Topic 3: Quantitative Chemistry

Question 1

What is the law of conservation of mass?

Answer 1

The law of conservation of mass states that no atoms are lost or made during a chemical reaction so the mass of the products equals the mass of the reactants.

Question 2

Write a balanced equation of magnesium reacting with hydrochloric acid.

Answer 2

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

Question 3

Define relative atomic mass.

Answer 3

RAM - average mass of atoms in an element taking into account masses and abundance of its isotopes, relative to 12C.

Question 4

Define relative formula mass.

Answer 4

RFM - sum of RAM’s of all atoms in the formula.

Question 5

What is the relative formula mass of:
A) CaF2
B) C6H12O6

Answer 5

CaF2 - (Ar values: Ca = 40, F = 19)
40 + 19 + 19 = 78

C2H12O6 - (Ar values: C = 12, H = 1, O = 16)
(12 x 6) + (1 x 12) + (16 x 6) = 180

Question 6

The following reaction occurs in a test tube under a Bunsen Burner:
4 MgO(s) + CH4(g) → 4 Mg(s) + 2 H2O(g) + CO2(g)

The carbon dioxide and water escape from the test tube.
Use the equation to explain why.

Answer 6

They are both gases

Question 7

The experiment was repeated three times. Calculate the mean mass of magnesium produced and suggest how you could increase the precision of the results.

Experiment
1 2 3
Mass of magnesium oxide used in g 4.0 4.0 4.0

Mass of magnesium produced in g 3.3 3.5 3.2
times.

Answer 7

(3.3 + 3.5 + 3.2) / 3 = 3.3

Measure to more decimal places or use a more sensitive balance / apparatus

Question 8

What is Avogadro’s constant?

Answer 8

The number of atoms, molecules or ions in a mole of a given substance. The value of the constant is 6.02 x 10²³.

Question 9

What is the formula that links mass, molecular mass and moles together?

Answer 9

Mass = Mr x Moles

Question 10

What is the mass of 20 moles of calcium carbonate, CaCO₃

Answer 10

Mass = Mr x Moles

Mr = 100

100 x 20 = 2000 g

Question 11

Calculate the amount of carbon dioxide in moles in 0.32 g of carbon dioxide.

Relative atomic masses (A_r): carbon = 12, oxygen = 16

Answer 11

Moles = Mass / Mr

0.32 / 44 = 0.007

Question 12

Nitrogen and hydrogen form ammonia shown by the following equation:

N₂(g) + 3 H₂(g) ⇌ 2 NH₂(g)

Calculate the mass of nitrogen needed to form 6.8 tonnes of ammonia.

Relative atomic masses (Ar): H = 1; N = 14

Answer 12

Step 1 - Work out the number of number of moles of ammonia (Mr of ammonia = 17)

6800000 / 17 = 400000 moles of ammonia

Step 2 - Use the balanced equation and number of moles of ammonia to work out the number of moles of nitrogen

The ratio of nitrogen to ammonia is 1:2

Therefore the number of moles of nitrogen is 400000/2 = 200000

Step 3 - Work out the mass of nitrogen (Mr of N₂ is 28)

200000 x 28 = 5600000 g = 5.6 tonnes.

Question 13

State what we mean by a limiting reactant in a chemical reaction

Answer 13

In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used. The reactant that is completely used up is called the limiting reactant because it limits the amount of products.

Question 14

Hydrogen peroxide decomposes in water to form water and oxygen. How many grams of oxygen gas will be given off from 40.8 g of hydrogen peroxide?

Answer 14

Step 1: Write the balanced equation 2 H₂O_2(l) → 2 H₂O + O_2(g) Mr of H₂O₂ = 34

Step 2: Number of moles in 40.8 g : 40.8/34 = 1.2 moles

Ratio in the balanced equation of H₂O₂ : O₂ = 2:1

Step 3 :Therefore number of moles of O₂ = 0.6 moles

Step 4: Mass of oxygen = 0.6 x 32 (Mr of O₂) = 19.2

Question 15

Write down the two formulae that link concentration, mole/mass and volume together.

Answer 15

Concentration (g per dm³) = Mass (g)/Volume (dm³)

Concentration (mol per dm³) = nr of moles/volume (dm³)

Question 16

31.0 cm3 of potassium hydroxide solution neutralised 25.0 cm3 of 2.0 moldm−3 nitric acid.

HNO₃ + KOH → KNO₃ + H₂O

Calculate the concentration of the potassium

hydroxide solution in moldm−3

Answer 16

Step 1: Calculate the moles of HNO₃ used = Concentration x volume

2 x 0.025 dm3 (25/1000 to convert the units) = 0.05 moles

Step 2 : Calculate the moles of KOH

Ratio is 1:1 therefore number of moles of KOH = 0.05

Step 3 : Calculate the concentration of KOH

Volume = Moles/concentration; 0.05 / 0.031 = 1.61

Question 17

What is the molar volume of a gas at room temperature and pressure?

Answer 17

1 mole of a gas at room temperature and pressure occupies 24 dm3

Question 18

Why is it not always possible to obtain the theoretical amount of product in a chemical reaction?

Answer 18

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 (side reactions may occur).