C3: Quantitative chemistry

Question 1

What is an isotope?

Answer 1

A variation of an element which occurs with different numbers of neutrons.

Question 2

Find the RFM of MgCl₂.

Ar of Mg = 24

Ar of Cl = 35.5

Answer 2

1. Mg= 24
2. Cl₂= 2 x 35.5 = 71
3. 24 + 71 = 95
4. RFM = 95

Question 3

How is the M_r of a substance calculated?

Answer 3

RAM = (sum of: (isotopic mass x % abundance)) / 100

Question 4

How do you find the percentage mass of an element in a compound?

Answer 4

(Ar/Mr) x 100

Question 5

A mixture’s mass is 20% iron. What mass of iron chloride (FeCl₂) would you need to provide the iron ions in 50g of the mixture?

Ar of Fe = 56

Ar of Cl = 35.5

Answer 5

1) Find the mass of iron in the mixture:

20% of 50g = 10g of iron needed

2) Calculate mass of Fe in FeCl₂

56 / (56 + 2 x 35.5) = 56/127 (decimal)

3) Find the mass of FeCl₂ which contains 10g of iron.

Mass of FeCl2 is 56/127 iron, and there is 10g of iron.

10g / (56/127) = 22.68g of iron chloride needed.

Question 6

What is a mole?

Answer 6

One mole is the amount of a substance which contains Avogadro’s number of particles.

^{The size or weight of particles does not change this value.}

Question 7

What is Avogadro’s number?

Answer 7

6.02 x 10²³

Question 8

Why is Avogadro’s number used?

Answer 8

One mole (6.02 x 10²³ particles) of any substance has a mass, in grams, equal to the A_<em>r</em> or M_r of that substance.

Question 9

What is relative atomic mass?

Answer 9

The number of grams which one mole of an element weighs.

E.g. M_r of Carbon = 12, so 1 mole of carbon weighs 12g.

Question 10

1. Carbon has an A_r of 12, so one mole of carbon weighs __g.
2. Oxygen has an A_r of 16, so CO₂ has an M_r of __. Therefore, one mole of CO₂ weighs __g.

Answer 10

1. 12g
2. 44
3. 44g

Question 11

The number of moles in a substance =

Answer 11

mass (g) / M_r

Question 12

M_r of a substance =

Answer 12

Mass / moles

Question 13

The mass of a substance =

Answer 13

Moles x M_r

Question 14

1. What is the rule of conservation of mass in a chemical reaction?
2. What does this mean about the atoms on each side of the equation?

Answer 14

1. No atoms are created or destroyed- mass is conserved.
2. The same number of atoms, of each element, should be on each side of the equation.

Question 15

Why could the total mass of the products of a reaction seem to differ from that of the reactants?

Answer 15

There would be a gas involved, and the reaction would have been in an unsealed vessel.

Question 16

If the measured mass of the products of a reaction is smaller than that of the reactants, it will be because one of the ___ is a ___ which has ___ ___ ___ ___, whose mass can’t accounted for because it isn’t___ in the ___ ___.

Answer 16

1. Products
2. Gas
3. Diffused into the air
4. Contained
5. Reaction vessel

Question 17

If the measured mass of the products of a reaction is larger than that of the reactants, it will be because one of the ___ is a ___ in the ___, whose mass was not initially accounted for because it wasn’t ___ in the ___ ___.

Answer 17

1. Reactants
2. Gas
3. Air
4. Contained
5. Reaction vessel

Question 18

Mg + 2HCl → MgCl₂ + H₂

What is the mole ratio of this reaction?

Answer 18

1 : 2 : 1 : 1

Question 19

Mg + 2HCl → MgCl₂ + H₂

For every _ ___of magnesium and _ ___ of hydrochoric acid which react together, _ ___ of magnesium chloride and _ ___ of hydrogen are produced.

Answer 19

1. 1 mole
2. 2 moles
3. 1 mole
4. 1 mole

Question 20

When given the masses of the reactants and products in a reaction, how would you work out the balanced symbol equation?

Answer 20

1. Find no. of moles of each substance: = mass/M_r
2. Simplify the numbers of moles if possible- e.g. 2:4:2:6 → 1:2:1:3
3. Write out the equation, putting the number of moles in front of the chemical formulae

Question 21

8. 1g of ZnO reacts with 0.6g of C to form 2.2g of CO₂ and 6.5g of Zn. Write a balanced symbol equation for this reaction.
   * A_r of Zn =65
   * A_r of O = 16
   * A*_r of C = 12

Answer 21

1. Find M_r of ZnO = 65 + 16 = 81
2. Find M_r of CO₂ = 12 + 2(16) = 44
3. Find no. of moles of each substance: = mass/M_r ZnO = 8.1 / 81 = 0.1 mol C = 0.6 / 12 = 0.05 mol CO₂ = 2.2 / 44 = 0.05 mol Zn = 6.5 / 65 = 0.1 mol
4. Simplify the numbers of moles: 0.1 : 0.05 : 0.05 : 0.1 → 2:1:1:2
5. Write out the equation: 2ZnO + C → CO₂ + 2Zn

Question 22

1. The reactant which is used up first in a reaction is referred to as what?
2. What are the other reactants classed as?

Answer 22

1. The limiting reactant.
2. Excess reactants.

Question 23

What is the releationship between the amount of product formed and the amount of limiting reactant?

Answer 23

The amount of product formed is directly proportional to the amount of limiting reactant.

Question 24

How would you calculate the mass of a product formed in a reaction, when given the mass of the limiting reactant?

Answer 24

1. Write out the balanced equation if not done for you.
2. Work out the M_r of the limiting reactant and the specified product.
3. Find out the no. moles of the reactant whose mass is given.
4. Work out how many moles there’ll be of the substance whose mass you need to find, using the balanced equation.
5. Use the no. moles to calculate the mass.

Question 25

Calculate the mass of aluminium oxide formed when 135g of aluminium is burned in air. * A*_r Al = 27 * A*_r O = 16

Answer 25

1. **Write out the balanced equation**: 4Al + 3O₂ →2Al₂O₃ 2. **Calculate the relative formula masses:** Al = 27 Al₂O₃ = (2 x 27) + (3 x 16) = 102 (Don't need to find Mr of oxygen because it's the excess reactant.) 3. **Find out the no. moles of aluminium in 135g.** = mass/*M*_r = 135/27 = 5 4. **Look at the ratio of moles in the equation to find no. moles of Al₂O₃.** 4 moles of Al react to make 2 moles of Al₂O_3. So ratio = 2 : 1 = 5 : _2.5_ 5. **Use the no. moles to calculate the mass.** Mass = moles x *M*_r = 2.5 x 102 = **_255_**g.

Question 26

At the same ___ and \_\_\_, equal numbers of ___ of any gas will occupy the same \_\_\_.

Answer 26

1. temperature 2. pressure 3. moles 4. volume

Question 27

What volume does one mole of any gas occupy at room temperature and pressure?

Answer 27

24dm³

Question 28

What formula can be used to find the volume of a known mass of any gas, at room temperature and pressure?

Answer 28

Volume = (mass of gas (g) / *M*_r of gas) x 24 ## Footnote *1 mole of any gas occupies 24cm² at room temp. & pressure. Moles = mass/Mr.*

Question 29

What is the volume of 319.5g of chlorine at room temperature and pressure? *M*_r of Cl = 35.5.

Answer 29

Volume = (mass of gas (g) / *M*_r of gas) x 24 = (319.5 / 35.5 x 2) x 24 = 108dm³

Question 30

What volume of carbon dioxide is formed when 30dm³ of oxygen reacts with carbon monoxide? The equation for this reaction is: 2CO + O₂ →2CO₂

Answer 30

Only need to focus on oxygen (because volume is given) and carbon dioxide (because you need to find its volume). 1. **Mole ratio** of oxygen : carbon dioxide = 1:2 2. **Equal no. moles of gases occupy the same volume** (assume same temperature and pressure) 3. **So 1 volume O₂ : 2 volumes CO₂** 4. So 30dm³ : 2 x 30dm³ 5. 2 x 30dm³ = **60dm³ of CO₂**

Question 31

What is concentration a measure of?

Answer 31

How much of a substance there is within a certain volume.

Question 32

When given mass and volume, how can concentration be calculated?

Answer 32

Concentration = mass of _solute_ / volume of _solvent_

Question 33

When given moles and volume, how can concentration be calculated?

Answer 33

Concentration = no. moles of _solute_ / volume of _solvent_

Question 34

What is concentration measured in?

Answer 34

g/dm³ **OR** mol/dm³

Question 35

What is volume of _liquids_ measured in?

Answer 35

dm³

Question 36

How do you convert cm³ into dm³?

Answer 36

÷ 1000

Question 37

How do you convert dm³ into cm³?

Answer 37

x 1000

Question 38

How do you convert m³ into dm³?

Answer 38

x 1000

Question 39

What is the concentration, in mol/dm³, of a solution with 2 moles of salt in 500cm³?

Answer 39

1. **Convert cm³ to dm³:** 500 ÷ 1000 = 0.5dm³ 2. **Concentration = no. moles / volume** = 2 ÷ 0.5 = 4 mol/dm³

Question 40

What are titrations?

Answer 40

Experiments which allow you to find the exact volumes of two solutions needed for them to react completely.

Question 41

What's the concentration, in g/dm³, of a 0.417 mol/dm³ solution of sulfuric acid (H₂SO₄)

Answer 41

1. *M*_r H₂SO₄ = (1 x 2) + 32 (16 x 4) = 98 2. In 1 dm³: mass = moles x *M*_r = 0.417 x 98 = 40.866g 3. so concentration = 40.87g/dm³

Question 42

Having done a titration, what would you be able to do if you knew the concentration of one of the reacting solutions?

Answer 42

You could use the volumes found in the titration and the reaction equation to find the concentration of the other reacting solution.

Question 43

A student started with a 30cm³ solution of sulfuric acid (H₂SO₄), of unknown concentration. She found by titration that it took an average of 25cm³ of 0.1mol/dm³ sodium hydroxide (NaOH) to neutralise the sulfuric acid. Find the concentration of the sulfuric acid in mol/dm³. The balanced symbol equation for the reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Answer 43

1. **Find no. moles of known substance:** = concentration x volume = 0.1 x (25 ÷ 1000) = 0.0025mol of NaOH 2. **Find no. moles of unknown substance** using the reaction equation: Mole ratio NaOH: H₂SO₄ = 2:1 = 0.0025 : _0.00125_ mol H₂SO₄ 3. **Find concentration of unknown substance:** = moles ÷ volume = 0.00125mol ÷ (30 ÷ 1000) = 0.416^r mol/dm³ = 0.417mol/dm³

Question 44

What is the atom economy/utilisation of a reaction?

Answer 44

The percentage mass of the reactants which form useful products when manufacturing a chemical.

Question 45

What is the formula for atom economy?

Answer 45

atom economy = 100 x (*M*_r of desired product ÷ *M*_r of all reactants) ## Footnote *Like efficiency equation.*

Question 46

What are the disadvantages of reactions with low atom economies?

Answer 46

1. Quickly use up resources 2. Make lots of waste 3. May not be profitable: raw materials are expensive and waste is expensive to dispose of

Question 47

What is the formula for percentage yield?

Answer 47

Percentage yield = 100 x (mass of product made ÷ max. theoretical mass of product made)

Question 48

Give the 3 main reasons why yields are always less than 100%.

Answer 48

1. **In reversible reactions**, products turn back into reactions- so it never goes to completion. 2. **Side reactions**- reactants could react with gases in the air or impurities in the reaction mixture. 3. **Product is lost when separated from others-** i.e. when filtering/transferring a liquid.