Download Topic 1: Quantitative chemistry

Topic 1: Quantitative chemistry
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.
In the beginning: The following laws and ideas underpin our knowledge about the relationships between
chemicals involved in reactions:
1. Law of Conservation of Mass: Lavoisier (18th 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 CO2) if
the mass of 1 element is the same (e.g. carbon in CO and CO2) than the ratio of the mass of the other
element (in our example oxygen) in each compound must be a whole number.
Fast forward:
1.1.1
1.1.2
1. 1 The mole concept and Avogadro’s constant
Apply the mole concept to substances.
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 1023 a number which is know as Avogadro’s constant.
So 1 mole of a substance = 6.02 x 1023 units of that substance; these units can be atoms, molecules, ions,
electrons or formula units.
1 mole = 6.02 x 1023 atoms, ions, molecules, electrons, …
Determining number of moles from a number of particles
Use the expression:
number of moles =
number of particles
6.02 x 1023
How many particles are in 3 moles,
0.5 moles and 0.001 moles?
Ratios in compounds or formula units:
Examples:
 in 1 mole of Fe2 O3 there are 2 moles of iron atoms and 3 moles of oxygen;
 in 3 moles of Na2CO3 there are 3 moles of CO32- and 6 moles of Na+;
 in 2 moles of (NH4)3 PO4 there are 6 moles of NH4+ and 12 moles of hydrogen atoms.
Exercises
1. If there are 2 x 1022 C-atoms in a sample of ethane, C2H6, how many hydrogen atoms are there in that
same sample?
2. if you have 4.5 x 1022 atoms of hydrogen in a sample of ammonia, how many molecules of ammonia do
you have?
3. How many moles of oxygen are there in 2 moles of Al2O3?
4. How many moles of NO3- are there in 1 mole of Ga(NO3)3?
Topic 1 notes
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5. What is the total number of atoms of 0.05 moles of NH4NO3?
1.2. Formulas
1.2.1
1.2.2
1.2.3
1.2.4.
1.2.5
1.2.6
Define the terms relative atomic mass (Ar) and relative molecular mass (Mr).
Calculate the mass of one mole of a species from its formula.
Solve problems involving the relationship between the amount of substance in moles, mass and molar mass.
Distinguish between the terms empirical formula and molecular formula.
Determine the empirical formula from the percentage composition or from other experimental data.
Determine the molecular formula when given both the empirical formula and experimental data.
Relative isotopic mass
The mass of a specific isotope of an element compared with 1/12th 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.
relative isotopic mass =
actual mass of isotope
1/12th of the mass of a C-12 atom
Relative atomic mass ( Ar)
The relative atomic mass is the weighted average mass of all isotopes of an element compared to 1/12th of
the mass of one 12C atom which has a mass of 12. It is calculated as shown below:
relative atomic mass =
weighted average mass of isotope
1/12th of the mass of a C-12 atom
It is calculated as shown below:
Ar =
(isotopic mass 1 x %) + (isotopic mass 2 x %) + (isotopic mass 3 x %)
100
Or Ar = (isotopic mass 1 x %/100) + (isotopic mass 2 x %/100) + (isotopic mass 3 x %/100)
It is the relative atomic mass that is found in the Periodic Table or data booklet.
Complete the following table:
element
Cu
Eu
Ag
isotopes
abundance
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%
relative
atomic mass
151.9
48.17%
Relative molecular mass ( Mr )
The average mass of a molecule compared to 1/12 th of the mass of one 12C atom.
To be able to calculate the Mr 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 H2O is 18.02 because (2 x 1.01) + (1 x 16.00) = 18.02
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
relative molecular mass of H2SO4 is 98.08 because (2 x 1.01) + 32.06 + (4 x 16.00) = 98.08
Relative formula mass
The Mr only really applies to molecules (a group of covalently bonded atoms); the concept of formula mass is
used for giant structures like ionic substances (e.g. sodium chloride, copper sulphate) and macromolecules
(e.g. silicon dioxide)
Examples:
 relative formula mass of NaOH is 40 because 22.99 + 16.00 + 1.01 = 40.00
 relative molecular mass of Al2(SO4)3 is because (2 x 26.92 ) + (3 x 32.06) + (12 x 16.00) = 342.02
Molar mass
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.
The molar mass is the mass of 6.02 x 1023 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.
Problems involving moles, mass and molar 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
number of moles =
mass of substance (g)
molar mass (g)
Calculate actual masses of atoms or molecules
Find the mass of 1 molecule of ethanol, C2H5OH.
The only actual mass you have is the molar mass of ethanol that can be calculated using a periodic table.
molar mass

mass of 1 atom/molecule
divide by 6.02 x 1023
Calculate number of atoms in a sample
Find the number of carbon atoms in 0.1 mole of C2H5OH.
number of moles

number of molecules
multiply
by 6.02 x 1023

number of atoms
multiply by
subscript of atom in
formula
Calculate the number of particles from a mass
Which sample contains the greatest number of particles? 5g of NH3 or 5 g of H2O
given mass
Topic 1 notes

divide
number of moles
12.5 hours

multiply by 6.02 x
or
5 g of CaO?
number of particles
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1023
by molar mass
Exercises
1. Find the mass of 1 molecule of CH3COOH?
2. Find the mass of 1 atom of Br?
3. Find the mass of 0.05 moles of NH4NO3.
4. Calculate the number of atoms in:
(a) 18.0 g of copper, Cu.
(b) 7.20 g of sulphur, S8.
(c) 1.6 g methane, CH4.
5. Calculate the number of molecules in:
(a) 1.00 g of ammonia, NH3.
(b) 3.28 g of sulphur dioxide, SO2.
6. Calculate the number of ions present in:
(a) 0.500 mol of sodium chloride, NaCl (Na+, Cl-).
(b) 14.6 g of sodium chloride, NaCl.
(c) 18.5 g of calcium chloride, CaCl2.
7. Which sample contains the greatest number of molecules? 10g of C2H4
10g of C2H6
10g of C6H6
8. Which sample contains the lowest number of molecules?
2g of NH3
2g of CO2
2g of CH4
9. How many molecules are there in 0.23 g of ethanol?
10. Which one of the following has the greatest mass?
1.0 x 10-3 moles of CH3COOH
1.2 x 1021 atoms of argon
2.4 x 1020 molecules of C8H18
11. How many glucose molecules are there in 5.23 g of C6H12O6?
Using the mole concept and molar ratios for analytical purposes: empirical
and molecular formulae.
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)
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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 calculate the mass of the gas that was liberated.
Percentage composition of a compound
Experimental analysis usually results in the amount of an element in a compound being expressed as a
percentage composition which can be calculated using the formula below:
% composition =
molar mass of element x coefficient in compound
molar mass of compound
x 100
Calculate
1. Percentage by mass of C in CH4, C2H6 and C2H4
2. Percentage by mass of Cu in CuSO4, CuO and Cu2O
Example calculation of empirical formula: analysis of methane:
elements in compound
carbon
hydrogen
mass in grams (or in %)
number of moles
47.98 (74.9%)
47.98/12 = 3.99 mole
(74.9/12 = 6.24 moles)
16.02
(25.1%)
16.11/1 = 16.02 moles
(25.1/1 = 25.1 moles)
ratio of moles
most simple ratio of moles
empirical formula
4 (6.25)
4/4 = 1 (6.25/6.25 = 1)
16 (25)
16/4 = 4 (25/6.25 = 4)
CH4
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, SiO2, this formula tells us that there are
twice as many oxygen atoms within the lattice as silicon atoms. The formula CuCl2, 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 CH2.
Answer:
 mass empirical formula: 14g
 molar mass of formula: 84g
 whole number ratio (n) of molar mass/empirical mass = 84 g/14g = 6
Topic 1 notes
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
molecular formula = n x empirical formula: (CH2 ) x 6 = C6H12
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?
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
most simple ratio
7.8/ 6.29 = 1.25
6.29/6.29 = 1
AGAIN DO NOT ROUND UP OR DOWN BUT FIND LOWEST
WHOLE NUMBER RATIO
lowest whole number ratio
5
empirical formula:
ratio molecular formula /empirical
formula:
4
C5H4
128g/ 64g = 2
molecular formula = 2 x C5H4 =
C10H8
Exercises:
For each of the following substances, calculate the empirical formula and the molecular formula
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.
1.3. Balanced equations
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).
Topic 1 notes
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As seen in 1.1 the stoichiometric amounts or molar ratios for any reaction can be obtained from a balanced
equation.
Example:
Fe2 O3 (s)
This means that
1 particle
of Fe2 O3
+

3CO (g)
+

3 particles
of CO
2Fe (s)
2 particles
of Fe
+
+
3 CO2 (g)
3 particles
of CO
Alternatively this means that there are 3 times as much particles of CO as Fe2 O3 particles; for the reaction
to go to completion (=both reactants are completely used up) 3 times as much CO is needed as Fe2 O3.
By multiplying each number of particles by 6.02 x 1023, the above balanced equation states the same ratios
but expressed in moles:
1 mole
of Fe2 O3
reacts with
3 moles
of CO
to give
2 moles
of Fe
and
3 moles
of CO2
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 CO2 are needed than Fe2O3
Complete the following table showing other molar ratio’s in the above equation;
Fe2 O3 : Fe
Fe : CO2
CO : CO2
Fe2 O3 : CO2
:
:
:
:
Conservation of mass
When applying the molar mass concept to the equation below the Law of Conservation of Mass can be
proven:
Fe2 O3 (s)
+
3CO (g)

2Fe (s)
+
3 CO2 (g)
1 x 160 g
+
3 x 28g
2 x 56 g
244 g
+
3 x 44 g
244 g
the mass before
=
Any balanced equation should include state symbols.
(s) = solid
(l) = liquid
(g) = gas
the mass after the reaction!!!!!
(aq) = dissolved in water
1. 4. Mass and gaseous volume relationships in chemical reactions
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
Fe2 O3 (s)
+
3CO (g)

2Fe (s)
+
3 CO2 (g)
Step 2: convert data in question into moles
data in question = 28 g or iron
Topic 1 notes
number of moles of iron = 28g / 56g = 0.5 mole of iron
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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 Fe2 O3 (= molar ratio)
 0.5 mole of iron needs 0.25 moles of Fe2 O3 .
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!!!
Theoretical yield and percentage yield
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.
percentage yield =
actual/experimental yield
theoretical yield
X 100
Limiting reagent
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.
N2(g)
+
3H2(g)
2NH3(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.
Topic 1 notes
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1. Find the actual number of moles of each reactant from the mass of reactant given.
for N2
n
=
m
M
=
25000
28
=
892.8 mol
for H2
n
=
m
M
=
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
x
ratio of A
ratio of B
Required moles of H2
=
892.8
x
3
1
=
2678.6 mol
Required moles of N2
=
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
2500 mol  2678.6 mol
So for H2
(for N2 892.8 mol 
so H2 is the limiting reactant
833.3 mol
so N2 is not limiting)
4. 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
e.g. to find the number of moles and the mass of NH3 formed
From the mole ratio’s in the equation
H2
: NH3
3
2
If the actual number of moles of the limiting reagent H2 = 2500 mol then using the mole ratio in the equation
the actual number of moles of NH3 = 2500 x 2 = 1666.6 mol
H2
3
2500
mass NH3
m
:
=
n
x
M
=
1666.6
x
17
=
28332
g
NH3
2
1666.6
5. Find the Percentage Yield when given the mass of product
e.g. find the percentage yield of 10390 g of ammonia
Topic 1 notes
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Experimental Yield %
=
Experimental Yield (g)
Theoretical Yield (g)
x
100
=
10390
28332
x
100
=
36.7 %
Exercises
1.
2.
Consider the reaction:
2 Al
+
3 I2

2 AlI3
Determine the limiting reagent when the following are reacted:
(a) 1.2 mol Al and 2.4 mol I2
(b) 1.2 g Al and 2.4 g I2
Freon-12 is used as a coolant in refrigerators. It is formed in the following reaction:
3 CCl4
+
2 SbF3

3 CCl2F2
+
2 SbCl3
150 g CCl4 (Mr = 154) is combined with 100 g SbF3 (Mr = 178.8) to give freon-12, CCl2F2 (Mr = 120.9).
(a)
Identify the limiting and excess reagents
(b)
How many grams of freon-12 can be formed?
3 Aspirin, C9H8O4 is made by reacting ethanoic anhydride, C4H6O3 (Mr = 102.1) with 2-hydroxybenzoic
acid C7H6O3 (Mr = 138.1) according to the equation:
C4H6O3
+
2 C7H6O3

2 C9H8O4
+
H2O
(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
Avogadro’s law to calculate reacting volumes of gases
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
and the products formed.
For instance,
 1 volume of hydrogen reacts with an equal volume of chlorine to form 2 volumes of hydrogen chloride
 1 volume of oxygen reacts with 2 volumes of hydrogen to form 2 volumes of water vapour.
To explain these results, Avogadro proposed that for Gay-Lussac’s observations to make sense, equal
volumes of gases (if measured in the same conditions i.e. same pressure and temperature) must have the
same number of molecules.
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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)
In other words, if there are x molecules of O2 in 10 cm3 of oxygen gas, then there are x molecules of N2 in
10 cm3 of nitrogen gas and there are 2x molecules of CO2 in 20 cm3 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 H2 gas is double the volume of O2 than there are twice as
many moles of O2 as H2.
Avogadro’s Law allows equations involving gases to be interpreted directly in terms of volumes.
For example, in the equation
N2
for 2.4 dm3 of N2,
+
3H2

2NH3
7.2 dm3 of H2 is needed to give 4.8 dm3 of NH3.
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 m3 mol-1 or 22.4 dm3 mol-1 and this is called the
gas molar volume.
At s.t.p. (which is 273K and 1 atm)
volume of 1 mole of a gas = 2.24 x 10-2 m3 mol-1 or 22.4 dm3 mol-1
Ideal gas equation
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 (this is the pressure acting on the
gas – from the outside) 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 V/n = constant (Avogadro’s Law)
pV = nRT
Irrespective of the conditions the gas is in (pressure, temperature, volume), the ratio of pV over nT 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!!!
Topic 1 notes
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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 m3
( 1m3 = 1000 dm3
Temperature should be in K.
and 1 dm3 = 1000 cm3 or ml)
If the pressure is in atm then change to Pa (1 atm = 1.01 x 105 Pa) (see data booklet)
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 provided the 2 other properties are known
calculate the molar mass of a gas:
o
o
o

we need to know the conditions it is in and the mass of the sample of the gas
we can then calculate n and then using n = mass/Mr calculate the Mr
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 of gas remains 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:
V1 P1
V2 P2
=
T1
T2
Worked example 1
What volume will be occupied by 0.216 g of CO2 at 21 C and at a pressure of 1.34 x 105 Pa?
1. Calculate the number of moles of the gas: 0.216 g/44.0 g = 4.91 x 10-3 mol
2. Express all temperatures in Kelvin: 21 C = 294 K
3. Apply pV = nRT
1.34 x 105 x V = 4.91 x 10-3 x
V = 8.95 x 10-5 m3 (= 89.5 cm3)
8.31 x 294
Worked example 2
A gas occupies 127 cm3 at a pressure of 8.4 x 104 Pa and at 28C.
(a) What volume will the same amount of gas occupy at 1.01 x 105 and 0 C
(b) How many moles of gas are present
Answer
1. Express all temperatures in Kelvin: 28 C = 301 K
2.
Topic 1 notes
12.5 hours
and
0 C = 273 K
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V1 P1
T1
Apply:
V2
=
V1 x P1 x T2
P2 T1
V2 P2
T2
=
127 x 8.4 x 104 x 273
1.01 x 105 x 301
=
= 95.8 cm3
When using the above expression there is no need to change volume in dm3 (or even pressure in atm).
What needs changing though is temperature as this needs to be in Kelvin.
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
n = PV/RT
n=
8.4 x 104 Pa x 1.27 x 10-4 m3
8.31 J K-1 mol-1 x 301 K
= 4.3 x 10 -3 mol
Exercises
1. The greatest volume for one mole of O2 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.
B.
C.
D.
number of molecules increases.
molecules expand and occupy a greater volume.
molecules collide more frequently and energetically with the surface of the container
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
C. temperature, rate of diffusion
B. volume, number of moles
D. pressure, volume
5. A 250 dm3 sample of an unknown gas has a mass of 1.42 g at 35 C and 1.01 x 105 Pa. Which
expression gives its molar mass, Mr? (R = 8.31 J K-1 mol-1)
1.42 x 8.31 x 35
1.42 x 8.31 x 308
A.
B.
0.25 x 1.01 x 10
5
0.25 x 1.01 x 105
1.42 x 250 x 1.01 x105
1.42 x 8.31 x 308
C.
D.
8.31 x 308
250 x 1.01x105
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
Topic 1 notes
B. 333C
C. 393C
12.5 hours
D. 666C
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7. Which change will have the greatest effect on the pressure of a fixed mass of an ideal gas?
VOLUME
TEMPERATURE /K
A
doubles
halves
B
doubles
doubles
C
halves
halves
D
halves
remains constant
For the following MCs you can use your calculator:
8. A 1.25 dm3 sample of air at 25C has a pressure of 85.9 kPa. What is its volume, in dm3, if its
temperature and pressure are changed to 100C and 65.7 kPa ?
A. 0.76
B. 1.20
C. 1.31
D. 2.05
9. A sample of ethane occupies a volume of 120 cm3 at 30C and 1.0 atm (101kPa). What volume (in cm3)
will it occupy at 40C and 1.2 atm (121 kPa)?
A. 103
B. 133
C. 149
D. 192
10. A 225 cm3 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 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 cm3 at 35C and 98.6 kPa. 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. Solutions
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 subtance
solution = a mixture made up from a solute and solvent
amount of solute in moles or grams
volume of solution in dm3
concentration =
Example: Calculate the volume of H2 SO4 with a concentration of 2 mol dm3 to neutralise 250 cm3 of
sodium hydroxide with a concentration of 1 mol dm3.
Answer:
Step 1: write balanced equation
2 NaOH (aq)
+
H2 SO4 (aq)
Step 2: convert data into moles
Topic 1 notes

12.5 hours
Na2 SO4 (aq)
+
H2 O (l)
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use :
concentration (mol dm3)
number of moles
volume (dm3)
=
number of moles of NaOH = 1 mol dm3 x 0.25 dm3 = 0.25 mole of NaOH
Step 3: find answer in equation in moles
equation tells us: molar ratio NaOH : H2 SO4 = 2 : 1
 0.25 mole of NaOH needs to 0.125 moles of H2 SO4.
Step 4: convert answer into relevant unit
volume of H2 SO4 = number of moles/concentration = 0.125 /2 mol dm3 = 0.0625 dm3
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.
Topic 1 notes
12.5 hours
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