needed to
react with 3.0 mol of Ca3(PO4)2?
 3.
42.0 mol of C produces how many grams of
CaSiO3?
Stoich Problems – Mass to Mole
 Given
grams – looking for moles
2H2 + O2
2H2O
How many moles of water are formed from 16.0 grams of H2?
Step 1: Label the equation correctly.
Step 2: Put down the given and divide by molar mass of the given to
get moles.
Step 3: Multiply by mole ratio – unknown on top – to get moles.
MASS TO MOLE EXAMPLES
2H2 + O2
 1.
2H2O
How many moles of water are
produced when 4.84 grams of hydrogen
reacts?
 2. How many moles of oxygen are need
to react with 54.2 grams of hydrogen?
 3. How many moles of hydrogen are
needed to produce 1.634 grams of water?
PRACTICE – MASS TO MOLE
Ca3(PO4)2 + 3SiO2 + 5C 3CaSiO3 + 5CO + 2P
 1. How many moles of SiO2 are needed to
produce 420 g of CO?

 2.
How many moles of CaSiO3 are produced
when 100.0 g of Ca3(PO4)2 are reacted?
 3.
How many moles of Ca3(PO4)2 are needed to
produce 280.0 g of CO?
Stoich Problems – Mass to Mass
 Given
grams – looking for grams
2H2 + O2
2H2O
How many grams of O2 are needed to react with 20.0 g of H2 ?
Step 1: Label the equation.
Step 2: Write down the given and divide by the molar mass of
the given.
Step 3: Multiply by the mole ratio to find moles of unknown.
Step 4: Multiply by molar mass of unknown to find grams of
unknown.
MASS TO MASS EXAMPLES
2H2 + O2
 1.
2H2O
How many grams of water are
produced when 4.84 grams of hydrogen
reacts?
 2. How many grams of oxygen are need
to react with 54.2 grams of hydrogen?
 3. How many grams of hydrogen are
needed to produce 1.634 grams of water?
PRACTICE – MASS TO MASS

N2 + 3H2
2NH3

1. How many grams of ammonia are formed from
15.0 g N2?

2. How many grams of H2 are needed to react
with 48.3 g N2?

3. How many grams of H2 are needed to produce
.914 g NH3?
UNDERSTANDINGS/KEY IDEA
1.3.A
Reactants can be either limiting or
excess.
APPLICATION/SKILLS
Be able to solve problems relating
to reacting quantities, limiting and
excess reactants, theoretical,
experimental and percentage
yields.
Limiting Reactant
 The
reactant that will run out first in a
reaction.
 The
excess reactant is the reactant that
is not used up completely in reaction.
Limiting Reactant
 You
will recognize a limiting reactant
problem because there will be 2 “givens”
in the problem.
2H2 + O2
2H2O
If 4 moles of H2 react with 8 moles of O2 ,
how much water will be formed?
Limiting Reactant - Calculations
Step 1: Write down and convert both givens to
moles (if not already in moles).
Step 2: Set up two stoichiometric problems with the opposite
reactant as the mole ratio.
1 mol O2
4 mols H2 given X
= 2 mols O2needed
2 mol H2
8 mols O2 given X
2 mol H2
1 mol O2
= 16 mols H2needed
Step 3: Interpret the equations by crossing and comparing to
determine the limiting reactant.
You need 2 moles of O2 and you have 8 mols O2 given
You need 16 moles of H2, but you were only given 4 moles H2.
Since you do not have enough H2 given, it will run out first.
Therefore, H2 is the limiting reactant.
Step 4: Use the limiting reactant, calculate the answer.
2H2(g) + O2(g)  2H2O(l)

How many grams of water are formed when 12.0 g of
hydrogen reacts with 17.0 g of oxygen?




First of all, I recognize that I am given an amount of hydrogen
and an amount of oxygen so I need to figure out which one is
going to run out first and stop the reaction. This one is called the
limiting reactant. The other reactant will be the excess reactant.
12.0 g H2 x mol = 5.94 mol H2 given
2.02g
17.0 g O2 x mol = .531 mol O2 given
32.0g
Notice that my first step was to convert to moles and to actually
stop after the conversion to moles and write the word “given”.
2H2(g) + O2(g)  2H2O(l)

The next step is to find out how much of each reactant is needed to
react with each other. You do this by multiplying each reactant by the
mole ratio with each other.
12.0 g H2 x mol = 5.94 mol H2 given x 1 mol O2 = 2.97 mol O2 needed
2.02g
2 mol H2
17.0 g O2 x mol = .531 mol O2 given x 2 mol H2 = 1.06 mol H2 needed
32.0g
1 mol O2

Now cross compare to find the reactant that you do not have enough
of. You need 1.06 mol H2 and you are given 5.94 mol H2 so you have
more than enough. You need 2.97 mol O2 and are only given .531
mol O2 so you do not have nearly enough. This means that oxygen is
your limiting reactant; therefore, hydrogen is your excess reactant.
2H2(g) + O2(g)  2H2O(l)

To solve the problem, you use the limiting reactant.
(Important! Be sure to use the given moles, not the
needed moles.)
 The problem asked you for the mass of water produced.
You now have a mole – mass problem.
 .531 mol O2 x 2 mol H2O x 18.02g = 19.1 g H2O
1 mol O2
mol
 The answer is your theoretical yield.
 If you were asked how much excess reactant remained,
simply subtract the moles H2 needed by the moles given.
 5.94 mol H2 given – 1.06 mol H2 needed = 4.88 mol H2 in
excess.
LIMITING REACTANT
EXAMPLE
 If
you have 6.70 mol Na reacting with 3.20
mol Cl2, what is your limiting reactant, how
many moles of product will be formed and
how much excess reactant remains?

2Na + Cl2
2NaCl
UNDERSTANDINGS/KEY IDEA
1.3.B
The experimental yield can be
different from the theoretical yield.
Theoretical Yield –
the maximum amount of product that can
be produced from a given amount of
reactant (found by using stoichiometry).
Experimental Yield –
the measured amount of product
obtained from a reaction (what you got
in the lab)
Percent Yield
 You
will recognize this type of
problem when they ask you to find the
percent yield.
 The experimental value will be given
in the problem and you have to
calculate the theoretical yield by using
stoichiometry.
EXAMPLE
2H2(g) + O2(g)  2H2O(l)
 If 17.0 g of oxygen reacts to form 18.7 g of
water, what is the percent yield?


First of all, you recognize that this is a percent yield
problem. Circle 18.7 g and label it “experimental”. Next
work the problem to solve for the theoretical yield.
17.0 g O2 x mol x 2 mol H2O x 18.02 g = 19.1 g
32.0 g 1 mol O2
mol
 % yield = exp x 100% = 18.7g x 100% = 97.9%
theo
19.1g
UNDERSTANDINGS/KEY IDEA
1.3.C
Avogadro’s law enables the mole
ratio of reacting gases to be
determined from volumes of the
gases.
Avogadro’s hypothesis states that
equal volumes of different gases
contain equal numbers of particles at
the same temperature and pressure.
This means that if you are given gas
volumes and asked for gas volumes at
the same conditions, you can find your
answer using the mole ratios without
doing any conversions.
APPLICATION/SKILLS
Be able to calculate reacting
volumes of gases using Avogadro’s
law.
Example Problem
40 cm3 of carbon monoxide is reacted with 40
cm3 of oxygen in the following reaction.
2CO + O2 →
2CO2
What volume of carbon dioxide is produced?
This is actually a limiting reactant “volumes of
gases” problem and is quite often assessed on
the IB exam.
40 cm3 CO given x 1 mol O2 = 20 cm3 O2 needed
2 mol CO
40 cm3 O2 given x 2 mol CO = 80 cm3 CO needed
1 mol O2
Notice you can use the mole ratio directly
without converting cm3 to moles first.
You need 80 cm3 CO and only have 40cm3
so CO is the limiting reactant.
40 cm3 CO given x 2 mol CO2 = 40 cm3 CO2 produced
2 mol CO
Oxygen is in excess by 20 cm3.
UNDERSTANDINGS/KEY IDEA
1.3.D
The molar volume of an ideal gas is
a constant at specified temperature
and pressure.
 Standard
Temperature and Pressure STP
= 273 K (0°C) and 101.3 kPa (1atm)
(Use 100 kPa when you can’t use a calculator.)
 Room
Temperature and Pressure
RTP = 298 K (25°C) and 101.3 kPa
 The molar volume of a gas at STP is 22.4
dm3/mol.
 The molar volume of a gas at RTP is 24
dm3/mol.
APPLICATION/SKILLS
Be able to solve problems and
analyze graphs involving the
relationship between temperature,
pressure and volume for a fixed
mass of an ideal gas.
KINETIC THEORY OF MATTER





1. The volume of the gas particles is assumed
to be zero.
2. The gas particles are in constant motion.
3. The collisions of the gas particles with the
sides of the container cause pressure.
4. The particles exert no forces on each other.
5. The average kinetic energy is directly
proportional to the Kelvin temperature of the
gas.
PROPERTIES OF GASES
 No
definite shape
 No definite volume
 Very easily compressed
 High rate of diffusion
 Gas particles exert pressure on their
surroundings.
 It
is important to remember that you
ALWAYS use Kelvin temperatures when
working gas law problems.
K = oC + 273

Zero K (-273 oC) is called absolute zero
and this is the temperature when all
motion theoretically ceases to exist.
BOYLE’S LAW
 Robert
Boyle, an Irish chemist, discovered
that the volume of a gas was inversely
proportional to the pressure applied. In
other words, as pressure is increased,
volume decreases.
P1V1 = P2V2
volume
pressure
Graph of pressure vs volume
pressure
1/volume
Boyle’s Law: The pressure of a gas is inversely proportional to the volume.
CHARLES’S LAW

Jacques Charles, a French physicist, discovered
the relationship between temperature and
volume at constant pressure. He was the first
person to make a solo balloon flight and it was
on his flight that he discovered that the volume
of a gas was dependent upon the temperature.
 He discovered that the volume of a gas is
directly proportional to the Kelvin temperature.
In other words, if you increase the temperature,
the volume increases.
T1V2 = T2V1
volume
Graph of volume vs temperature
Temperature (K)
Charles’ Law – the volume of a gas is proportional to the Kelvin temperature.
GAY LUSSAC’S LAW

Joseph Gay Lussac, a French physicist and
chemist, another avid balloonist discovered the
relationship between temperature and pressure.
 He discovered that at constant volume, if you
increase the temperature, the pressure
increases.
 He also invented many types of chemical
glassware that is still in use today.
T1P2 = P1T2
pressure
pressure
Graph of pressure vs temperature
Temperature (°C)
Temperature (K)
Gay Lussac’s Law – the pressure of a gas is directly proportional
to the Kelvin temperature.
Effect of temperature on gas
volume
 1.
If you double the temperature and keep
volume constant, the pressure doubles.
 2. If you then double the volume and keep
the temperature constant, the pressure is
halved.
 3. Therefore, the net overall effect of
doubling the temperature and doubling the
volume is that there is no overall change in
pressure.
AVOGADRO’S LAW

Amadeo Avogadro, an Italian chemist,
postulated that equal volumes of gases at the
same temperature and pressure contained the
same number of particles.
 So for a gas at constant temperature and
pressure, the volume is directly proportional to
the number of moles of a gas.
V1n2 = n1V2
COMBINED GAS LAW
 You
can combine the first three gas laws
into one equation. Remember to always
use Kelvin temperature.
 In this equation, you will be given 5
knowns and solve for 1 unknown.
P1V1T2 = P2V2T1
APPLICATION/SKILLS
Be able to solve problems relating
to the ideal gas equation.
IDEAL GAS LAW

Use the ideal gas law when given moles or mass
in a gas law problem or if you are asked to solve
for moles or mass.
PV = nRT
 Where P is pressure, V is volume, n is moles, T
is Kelvin temp, and R is the universal gas law
constant.
 R = 8.31 J K-1 mol-1 or 8.31 m3 Pa / K mol
(A J is a N-m and a pascal is a Nm-2)
 You must use the correct units. P has to be in
Pa and V must be in m3.
APPLICATION/SKILLS
Be able to explain the deviation of
real gases from ideal behavior at
low temperature and high pressure.
REAL GASES
 An
ideal gas exactly obeys the gas laws.
 Real gases actually have attractive forces
between them.
 Real gases actually take up some volume.
 A gas most behaves like an ideal gas at
high temperatures and low pressures.
APPLICATION/SKILLS
Be able to obtain and use
experimental values to calculate
the molar mass of a gas from the
ideal gas equation.
MOLAR MASS OF A GAS
Using density and the ideal gas law, the
molar mass can be derived.
M = dRT or M = mRT
P
VP
where d is density and M is molar mass.
UNDERSTANDINGS/KEY IDEA
1.3.E
The molar concentration of a
solution is determined by the
amount of solute and the volume of
solution.





The concentration is the composition of the
solution expressed in moles or grams solute
over volume of the total solution.
When no more solute can be dissolved in a
solution, the solution is saturated.
Square brackets [ ] are used to represent
concentration or molarity.
Concentration [ ] = moles of solute/volume of
solution
The unit of volume in [conc] is usually dm3.
***To solve for moles, multiply the concentration
by the volume given. This is a very common
calculation.***
you are given a volume of cm3 and
asked to find moles with a concentration
using mol/dm3, you must divide by 1000.
 1 dm3 = 1000 cm3
 If
GUIDANCE
Units of concentration include:
g/dm3, mol/dm3 and parts per
million (ppm).
GUIDANCE
Use square brackets to denote
molar concentration.
UNDERSTANDINGS/KEY IDEA
1.3.F
A standard solution is one of known
concentration.
UNDERSTANDINGS/KEY IDEA
Strong electrolytes are assumed to
completely break down into ions in
solution.
ELECTROLYTES
 Electrolytes
are substances that when in
solution break into ions and conduct
electricity due to the presence of said ions.
 All ionic compounds are electrolytes
meaning that when in solution, they
dissociate into the ions making up the
compound.
 Strong acids and strong bases dissociate
completely into their ions.
APPLICATION/SKILLS
Be able to determine the ions
produced by an electrolyte in
aqueous solution and calculate the
concentration of those ions.
Sample Problems
mol/dm3 HCl breaks into 1.0 mol/dm3 H+
ions and 1.0 mol/dm3 Cl- ions
 2.0 mol/dm3 Ba(OH)2 breaks into 2.0
mol/dm3 Ba2+ ions and 4.0 mol/dm3 OH- ions
 1.0
 You
should be able to break any ionic
compound into its ions and be able to
determine the concentration of said ions by
the original concentration of the ionic cmpd.
APPLICATION/SKILLS
Be able to solve problems involving
molar concentration, amount of
solute and volume of solution.

A solution is a homogeneous mixture of the
solvent with the solute.



The solute is the substance being dissolved in the
solvent and it is usually the less abundant
component.
The solvent is the substance doing the dissolving. It
is usually a liquid and is the more abundant
component.
An aqueous solution is a solution with water as
the solvent.
 Solutes can be solids, liquids or gases, but the
solvent is generally a liquid.
EXAMPLE

Calculate the mass of copper II sulfate
pentahydrate, CuSO4.5H20, required to prepare
500 cm3 of a 0.400 mol/dm3 solution.



n = 500 cm3 x 1 dm3/1000cm3 x 0.400 mol/dm3 = .200 mol


First of all, they are asking for mass so you know you
need to solve for moles.
You can solve for moles by multiplying volume by the
concentration – watch for units.
To solve for mass when you have moles, multiply by the molar
mass.
Mass = .200 mol x 249.61g/mol = 49.9 g
APPLICATION/SKILLS
Be able to use the experimental
method of titration to calculate the
concentration of a solution by
reference to a standard solution.
 Standard
solutions are used to find
concentrations of other solutions.
 The volumetric technique “titration” is most
commonly used to find the concentration
of an unknown solution.
EXAMPLE

What volume of 2.00 mol/dm3 hydrochloric acid would have to be added to
25.0 cm3 of a 0.500 mol/dm3 sodium carbonate solution to produce a
neutral solution of sodium chloride?

First you need a balanced equation:
2HCl + Na2CO3  2NaCl + H2O + CO2

Next find moles of sodium carbonate:
25.0cm3 x 1dm3/1000cm3 x 0.500mol/dm3 = .0125mol

Use the mole ratio to find moles of HCl needed:
.0125 mol Na2CO3 x 2 mol HCl = .0250 mol HCl
1 mol Na2CO3

Use the concentration equation to solve for volume since you have
moles and concentration.
[Conc] = mol/volume so volume = mol/[conc]
volume = .0250mol/2.00mol/dm3 = .0125dm3 x 1000cm3/dm3 = 12.5cm3
EXAMPLE

Calculate the volume of carbon dioxide produced at
STP when 1.00g of calcium carbonate reacts with
25.0cm3 of 2.00 mol/dm3 hydrochloric acid.
CaCO3 + 2HCl  CaCl2 + H2O + CO2
Wow! Another limiting reactant problem with mixed units
given. Both givens have to be converted to moles:
1.00gCaCO3 x mol/100.09g = 0.0100mol CaCO3 given
25.0cm3 x 1dm3/1000cm3 x 2.00mol/dm3 = 0.0500mol HCl

0.0100mol CaCO3 given x 2 mol HCl = 0.0200 mol HCl needed
1molCaCO3
0.0500mol HCl given x 1 mol CaCO3 = 0.025mol CaCO3 needed
2 mol HCl

You need 0.0250 mol CaCO3 and are only given 0.0100
mol so this is your limiting reactant.
0.0100 mol CaCO3 x 1 mol CO2 = 0.0100 mol CO2
1mol CaCO3
0.0100 mol CO2 x 22.4 dm3/mol = .224 dm3
.224 dm3 x 1000 cm3/dm3 = 224 cm3
Citations
Brown, Catrin, and Mike Ford. Higher Level
Chemistry. 2nd ed. N.p.: Pearson Baccalaureate,
2014. Print.
Most of the information found in this power point
comes directly from this textbook.
The power point has been made to directly
complement the Higher Level Chemistry textbook by
Catrin and Brown and is used for direct instructional
purposes only.