2 - Amount of substance

Question 1

Define relative atomic mass (Ar)

Answer 1

The average mass of an atom of an element relative to 1/12 of the mass of a carbon-12 atom

Question 2

Define relative molecular mass (Mr)

Answer 2

The average mass of a molecule relative to 1/12 of the mass of a carbon-12 atom

Question 3

What is a ‘mole’?

Answer 3

The amount of a substance that contains the same number of particles as the number of atoms in 12 grams of carbon-12

Question 4

1 mole of any substance contains 6.02x1023…

Answer 4

atoms / molecules / electrons / particles

Question 5

What equation do we use to calculate the number of particles?

Answer 5

number of particles = number of moles x Avogadro’s number

no of particles = n x L

Question 6

Example question

How many particles make up 0.67 moles of ammonia (NH3)?

Answer 6

Number of particles = 6.02x10²³ x 0.67

number of particles = 4.03x10²³

Question 7

Example question

How many atoms are in 57.0g of NH₃?

Answer 7

Mr of NH₃ = 14.0 + (3 x 1.0) = 17.0 g mol^-1

n = m / Mr
n = 57.0 / 17.0 = 3.35 mol

no of particles = n x L
3.35 x 6.022 x 10²³ = 2.02 x 10²⁴

There are 4 atoms present in a molecule of NH₃ so therefore we times 2.02 x 10²⁴ by 4 which equals 8.08 x 10²⁴

Question 8

Example question

What is the mass of 1 atom of Carbon?

Answer 8

Ar = 12.0

1 mole of C = 12g
6.022 x 10²³ atoms of C = 12.0g

Divide both sides by Avogadro’s number

1 atom of C = 1.9926 x 10^-23 g
3 sig fig = 1.99 x 10^-23 g

Question 9

How can we calculate the number of moles from the mass and Mr?

Answer 9

Number of moles = mass (g) / Mr or Ar

n = m / Mr or Ar

Question 10

Example question

Calculate the number of moles of 23g of gold. Give your answer to 2 significant figures

Answer 10

Number of moles = mass / Ar

Number of moles = 23 / 197

Number of moles = 0.12mol (2 s.f)

Question 11

Example question

Which contains the greatest number of atoms?
10.8g of water or 4.8g of helium

Answer 11

H₂O
n = m / Mr
n = 10.8g / 18.0 = 0.6 moles of H₂O
There is 3 atoms present in a H₂O molecule so we therefore times the number of moles by the amount of atoms
0.6 x 3 = 1.8 moles

He
n = m / Ar
n = 4.8 / 4 = 1.2 moles

Water contains the greatest number of atoms

Question 12

How can we calculate the number of moles in solutions?

Answer 12

Number of moles = concentration (moldm^-3) x volume (dm³)

n = c x v

Question 13

Example question - moles in solution

Calculate the number of moles of 200cm³ of 0.35moldm^-3 HCl

Answer 13

Number of moles = concentration x volume

200cm³ must be in dm³: 200 / 1000 = 0.2 dm³

Number of moles = 0.35 x 0.2 = 0.07moldm^-3

Question 14

What is the ideal gas equation?

Answer 14

pV = nRT

Question 15

What are the units of the ideal gas equation?

Answer 15

p = pressure (Pa)
V = volume (m³)
n = number of moles (mol)
R = gas constant (given in exam)
T = temperature (K)

Question 16

What are the standard conditions of temperature and pressure?

Answer 16

Temperature = 298K
Pressure = 100kPa

Question 17

Example questions - ideal gas equation

Calculate the volume in cm³ of 0.36 moles of a gas at 100kPa and 298K

Answer 17

pV = nRT rearrange to give V = nRT / p

100kPa to 100,000 Pa

V = 0.36 x 8.31 x 298 / 100000

V = 8.91x10^-3 m³

V = 8.91x10^-3 x 1,000,000 = 8910cm³

Question 18

Define ‘empirical formula’

Answer 18

The simplest whole number ratio of each element in a compound

Question 19

How do we work out the empirical formula?
(Example) A compound contains 23.3% Magnesium, 30.7% Sulfur and 46.0% Oxygen. What is the empirical formula of this compound?

Answer 19

1) Write out the elements involved
2) Write the percentages as masses
3) Divide these by their relative atomic masses to get the number of moles
4) Divide all the numbers by the smallest number of moles

Mg = 23.3g / 24.3 (Ar) = 1

S = 30.7g / 32.1 (Ar) = 1

O = 46.0g / 16.0 (Ar) = 3

Empirical formula = MgSO₃

Question 20

Define ‘molecular formula’

Answer 20

The molecular formula is the actual number of atoms of each element in the compound

Question 21

How do we work out the molecular formula from the empirical formula?
(Example) The empirical formula of a molecule is CH₂O. It has a relative molecular mass (Mr) of 180. What is its molecular formula?

Answer 21

1) Firstly work out the empirical mass of the molecule
2) Divide the molecular mass by the empirical mass
3) Multiply the empirical formula by the number of empirical units

Empirical mass = 12 + (1 x 2) + 16 = 30
Mr = 180

Mr / empirical mass = 180 / 30 = 6 (empirical units)

CH₂O x 6 = C₆H₁₂O₆

Question 22

What is the purpose of a ionic equation?

Answer 22

Ionic equations show the ions that are formed in solution and show which particles are reacting

Question 23

What form ions in solutions?

Answer 23

Acids, bases and salts

Question 24

How can we use equations to work out masses?
(Example) How much CaO can be made when 34g of Ca is burnt completely in oxygen?

Answer 24

1) Write out the equation and balance it
2) Work out the Mr / Ar of species involved then write these as mass in grams
3) Divide the Ca side by 80 to find 1g then multiply by 34 to get 34g and do the same for the CaO side

2Ca _(g) + O_{2 (g)} → 2CaO _(s)

Ar of Ca = 40
40 x 2 = 80
80g / 80 = 1g
1g x 34 = 34g

Mr of CaO = 56
56 x 2 = 112
112g / 80 = 1.4g
1.4g x 34 = 47.6g of CaO (Theoretical mass)

Question 25

How can we use equations to work out masses of gases?
(Example) What volume of H₂ is produced when 12g of potassium reacts with water at 100kPa of pressure and 298K? Gas constant is 8.31JK^-1mol^-1.

Answer 25

1) Write out the equation and balance it
2) Work out the number of moles of potassium
3) Use the equation to find out the molar ratio of K:H₂
4) Use ideal gas equation to work out the volume

2K _(s) + 2H₂O _(l) → 2KOH _(aq) + H_{2 (g)}

Moles = mass / Ar
moles = 12g / 39 = 0.31 moles of K

2 moles of K react to produce 1 mole of H₂
moles of H2 is 0.31 / 2 = 0.155 moles of H₂

pV = nRT but rearrange to get V = nRT / p
V = 0.155 x 8.31 x 298 / 100,000
V = 3.84 x 10^-3 m³

Question 26

What is the theoretical yield?

Answer 26

The amount of a product produced assuming no products are lost and all reactants react fully

Question 27

How do we calculate percentage yield?

Answer 27

Percentage yield = (actual yield / theoretical yield) x 100

Question 28

Example question

In a reaction involving the complete combustion of Calcium 32.6g of Calcium Oxide was produced. The theoretical mass is 47.6g. Calculate the percentage yield of this reaction.

Answer 28

Percentage yield = (actual / theoretical) x 100

Percentage yield = (32.6 / 47.6) x 100 = 68.5%

Question 29

Why is a high percentage yield important?

Answer 29

Efficient conversion of reactants to products

Question 30

What is atom economy?

Answer 30

How efficient a reaction is

Question 31

How do we calculate the percentage atom economy?

Answer 31

% Atom economy = (Mr of desired product / sum of Mr of all reactants) x 100

Question 32

Example question

Iron oxide (Fe₂O₃) can be reduced using carbon dioxide to make pure iron and carbon dioxide. Calculate the atom economy in the extraction of iron.

Answer 32

Fe₂O₃ + 1.5C → 2Fe + 1.5CO₂

Atomic mass of desired product = 2 x 55.8 = 111.6
Sum of molecular masses of all reactants = (2 x 55.8) + (3 x 16.0) + (1.5 x 12.0) = 177.6

% atom economy = 111.6 / 177.6 x 100 = 62.8%

Question 33

Why is a high atom economy important?

Answer 33

• High atom economies produce less waste and will therefore benefit the environment
• High atom economies mean that raw materials are used more efficiently which is more sustainable
• Higher atom economy means less by-products so less time and money will be spent separating these from the desirable product
• Companies will try to use reactions that tend towards 100% atom economy