Topic 3 Quantitative Chemistry

Question 1

1) True or false? The M_r of a compound is always greater than the A_r of any of the
elements in that compound.

Answer 1

1) True - you find the M_r by adding up the A of each element, so it must be greater.

Question 2

2) The A of oxygen is 16 and the A_r of nitrogen is 14. Find the M_r of nitric oxide, NO.

Answer 2

2) 14 + 16 = 30. There’s only one atom of each element in the formula, so just add the A_r values together.

Question 3

3) How do you calculate the percentage mass of an element in a compound?

Answer 3

3) Multiply the A_r of the element by the number of atoms of that element in the formula of the compound. Then divide by the compound’s M_r, and multiply by 100.

Question 4

4) The formula of the compound sulfur hexafluoride is SF. Calculate the M_r of sulfur hexafluoride. A, of F = 19 and A_r of S = 32.

Answer 4

4) 32 (19 x 6) = 32 + 114 = 146

Question 5

5) The M, of methane, CH, is 16. The A, of carbon is 12.
What is the percentage mass of carbon in methane?

Answer 5

5) (12/16) x 100 = 0.75 x 100 = 75%
12+ 16 might look tricky, but it cancels down to 3/4, and 3/4 = 0.75= 75%.

Question 6

6) Find the M, of calcium hydroxide, Ca(OH)₂. A_r of Ca = 40, A_r of O = 16, A_r of H = 1.

Answer 6

6) 40+ [2 x (16+1)] = 40 + [2 x 17] = 40 + 34 = 74.

Question 7

7) Compound X has M_r = 30 and contains 10% hydrogen by mass. A_r of H = 1.
How many hydrogen atoms are there in the molecular formula of compound X?

Answer 7

7) Mass of H = 10% of M_r of X = 10% of 30 = 3
So number of H atoms = 3+ (A_r of H) = 3 + 1 = 3

Question 8

1) What is the definition of a mole?

Answer 8

1) One mole is 6.02 x 10²³ (the Avogadro constant) particles of a substance.

Question 9

2) How is the A_r or M_r, of a substance related to the idea of moles?

Answer 9

2) The value of A_r of an element or the M_r of a compound is equal to the mass in grams of 1 mole of the substance. E.g. the M_r of CO₂ is 44, so 1 mole of CO₂ has a mass of 44 g.

Question 10

3) Which contains more particles, a mole of water or a mole of oxygen gas?

Answer 10

3) Neither they both contain the same number of particles. 1 mole of water contains more atoms, but there are 6.02 x 10²³ molecules of H₂O or O₂ in 1 mole.

Question 11

4) What’s the formula for the number of moles in a given mass?

Answer 11

4) number of moles = mass in grams/M_r

Question 12

5) The M, of sodium hydroxide, NaOH, is 40. How many moles are there in 500 g of NaOH?

Answer 12

5) number of moles = mass = M = 500/ 40 = 12.5 moles

Question 13

6) 3 moles of compound A have a mass of 126 g. What is the M_r of compound A?

Answer 13

6) M_r = mass number of moles = 126 / 3 = 42

Question 14

7) What’s the mass of 4 moles of potassium fluoride, KF? A_r of K = 39, A_r
of F?

Answer 14

7) mass = number of moles x M_r = 4x (39+19) = 4 x 58 = 232 g

Question 15

8) The formula of magnesium bromide is MgBr₂. A_r of Mg = 24, A_r of Br = 80.
What mass of bromine is there in 0.25 moles of MgBr₂?

Answer 15

8) 1 mole of MgBr₂, contains 2 moles of Br atoms, so mass of Br in 1 mole of MgBr₂ = 2 × 80 = 160 g. So mass of Br in 0.25 moles = 160 ÷ 4 = 40 g.

Question 16

1) What is meant by the term ‘conservation of mass’?

Answer 16

1) Conservation of mass means the total mass of the products in a reaction will always be equal to the total mass of the reactants. This is because no atoms are ever destroyed or created during a reaction.

Question 17

2) How does the balanced symbol equation for a reaction to show that mass is conserved in that reaction?

Answer 17

2) In a balanced symbol equation, the sum of the relative masses of the reactants will equal the sum of the relative masses of the products. There will be the same number of each
type of atom on both sides of the equation.

Question 18

3) A scientist places a lump of a metal in an open container on a mass balance, then forgets to put it away before locking up the lab and going on holiday for two weeks. When the scientist returns, the reading on the mass balance has gone up. Suggest an explanation for this.

Answer 18

3) The metal must have reacted with one or more of the gases in the air, to form a solid or liquid product. The mass of the gas that reacted was not recorded by the balance before the reaction because the system wasn’t closed.

Question 19

The equation for the reaction between sulfuric acid and sodium carbonate is:

H₂SO_{4 (aq)} + Na₂CO_{3 (aq)}→ Na₂SO_{4 (aq)} + H₂O_(l) + CO₂_(g)

This reaction was carried out in an open beaker. Predict whether the mass of the beaker and its contents will increase, decrease or stay the same during the course of the reaction. Explain your answer.

Answer 19

4) The mass will decrease. Both reactants are in solution, so both their masses will be included in the mass of the beaker at the start of the experiment. However, one of the
products is carbon dioxide gas, which will escape the open beaker.

Question 20

1) The reaction between sodium and water is: 2Na+ 2H₂O → 2NaOH + H₂ What is the ratio of the total number of moles of reactants to the total number of moles of products in this reaction?

Answer 20

4:3. Every 2 + 2 = 4 moles of reactants makes 2 + 1 = 3 moles of products.

Question 21

2) Zinc reacts with oxygen to form zinc oxide, ZnO, as follows: 2Zn + O₂ → 2ZnO
Find the total number of moles of reactants needed to make 8 moles of ZnO.

Answer 21

2) The equation makes 2 moles of ZnO, so to make 8 moles of ZnO, multiply by 8/2 = 4:
8Zn + 40₂ →8ZnO, so making 8 moles of ZnO uses 8 + 4 = 12 moles of reactants.

Question 22

3) 0.9 moles of compound A react completely with 0.6 moles of compound B to form 0.6 moles of compound C, which is the only product of this reaction.
Find the balanced equation for this reaction in terms of A, B and C.

Answer 22

3) Divide all the numbers of moles by the smallest number of moles (0.6):

A: 0.9/0.6=1.5 B: 0.6/0.6=1 C: 0.6/0.6=1
Now multiply to get them all to be whole numbers. So in this case, multiply by 2:
A: 1.5x2=3 B: 1x2=2 C: 1x2=2.

So the balanced equation is: 3A + 2B → 2C

Question 23

4) Compound Z decomposes to make nitrogen gas, N₂, and sodium, Na. When 2 moles of Z decomposes, it produces 46 g of sodium and 84 g of nitrogen. Deduce the balanced equation for the decomposition of Z, in terms of Z, N₂ and Na.

A_r of Na = 23, A_r of N = 14.

Answer 23

46 g of Na = 46 ÷ 23 = 2 moles
84 g of N₂ = 84 (2 x 14) = 84 ÷ 28 = 3 moles

The numbers of moles are all whole numbers, so the balanced equation is:

2Z → 2Na+ 3N₂

Question 24

1) What’s the definition of a limiting reactant?

Answer 24

1) A limiting reactant is the reactant that gets completely used up in a given reaction.

Question 25

2) What does it mean if a reactant is said to be in excess?

Answer 25

2) E.g. a reactant is in excess if there is more than enough of it present to allow the limiting reactant to be fully used up.

Question 26

3) Magnesium reacts with hydrochloric acid as follows: Mg + 2HCI→ MgCl₂ + H₂ How many moles of magnesium chloride, MgCl₂, form when 0.4 moles of magnesium react with an excess of hydrochloric acid?

Answer 26

3) 0.4 moles. The equation shows you get 1 mole of MgCl, for each mole of Mg that reacts. The acid is in excess, so all 0.4 moles of magnesium will react to make 0.4 moles of MgCl₂.

Question 27

4) Methane, CH₄, burns accor to the
wing equation: CH₄+ 20₂ → CO₂ + 2H₂O.
What mass of water is formed when 64 g of methane burns in air? M, of CH₄ = 16, M_r of H₂O = 18.

Answer 27

4) Number of moles in 64 g of CH = mass M₁ = 64 ÷ 16 = 4

From the equation, each mole of CH₄ makes 2 moles of H₂O, so 2 x 4 = 8 moles of H₂O are made.

So mass of H₂O produced =
moles x M_r = 8 x 18 = 144g

Question 28

5) Fluorine reacts with water to form hydrogen fluoride, HF. The reaction equation is 2F₂ + 2H₂O → O₂ + 4HF. When 76 g of F₂ reacts with an excess of water, what mass of hydrogen fluoride is produced? A_r of F = 19, A_r of H = 1.

Answer 28

5)

M_r of F₂ = 2 x 19 = 38, so moles of F₂ reacting = mass + M_r = 76 ÷ 38 = 2

From the equation, 2 moles of F₂ produces 4 moles of HF.

M_r of HF = 1 + 19 = 20

so mass of HF produced = moles x M_r = 4 × 20 = 80 g

Question 29

1) What’s the volume of one mole of any gas at room temperature and pressure?

Answer 29

1) 24 dm³. The same number of moles of any two gases will always occupy the same volume as each other if they’re at the same temperature and pressure.

Question 30

2) What’s the formula for calculating the volume of a gas at r.t.p. from its mass?

Answer 30

volume of gas =

(mass of gas/ Mr of gas) x 24

Question 31

3) What volume does 1.6 kg of bromine gas, Br₂, occupy at r.t.p.? A_r of Br = 80.

Answer 31

The mass needs to be in grams.

1.6 x 1000 = 1600g. Mr of Br₂ = 2 x 180 = 160.

So volume = (mass of gas / M_r of gas) x 24 = 10 x 24 = 240dm³

Question 32

4) The reaction between nitric oxide gas, NO, and oxygen is: 2NO+O₂ → 2NO₂

What mass of nitrogen dioxide gas, NO₂, is produced when 120 dm³ of NO reacts completely with oxygen at r.t.p.? A, of N = 14, A_r of O = 16

Answer 32

4) From the equation, 120 dm³ of NO will produce 120 dm³ of NO₂.

So mass of NO₂ formed = (volume/24) x M_r of NO₂

= (120/24) x (14+(2x16)) = 5 x 46 = 230 g

Question 33

5) The equation for the decomposition of calcium carbonate is:

CaCO₂ → CaO + CO₂

What is the volume at r.t.p. of the carbon dioxide produced when 25 g of calcium carbonate fully decomposes?

A_r of Ca = 40, A_r of C = 12, A_r of O = 16.

Answer 33

5) Moles of CaCO₂ = mass/M_r

= 25 ÷ (40 + 12 + (3 x 16))

= 25 ÷ 100 = 0.25

From the equation, 0.25 moles of CaCO₃, produces 0.25 moles CO₂.

Volume at r.t.p. of 0.25 moles of CO₂ = 0.25 x 24 = 6 dm³.

Question 34

1) How do you find the concentration of a solution in g/dm³?

Answer 34

1) Divide the mass of solute in grams by the volume of the solution in dm³.

Question 35

2) A large bottle of a popular brand of soft drink contains 130 g of sugar. The bottle holds 2 dm³ of the drink. The drink can’t be named, for legal reasons.
What is the sugar concentration of the drink in g/dm³?

Answer 35

2) concentration = mass + volume = 130 ÷ 2 = 65 g/dm³

Question 36

3) What mass of sodium sulfate is there in 100 cm³ of sodium sulfate solution if the solution has a concentration of 5 g/dm³?

Answer 36

3) 100 cm³= 100 ÷ 1000 = 0.1 dm³
mass = concentration x volume

= 5 x 0.1 = 0.5 g

Question 37

4) M dissolved some sodium chloride in water to form a solution. She then made a second solution, using the same mass of sodium chloride, but only half as much water. How will the concentration of the second solution compare with the first?

Answer 37

4) Halving the volume of the solution while keeping the mass of sodium chloride the same will double the concentration.

If that’s not clear, try it with some numbers - for example, 50g dissolved in 2dm³ is a concentration of 25g/dm³, but if you halve the volume and keep the mass the same, the concentration doubles to 50 g/dm³.

Question 38

5) Find the concentration, in mol/dm³, of the solution formed when 0.2 moles of
potassium nitrate are dissolved in 250 cm³ of water.

Answer 38

5) First convert the volume to dm³: 250 1000 = 0.25 dm³
So concentration = moles volume = 0.2 0.25 = 0.8 mol/dm³

Question 39

1) What’s the formula for concentration in mol/dm³?

Answer 39

1) concentration = number of moles/ volume in dm³

Question 40

2) How many moles of solute are there in 4.0 dm³ of a 0.3 mol/dm³ solution?

Answer 40

2) number of moles concentration x volume = 0.3 x 4.0 = 1.2 moles.

Question 41

3) A solution of potassium bromide, KBr, has a concentration of 0.2 mol/dm³.
What is this in g/dm³? M_r of KBr = 119.

Answer 41

3) mass = moles x M_r = 0.2 x 119 = 23.8 g, so concentration = 23.8 g/dm³

Question 42

4) The reaction of sodium hydroxide, NaOH, and hydrochloric acid, HCl, is: NaOH + HCI → NaCl + H₂O.

What volume, in dm³, of 0.30 mol/dm³ HCI is needed to react exactly with 0.15 dm³ of 0.2 mol/dm³ NaOH?

Answer 42

4) moles of NaOH =

concentration x volume = 0.15 x 0.2 = 0.03 moles

So 0.03 moles of HCI are needed to react fully with the NaOH.

So volume of HCI = no. of moles concentration = 0.03 0.30 = 0.1 dm³

Question 43

5) How many grams of beryllium chloride, BeCl₂, are there in 4.0 dm³ of a solution with a concentration of 0.50 mol/dm³? A_r of Be=9, A_r of Cl = 35.5.

Answer 43

5) moles of BeCl₂ = concentration x volume = 0.50 x 4.0 = 2.0 moles

mass of BeCl₂ = no. of moles x M_r = 2.0 x [9+ (2x 35.5)] = 2.0 × 80 = 160

Question 44

6) 1 mole of calcium will react exactly with 1 mole of sulfuric acid. If 30 g of calcium
reacts with 3 dm³ of sulfuric acid, what is the concentration of the acid? A of Ca = 40.

Answer 44

6) moles of Ca = mass + A, = 30 + 40 = 0.75
0.75 moles of Ca react with 0.75 moles of sulfuric acid.
So concentration of acid = no. of moles volume = 0.75 + 3 = 0.25 mol/dm³

Question 45

1) What is atom economy?

Answer 45

1) The atom economy of a reaction is the percentage of the mass of the reactants that ends up as part of the desired product.

Question 46

2) What does it tell you if a reaction has an atom economy of 100%?

Answer 46

2) All the atoms in the reactants become part of the desired product(s), so no waste is produced.

Question 47

3) What’s the formula for calculating atom economy?

Answer 47

3) atom economy =
(relative formula mass of desired products/ relative formula mass of all reactants)
x 100

Question 48

4) The following reaction can be used to produce calcium oxide, CaO:
CaCO₃ → CaO + CO_2.

Find the atom economy of this reaction. A_r of Ca = 40, A_r of C = 12, A_r of O = 16.

Answer 48

4) atom economy =

[M_r of CaO / M_r of CaCO3] x 100

= [(40 + 16) + (40 + 16 + (12 x 3))] x 100 =

[56100] x 100 = 56%

Question 49

5) Explain two advantages of using a reaction with a high atom economy in industry.

Answer 49

5) E.g. high A.E. reactions use less raw materials so are more sustainable than low A.E. reactions.

There is less waste to be disposed of with high A.E. reactions than low A.E.reactions.

High A.E. reactions tend to be more profitable than low A.E. reactions as costs of buying materials and disposing of waste are lower.

Question 50

6) Silicon can be produced by reacting silica, SiO₂, with carbon: SiO₂ + 2C → Si + 2CO
Find the atom economy of this reaction. A_r of Si = 28, A_r of O=16, A_r of C = 12

Answer 50

6) atom economy =

A_r of Si / (M_r of SiO₂ + 2 x A_r of C) x 100

= [28 / (28 + (2 x 16))+ (2x12)] x 100

= [28 / (60 +24)] x 100

= [28/84] × 100

= ⅓ x 100 = 33.33%

Question 51

1) What is the formula for working out the percentage yield of a reaction?

Answer 51

1) percentage yield =
(mass of product actually made/
maximum theoretical mass of product) x 100

Question 52

2) True or false? A reaction with only one product always has a 100% yield.

Answer 52

2) False. The percentage yield doesn’t depend on how many products there are, but on how much of the product you actually produce. (And it’s never 100%
in real life.)

Question 53

3) Explain why you can never get a 100% yield from a reversible reaction.

Answer 53

3) In a reversible reaction, both the forward and backwards reactions happen at the same time, so some of the products always get converted back to the reactants and the reaction
never goes to completion.

Question 54

4) Marlon carried out a reaction in the lab. He calculated his maximum yield would be 3.2 g of product. He measured the actual yield as 3.5 g. How can you tell Marlon has made a mistake? Suggest two possible mistakes that could have led to this error.

Answer 54

4) Marlon’s actual yield is greater than his maximum theoretical yield, so his percentage yield is greater than 100%. This is impossible, so he must have made a mistake. E.g. he could have calculated his maximum yield rectly, measured out more of his reactants than he intended, or failed to separate out some impurities his product.

Question 55

5) The equation for the combustion of magnesium is 2Mg + O₂ → 2MgO. 2.4 g of magnesium was burned in air and the actual yield was found to be 3.6 g of magnesium oxide, MgO. Find the percentage yield. A_r of Mg = 24, A_r of O = 16.

Answer 55

5) Moles of magnesium burned =

mass / A_r = 2.4 ÷ 24 = 0.1.

From the equation, each mole of Mg produces 1 mole of MgO, so maximum theoretical yield = 0.1 moles of MgO.

So max. theoretical mass of product = moles x M_r = 0.1 x (24 + 16) = 0.1 x 40 = 4.0 g.

So percentage yield = (3.6 / 4.0) x 100 = 0.9 x 100 = 90%.