Separate Chemistry - C3 Quantitative Chemistry

Question 1

What does the law of the conservation of mass state?

Answer 1

No atoms are lost or made during a chemical reaction

Question 2

A piece of magnesium is added to acid. What would happen to the mass reading on the balance and why?

Answer 2

It would stay the same because there is a lid on the conical flask so no gas can escape.

Question 3

Count the number of elements in the following compounds:

a) CaO
b) Al₂O₃
c) H₂SO₄

Answer 3

a) CaO - 2 elements (Ca and O)
b) Al₂O₃ - 2 elements (Al and O)
c) H₂SO₄ - 3 elements (H, S and O)

Question 4

Count the number of atoms in the following compounds:

a) CaO
b) Al₂O₃
c) H₂SO₄

Answer 4

a) CaO - 2 atoms (1 x Ca and 1 x O)
b) Al₂O₃ - 5 atoms (2 x Al and 3 x O)
c) H₂SO₄ - 7 atoms (2 x H, 1 x S and 4 x O)

Question 5

Calculate the relative formula mass (Mr) of each of the compounds (a, b & c) below.

AR: Mg = 24; Cl = 35.5; Ca = 40; O = 16; Al = 27

a) CaO
b) MgCl₂
c) Al₂O₃

*this is just an example, you need to be able to do this for any compound.

Answer 5

a) CaO: 40 + 16 = 56
b) MgCl₂: 24 + (35.5x2) = 95
c) Al₂O₃: (27x 2) + (16 x 3) = 102

Question 6

A piece of magnesium is added to acid. What would happen to the mass reading on the balance and why?

Answer 6

It would decrease because there is no lid on the conical flask so gas can escape and therefore mass is ’lost’ from the flask.

Question 7

Calculate the uncertainty in the following sets of results

a) 1, 3, 6, 7, 10
b) 22, 25, 26, 25, 24

*this is just an example, you need to be able to do this for any set of data given.

Answer 7

Use the equation: uncertainty = range / 2

a) 1, 3, 6, 7, 10

Range = 10 – 1 = 9. Uncertainty = 9/2

Uncertainty = ± 4.5

b) 22, 25, 26, 25, 24

Range = 26 – 22 = 4. Uncertainty = 4/2

Uncertainty = ± 2

Question 8

Calculate the percentage mass of hydrogen in:

Ar: N = 14 ; H = 1 ; S = 32 ; O = 16.

a) NH₃
b) H₂SO₄

Give your answer to the nearest whole number.

*this is just an example, you need to be able to do this for any compound when given the appropriate information.

Answer 8

% mass = (Ar x number of atoms of the element) / (Mr of compound) x 100

a) NH₃ : (3x1) / 17 = 0.176
0. 176 x 100 = 18 %
b) H₂SO₄ : (2x1) / 98
0. 02 x 100 = 2 %

Question 9

Higher Q. How many molecules, atoms or ions are in one mole of a substance?

Answer 9

6.02 x 10²³

Question 10

Higher Q. How many moles are in the following?

a) 46 g of CaO?
b) 50 g of O₂?

Answer 10

Number of moles = (mass in g)/Mr

a) Mr of CaO = 40 + 16 = 56.

Number of moles = 46 g /56 = 0.82 moles

b) Mr of O₂ =16

Number of moles = 50g / 32 = 1.6 moles

Question 11

Higher Q. Describe the equations below in words in terms of the number of moles of each substance reacting.

1) 2Mg + O₂ → 2MgO
2) H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Answer 11

1) 2 moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide.
2) One mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate and two moles of water.

Question 12

Higher Q. Calculate the mass of magnesium needed to produce 4.6 tonnes of magnesium oxide.

2Mg + O₂ → 2MgO

*this is just an example, you need to be able to do this for any equation when given a mass of a reactant or product.

Answer 12

Step 1 – Make sure the equation is balanced - This equation is already balanced.

Step 2 – Calculate the Mr of 2Mg and the Mr of 2MgO.

Mr of 2Mg = 48, Mr of 2MgO =80

Step 3 – You now know 48 tonnes of magnesium would produce 80 tonnes of magnesium oxide.

Step 4 – Find how much magnesium you would need to produce 1 tonne of magnesium oxide: (48/80) magnesium and (80/80) magnesium oxide.

Therefore 0.6 tonnes of magnesium would produce 1 tonne of magnesium oxide.

Step 5 – Multiply both sides by 4.6 to work out how much magnesium would be needed to make 4.6 tonnes of magnesium oxide.

(0.6 x 4.6) = 2.76 tonnes

Question 13

Higher Q. Calculate the mass of CO₂ produced when 50 g of CaCO₃ is thermally decomposed.

CaCO₃ → CaO + CO₂

*this is just an example, you need to be able to do this for any equation when given a mass of a reactant or product.

Answer 13

Step 1 – Make sure the equation is balanced - This equation is already balanced.

Step 2 – Calculate the Mr of CaCO₃ and the Mr of CO₂.

Mr of CaCO₃ = 100, Mr of CO₂ = 44

Step 3 – You now know 100 g of CaCO₃ would produce 44 g of CO₂.

Step 4 – Find what 1 g of CaCO₃ would produce.

(100/100) CaCO₃ and 44/100 CO₂

1g of CaCO₃ would produce 0.44 g of CO₂

Step 5 – Multiply both sides by 50 to work out what 50 g of CaCO₃ would make.

50 g of CaCO₃ would make (0.44 x 50g) of CO₂

50 g of CaCO₃ would make 22g of CO₂

Question 14

Higher Q. Write a balanced symbol equation for the following reaction:

48 g of magnesium (Mg) reacts with 32 g oxygen (O₂) to produce 80 g of magnesium oxide (MgO).

*this is just an example, you need to be able to do this for any equation when given the appropriate information.

Answer 14

Find ratio of moles in equation by finding number of moles of each substance using the equation:

Number of moles=(mass in g)/Mr

Mg: 48/24 = 2 ; O₂: 32/32 =1 ; MgO: 80/40 = 2

Ratio = 2Mg: 1O2 : 2MgO

Balanced equation: 2Mg + O₂ → 2MgO

Question 15

Higher Q. What is a ‘limiting reactant’?

Answer 15

The first reactant that gets used up in a reaction and causes the reaction to stop.

Question 16

Higher Q. Why is one reactant often added in excess to a reaction?

Answer 16

To ensure that all of the other reactant is used up.

Question 17

Higher Q. What are the two possible units for concentration?

Answer 17

g/dm³ or mol/dm³

Question 18

Higher Q. What equation links concentration, mass of solute and volume of solvent?

Answer 18

Concentration =(mass of solute)/(volume of solvent)

Question 19

Higher Q. What equation links concentration, number of moles of solute and volume of solvent?

Answer 19

Concentration =(number of moles of solute)/(volume of solvent)

Question 20

Higher Q. What is the mass of one mole of a substance, in grams, equal to?

Answer 20

The relative formula mass of the substance

Question 21

Higher Q. Calculate the concentration of sodium hydroxide solution when 44 g of sodium hydroxide is dissolved in 0.5 dm³ of water.

*this is just an example, you need to be able to do this for any equation when given a mass of a reactant or product.

Answer 21

44 g / 0.5 dm³

= 88 g / dm³

Question 22

Higher Q. Calculate the concentration of sodium hydroxide solution when 3.5 moles of sodium hydroxide is dissolved in 100 cm³ of water. Give your answer in mol/dm³.

*this is just an example, you need to be able to do this for any equation when given a mass of a reactant or product.

Answer 22

Convert volume of water to dm³

100 cm³ / 1000 = 0.1 dm³

3.5 moles / 0.1 dm³ = 35 mol/dm³

Question 23

Higher Q. What is the value of the Avogadro Constant?

Answer 23

6.02 x 10²³ per mole

Question 24

Separate Q. Give three reasons why you might not achieve the amount of product you have calculated (theoretical yield).

Answer 24

• The reaction may not go to completion because it is reversible.
• Some of the product may be lost when it is separated from the reaction mixture.
• Some of the reactants may react in ways different to the expected reaction.

Question 25

Separate Q. What does ‘yield’ mean?

Answer 25

The amount of product that is obtained

Question 26

Separate Q. What equation would you use to work out the percentage yield of a reaction?

Answer 26

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

Question 27

Separate Higher Q. What volume of carbon dioxide is formed when 20 dm³ of carbon monoxide reacts with oxygen?

2CO (g) + O₂ (g) → 2CO₂ (g)

Answer 27

2CO (g) + O₂ (g) → 2CO₂ (g)

Ratio of moles of carbon monoxide to carbon dioxide is 2:2 or 1:1.

Therefore 20 dm³ of carbon monoxide will produce…

20 dm³ of carbon dioxide

Question 28

Separate Higher Q. What’s the volume of the following gases at room temperature and pressure?

a) 24.5 g of Oxygen
b) 53.6 g of Chlorine

*these are just examples, you need to be able to do this for any gas.

Answer 28

Volume of gas= (Mass of gas) / (Mr of gas) x 24

a) Oxygen: Volume of gas = 24.5/32 x 24 = 18.4 dm³
b) Chlorine: Volume of gas= 53.6/71 x 24 = 18.1 dm³

Question 29

Separate Q. What does atom economy mean and why is it important?

Answer 29

• Atom economy is a measure of the amount of starting materials that end up as useful products.
• It’s important for sustainable development and economical reasons.

Question 30

Separate Q. How would you calculate percentage atom economy?

Answer 30

% atom economy = (Mr of desired product) / (Sum Mr of all reactants) x 100

Question 31

Separate Higher Q. If one mole of oxygen occupies 24 dm³ at room temperature, how much volume would Nitrogen occupy?

Answer 31

• 24 dm³

*one mole of any gas at 20 ^oC and 1 atm of pressure occupies a volume of 24 dm³