Stoichiometry is the study of the relationship or ratio’s between the amounts of reactants and products in a chemical reaction.
Stoichiometric amounts or ratio’s are the amounts of reactant or product as given by the coefficients in the balanced equation.
The following laws and ideas underpin our knowledge about the relationships between chemicals involved in reactions (see PowerPoint):
1. Law of Conservation of Mass: Lavoisier (18 th Century) concluded, from studying reactions involving oxygen, that during any chemical reaction no atoms were destroyed.
2. Law of Definite Proportions: analytical tests showed that any compound consists of the same elements in definite proportions e.g. in every water molecule 88.88% of its mass is oxygen.
3. Law of Combining Volumes of Gases: Gay-Lussac (1803) concluded that when gases react they do so in whole number ratios in terms of volume e.g. 3 volumes of hydrogen react with 1 volume of nitrogen to form 2 volumes of ammonia.
4 . Avogadro’s Theory (1811): Avogadro stated that equal volumes of gases must have equal number of particles (provided temperature and pressure are the same)
5. Law of Multiple Proportions: in 2 different compounds of the same elements (e.g. CO and CO
2
) if the mass of 1 element is the same (e.g. carbon in CO and CO
2
) than the ratio of the mass of the other element (in our example oxygen) in each compound must be a whole number.
1.1.1 Apply the mole concept to substances.
1.1.2 Determine the number of particles and the amount of substance (in moles).
As particles are extremely small chemists measure amounts of substance using a quantity called the
MOLE which stands for 6.02 x 10 23 a number which is know as Avogadro’s constant.
So 1 mole of a substance = 6.02 x 10 23 units of that substance; these units can be atoms, molecules, ions, electrons or formula units.
1 mole = 6.02 x 10 23 atoms, ions, molecules, electrons, …
Determining number of moles and the number of particles
Use the expression: number of particles
number of moles =
6.02 x 10 23
Topic1notes 12.5 hours Page 1 of 31

Molar ratios in compounds or formula units:
Examples:

in 1 mole of Fe
2
O
3 there are 2 moles of iron atoms and 3 moles of oxygen;

in 3 moles of Na
2
CO
3
there are 3 moles of CO
3
2 and 6 moles of Na + ;

in 2 moles of (NH
4
)
3
PO
4
there are 6 moles of NH
4
+ and 12 moles of hydrogen atoms and so on…

if there are 2 x 10 22 C-atoms in a sample of ethane how many hydrogen atoms are there in that same sample?

if you have 6 x 10 22 atoms of hydrogen, how many molecules of ammonia do you have?
Exercises
1. How many moles of oxygen are there in 2 moles of Al
2
O
3?
2. How many moles of NO
3
are there in 1 mole of Ga(NO
3
)
3
?
3. What is the total number of atoms of 0.05 moles of NH
4
NO
3
?
1.2.1 Define the terms relative atomic mass ( A r) and relative molecular mass ( M r).
1.2.2 Calculate the mass of one mole of a species from its formula.
1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar
mass.
1.2.4. Distinguish between the terms empirical formula and molecular formula.
1.2.5 Determine the empirical formula from the percentage composition or from other experimental data.
1.2.6 Determine the molecular formula when given both the empirical formula and experimental data.
The mass of a specific isotope of an element compared with 1/12 th of the mass of the most common carbon isotope which itself has been given a relative mass of 12. As this is a relative quantity no unit is needed.
actual mass of isotope
relative isotopic mass =
1/12 th of the mass of a C-12 atom
r
The relative atomic mass is the average mass of all isotopes of an element again compared to the same standard which is 1/12 th of the mass of one 12 C atom. It is calculated as shown below:
(isotopic mass 1 x %) + (isotopic mass 2 x %) + (isotopic mass 3 x %)
Topic1notes 12.5 hours Page 2 of 31

A r
=
100
It is the relative atomic mass which is found in the Periodic Table or data booklet.
Example:
The average magnesium atom has a relative atomic mass of 24 which means that a magnesium atom is 2 times heavier than a carbon atom which has a mass of 12 and 24 times heavier than a hydrogen atom which has a mass of 1.
Complete the following table: element isotopes abundance relative atomic mass
Cu
Eu
Ag
Cu-63 (relative isotopic mass = 63)
Cu-65 (relative isotopic mass = 65)
Eu-151
Eu-153
Ag-106.9041
Ag-108.9047
69.09%
30.91%
48.17%
151.9
r
The average mass of a molecule compared to 1/12 the of the mass of one 12 C atom.
To be able to calculate the M r
correctly, the formula of the compound must be correct and the relative atomic masses of the elements in the compound need to be used.
Examples:

relative molecular mass of H
2
O is 18 because (2 x 1.01) + (1 x 16.00) = 18.02

relative molecular mass of H
2
SO
4
is 98 because (2 x 1.01) + 32.06 + (4 x 16.00) = 98.08
The M r
only really applies to molecules (a group of covalently bonded atoms); the concept of formula mass is used for giant structures like ionic substances (eg sodium chloride, copper sulphate) and macromolecules (eg silicon dioxide)
Examples:

relative formula mass of NaOH is 40 because 22.99 + 16.00 + 1.01 = 40.00

relative molecular mass of Al
2
(SO
4
)
3
is because (2 x 26.92 ) + (3 x 32.06) + (12 x 16.00) = 342.02
How do we measure 1 mole of a substance?
The mass of 1 mole of a substance = molar mass which is measured in g mol -1 .
Topic1notes 12.5 hours Page 3 of 31

The molar mass is the mass of 6.02 x 10 23 units of a substance; it is the relative atomic, molecular or formula mass but expressed in grams as the molar mass is an absolute mass.
Below are examples of the type of calculations you should be able to solve using the mole concept
Calculate the number of moles (using molar mass) when you are given the mass of the substance
mass of substance
number of moles =
molar mass
Calculate actual masses of atoms or molecules
Find the mass of 1 molecule of ethanol, C
2
H
5
OH
The only actual mass you have is the molar mass of ethanol which can be calculated using a periodic table molar mass  mass of 1 atom/molecule
d ivide by 6.02 x 10 23
Calculate number of atoms in a sample
Find the number of carbon atoms in 0.1 mole of C
2
H
5
OH. number of moles  number of molecules  number of atoms
multiply multiply
by 6.02 x 10 23 by subscript of atom in formula
Calculate the number of particles from a mass
Which sample contains the greatest number of particles? 5g of NH
3
or 5 g of H
2
O or 5 g of CaO? given mass  number of moles  number of particles
divide by molar mass multiply by 6.02 x 10 23
Exercises
1. Find the mass of 1 molecule of CH
3
COOH?
Topic1notes 12.5 hours Page 4 of 31

2. Find the mass of 1 atom of Br?
3. Find the mass of 0.05 moles of NH
4
NO
3.
4. Which sample contains the greatest number of molecules?
10g of C
2
H
4
10g of C
2
H
6
10g of C
6
H
6
5. Which sample contains the lowest number of molecules?
2g of CH
4
2g of NH
3
2g of CO
2
6. How many molecules are there in 0.23 g of ethanol?
7. Which one of the following has the greatest mass?
1.0 x 10 -3 moles of CH
3
COOH 1.2 x 10 21 atoms of argon 2.4 x 10 20 molecules of C
8
H
18
8. How many glucose molecules are there in 5.23 g of C
6
H
12
O
6
?
Empirical formula
To determine the empirical formula (=the most simple molar ratio) of a compound, a chemist needs to find out how much (a mass or a percentage) of each element there is in a certain mass of the compound.
This raw data (how much there is of an element in a compound) can be obtained in various ways:

Combustion analysis (usually used with organic compounds)
This involves combusting a known amount of the organic compound and finding the masses of carbon dioxide and water (=raw data). Using the percentage composition of hydrogen and carbon in both compounds allows the chemist to calculate the starting amount of hydrogen and carbon in the compound.

Direct determination (used for binary compounds)
Involves reacting the known mass of one element with a second element, measuring the mass of the compound and then finding the mass of the second element.

Decomposition (usually used to find the formula of a hydrated salt)
Decompose (usually using heat) a sample with a known mass, measure the mass of the final product and use this data to
From this mass analysis of a compound the amount in mass of each element or the percentage composition of the compound can be determined; from this information the most simple whole number molar ratio can be found and this then gives us the empirical formula of the compound.
Percentage composition of a compound
Topic1notes 12.5 hours Page 5 of 31

Experimental analysis usually results in the amount of an element in a compound being expressed as a percentage composition
molar mass of element x coefficient in compound
% composition = x 100
molar mass of compound
Calculate
1. Percentage by mass of C in CH
4
, C
2
H
6
and C
2
H
4
2. Percentage by mass of Cu in CuSO
4,
CuO and Cu
2
O
Example calculation of empirical formula: analysis of methane:
elements in compound carbon hydrogen
mass in grams (or %) 47.98 (74.9%) 16.02 (25.1%)
number of moles 47.98/12 = 3.99 mole 16.11/1 = 16.02 moles
(74.9/12 = 6.24 moles) (25.1/1 = 25.1 moles)
ratio of moles 4 (6.25) 16 (25) most simple ratio of moles 1 4 empirical formula: CH
4
Molecular formula
The actual formula of the compound i.e. the molecular formula - which shows the actual number of atoms and ions present in a compound - is always a multiple of the empirical formula both in terms of mass and particles; it is up to the researcher to determine how many times heavier than the empirical formula the actual molecule is.
Strictly speaking we can not use the term molecular formula when dealing with giant structures as they do not consist of separate molecules; in that case any formula is always an empirical formula indicating the most simple ratio of ions or atoms e.g. in the case of silicon dioxide, SiO
2
, this formula tells us that there are twice as many oxygen atoms within the lattice as silicon atoms. The formula CuCl
2
, tells that there are twice as many Cl ions in the giant ionic lattice as copper ions.
There are many possible molecular formulae that can be derived from one empirical formula. However, if the molecular formula is known, the molecular formula can be obtained.
Example : Calculate the molecular formula of a compound with a molecular mass of 84g and an empirical formula of CH
2
.
Answer:

mass empirical formula: 14g
Topic1notes 12.5 hours Page 6 of 31


molar mass of formula: 84g

ratio of molar mass/empirical mass = 84/14 = 6

molecular formula = (CH
2
) x 6 = C
6
H
12
Exercise : Naphthalene, best known as ‘mothballs’, is composed of carbon (93.71%) and hydrogen
(6.29%). If the molar mass of the compound is 128g, what is the molecular formula of naphthalene?
Answer: carbon hydrogen
mass of element 93.71 6.29
number of moles 93.71/12 = 7.8 6.29/ 1 = 6.29
DO NOT ROUND UP OR DOWN AT THIS TIME !!!!!!!!!
most simple ratio 7.8/ 6.29 = 1.25 6.29/6.29 = 1
AGAIN DO NOT ROUND UP OR DOWN !!!!!!
lowest whole number ratio 5 4
empirical formula:
C
5
H
4
ratio molecular formula /empirical formula: 128g/ 64g = 2
molecular formula = 2 x C
5
H
4
= C
10
H
8
Exercises:
Calculate i. The empirical formula ii. The molecular formula
For each of the following substances:
1. Water (one mole = 18g) that contains 11.1% of hydrogen and 88.9% of oxygen.
2. Ammonia (one mole = 17g) that contains 82.4% of nitrogen and 17.6% of hydrogen.
3. Potassium carbonate (one mole = 138g) that contains 56.5% of potassium, 8.7% of carbon and 34.8% of oxygen.
4. Hydrazine (one mole = 32g) that contains 87.5% of nitrogen and 12.5% of hydrogen.
5. Glucose (one mole = 180g) that contains 40.0% of carbon, 6.67% of hydrogen and 53.3% of oxygen.
6. Ethane (one mole = 30g) that contains 80% of carbon and 20% of hydrogen.
7. Phosphorus(III) chloride (one mole = 137.5g) that contains 22.5% of phosphorus and 77.5% of chlorine.
8. Butane (one mole = 58g) that contains 82.8% carbon and 17.2% of hydrogen.
9. A compound X (one mole = 342g) that contains 15.8% of aluminium, 28.1% of sulfur and 56.1% of oxygen. The compound forms a white precipitate when it reacts with barium nitrate solution.
Topic1notes 12.5 hours Page 7 of 31

1.3.1 Deduce chemical equations when all reactants and products are given.
1.3.2 Identify the mole ratio of any two species in a chemical equation.
1.3.3 Apply the state symbols (s), (l), (g) and (aq).
As seen in 1.1 the stoichiometric amounts or molar ratios for any reaction can be obtained from a balanced equation.
Example:
Fe
2
O
3
(s) + 3CO (g)

2Fe (s) + 3 CO
2
(g)
This means that
1 particle + 3 particles

2 particles + 3 particles
of Fe
2
O
3
of CO of Fe of CO
Alternatively this means that there are 3 times as much particles of CO as Fe
2
O
3
particles; for the reaction to go to completion (=both reactants are completely used up) 3 times as much CO is needed as Fe
2
O
3
.
By multiplying each number of particles by 6.02 x 10 23 , the above balanced equation states the same ratios but expressed in moles:
1 mole reacts with 3 moles to give 2 moles and 3 moles
of Fe
2
O
3
of CO of Fe of CO
2
In the above equation the molar ratio of iron oxide to carbon monoxide is 1: 3 i.e. for the reaction to go to completion 3 times more moles of CO
2 are needed than Fe
2
O
3
Complete the following table showing other molar ratio’s in the above equation;
Fe
2
O
3
: Fe
Fe : CO
2
:
:
CO : CO
2
Fe
2
O
3
: CO
2
:
:
Balancing equations
In getting an equation to balance , the number of moles of each substance is changed, until there are the same number of atoms of each element on each side of the equation:
Unbalanced equation:
Balanced equation:
C
3
H
8
+ O
2
---> CO
2
+ H
2
O
C
3
H
8
+ 5 O
2
---> 3 CO
2
+ 4 H
2
O
Conservation of mass
When applying the molar mass concept to the equation below the Law of Conservation of Mass can be proven:
Fe
2
O
3
(s) + 3CO (g)

2Fe (s) + 3 CO
2
(g)
1 x 160 g + 3 x 28g 2 x 56 g + 3 x 44 g
244 g 244 g
Topic1notes 12.5 hours Page 8 of 31

Any balanced equation should include state symbols .
When possible you should always apply the state symbols:
(s) = solid (l) = liquid (g) = gas (aq) = dissolved in water
1.4.1 Calculate theoretical yields from chemical equations.
Example : How much iron oxide is needed to produce 28g of iron during the reduction of iron oxide by carbon monoxide?
Step 1: write balanced equation
Fe
2
O
3
(s) + 3CO (g)

2Fe (s) + 3 CO
2
(g)
Step 2: convert data in question into moles data in question = 28 g or iron
number of moles of iron = 28g / 56g = 0.5 mole of iron
Step 3: find answer in equation i.e. use molar ratios the equation tells us that to get 2 moles of Fe we need 1 mole of Fe
2
O
3
(= molar ratio)

0.5 mole of iron needs 0.25 moles of Fe
2
O
3
.
Step 4: convert answer into relevant units mass = number of moles x molar mass = 0.25g x 160 g = 40 g
To get 28g or iron, 40g of iron oxide is needed to get the reaction to go to completion!!!
1.4.3 Solve problems involving theoretical, experimental and percentage yield.
The quantity of product that you should obtain according to the amounts of reactants available and the molar ratios of reactants is called the theoretical yield.
However, when the reaction is carried out the actual yield is nearly always less than the theoretical yield.
Comparing the actual and theoretical yield by calculating their ratio allows us to determine the efficiency of a reaction; the ratio is called the percentage yield.
Topic1notes 12.5 hours Page 9 of 31

actual/experimental yield
percentage yield = x 100
theoretical yield
1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are
given.
If in a chemical reaction the molar ratios of the equation are not followed, usually one of the reactants becomes the limiting reagent whilst all the others are in excess. The limiting reagent controls or limits the amount of product that is formed. At the end of such a reaction in which there is a limiting reagent, besides product, some of the reactants that are in excess will be left.
To determine which reactant is the limiting reagent:

Calculate the number of moles of each reactant;

Determine the most simple ratio of the moles

From the equation find the required molar ratio so that the reaction can go to completion (i.e. all reactants turned into products).

The reactant which has a lower number of moles than what is required according to the equation is the limiting reagent.
The number of moles of the limiting reagent is the number of moles used to calculate the theoretical yield of the reaction.
Sample exercise
Consider the Haber process, when 25000 g of nitrogen is reacted with 5000 g of hydrogen to produce ammonia.
N
2
(g) + 3H
2
(g)

2NH
3
(g)
Determine which reactant nitrogen or hydrogen is the limiting reactant? Calculate the theoretical yield of ammonia in grams and the experimental yield as a % of 10390 g of ammonia.
1. Find the actual number of moles of each reactant from the mass of reactant given. for N
2 n = m = 25000 = 892.8 mol for H
2 n =
M m
M
=
28
5000
2
= 2500 mol
2. Find the required number of moles of each reactant from the moles ratio in the reaction.
Required moles of reactant A = Actual moles of reactant B
Required moles of H
2
= 892.8 x x Ratio of A
Ratio of B
3
1
= 2678.6 mol
Topic1notes 12.5 hours Page 10 of 31

Required moles of N
2
= 2500 x 1
3
= 833.3 mol
3. Find the limiting reagent
If the: actual moles

required moles that reactant is the one that is limiting
So for H
2
2500 mol

2678.6 mol so H
2
is the limiting reactant
4.
(for N
2
892.8 mol

833.3 mol so N
2
is not limiting)
Find the mass of one of the products formed (also called the theoretical yield)
Use the actual number of moles of the limiting reagent and compare the mole ratios
5. e.g. to find the number of moles and the mass of NH
3
formed
From the mole ratio’s in the equation
H
2
: NH
3 2
3 if the actual number of moles of the limiting reagent H
2
= 2500 mol then using the mole ratio in the equation the actual number of moles of NH
3
= 2500 x 2 = 1666.6 mol
3 mass NH
3
H
2
:
3
M
17
NH
3
2
2500 1666.6 m = n x
= 1666.6 x
= 28332 g
Find the Percentage Yield when given the mass of product e.g. find the percentage yield of 10390 g of ammonia
Experimental Yield % = Experimental Yield (g)
=
=
Theoretical Yield (g)
10390
28332
36.7 % x 100 x 100
Topic1notes 12.5 hours Page 11 of 31

3.
Exercises
1. Consider the reaction:
2 Al + 3 I
2

2 AlI
3
Determine the limiting reagent when the following are reacted:
(a) 1.2 mol Al and 2.4 mol I
2
(b) 1.2 g Al and 2.4 g I
2
2. Freon-12 is used as a coolant in refrigerators. It is formed in the following reaction:
3 CCl
4
+ 2 SbF
3

3 CCl
2
F
2
+ 2 SbCl
3
150 g CCl
4
(M r
= 154) is combined with 100 g SbF
3
(M r
= 178.8) to give freon-12, CCl
2
F
2
(M r
= 120.9).
(a) Identify the limiting and excess reagents
(b) How many grams of freon-12 can be formed?
IB Chemistry Exam November 1998
Aspirin, C
9
H
8
O
4
is made by reacting ethanoic anhydride, C acid C
7
H
6
O
3
(M r
= 138.1) according to the equation:
4
H
6
O
3
(M r
= 102.1) with 2-hydroxybenzoic
C
4
H
6
O
3
+ 2 C
7
H
6
O
3

2 C
9
H
8
O
4
+ H
2
O
(a) if 15.0 g of 2-hydroxybenzoic acid is reacted with 15.0 g of ethanoic anhydride, determine the limiting reagent in this reaction.

2 marks

(b) Calculate the maximum mass of aspirin that could be obtained in this reaction.

2 marks

(c) If the mass obtained in this experiment was 13.7 g calculate the percentage yield of aspirin.

1 mark

1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases.
1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations.
1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an
ideal gas.
1.4.7 Solve problems using the ideal gas equation, PV = nRT
1.4.8 Analyse graphs relating to the ideal gas equation.
Gay-Lussac noticed that there is always a very simple ratio between volumes of gases that react together.
For instance,
Topic1notes 12.5 hours Page 12 of 31


1 volume of hydrogen + an equal volume of chlorine

2 volumes of hydrogen chloride

1 volume of oxygen + 2 volumes of hydrogen

2 volumes of water vapour
To explain these results, Avogadro proposed that equal volumes of gases (if measured in the same conditions) must have the same number of molecules.
As a result the above volume observation can also be considered as
1 molecule of hydrogen + 1 molecule of chlorine

2 molecules of hydrogen chloride
Avogadro’s Law states that:
Equal volumes of different gases at the same temperature and pressure contain the same number of molecules (or same number of moles)
If the same volume has the same number of particles than the same number of particles must have the same volume so 1 mole at s.t.p. has a volume of 2.24 x 10 -2 m 3 mol -1 or 22.4 dm 3 mol -1 and this is called the gas molar volume.
At s.t.p. (which is 273K and 1 atm)
-2
3
-1
3
-1
In other words, if there are x molecules of O
2
in 10 ml of oxygen gas, then there are x molecules of N
2
in 10 ml of nitrogen gas and there are 2x molecules of CO
2
in 20 ml of carbon dioxide gas.
Another way of putting this is that the volume of a gas depends on the amount of moles, n (provided the conditions are the same). Or, if the volume of H
2
gas is double the volume of O
2
than there are twice as many moles of O
2
as H
2
.
Avogadro’s Law allows equations involving gases to be interpreted directly in terms of volumes.
For example, in the equation N
2
+ 3H
2

2NH
3
for 2.4 dm 3 of N
2
, 7.2 dm 3 of H
2
is needed to give
4.8 dm 3 of NH
3
.
Question: Cl
2
+ H
2

2HCl
Hydrogen and chlorine react according to the equation above. What will be the result of the reaction of 2.0 moles of H
2
and 1.5 moles of Cl
2
?
A. 3.5 mol of HCl B. 1.5mol of HCl and 0.5 mol of H
2
C. 2.0 mol of HCl and 0.5 mol of Cl
2
D. 3.0mol of HCl and 0.5 mol of H
2
Ideal gas equation
Topic1notes 12.5 hours Page 13 of 31

What now happens when gases are not at s.t.p.? Is there still a relationship between the number of moles and volume of the gas?
We can use the gas laws and the ideal gas equation which summarises the gas laws.
The gas laws describe the effects of changes in temperature, pressure and volume on a fixed mass of an ideal gas.
Simulations showing the gas laws can be found on:
 Charles’s Law: http://www.chm.davidson.edu/ChemistryApplets/GasLaws/CharlesLaw.html
 Boyle’s Law: http://www.chm.davidson.edu/ChemistryApplets/GasLaws/BoylesLaw.html
You can also use the following simulations to study the effects of the above factors.
 http://mc2.cchem.berkeley.edu/Java/molecules/index.html
 http://www2.biglobe.ne.jp/%7Enorimari/science/JavaApp/Mole/e-gas.html
The ideal gas equation
When combining the above gas laws, the ideal gas equation is obtained
 pV = constant (Boyle’s Law)
 V/T = constant (Charles’s Law)

P/T = constant

If n /p = constant (Avogadro’s Law) pV = n RT
Irrespective of the conditions the gas is in (pressure, temperature, volume), the ratio of pV over n T is always R!!!
The gas constant can have a different value depending on the units it is expressed in. The value you chose depends on the units the raw data is expressed in although we prefer to use the SI units!!!
See: http://www.chm.davidson.edu/ChemistryApplets/GasLaws/GasConstant.html
for determining the gas constant.
See http://resources.schoolscience.co.uk/BAMA/16plus/aerosch3pg1.html
When carrying out calculations using the ideal gas equation, we need to ensure we can cancel units!!! Better to use SI units!!
Therefore:

R = 8.31 J K -1 mol -1 ( J = Nm)

Pressure is in Pa = N m -2 ,

Volume should be in m 3 ( 1m 3 = 1000 dm 3 and 1 dm 3 = 1000 cm 3 or ml)

Temperature should be in K
If the pressure is in atm then:

Either change to Pa (1 atm = 1.013 x 10 5 Pa)

Or use 0.083 L atm K -1 mol -1
The ideal gas equation can be used to solve a variety of problems such as:

calculate the p,V or T values of a gas in a particular condition provided 2 of the 3 quantities are known
Topic1notes 12.5 hours Page 14 of 31


calculate the molar mass of a gas: o we need to know the conditions it is in and the mass of the sample of the gas o we can then calculate n and then using n = mass/M r
calculate the M r o we can also use this calculation to identify an unknown gas experimentally !!!

calculate the new value of either p, V or T provided we know what the other 2 have changed into. If we are assuming that when we change 1 or 2 conditions and the mass or amount gas stays the same, then in pV = nRT both n and R are constants and n = pV/T should be the same for both sets of conditions.
So
V
1
P
1
V
2
P
2
=
T
1
T
2
Worked example
What volume will be occupied by 0.216 g of CO
2 at 21

C and at a pressure of 1.32 atm?
1. Calculate the number of moles of the gas: 0.216g/44.0g = 4.91 x 10 -3 mol
2. Express all temperatures in Kelvin: 21

C = 294 K
3. Convert all other units in SI units: 1.32 atm = 1.32 x 1.013 x 10 5 Pa = 1.34 x 10 5 Pa
4. Apply pV = nRT
1.34 x 10 5 x V = 4.91 x 10 -3 x 8.314 x 294 V = 8.96 x 10 -5 m 3 (= 89.6 cm 3 )
Worked example
A gas occupies 127 cm 3 at a pressure of 0.830 atm and at 28

C.
(a) What volume will the same amount of gas occupy at 1.00 atm and 0

C
(b) How many moles of gas are present
1. Express all temperatures in Kelvin: 28

C = 301 K and 0

C = 273 K
P
1
V
1
P
2
V
2
2. Apply: =
T
1
T
2
(no need to change the units to SI units!! when using the above formula)
V
1 x P
1
x T
2
127 x 0.830 x 273
V
2 =
= = 95.6 cm 3
P
2
T
1
1.00 x 301
3. The number of moles can be calculated by using the ideal gas equation. When using the ideal gas equation, SI units need to be used
Topic1notes 12.5 hours Page 15 of 31

n = PV/RT
8.41 x 10 4 Pa x 1.27 x 10 -4 m 3
n = = 4.27 x 10 -3 mol
8.314 J K -1 mol -1 x 301 K
Exercises
1. The greatest volume for one mole of O
2
would be expected at
A. 273 K and 1 atm B. 273 K and 2 atm C. 546 K and 1 atm D. 546 K and 2 atm
2. The temperature (in K) is doubled for a sample of gas in a flexible container while the pressure on it is doubled. The final volume of the gas compared with the initial volume will be
A. the same B. twice as large C. four times as large D. half as large
3. When the temperature of a fixed mass and volume of nitrogen is raised from 25

C to 50

C, the pressure increases because the
A. number of molecules increases.
B. molecules expand and occupy a greater volume.
C. molecules collide more frequently and energetically with the surface of the container
D. molecules dissociate into separate atoms.
4. For an ideal gas, the variables in one of the pairs below are inversely related to one another (i.e. one increases as the other decreases). Which pair is it?
A. temperature, kinetic energy B. volume, number of moles
C. temperature, rate of diffusion D. pressure, volume
5. A 250 cm 3 sample of an unknown gas has a mass of 1.42 g at 35

C and 0.85 atmospheres.
Which expression gives its molar mass, Mr? (R = 82.05 atm cm 3 K -1 mol -1 )
1.42 x 82.05x 35 1.42 x 82.05 x 308 1.42 x 250 x 0.85 1.42 x 82.05 x 308
A. B. C. D.
0.25 x 0.85 0.25 x 0.85 82.05 x 308 250 x 0.85
6. A sample of gas has a certain volume at a temperature of 60

C. What must the temperature be in order to double the volume if the pressure is kept constant?
A. 120

C B. 333

C C. 393

C D. 666

C
7. 125 cm 3 of an unknown gas has a mass of 0.725 g at 25

C and 0.97 atmospheres. Which expression will give the relative molar mass of the gas? (R = 82.05 cm 3 atm K -1 mol -1 )
0.725 x 82.05x 25 125 x 0.97 0.725 x 82.05 x 298 0.725 x 82.05 x 298
A. B. C. D.
0.97 x 125 0.725 x 82.05 x 298 0.97 x 0.125 0.97 x 125
For the following MCs you can use your calculator:
8. A 1.25 dm
(in dm 3
3 sample of air at 25

C has a pressure of 85.9 kPa (0.85 atmospheres). What is its volume
) if its temperature and pressure are changed to 100

C and 65.7 kPa (0.65 atm)?
Topic1notes 12.5 hours Page 16 of 31

A. 0.76 B. 1.20 C. 1.31 D. 2.05
9. A sample of ethane occupies a volume of 120 cm 3 at 30

C and 1.0 atm (101kPa). What volume (in cm 3 ) will it occupy at 40

C and 1.2 atm (121 kPa)?
A. 103 B. 133 C. 149 D. 192
10. A 225 cm 3 sample of a particular gas weighs 0.774 g. What is the approximate molar mass of this gas if the volume was determined at 98.7 kPa (0.974 atm) and 30

C?
A. 120 g mol -1 B. 88 g mol -1 C. 86 g mol -1 D. 77 g mol -1
11. A 0.365g sample of a common anaesthetic has a volume of 225 cm 3 at 35

C and 98.6 kPa (740 mm
Hg). What is its molar mass?
A. 42.1 g mol -1 B. 40.8 g mol -1 C. 39.9 g mol -1 D. 4.79 g mol -1
1.5.1 Distinguish between the terms solute , solvent , solution and concentration (g dm
–3
and mol dm
–3
).
1.5.2 Solve problems involving concentration, amount of solute and volume of solution. solute = a solid, liquid or gas that is dissolved in another substance solvent = substance, usually a liquid, that will dissolve another substance solution = a mixture made up from a solute and solvent concentration =
amount of solute in moles or grams
concentration =
volume of solution in dm 3
Example: Calculate the volume of H
2
SO
4 with a concentration of 2 mol/L to neutralise 250 ml of sodium hydroxide with a concentration of 1 mol/L.
Answer:
Step 1: write balanced equation
2 NaOH (aq) + H
2
SO
4
(aq)

Na
2
SO
4
(aq) + H
2
O (l)
Step 2: convert data into moles use : number of moles
concentration =
(mol/L) volume (in L)
Topic1notes 12.5 hours Page 17 of 31

number of moles of NaOH = 1 mol/L x 0.25 L = 0.25 mole of NaOH
Step 3: find answer in equation in moles equation tells us: molar ratio NaOH : H
2
SO
4
= 2 : 1

0.25 mole of NaOH needs to 0.125 moles of H
2
SO
4
.
Step 4: convert answer into relevant unit volume of H
2
SO
4
= number of moles/concentration = 0.125 /2 mol/L = 0.0625 L
Tip: Whenever you are given a question in which amounts of all or nearly reactants are given you will very likely need to decide which one is the limiting reagent.
1. (M07) On complete combustion, a sample of a hydrocarbon compound produces 1.5 mol of carbon dioxide and 2.0 mol of water. What is the molecular formula of this hydrocarbon?
A. C
2
H
2
B. C
2
H
4
C. C
3
H
4
D. C
3
H
8
2. (M07) When excess BaCl
2
(aq) was added to a sample of Fe (NH
4
)
2
(SO
4
)
2
(aq) to determine the amount in moles of sulfate present, 5.02

10
_ mol of BaSO
4 was obtained. How many moles of sulfate ions and iron ions were in the sample of Fe (NH
4
)
2
(SO
4
)
2
?
3. (M07) What volume of 0.500 mol dm
-3 sulfuric acid solution is required to react completely with 10.0 g of calcium carbonate according to the equation below?
CaCO
_
(s)

H
2
SO
4
(aq)

CaSO
4
(aq)

H
2
O (l)

CO
2
(g)
A.
100 cm
-3
B. 200 cm
-3
C. 3 00 cm
-3
D. 400 cm -3
4. (M06) Which of the following quantities has units?
Topic1notes 12.5 hours Page 18 of 31

A. Relative atomic mass B. Relative molecular mass
C. Molar mass D. Mass number
5. (M06) A reaction occurring in the extraction of lead from its ore can be represented by this unbalanced
equation:
__ PbS + __ O
2
→ __ PbO + __SO
2
When the equation is balanced using the smallest possible whole numbers, what is the coefficient
for O
A. 1 B. 2 C. 3 D. 4
6. (M06) The equation for a reaction occurring in the synthesis of methanol is
What is the maximum amount of methanol that can be formed from 2 mol of carbon dioxide and 3 mol
of hydrogen?
A. 1 mol B. 2 mol C. 3 mol D. 5 mol
7. (N05) The complete oxidation of propane produces carbon dioxide and water as shown below.
What is the total of the coefficients for the products in the balanced equation for 1 mole of propane?
A. 6 B. 7 C. 12 D. 13
8. (N05) The relative molecular mass ( M r) of a compound is 60. Which formulas are possible for this
compound?
A. I and II only B. I and III only C. II and III only D. I, II and
9. (N05) Which aqueous solution contains the most hydrogen ions?
10. (N06) A 4 g sample of sodium hydroxide, NaOH, is dissolved in water and made up to 500 cm 3 of
Topic1notes 12.5 hours Page 19 of 31

aqueous solution. What is the concentration of the resulting solution?
A. 0.1 mol dm -3 B. 0.2 mol dm -3 C. 0.5 mol dm -3 D. 1.0 mol dm -3
11 (N06) Calcium carbonate decomposes on heating as shown below.
CaCO
3
→ CaO + CO
2
When 50 g of calcium carbonate are decomposed, 7 g of calcium oxide are formed. What is the
percentage yield of calcium oxide?
A. 7 % B. 25 % C. 50 % D. 75 %
12. (N06) Sodium reacts with water as shown below.
__ Na + __ H
2
O → __NaOH + __H
2
What is the total of all the coefficients when the equation is balanced using the smallest possible
whole numbers?
A.3 B. 4 C. 6 D. 7
13. (M05) The equation for the complete combustion of butane is
What is the amount (in mol) of carbon dioxide formed by the complete combustion of three moles of
butane?
A. 4 B. 8 C. 12 D. 24
14. (M05) Which solution contains the greatest amount (in mol) of solute?
A. 10.0 cm 3 of 0.500 mol dm
−3
NaCl B. 20.0cm
3 of 0.400 moldm
−3
NaCl
C. 30.0cm
3 of 0.300 moldm
−3
NaCl D. 40.0cm
3 of 0.200 moldm
−3
NaCl
15. (M05) How many oxygen atoms are present in 0.0500 mol carbon dioxide?
A.
3.01

10 22 B. 6.02

10 22 C. 6.02

10 23 D. 1.20

10 24
16. (M05) How many electrons are there in one 24
12
Mg 2+ ion?
A. 10 B. 12 C. 14 D. 22
17. (N04) Consider the following equation.
2C
4
H
10
(g) + 13O
2
(g) → 8CO
2
(g) + 10H
2
O(l)
How many moles of CO
2
(g) are produced by the complete combustion of 58 g of butane, C
4
H
10
(g) ?
A. 4 B. 8 C. 12 D. 16
Topic1notes 12.5 hours Page 20 of 31

18. (N04) 6.0 moles of Fe2O3 (s) reacts with 9.0 moles of carbon in a blast furnace according to the
equation below.
Fe2O3 (s)+3C(s) →2Fe(s)+3CO(g)
What is the limiting reagent and hence the theoretical yield of iron?
19. (N04) What volume of 0.500 moldm
−3
HCl (aq) is required to react completely with 10.0 g of calcium
carbonate according to the equation below?
CaCO
3
(s) + 2HCl(aq) → CaCl
2
(aq) + H
2
O(l) + CO
2
(g)
A. 100 cm 3 B. 200 cm 3 C. 300 cm 3 D. 400 cm 3
20. (M04) What amount of NaCl (in moles) is required to prepare 250 cm 3 of a 0.200 mol dm -3 solution?
A. 50.0 B. 1.25 C. 0.800 D. 0.0500
21. (M02) A compound that contains only carbon, hydrogen and oxygen has the following percentage by mass:
carbon: 60% hydrogen: 8% oxygen: 32%
What is a possible molecular formula?
A. C
5
H
8
O
2
. B. C
5
H
4
O C. C
6
HO
3
D. C
7
HO
4
22. (M02) Which sample contains the smallest amount of oxygen?
A. 0.3 mol H
2
SO
4
B. 0.6 mol O
3
C. 0.7 mol HCOOH D. 0.8 mol H
2
O
23. (M02) 6.4 g of copper wire is added to 0.10 dm 3 of 1.0 mol dm -3 aqueous AgNO
3
to form
metallic silver and aqueous copper (II) nitrate. When the reaction is complete, a. excess copper wire remains. b. all the copper wire dissolves and some silver ions are left in solution. c. All the copper wire dissolves and no silver ions are left in solution. d. The mass of metallic silver formed is equal to the mass of copper wire that reacts.
24. (M02) 2.02 g of KNO
3
(M r
= 101) is dissolved in sufficient water to prepare 0.500 dm 3 of solution. What is the concentration of this solution in mol dm -3 ?
Topic1notes 12.5 hours Page 21 of 31

A. 0.02 B. 0.04 C. 0.10 D. 0.20
25. (M03) What amount of oxygen, O
2
, (in moles) contains 1.8 x 10 22 molecules?
A. 0.0030 B. 0.030 C. 0.30 D. 3.0
26. (M03) 3.0 dm 3 of sulfur dioxide are reacted with 2.0 dm 3 of oxygen according to the equation:
2SO
2
(g) + O
2
(g)

2SO
3
(g)
What volume of sulfur trioxide (in dm 3 ) is formed? (Assume the reaction goes to completion and all
gases are measured at the same temperature and pressure.
A. 5.0 B. 4.0 C. 3.0 D. 2.0
27. (M03) _ C
2
H
2
(g) + _ O
2
(g)

_ CO
2
(g) + _ H
2
O (g)
When the equation is balanced, what is the coefficient of oxygen?
A. 2 B. 3 C. 4 D. 5
28. (N03) Copper can react with nitric acid as follows
3Cu + _ HNO
3

_ Cu(NO
3
)
2
+ _ H
2
O + _NO
What is the coefficient for HNO
3
when the equation is balanced?
A. 4 B. 6 C. 8 D. 10
29. (N03) A hydrocarbon contains 90% by mass of carbon. What is its empirical formula?
A. CH
2
B. C
3
H
4
C. C
7
H
10
D. C
9
H
10
30. (N03) Lithium hydroxide reacts with carbon dioxide as follows
2LiOH + CO
2

Li
2
CO
3
+ H
2
O
What mass (in grams) of lithium hydroxide is needed to react with 11g of carbon dioxide?
A. 6 B. 12 C. 24 D. 48
31. (M04) How many hydrogen atoms are contained in one mole of ethanol, C
2
H
5
OH?
A. 5 B. 6 C. 1.0 x 10 23 D. 3.6 x 10 24
32. (M04) The percentage by mass of the elements in a compound is
C = 72% H = 12% O = 16%
What is the mole ratio C : H in the empirical formula of its compound?
Topic1notes 12.5 hours Page 22 of 31

A. 1 : 1 B. 1 : 2 C. 1 : 6 D. 6 : 1
33. (M04) What is the coefficient for O
2
(g) when the equation below is balanced?
_ C
3
H
8
(g) + _ O
2
(g)

_ CO
2
(g) + _ H
2
O (g)
A. 2 B. 3 C. 5 D. 7
34. According to the equation
2SO
2
(g) + O
2
(g)

2SO
3
(g)
what volume of air (20% O
2
) is required to react with 10 dm 3 of SO
2
?
A. 2 dm 3 B. 5 dm 3 C. 10 dm 3 D. 25 dm 3
35. (M03) Which of the following compounds has the greatest empirical formula mass?
A. C
6
H
6
B. C
4
H
10
C. C
5
H
10
D. C
2
H
6
36. CaCO
3
(s)

CaO(s) + CO
2
(g)
When heated, CaCO
3
(M r
= 100) decomposes as shown above. When 20 g of impure CaCO
3
is
heated, 0.15 moles of CO
2
are obtained. What is the % purity of the CaCO
3
?
A. 15 B. 25 C. 55 D. 75
Paper 2
1. (M07)
(a) Propane and oxygen react according to the following equation.
C
3
H
8
(g) + 5O
2
(g) → 3CO
2
(g) + 4H
2
O(g)
Calculate the volume of carbon dioxide and water vapour produced and the volume of oxygen
remaining, when 20.0 dm 3 of propane reacts with 120.0 dm 3 of oxygen. All gas volumes are
measured at the same temperature and pressure. [3]
(b) State and explain what would happen to the pressure of a given mass of gas when its absolute
temperature and volume are both doubled.
[3]
2. (N05) An organic compound, A , containing only the elements carbon, hydrogen and oxygen was
analysed.
(a) A was found to contain 54.5 % C and 9.1 % H by mass, the remainder being oxygen.
Determine the empirical formula of the compound. [3]
(b) A 0.230 g sample of A , when vaporized, had a volume of 0.0785 dm 3 and 102 kPa at 95 C °.
Determine the relative molecular mass of A . [3]
Topic1notes 12.5 hours Page 23 of 31

(c) Determine the molecular formula of A using your answers from parts (a) and (b). [1]
3. (M06) The reaction below represents the reduction of iron ore to produce iron.
A mixture of 30 kg of and 5.0 kg of C was heated until no further reaction occurred.
Calculate the maximum mass of iron that can be obtained from these masses of reactants. [5]
4. (N06) An organic compound A contains 62.0 % by mass of carbon, 24.1 % by mass of nitrogen,
the remainder being hydrogen.
(i) Determine the percentage by mass of hydrogen and the empirical formula of A . [3]
(ii) Define the term relative molecular mass . [2]
(iii) The relative molecular mass of A is 116. Determine the molecular formula of A . [1]
5. (M05) The percentage composition by mass of a hydrocarbon is C = 85.6 % and H = 14.4 %.
(a) Calculate the empirical formula of the hydrocarbon. [2]
(b) A 1.00g sample of the hydrocarbon at a temperature of 273K and a pressure of 1.01
×105 Pa (1.00
atm) has a volume of 0.399 dm 3 .
(i) Calculate the molar mass of the hydrocarbon. [2]
(ii) Deduce the molecular formula of the hydrocarbon. [1]
6. (M03)
(a) The relative molecular mass of aluminium chloride is 267 and its composition by mass is 20.3% Al
and 79.7% chlorine. Determine the empirical formula and molecular formulas of aluminium
chloride. [4]
(b) Sodium reacts with water as follows. 2Na(s) + 2H
2
O (l)

2NaOH (aq) + H
2
(g)
1.15 g of sodium is allowed to react completely with water. The resulting solution is diluted to
250 cm 3 . Calculate the concentration, in mol dm -3 , of the resulting sodium hydroxide solution.
[3]
7. (N03)
(a) Aqueous XO
4
3 ions form a precipitate with aqueous silver ions, Ag + . Write a balanced equation for
the reaction, including state symbols.
[2]
(b) When 41.18 cm 3 of a solution of aqueous silver ions with a concentration of 0.2040 mol dm -3 is
added to a solution of XO
4
3 ions, 1.172 g of the precipitate is formed.
(i) Calculate the amount (in moles) of Ag + ions used in the reaction. [1]
(ii) Calculate the amount (in moles) of the precipitate formed. [1]
Topic1notes 12.5 hours Page 24 of 31

(iii) Calculate the molar mass of the precipitate. [2]
(iv) Determine the relative atomic mass of X and identify the element. [2]
8. (M02) A student was asked to make some copper (II) sulfate-5-water (CuSO
4
.5H
2
O) by reacting
copper (II) oxide (CuO) with sulfuric acid.
(a) Calculate the molar mass of copper (II) sulfate-5-water. [1]
(b) Calculate the amount (in mol) of copper (II) sulfate-5-water in a 10.0 g sample. [1]
(c) Calculate the mass of copper (II) oxide needed to make this 10.0 g sample. [1]
9. (M03)
(a) Write an equation for the reaction between hydrochloric acid and calcium carbonate.
[2]
(b) Determine the volume of 1.50 mol dm -3 hydrochloric acid that would react with exactly 1.25 g of calcium carbonate. [3]
(c) Calculate the volume of carbon dioxide, measured at rtp, which would be produced when 1.25 g of calcium carbonate reacts completely with the hydrochloric acid. [2]
10. (M01)
(a) An anti-cancer drug called Cisplatin has the following percentage composition by mass:
Pt = 65.01% Cl = 23.63% N = 9.340 % H = 2.020 %
Calculate the empirical formula of Cisplatin. [3]
(b) The molecular and empirical formulas of Cisplatin are the same. Analysis of the molecule shows
platinum to be the central atom, being bonded to four separate atoms; the hydrogen is bonded to
nitrogen. Draw the representation of the molecule. [1]
(c) 16.20 x 10 -3 dm 3 of 0.1020 mol dm -3 aqueous AgNO
3
is added to 14.80 x 10 -3 dm 3 of 0.125 mol
dm -3 aqueous NaCl. Calculate the maximum mass (g) of AgCl which could be obtained from this
reaction. [4]
11. (N01 ) Indigo is a blue dye which contains only carbon, nitrogen, hydrogen and oxygen.
(a) 2.036 g of indigo was completely oxidized to produce 5.470 g of carbon dioxide and 0.697 g of
water. Calculate
(i) the percentage of mass of carbon in indigo. [2]
(ii) the percentage of mass of hydrogen in indigo. [2]
(b) If the percentage by mass of nitrogen in the indigo sample is 10.75%, determine the empirical
formula of indigo. [3]
(c) If the molar mass is approximately 260 g mol -1 , determine the molecular formula of indigo. [2]
Topic1notes 12.5 hours Page 25 of 31

12. (N02) A balloon, which can hold a maximum of 1100 cm 3 of air before bursting, contains 955 cm 3 of air at 5ºC . Determine whether the balloon will burst if the temperature is increased to 25 C. Assume
that the pressure of the gas in the balloon remains constant. [3]
13. (N02) An element X reacts with oxygen to form the oxide X
2
O
3
.
(a) Write a balanced equation for the reaction. [1]
(b) If 2.199 g of the oxide was obtained from 1.239 g of X, calculate the relative atomic mass
of X and identify the element. [5]
(c) Nitrogen also forms an oxide on reaction with oxygen. This oxide contains 25.9 % of
nitrogen and 74.1 % of oxygen by mass. Calculate the empirical formula of this second
oxide. [3]
14. (M02) A student is asked to prepare some copper(II) nitrate by reacting nitric acid with copper(II)
oxide.
(a) Write a balanced equation for this reaction. [1]
(b) The student carries out this reaction by adding 0.0345 mol of copper(II) oxide to 36.0 of cm 3
1.15 mol dm -3 nitric acid solution. Calculate the amount (in mol) of nitric acid. [1]
(c) Use the information in (a) and (b) to identify the limiting reagent and determine the amount
(in mol) of copper(II) nitrate formed. [2]
15. The value of X in NaAl(SO
4
)
2
.
X H
2
O can be found by determining the number of moles of sulphate in
the compound quantitatively. A 5.218g sample was dissolved and excess BaCl
2
was added. The
precipitate of BaSO
4
was separated, dried and found to weigh 5.315g. (use integer mass numbers).
(a) Calculate the number of moles of BaSO
4
in the 5.315g sample.
(b) How many moles of sulphate are there in the 5.218g of NaAl(SO
4
)
2
.XH
2
O?
(c) Calculate the number of moles of Al in the 5.218g of NaAl(SO
4
)
2
.XH
2
O.
(d) Determine the total mass of sulphate, sodium and aluminium that are present in the 5.218g sample of NaAl(SO
4
)
2
.XH
2
O
(e) Determine the number of moles of water in the 5.218g sample of NaAl(SO
4
)
2
.XH
2
O
(f) Determine the value of X in NaAl(SO
4
)
2
.XH
2
O
Topic1notes 12.5 hours Page 26 of 31

3
4
5
6
7
8
9
10
PAPER 2
1. (M07)
2. (N05)
Mark scheme
PAPER 1
1
2
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Topic1notes 12.5 hours Page 27 of 31

1. (M06)
2. (N06)
Topic1notes 12.5 hours Page 28 of 31

4.(M01)
Topic1notes 12.5 hours Page 29 of 31

5. (N01)
6. (N02)
Topic1notes 12.5 hours Page 30 of 31

7. (N02)
8.(M02)
Topic1notes 12.5 hours Page 31 of 31