1. Definitions
The relative atomic mass (Ar) of an element is the mass of one atom of
the element, relative to the mass of 1/12 the mass of one atom of carbon 12.
The relative molecular mass (Mr) is the mass of one molecule, relative
to the mass of 1/12 the mass of one atom of carbon 12.
2. Mole
The mole is used to represent the amount of particles. One mole refers to 6.02 x 10^{23} particles.
The Avogadro constant (L) has a value of 6.02 x 10^{23}.
Hence, number of particles = number of moles x Avogadro constant
Another formula for finding number of moles = mass in grams / molar mass (or mass you get from periodic table)
Example. Find the mass of 5 mol of chlorine atom.
Step 1: Since questions ask for chlorine atom, we
make use of Ar. From the periodic table, Ar of chlorine is 35.5.
Step 2: moles = mass/ molar mass ==> mass =
molar mass x moles ===> mass = 5 x 35.5 = 177.5g
Example. Find the mass fo 5 mol of chlorine gas.
Step 1: Chlorine gas exists as Cl_{2}.
Since the Ar of one chlorine is 35.5, the Mr of Cl_{2} is 35.5 x 2 = 71.
Step 2: moles = mass/ molar mass ==> mass = molar mass x
moles ===> mass = 5 x 71 = 355g
3. Calculating relative molecular mass
You will need to use the mass number of the elements, to determine
relative molecular mass.
Example: Find the relative molecular mass of carbon dioxide.
Carbon dioxide has a formula of CO_{2}. This means that it contains
1 carbon atom, and 2 oxygen atoms per molecule. From the periodic table, the
mass number of carbon is 12, and the mass number of oxygen is 16. Hence, the Mr
of carbon dioxide = 12 + 16 x 2 = 44.0
4. Calculating percentage mass (or percentage composition)
Example: Calculate the percentage composition of oxygen in carbon
dioxide.
Step 1 Find the Mr of carbon dioxide.
Mr of CO_{2} = 12 + 16 x 2 = 44.0
Step 2 Determine relative mass of oxygen per molecule.
Since there are 2 oxygen per molecule of carbon dioxide, and the mass
number of each oxygen is 16, hence the relative mass of oxygen per molecule is
16 x 2 = 32.
Step 3 Calculate percentage composition, by taking answer from step 2
divided by answer from step 1, and multiply by 100%.
Percentage composition of oxygen = 32/44 x 100% = 72.7% (3 significant
figures)
Example: Calculate the mass of aluminium, in 25g of aluminium oxide.
Step 1 Write out the formula of aluminium oxide, since it is not given.
Al_{2}O_{3}.
Step 2 Calculate the Mr of Al_{2}O_{3}.
Mr of Al_{2}O_{3 }= 2 x 27 + 3 x 16 = 102
Step 3  Determine the relative mass of aluminium in one compound of
aluminium oxide.
There are 2 aluminium atoms per Al_{2}O_{3}. Hence,
relative mass of Al in Al_{2}O_{3 }= 2 x 27 = 54.
Step 4 Step 2 and 3 tell you that for every 102 g of Al_{2}O_{3},
there is 54g of Al. Hence, in 25g of Al_{2}O_{3}, mass of
Al = 54/102 x 25 = 13.2 g (3 significant figures).
5. Calculating empirical and molecular formulae
Empirical formula gives the simplest whole number ratio of each element
in a compound.
Molecular formula shows the exact number of each element in a compound.
5.1. Calculating empirical and molecular formulae, given mass of each element.
Example. 50 g of a hydrocarbon is make up of 42.9g carbon. Find the
empirical formula of this compound. Given that the molecular mass of this
hydrocarbon is 42.0, determine its molecular formula.
Step 1. Since this is a hydrocarbon, it means that the compound consists
of hydrogen and carbon only. Since it contains 42.9g of carbon, mass of
hydrogen = 50  42.9 = 7.10 g
Step 2. Set up the table.
C

H


mass/g (1)

42.9

7.10

Ar (2)

12

1

Number of moles per 100 g (3)
[Take (1) divided by (2)]

3.58

7.1

Ratio (round off to nearest whole number).
From (3), notice that 3.58 is the smallest
number. Hence, divide all numbers in (3) by 3.58.

1

2

Empirical Formula

CH_{2}

Step 3: Determine molecular formula.
Let the molecular formula be (CH_{2})_{n}, where n is an
integer.
Molecular mass = (12 + 2 x 1) x n = 42
14 n = 42
n = 3
Hence, molecular formula = C_{3}H_{6}.
5.2 Calculating empirical and molecular formulae, given percentage
composition of each element.
Example. A hydrocarbon consists of 86% by mass of carbon. Find the
empirical formula of this compound. Given that the molecular mass of this
hydrocarbon is 42.0, determine its molecular formula.
Step 1. Since this is a hydrocarbon, it means that the compound consists
of hydrogen and carbon only. Since it contains 86% carbon, percentage
composition of hydrogen = 100%  86% = 14%
Step 2: Set up the table
C

H


% composition (1)

86

14

Ar (2)

12

1

Number of moles per 100 g (3)
[Take (1) divided by (2)]

7.17

14

Ratio (round off to nearest whole number).
From (3), notice that 7.17 is the smallest
number. Hence, divide all number in (3) by 7.17.

1

2

Empirical Formula

CH_{2}

Step 3: Determine molecular formula.
Let the molecular formula be (CH_{2})_{n},where n is an
integer.
Molecular mass = (12 + 2 x 1) x n = 42
14 n = 42
n = 3
Hence, molecular formula = C_{3}H_{6}.
6. Gases
number of moles of gases = (volume of gas in dm^{3})/
(molar volume)
molar volume at room temperature and pressure = 24 dm^{3}
molar volume at standard temperature and pressure = 22.4 dm^{3}.
Note: 1dm^{3} = 1000 cm^{3}
Example.Find the number of moles of carbon dioxide in 250 cm^{3}of
carbon dioxide gas.
Step 1. 250 cm^{3}= 250/1000 dm^{3} = 0.250 cm^{3}
Step 2. moles = volume/ molar volume = 0.25/24 =0.0104 mol
7. Solutions
Concentration of solutions can be represented by mol/dm^{3} or
g/dm^{3}.
Concentration in mol/dm^{3}= (number of
moles) / (volume in dm^{3})
Concentration in g/dm^{3}= (mass in grams)
/ (volume in dm^{3})
Concentration in g/dm^{3}= Concentration in
mol/dm^{3 }x Mr
8. Chemical equations and mole ratios
2NaOH (aq)+ H_{2}SO_{4} (aq)> Na2SO4 (aq) + 2H_{2}O (l)
From the above equation, it means that 2 moles of NaOH react with 1 mole
of H_{2}SO_{4} to form 1 mole of Na_{2}SO_{4} and 2 moles of H_{2}O.
Example. 20.0 cm^{3 }of 0.250 mol/dm^{3} sodium
hydroxide reacts completely with 15 cm^{3} of sulfuric acid.
Find the concentration of the sulfuric acid in (i) mol/dm^{3} (ii)
g/dm^{3}.
Step 1. Write the equation for the reaction, if it is not given. In this
case, the equation is:
2NaOH (aq)+ H_{2}SO_{4} (aq)> Na_{2}SO_{4} (aq)
+ 2H_{2}O (l)
Step 2. Find the number of moles of NaOH.
no. of mol of NaOH = 20/1000 x 0.250 = 0.005 mol
Step 3. Use equation to find number of moles of H_{2}SO_{4}.
From equation, 2 mol of NaOH reacts with 1 mol of H_{2}SO_{4}.
Hence, number of moles of H_{2}SO_{4}= 0.005/2 = 0.00250
mol
Step 4. Find concentration of H_{2}SO_{4}.
15 cm^{3}= 15/1000 dm^{3}= 0.015 dm^{3}.
Concentration (mol/dm^{3}) of H_{2}SO_{4 } =
0.00250/ 0.015 = 0.167 mol/dm^{3}
Mr of = 2 x 1 + 32 + 16 x 4= 98.0
Concentration (g/dm^{3}) of H_{2}SO_{4 }=
0.167 x 98 = 16.4 g/dm^{3}.
9. Limiting and excess reactant
The limiting reactant is one that will be used up in the reaction. The
excess reactant is one that will not be used up in the reaction.
Example. 20.0 cm^{3 }of 0.250 mol/dm^{3} sodium
hydroxide is added to 20.0 cm^{3} of 0.2 mol/dm^{3}
sulfuric acid. Find the mass of sodium sulfate formed.
Step 1: Write the equation for the reaction, if it is not given. In this
case, the equation is:
2NaOH (aq)+ H_{2}SO_{4} (aq)> Na_{2}SO_{4} (aq)
+ 2H_{2}O (l)
Step 2: The question gives sufficient information to find number of
moles of both reactants  H_{2}SO_{4} and NaOH.
This means that you need to find limiting and excess reactant.
Step 3: Find number of moles of both reactant.
No. of mol of NaOH = 20/1000 x 0.250 = 0.005 mol
No. of mole of H_{2}SO_{4} = 20/1000 x 0.2 =
0.004 mol
Step 4. Use values from step 3, and equation to determine limiting and
excess reactant.
From equation, 2mol of NaOH will react with 1 mol of H_{2}SO_{4}.
Hence, 0.005 mol of NaOH will react with 0.005/2 = 0.0025 mol of H_{2}SO_{4}.
However, 0.004 mol of H_{2}SO_{4 }is added. This means
that H_{2}SO_{4 }is in excess, and NaOH is the limiting
reagent.
Alternatively
From equation, 1 mol of H_{2}SO_{4 }will
react with 2 mol of NaOH
Hence, 0.004 mol of H_{2}SO_{4} will react with
0.004 x 2 = 0.008 mol of NaOH. However, only 0.005 mol of NaOH_{ }is
added. This means that H_{2}SO_{4 }is in excess, and NaOH
is the limiting reagent.
Step 5. Once you found the limiting reagent, use its number of moles for
calculating amount or mass of other reactant/ products, because it is the one
that reacts completely.
Since NaOH is the limiting reactant, we use it to calculate number of
moles of Na_{2}SO_{4}.
No. of mol of NaOH = 20/1000 x 0.250 = 0.005 mol
From the equation, 2 moles of NaOH form 1 mole of Na_{2}SO_{4}.
No. of mol of Na_{2}SO_{4 }= 0.005 / 2 =
0.00250
Mr of Na_{2}SO_{4 }= 23 x 2 + 32+ 16 x 4 =
142
Mass of Na_{2}SO_{4}= 142 x 0.00250 = 0.355 g
10. Percentage yield and purity
% yield = (actual mass)/ (theoretical mass) x 100%
% purity = (calculated mass)/ (mass of impure substance) x 100%
Example. When 100 g of calcium carbonate is heated, 32 g of calcium
oxide is obtained. Calculate the percentage yield.
Step 1. Write the equation for the reaction, if it is not given. In this
case, the equation is:
CaCO_{3}_{ }> CO_{2} + CaO
Step 2. Find the Mr of CaCO_{3 } (what is given)
and CaO (what you need to caculate).
Mr of CaCO_{3 }= 40 + 12 + 16 x 3 = 100
Mr of CaO = 40 + 16 = 56.0
Step 3. Find mass of CaO produced from 100g of CaCO_{3 }.
No. of mol of CaCO_{3 }= 100/ 100 = 1
From equation, 1 mol of CaCO_{3 }forms 1 mol of CaO.
No. of mol of CaO = 1
Theoretical mass of CaO = 1 x 56 = 56.0 g
Step 4. Calculate percentage yield.
percentage yield = actual mass/ theoretical mass x 100% = 32/56 x 100% =
57.1%
Example. Limestone is consist of calcium carbonate and other impurities.
When 100 g of calcium carbonate is heated, 32 g of calcium oxide is obtained.
Calculate the percentage purity.
Step 1. Write the equation for the reaction, if it is not given. In this
case, the equation is:
CaCO_{3 }> CO_{2} + CaO
Step 2. Find the Mr of CaCO_{3 } and CaO.
Mr of CaCO_{3 }= 40 + 12 + 16 x 3 = 100
Mr of CaO = 40 + 16 = 56.0
Step 3. Find mass of CaCO_{3 }required to produce 32 g of
CaO.
No. of mol of CaO = 32/ 56 = 0.571 mol
From equation, 1 mol of CaCO_{3 }forms 1 mol of CaO.
No. of mol of CaCO_{3 }reacted = 0.571 mol
Step 4. Calculate mass of CaCO_{3 }present in limestone.
Mass of CaCO_{3} = 0.571 x 100 = 57.1 g
Step 4. Calculate percentage yield.
percentage yield = theoretical mass/ mass of impure substance x 100% =
57.1/100 x 100% = 57.1%
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