Please Return This Booklet to Your Teacher at the End of the Course

National 5 Chemistry

Calculations Booklet

Please return this booklet to your teacher at the end of the course.

WRITING FORMULAE

Example 1 Hydrogen Oxide

Elements present

Valency

Formula H2O

Example 2 Calcium Nitrate

Elements Present

Formula Ca(NO3)2

1. Write the formula for the following compounds:

a) sodium fluoride

b) hydrogen sulphide

c) calcium oxide

d) sulphur dioxide

e) sodium chloride

f) iron(II) chloride

g) copper(II) oxide

h) potassium sulphate

i) calcium nitrate

j) ammonium chloride

k) magnesium carbonate

l) ammonium hydroxide

m) iron (III) sulphate

You should be able to remember the formulae for these common compounds.

2. What is the formula for:

a) Oxygen

b) Carbon dioxide

c) Water

d) Hydrogen

e) Carbon monoxide

f) Bromine

g) Hydrochloric acid

h) Sulphur dioxide

i) Nitric acid

j) Chlorine

FORMULA MASS

Example 1 sodium chloride

Formula NaCl

Elements present Na Cl

Number of atoms present 1 1

Relative Atomic Mass 23 35.5

Totals 23 + 35.5 = 58.5 amu

Example 2 hydrogen sulphate (sulphuric acid)

Formula H2SO4

Elements present H S O

Number of atoms present 2 1 4

Relative Atomic Mass 1 32 16

Totals 2 + 32 + 64 = 98 amu

3. Calculate the formula mass of the following:

a. CO2

b. MgCl2

c. C2H6

d. CaSO4

e. O2

f. sodium carbonate

g. aluminium oxide

h. bromine

i. sodium oxide

j. sulphur dioxide

MOLES

A mole of a substance is the formula mass in grams. This is also known as the gram formula mass (GFM).

Example 1 calcium chloride

Formula CaCl2

Elements present Ca Cl

Number of atoms present 1 2

Relative Atomic Mass 40 35.5

Totals 40 + 71 GFM = 111g

Example 2 magnesium nitrate

Formula Mg(NO3)2

Elements present Mg N O

Number of atoms present 1 2 6

Relative Atomic Mass 23·5 14 16

Totals 23·5 + 28 + 96 GFM = 147·5g

4. Calculate the GFM of each of the following examples:

a) C

b) H2

c) HCl

d) H2O

e) C4H8

f) C2H4

g) C2H6

h) C2H5OH

i) Mg(OH)2

j) calcium oxide

k) lithium chloride

l) sulphuric acid

m) magnesium oxide

n) potassium hydroxide

o) ammonium nitrate

p) magnesium sulphate

q) ammonium sulphate
Example 3 sodium hydroxide

Formula NaOH

Elements present Na O H

Number of atoms present 1 1 1

Relative Atomic Mass 23 16 1

Totals 23 + 16 + 1 GFM = 40g

2 moles of NaOH = 2 × 40 = 80g

3 moles of NaOH = 3 × 40 = 120g

10 moles of NaOH = 10 × 40 = 400g

0.1 moles of NaOH = 0.1 × 40 = 4 g

0.25 moles of NaOH = 0.25×40 = 10g

Example 4 Calculate the mass of 0·1 mole of lithium chloride, LiCl

GFM = 7 + 35·5 = 42·5g

0·1 mole = 0·1 × 42·5g = 4·25g

5. Calculate the mass of each of the following substances:

a) 2 moles of copper

b) 0.1mole of carbon monoxide

c) 0.2moles of NaOH

d) 2 moles of lithium chloride

e) 3 moles of propane

f) 0.8 moles of (NH4)2SO4

g) 0.3 moles of water

h) 10 moles of copper sulphate

i) 0.5 moles of potassium hydroxide

j) 0.3 moles of iron (II) chloride

MASS TO MOLES

Example 1 How many moles are there in 2g of NaOH? (GFM = 40 grams).

You use the formula from page 3 of the data booklet.

n =

= 0·05 moles

Example 2 How many moles are there in 20g of calcium carbonate, CaCO3? (GFM = 100g)

n =

= 0·2 moles

6. Calculate the number of moles in the following substances:

a) 20g calcium

b) 54g water

c) 12g helium

d) 14g nitrogen gas

e) 10g calcium carbonate

f) 280g carbon monoxide

g) 10g hydrogen gas

h) 10·1 g potassium nitrate

i) 8g sulphur

j) 6·6g ammonium sulphate

CONCENTRATION

The concentration (C) of a solution is usually expressed as the number of moles (n) of solute dissolved in one litre of solution and the units are mol/litre (mol l-1).

You start with the formula from page 3 of the data booklet n = CV. (NB V must be in litres)

Example One A solution of sodium chloride (NaCl) has 2 moles dissolved in 500cm3 of solution. The equation is rearranged to calculate the concentration:-

C =

C = = 4 moles/ litre (4 mol l-1)

Example Two Calculate the concentration of a solution which contains 1g of NaOH dissolved in 250 cm3 of solution.

Step one — calculate the number of moles of solid

GFM (mass of 1 mole) of NaOH = 40g

n =

= 0·025

Step two – calculate the concentration

C = = = 0·1 moles/ litre (0.1 mol l-1)

7. Calculate the concentrations of the following solutions.

a) 4g of sodium hydroxide dissolved in 1 litre of solution

b) 5.55g of calcium chloride dissolved in 2 litres of solution

c) 16 g of copper (II) sulphate dissolved in 25cm3 of solution

d) 10g of ammonium nitrate dissolved in 100cm3 of solution

e) 3·65g of hydrogen chloride dissolved in 250cm3 of solution

f) 27.6g of potassium carbonate dissolved in 200cm3 of solution

g) 23.4g of sodium chloride dissolved in 100cm3 of solution

You can rearrange the equation to calculate the volume of solution: V =

8. Calculate the volumes required in order to make the following solutions:

a) What volume of a 1 mole/litre solution contains 2 moles solute?

b) What volume of a 2 mole/litre solution contains 0·4 moles solute?

c) What volume of a 0·5 mole/litre solution contains 0·1 moles solute?

d) What volume of a 1 mole/litre solution contains 0·2 moles solute?

e) What volume of a 2 mole/litre solution contains 0·5 moles solute?

f) What volume of a 4 mole/litre solution contains 0·4 moles solute?

g) What volume of a 0·5 mole/litre solution contains 0·15 moles solute?

To calculate the mass of solute you firstly use n = CV to calculate the number of moles (n) then calculate the mass of solute (m) required by rearranging:

n = to m = n ÍGFM

9. Calculate the mass of solute required in the following solutions:

a) 0·8 litre of a 0·5 mol/litre solution of ammonium nitrate, NH4NO3, has to be made up; what mass of solid ammonium nitrate would be required?

b) 25cm3 of a 0·4 mol/litre solution of ammonium sulphate (NH4)2SO4 is made up. What mass of solid must have been dissolved?

c) What mass of anhydrous copper (II) sulphate, CuSO4, would be required to make 80cm3 of a 0·15 mol/litre solution?

d) What mass of pure ethanoic acid, CH3COOH, would be required to make 40cm3 of a 4 mol/litre solution?

e) How many moles of potassium hydroxide are required to make 200cm3 of solution, concentration 0·5mol/litre?

f) How many moles of a substance would be dissolved to make 0·2 litre of a 1 mol/litre solution?

g) What mass of sodium hydroxide would be required to make 100cm3 of a 0·25 mol/litre solution?

VOLUMETRIC TITRATIONS

Example 1 20cm3 of sodium hydroxide was neutralised by 15cm3 of 0·1 mol/litre sulphuric acid. Calculate the concentration of the sodium hydroxide.

H2SO4 + 2NaOH → Na2SO4 + H2O

Step 1 – calculate the number of moles of acid reacting

m = c × v

m = 0·1 x 0·015

= 0.0015

Step 2 – relate the number of moles of acid reacting to the appropriate number of moles of base, then calculate the concentration using c=

From the equation, 1 mole H2SO4 reacts with 2 moles NaOH

So 0·0015 moles of H2SO4 will react with 0·003 moles NaOH

concentration of NaOH c =

c =

c = 0.15 mol/litre

Alternatively the following equation from page 3 of the data booklet can be used:

= where: C1 = concentration of acid (reactant 1)

C2 = concentration of alkali or soluble base (reactant 2)

V1 = volume of acid

V2 = volume of alkali or soluble base

n1 = number of moles of acid in balanced equation

n2 = number of moles of alkali or soluble base in equation

H2SO4 + 2NaOH → Na2SO4 + H2O

1 mole 2 mole

C2 =

C2 = 0·15 mol/litre

10. Calculate the answers to the following volumetric titrations:

a) What volume of HCl (0·1 mol/litre) is required to neutralise 100cm3 of NaOH solution (0·5 mol/litre)?

b) What volume of potassium hydroxide, KOH, (2 mol/litre) is required to neutralise 50cm3 of H2SO4 (1 mol/litre) solution

c) What volume of nitric acid (0·5 mol/litre) is required to neutralise 25cm3 NaOH (4 mol/litre)?

d) What volume of sulphuric acid (2 mol/litre) is required to neutralise 25cm3 of KOH solution (4 mol/litre)?

e) If 25cm3 of HCl is neutralised by 40·7 cm3 of NaOH solution (2 mol/litre), what is the concentration of the acid?

f) If 50cm3 KOH solution is neutralised by 17·8cm3 of H2SO4 (2 mol/litre), what is the concentration of the alkali?

g) What mass of solid calcium hydroxide, Ca(OH)2, would exactly neutralise 25cm3 of 0·4 mol/litre nitric acid?

PERCENTAGE COMPOSITION

The percentage composition is the percentage by mass of one or more of the elements in a compound.

It is calculated using the equation form page 3 of the data booklet:

% by mass =

Example 1 What is the percentage of copper in copper (II) oxide, CuO?

Elements present Cu O

Number of atoms present 1 1

RAM 63·5 16

GFM 63·5 + 16 = 79·5g

Percentage of Cu present = × 100

= × 100%

= 80%

Example 2 What is the percentage of iron in iron (III) oxide, Fe2O3?

Elements present Fe O

Number of atoms present 2 3

RAM 56 16

GFM 112 + 48 = 160 g

Percentage of Fe present = × 100%

= × 100%

= 70%

11. Calculate the percentage compositions of the following:

a) Calculate the percentage by mass of carbon in carbon dioxide, CO2.

b) Calculate the percentage by mass of carbon in sodium carbonate.

c) Calculate the percentage by mass of iron in Fe2O3

d) Calculate the percentage by mass of oxygen in water.

e) Calculate the percentage by mass of nitrogen by mass in NH4NO3

f) Calculate the percentage by mass of copper in the ore malachite (CuCO3)

g) Calculate the percentage by mass of each element in sucrose (C12H22O11)

BALANCING EQUATIONS

Example one

H2 + O2 → H2O

LHS H 2 RHS H 2

O 2 O 1

The hydrogen atoms are balanced but the oxygen atoms are not. There are two oxygen atoms on the LHS but only 1 on the RHS, so we put a 2 in front of H2O

H2 + 02 → 2H2O

However, this now means there are two hydrogen atoms on the left, but four on the right. In order to have four hydrogen atoms on the left, we put a 2 in front of the H2

2H2 + O2 → 2H2O

LHS H 4 RHS H 4

O 2 O 2

The equation is now balanced.

Example Two

C2H6 + O2 → CO2 + H2O

LHS C 2 RHS C 1

H 6 H 2

O 2 O 3

Balancing carbon atoms gives:

C2H6 + O2 → 2CO2 + H2O

Balancing hydrogen atoms gives:

C2H6 + O2 → 2CO2 + 3H2O

Balancing oxygen atoms requires 7 oxygen atoms on the LHS so we multiply the whole equation by two in order to avoid using fractions.

2C2H6 + 7O2 → 4CO2 + 6H20

LHS C 4 RHS C 4

H 12 H 12

O 14 O 14

The equation is now balanced.

12. Balance the following equations:

a) C + Br2 → CBr4

b) H2 + Cl2 → HCl

c) Ca + O2 → CaO

d) N2 + H2 → NH3

e) C2H4 + O2 → CO2 + H2O

f) C4H10 + O2 → CO2 + H2O

g) C5H12 + O2 → CO2 + H2O

h) ZnCO3 → ZnO + CO2

i) Mg + HNO3 → Mg(NO3)2 + H2

j) Na + H2O → NaOH + H2

k) Al + H2SO4 → Al2(SO4)3 + H2

l) Al + Fe2O3 → Fe + Al2O3

m) K2CO3 + H2SO4 → K2SO4 + H2O + CO2

n) H2SO4 + NaOH → Na2SO4 + H2O

o) C6H12O6 → C2H5OH + CO2

REACTING MASSES

Step 1 Write a balanced equation

Step 2 Select the two chemicals

Step 3 From the equation, workout the number of moles of each of these chemicals

Step 4 Convert moles into masses

Step 5 Divide down to 1g of the known substance

Step 6 Multiply to the correct mass of the known substance

Step 7 Calculate the mass of the other chemical by proportion

Example 1 What mass of copper would be obtained by heating 4g of copper(II) oxide with an excess of carbon?

2CuO + C → 2Cu + CO2

2 moles 2 moles

1 mole 1 mole

79·5g 63·5g

4g = 3·19g

Example 2 What mass of hydrogen gas would be evolved if 6g of magnesium reacted completely with dilute sulphuric acid?

Mg + H2SO4 → MgSO4 + H2

1 mole 1 mole

23·5g 2g

6g = 0·52g

13.

a) Mg + H2SO4 → MgSO4 + H2

What mass of hydrogen gas would be evolved if 6g of magnesium reacted completely with dilute sulphuric acid?

b) CuO + H2 → Cu + H2O

What mass of copper metal would be obtained by the complete reduction of 16g of copper(II) oxide by hydrogen gas?

c) 2CO + O2 → 2CO2

What mass of CO would required to be completely burned in oxygen to form 5·5g of CO2?