IB Chemistry Topic 1 Questions (SL & HL)

Topic 1: Stoichiometrics

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You must login or register for free to use our interactive syllabus checklist and save your progress! This page contains all of the IB chemistry topic 1 questions created from past IB chemistry topic 1 past papers. IB chem topic 1 covers the IB chemistry moles content from the IB chemistry course. The sub-topics included are shown below, covering the IB chemistry topic 1 areas of: atoms, mixtures, atomic mass, molecular mass, empirical formula, limiting reactants, Avogadro's law, and standard solutions.

Our IB chem topic 1 questions on IB chemistry stoichiometry test your topic 1 syllabus knowledge required for the IB chemistry topic 1 questions in the exam. They will also prepare you well for IB chemistry topic 1 past paper questions!

Topic 1 Sub-topics

1.1: Introduction to the particulate nature of matter & chemical change

1.2: The concept of moles

A Avogadro's constant A Empirical formula

1.3: Reacting masses & volumes

A Limiting reactants & yield A Volume & the ideal gas equation (PV=nRT) A Volume & concentration A Neutralisation A Percentage composition & hydrated salts

Question 1

[Total Marks: 1] Easy

When the equation below is balanced, the sum of all the coefficients is

(CH₃)₂NNH₂ + N₂O₄ → N₂ + CO₂ + H₂O

Select an answer from the options

Correct, the answer is C!

The equation balances to: (CH₃)₂NNH₂ + 2N₂O₄ → 3N₂ + 2CO₂ + 4H₂O

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Question 2

[Total Marks: 1] Easy

For the following equation, the sum of all the balanced coefficients is

CH₃CH₂COCH₃_(l) + O₂ _(g) → CO₂ _(g) + H₂O_(l)

Select an answer from the options

Correct, the answer is D!

The equation balances to: 2CH₃CH₂COCH₃_(l) + 11O₂ _(g) → 8CO₂ _(g) + 8H₂O_(l)

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Question 3

[Total Marks: 1] Easy

What is the molar ratio of O₂ to H₂O when the equation below is balanced?

NH₃_(g) + O₂ _(g) → NO_(g) + H₂O_(g)

Select an answer from the options

Correct, the answer is B!

The equation balances to: 4NH₃_(g) + 5O₂ _(g) → 4NO_(g) + 6H₂O_(g)

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Question 4

[Total Marks: 1] Medium

Write a balanced equation for the complete combustion of hexane, C₆H₁₄. The sum of the coefficients in this balanced equation would be

Select an answer from the options

Correct, the answer is D!

The equation would be: 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O.

Note: A useful tip is to balance C, H and O in that order. If you need to, use a decimal to make the O balance, then double everything after!

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Question 5

[Total Marks: 1] Medium

For the below equation, what is the value of b when a = 4?

aC₂H₃Cl_(g) + bO₂ → cCO₂ _(g) + dH₂O_(g) + eHCl_(g)

Select an answer from the options

Correct, the answer is A!

The equation would be: 4C₂H₃Cl_(g) + 10O₂ → 8CO₂ _(g) + 4H₂O_(g) + 4HCl_(g)

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Question 6

[Total Marks: 1] Medium

If the following reaction produced 3.15 mol of oxygen, how many moles of KCl would be produced?

2KClO₃ → 2KCl + 3O₂

Select an answer from the options

Correct, the answer is A!

O₂ and KCl are in a 3:2 ratio. So, moles of KCl is 3.15 ÷ 3 * 2 = 2.1

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Question 7

[Total Marks: 1] Medium

Calculate the mass of H₂ when 25 g of Al reacts with excess HCl as per the equation shown

2Al_(s) + 6HCl_(aq) → 2AlCl₃_(aq) + 3H₂_(g)

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_r. So, moles of Al is 25 ÷ 27 = 0.926
Al and H₂ are in a 2:3 ratio, so moles of H₂ is 0.926 ÷ 2 * 3 = 1.38
Mass = Moles * M_r. So, mass of H₂ is therefore 1.38 * 2 = 2.77

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Question 8

[Total Marks: 1] Hard

Write a balanced equation for the complete combustion of pentane, then calculate the mass of oxygen required to burn 14.4g of C₅H₁₂

Select an answer from the options

Correct, the answer is C!

The equation would be: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
Moles of C₅H₁₂ is 14.4 ÷ 72 = 0.2
C₅H₁₂ is in a 1:8 ratio with O₂ , so moles of O₂ is 0.2 * 8 = 1.6
Mass = Moles * M_r. So, mass of O₂ is therefore 1.6 * 32 = 51.2

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Question 9

[Total Marks: 1] Easy

For the reaction: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
How many moles of water would form from 28.8 g of pentane?

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_r. So, moles of pentane is 28.8 ÷ 72 = 0.4
Pentane and water are in a 1:6 ratio, so moles of water is 0.4 * 6 = 2.4

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Question 10

[Total Marks: 1] Medium

What is the minimum mass of O₂ required for combustion of 3.2 grams of methane according to the equation shown below?

CH₄ + 2O₂ → CO₂ + 2H₂O

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_r. So, moles of methane is 3.2 ÷ 16 = 0.2
Methane and O₂ are in a 1:2 ratio, so moles of O₂ is 0.2 * 2 = 0.4
Mass = Moles * M_r. So, mass of O₂ is therefore 0.4 * 32 = 12.8

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Question 11

[Total Marks: 1] Medium

Using the equation for the oxidation of NH₃ shown below, calculate the water produced (grams) from oxidising 3.4g of NH₃
4NH₃ + 5O₂ → 4NO + 6H₂O

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_r. So, moles of NH₃ is 3.4 ÷ 17 = 0.2
NH₃ and water are in a 4:6 ratio, so moles of water is 0.2 ÷ 4 * 6 = 0.3
Mass = Moles * M_r. So, mass of water is therefore 0.3 * 18 = 5.4

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Question 12

[Total Marks: 1] Hard

Al reacts with HCl to form hydrogen gas as per the equation below:

2Al_(s) + 6HCl_(aq) → 3H₂_(g) + 2AlCl₃_(aq)

Which expression calculates the number of moles of H₂ formed from 0.12 moles of Al? Assume the HCl is in excess

Select an answer from the options

Correct, the answer is A!

Al and H₂ are in a 2:3 ratio, so moles of H₂ is (0.12 ÷ 2) * 3. This is the same as 0.12 * (3 ÷ 2)

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Question 13

[Total Marks: 1] Medium

For the reaction shown, how much MnO₂ is required to produce 50 g of KMnO₄ ?

2MnO₂ + 4KOH + O₂ + Cl₂ → 2KMnO₄ + 2KCl + 2H₂O

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_r. So, moles of KMnO₄ is 50 ÷ 158 = 0.316
KMnO₄ and MnO₂ are in a 2:2 ratio, so, moles are identical
Mass = Moles * M_r. So, mass of MnO₂ is therefore 0.316 * 87 = 27.5

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Question 14

[Total Marks: 1] Medium

For the balanced equation: 3BaCl₂_(aq) + 2Na₃ PO₄ _(aq) → Ba₃(PO₄)₂_(s) + 6NaCl_(aq)
How many moles of sodium chloride could be collected using 20 moles of BaCl₂ in excess Na₃ PO₄ ?

Select an answer from the options

Correct, the answer is D!

BaCl₂ and sodium chloride are in a 3:6 ratio, so moles of sodium chloride is 20 ÷ 3 * 6 = 40

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Question 15

[Total Marks: 3] Medium

Using the equation: 2Al_(s) + 3Cl₂_(g) → Al₂Cl₆_(s)
Calculate the mass of Al required to produce 34.2 g of Al₂Cl₆

Write out your answer in the box

Moles = Mass ÷ M_r. So, moles of Al₂Cl₆ is 34.2 ÷ 267 = 0.128 ✓
Al₂Cl₆ and Al are in a 1:2 ratio, so moles of Al is 0.128 * 2 = 0.256 ✓
Mass = Moles * M_r. So, mass of Al is therefore 0.256 * 27 = 6.917g → 6.92g (3sf) ✓

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Question 16

[Total Marks: 3] Medium

Calculate the mass of Fe₂O₃ required to produce 3632 kg of Fe as per the equation shown below:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Write out your answer in the box

Moles = Mass ÷ M_r. So, moles of Fe is 3632 ÷ 56 = 64.857. ✓
Fe and Fe₂O₃ are in a 2:1 ratio, so moles of Fe₂O₃ is 64.857 ÷ 2 = 32.428. ✓
Mass = Moles * M_r. So, mass of Fe₂O₃ is therefore 32.428 * 160 = 5188.57kg → 5190kg (3sf) ✓

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Question 1

[Total Marks: 1] Medium

Which of these options has the greatest mass?

Select an answer from the options

Correct, the answer is D!

Mass = M_r * Moles. Substituting the relevant values from each option into this formula reveals that the last option has the greatest mass.

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Question 2

[Total Marks: 1] Easy

The number of moles in 250 g of water is:

Select an answer from the options

Correct, the answer is A!

Moles = Mass ÷ M_r. So, 250 ÷ 18 = 13.88

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Question 3

[Total Marks: 1] Medium

The mass of two molecules of water in grams would be:

Select an answer from the options

Correct, the answer is A!

Mass = M_r * Mole. So, 1 mole of water weighs 18g
1 mole of anything contains 6.02x10²³ molecules, so 1 molecule weighs 18 ÷ 6.02x10²³ = 2.99x10^-23

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Question 4

[Total Marks: 1] Medium

How many molecules are present in 18 g of H₂O?

Select an answer from the options

Correct, the answer is C!

1 mole of water weighs 18g. One mole of anything contains 6.02x10²³ molecules

Note: 1 Mole of anything weighs the same as its M_r!

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Question 5

[Total Marks: 1] Hard

What is the number of atoms in 0.20 mol of propyne, C₃H₄?

Select an answer from the options

Correct, the answer is C!

The number of molecules present is 0.2 * 6.02x10²³ = 1.2x10²³
There are 7 atoms in each molecule so 7 * 1.2x10²³ = 8.42x10²³ atoms

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Question 6

[Total Marks: 1] Medium

A single atom of an element has a mass of 1.06x10^-22 grams. The element is:

Select an answer from the options

Correct, the answer is A!

The total weight of 1 mole of atoms would be 1.06x10^-22 * 6.02x10^-23 = 63.8
Since 1 mole of anything = Mr, this would be the Mr

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Question 7

[Total Marks: 1] Easy

What is the mass of one molecule of propanol in grams?

Select an answer from the options

Correct, the answer is B!

1 mole weighs the same as Mr, which = 60
There are 6.02x10²³ molecules in this, so 60 ÷ 6.02x10²³ = 1x10^-22

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Question 8

[Total Marks: 1] Hard

The mass spectrum of Mg produced the data shown below. Using this information, the A_r of Mg would be

Mass/Charge	% Abundance
79	78.6
80	10.11
81	11.29

Select an answer from the options

Correct, the answer is B!

The formula to use here is (M_r * %) + (M_r * %) + (M_r * %)100 = 79.32

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Question 9

[Total Marks: 1] Easy

Which contains the largest number of molecules?

Select an answer from the options

Correct, the answer is B!

Whichever has the most moles will have the most molecules. Moles = Mass ÷ M_r. They all have the same mass, so whichever has the lowest M_r will be correct

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Question 10

[Total Marks: 1] Easy

The mass in grams of one molecule of C₃H₇Br is

Select an answer from the options

Correct, the answer is B!

1 mole weighs the same as Mr, which = 123
There are 6.02x10²³ molecules in this, so 123 ÷ 6.02x10²³ = 2.04x10^-22

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Question 11

[Total Marks: 1] Medium

Half a mole of carbon dioxide molecules would contain

Select an answer from the options

Correct, the answer is D!

1 mole of anything contains 6.02x10²³ molecules, so 0.5 mole contains 0.5 * 6.02x10²³ = 3.01x10²³
There are 3 atoms in each molecule, so 3.01x10²³ * 3 = 9.03x10²³

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Question 12

[Total Marks: 1] Medium

In 0.250 moles of CH₃CHClOH, there are

Select an answer from the options

Correct, the answer is C!

1 mole of anything contains 6.02x10²³ molecules, so 0.250 mole contains 0.250 * 6.02x10²³ = 1.505x10²³ molecules
There are 9 atoms in each molecule, so 1.505x10²³ * 9 = 1.35x10²⁴

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Question 13

[Total Marks: 1] Hard

Which of the answers contains the smallest number of molecules?

Select an answer from the options

Correct, the answer is B!

Whichever has the fewest moles will have the fewest molecules. Moles = Mass ÷ M_r. They all have the same mass, so whichever has the highest M_r will be correct

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Question 14

[Total Marks: 1] Hard

A sample of 6.44 g of O₃ contains the same number of atoms as

Select an answer from the options

Correct, the answer is A!

The number of moles is Mass ÷ Mr, so 6.44 ÷ 48 = 0.134
The number of molecules is 0.134 * 6.02x10²³ = 8.06x10²²
So, the number of atoms is 3 * 8.06x10²² = 2.42x10²³. Repeat this for each option

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Question 15

[Total Marks: 1] Easy

The mass in grams of one molecule of C₁₀H₁₄N₂ (Nicotine) would be

Select an answer from the options

Correct, the answer is B!

1 mole weighs the same as Mr, which = 162.26
There are 6.02x10²³ molecules in this, so 162.26 ÷ 6.02x10²³ = 2.70x10^-22

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Question 16

[Total Marks: 1] Medium

One mole of CO₂ molecules contains roughly

Select an answer from the options

Correct, the answer is D!

1 mole of anything contains 6.02x10²³ molecules
There are 3 atoms in each molecule, so 6.02x10²³ * 3 = 1.806x10²⁴ atoms in 1 mole

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Question 17

[Total Marks: 1] Medium

The option which would contain 1.0x10²³ atoms is

Select an answer from the options

Correct, the answer is B!

Mass ÷ M_r = Moles
Moles * 6.02x10²³ = molecules
Molecules * number of atoms in a molecule = total atoms in that many moles. Repeat this for each

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Question 18

[Total Marks: 1] Medium

The mass in grams of two molecules of CO₂ is

Select an answer from the options

Correct, the answer is C!

1 mole weighs the same as Mr, which = 44
There are 6.02x10²³ molecules in this, so 44 ÷ 6.02x10²³ = 7.308x10^-23 is the mass of one molecule

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Question 19

[Total Marks: 1] Hard

Which of the following has the greatest mass?

Select an answer from the options

Correct, the answer is A!

8 moles of oxygen have a mass of 8 * 32 = 256. 1 mole of gallium has a mass of 1 * 70 = 70
Since it is diatomic, the number of moles of helium is 8x10²⁵ * 2 ÷ 6.02x10²³ = 265.78
So, the mass is 4 * 265.78 = 1063.12. Repeat for iron

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Question 20

[Total Marks: 1] Hard

Which of the answers would contain the fewest atoms?

Select an answer from the options

Correct, the answer is C!

For A, the number of moles is 1.5 ÷ 8 = 0.1875
The number of molecules is thus 0.1875 * 6.02x10²³ = 1.13x10²³
There are 2 atoms in each molecule so 2 * 1.13x10²³ = 2.26x10²³ atoms. Repeat for the rest

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Question 21

[Total Marks: 1] Hard

Which of the following would contain the largest number of atoms?

Select an answer from the options

Correct, the answer is A!

For A, the number of moles is 1.5 ÷ 20 = 0.075
The number of molecules is thus 0.075 * 6.02x10²³ = 4.515x10²²
There is 1 atom in each molecule so 4.515x10²² atoms. Repeat for the rest

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Question 22

[Total Marks: 1] Hard

A sample of X with atomic mass 139.446 is made up of 60.4% of X-138 and 39.6% of X-142. If the mass of X-138 is 138.058, the mass of X-142 would be

Select an answer from the options

Correct, the answer is D!

The formula to use here is (M_r * %) + (M_r * %)100 = 139.446
So ((138.058 * 60.4) + (M_r * 39.6)) ÷ 100 = 139.446. Then rearrange

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Question 23

[Total Marks: 1] Medium

Which sample contains the smallest amount of oxygen?

Select an answer from the options

Correct, the answer is A!

Comparing the number of atoms would be easiest
So, moles * 6.02x10²³ = molecules
Molecules * number of O atoms in each molecule = total number of O atoms

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Question 24

[Total Marks: 1] Medium

How many moles of C₂H₆ are needed to obtain 6.0x10²³ hydrogen atoms?

Select an answer from the options

Correct, the answer is A!

Number of molecules = 6.0x10²² ÷ 6 = 1x10²²
Number of moles = 1x10²² ÷ 6.02x10²³ = 0.167

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Question 25

[Total Marks: 1] Easy

How many moles of O₃ would contain 3.6x10²² molecules?

Select an answer from the options

Correct, the answer is B!

Moles = 3.6x10²² ÷ 6.02x10²³ = 0.0598

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Question 1

[Total Marks: 1] Easy

Which of the answers is both an empirical and molecular formula?

Select an answer from the options

Correct, the answer is A!

Empirical formula cannot be divided, so B, C and D are wrong

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Question 2

[Total Marks: 1] Easy

Which of the following would have the greatest empirical formula mass?

Select an answer from the options

Correct, the answer is B!

A would be CH, B would be C₂H₅, C would be CH₂, and D would be CH₃

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Question 3

[Total Marks: 1] Medium

A compound containing only C, H, O has the percentage by mass of 60 %, 8 %, 32 % respectively. The empirical formula would be?

Select an answer from the options

Correct, the answer is A!

Empirical formula are best calculated in a table:

	C	H	O
Mass/Percentage	60	8	32
/\ ÷ M_r (Moles)	5	8	2
/\ ÷ Smallest (Ratio)	2.5	4	1
Final ratio	5	8	2

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Question 4

[Total Marks: 1] Medium

A compound has the empirical formula CH₂ and an M_r in the region of 120 to 160. What is its most likely molecular formula?

Select an answer from the options

Correct, the answer is B!

The empirical formula has M_r of 14. 120 ÷ 14 = 8.6 and 160 ÷ 14 = 11.4. So, the M_r is roughly 9-11 times larger

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Question 5

[Total Marks: 1] Hard

A compound consists of 87.5% N₂ and 12.5% H₂. The molecular mass is 32.0 g mol^-1. What would be the molecular formula?

Select an answer from the options

Correct, the answer is D!

First work out the empirical formula, which is NH₂, as below:

	N	H
Mass/Percentage	87.5	12.5
/\ ÷ M_r (Moles)	6.25	12.5
/\ ÷ Smallest (Ratio)	1	2
Final ratio	1	2

The empirical formula has an empirical mass of 16. 32 ÷ 16 = 2
The answer is therefore twice the empirical formula

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Question 6

[Total Marks: 1] Medium

The molecular formula of a compound containing 85.7 % C and 14.3 % H₂, by mass, could be?

Select an answer from the options

Correct, the answer is B!

Work out the empirical formula, which is CH₂, as below:

	C	H
Mass/Percentage	85.7	14.3
/\ ÷ M_r (Moles)	7.14	14.3
/\ ÷ Smallest (Ratio)	1	2
Final ratio	1	2

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Question 7

[Total Marks: 1] Medium

A compound contains 24 % Mg, 28 % Si and 48 % O₂ . What is its empirical formula?

Select an answer from the options

Correct, the answer is D!

The empirical formula is MgSiO₃ , as below:

Element	Mg	Si	O
Mass/Percentage	24	28	48
/\ ÷ M_r (Moles)	1	1	3
/\ ÷ Smallest (Ratio)	1	1	3
Final ratio	1	1	3

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Question 8

[Total Marks: 1] Easy

What is the empirical formula for a compound with the molecular formula C₁₂H₆Cl₆?

Select an answer from the options

Correct, the answer is C!

All of the elements can be divided by 6, giving 2, 1, 1 respectively

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Question 9

[Total Marks: 1] Easy

Which answer shows the % by mass of C in C₇H₅(NO₂)₃?

Select an answer from the options

Correct, the answer is B!

The total M_r is 227. C accounts for 84 of this. So as a %, (84 ÷ 227) * 100 = 37%

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Question 10

[Total Marks: 1] Hard

Thallium (0.203 g) is reacted completely with bromine gas to form 0.805 g of a compound containing only these two elements. What is its formula?

Select an answer from the options

Correct, the answer is C!

The mass of Br in the compound is 0.805 - 0.203 = 0.602. Calculate the empirical formula:

Element	Tl	Br
Mass/Percentage	0.203	0.602
/\ ÷ M_r (Moles)	9.93x10^-4	7.52x10^-3
/\ ÷ Smallest (Ratio)	1	7.56
Final ratio	1	7.5

Note: You can often round ratios from <1.3 down to 1, and >1.8 up to 2. However, keep ratios around .33, .5 and .75 and triple, double and quadruple the final ratio respectively

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Question 11

[Total Marks: 1] Medium

A iodide of tungsten contains 55.1% iodine by mass. What is its simplest formula?

Select an answer from the options

Correct, the answer is D!

The % of tungsten in the compound is 100 - 55.1 = 44.9. Calculate the empirical formula:

Element	W	I
Mass/Percentage	44.9	55.1
/\ ÷ M_r (Moles)	0.244	0.434
/\ ÷ Smallest (Ratio)	1	1.77
Final ratio	4	7

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Question 12

[Total Marks: 1] Medium

8 g of oxygen combine with a metal (A) of A_r 40.0 to give 18.0 g of product. What would be the empirical formula of the oxide formed?

Select an answer from the options

Correct, the answer is B!

The mass of A in the compound is 18 - 8 = 10. Calculate the empirical formula:

Element	A	O
Mass/Percentage	10	8
/\ ÷ M_r (Moles)	0.25	0.5
/\ ÷ Smallest (Ratio)	1	2
Final ratio	1	2

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Question 13

[Total Marks: 1] Easy

A compound has an M_r of 112 g mol^-1. The empirical formula could never be

Select an answer from the options

Correct, the answer is D!

The empirical formula must have an empirical mass that is a factor of 112 (can divide into it exactly)

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Question 14

[Total Marks: 1] Medium

A sample contains 0.500 g of H₂, 3.000 g of C and 17.75 g of Cl₂. What is the empirical formula?

Select an answer from the options

Correct, the answer is B!

The mass of A in the compound is 18 - 8 = 10. Calculate the empirical formula:

Element	C	Cl	H
Mass/Percentage	3	17.75	0.5
/\ ÷ M_r (Moles)	0.25	0.5	0.5
/\ ÷ Smallest (Ratio)	1	2	2
Final ratio	1	2	2

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Question 15

[Total Marks: 1] Hard

A 20.0 g sample copper oxide, when heated in hydrogen, forms 2.52 g of water. The % mass of Cu in copper oxide would be

Select an answer from the options

Correct, the answer is D!

The moles of water would be 2.52 ÷ 18 = 0.14. This is the moles of oxygen in this water, and thus in the copper oxide (as it was not heated in air). The mass of oxygen would be 0.14 * 16 = 2.24. The mass of copper is therefore 20 - 2.24 = 17.76. This as a % of 20 is 88.8%

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Question 16

[Total Marks: 1] Medium

When 11.864 g of lead oxide was heated with H₂, it was completely reduced to 10.758 g Pb. What is the empirical formula for the original lead oxide?

Select an answer from the options

Correct, the answer is D!

The mass of oxygen in the lead oxide is 11.864 – 10.758 = 1.106. Calculate the empirical formula:

Element	Pb	O
Mass/Percentage	10.758	1.106
/\ ÷ M_r (Moles)	0.0519	0.069
/\ ÷ Smallest (Ratio)	1	1.33
Final ratio	3	4

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Question 17

[Total Marks: 1] Medium

Determine the empirical formula of terbium peroxide given 0.140 mol of Tb and 0.245 mol of O₂ in a sample

Select an answer from the options

Correct, the answer is D!

Calculate the empirical formula:

Element	Tb	O
Mass/Percentage	-	-
/\ ÷ M_r (Moles)	0.140	0.245
/\ ÷ Smallest (Ratio)	1	1.75
Final ratio	4	7

Note: You can often round ratios from <1.3 down to 1, and >1.8 up to 2. However, keep ratios around .33, .5 and .75 and triple, double and quadruple the final ratio respectively

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Question 18

[Total Marks: 1] Hard

A hydrated salt contained 45.7% water, but the anhydrous salt contains: Na 29.8%; C 7.8%; O 73.4%. The formula of the hydrated salt would be

Select an answer from the options

Correct, the answer is D!

Calculate the empirical formula of the anhydrous salt, which is Na₂CO₇ , as below:

Element	Na	C	O
Mass/Percentage	29.8	7.8	73.4
/\ ÷ M_r (Moles)	1.295	0.65	4.64
/\ ÷ Smallest (Ratio)	1.99	1	7.14
Final ratio	2	1	7

If the hydrated salt is 45.7% water, the Na₂CO₇ (M_r 170) accounts for 54.3%
Thus, the water must have a total M_r of (170 ÷ 54.3) * 45.7 = 143.07
The number of water molecules that would make this M_r is 143.07 ÷ 18 = 7.95

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Question 19

[Total Marks: 1] Medium

Citrazinic acid contains 46.46% C, 3.25% H₂, 9.03% N₂ and 41.26% O₂ . If there is only one N atom in the molecule, the total number of atoms combined would be?

Select an answer from the options

Correct, the answer is B!

Calculate the empirical formula of the acid, which is C₆H₅NO₄ , as below:

Element	C	H	N	O
Mass/Percentage	46.46	3.25	9.03	41.26
/\ ÷ M_r (Moles)	3.87	3.25	0.645	2.578
/\ ÷ Smallest (Ratio)	6	5.03	1	3.99
Final ratio	6	5	1	4

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Question 20

[Total Marks: 1] Easy

A compound has an M_r of 264 and contains 54.2% oxygen by mass. The number of oxygen atoms in each molecule of this compound is

Select an answer from the options

Correct, the answer is C!

The oxygen must have a total M_r of 264 * 0.542 = 143.088
The number of oxygen atoms that would produce this is 143.088 ÷ 16 = 8.94

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Question 21

[Total Marks: 1] Medium

A compound contains 0.3 mol of hydrogen, 0.0600 mol of oxygen and 0.090 mol of nitrogen. The number of H atoms in the empirical formula of this compound is

Select an answer from the options

Correct, the answer is D!

Calculate the empirical formula:

Element	N	H	O
Mass/Percentage	-	-	-
/\ ÷ M_r (Moles)	0.09	0.3	0.06
/\ ÷ Smallest (Ratio)	1.5	5	1
Final ratio	3	10	2

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Question 22

[Total Marks: 1] Easy

The mass of hydrogen in grams in C₆H₁₂O₆ that contains 48.0 g of C is

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_r. The moles of C atoms is therefore 48 ÷ 12 = 4
The moles of C₆H₁₂O₆ is therefore 4 ÷ 6 = 0.666
The moles of H are therefore 0.666 * 12 = 8
The mass of H is therefore 8 * 1 = 8

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Question 23

[Total Marks: 1] Easy

Which one of the answers has the highest percentage N by mass?

Select an answer from the options

Correct, the answer is C!

The M_r for A is 114, of which N accounts for 14. So (14 ÷ 114) * 100 = 12.28%
Repeat for the others

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Question 24

[Total Marks: 1] Hard

A compound of M_r 160 contains 0.175 mol of C, 0.140 mol of H and 0.0350 mol of N. How many C atoms would be found in the empirical and molecular formula?

Select an answer from the options

Correct, the answer is C!

Calculate the empirical formula, which is C₅H₄N, as below:

Element	C	H	N
Mass/Percentage	-	-	-
/\ ÷ M_r (Moles)	0.175	0.140	0.0350
/\ ÷ Smallest (Ratio)	5	4	1
Final ratio	5	4	1

The empirical mass of this formula is 78. The molecular formula is therefore double the empirical formula (C₁₀H₈N₂)

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Question 25

[Total Marks: 1] Easy

If a compound consists of Na₃ (PO₄)₂ and Na(OH)₂ and a molar ratio of sodium to phosphorus of 5:3, which formula would fit this information?

Select an answer from the options

Correct, the answer is B!

This simply means there must be a ratio of 5 Na atoms to every 3 P atoms. The total number of atoms for each must therefore be in this ratio

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Question 1

[Total Marks: 1] Easy

Ammonia is manufactured as shown below:

N₂_(g) + 3H₂_(g) → 2NH₃_(g)

If 28.0 g of N₂ produces 17.0 g of NH₃. What is the percentage yield?

Select an answer from the options

Correct, the answer is A!

Moles = Mass ÷ M_r. So, moles of N₂ is 28 ÷ 28 = 1
N₂ and NH₃ are in a 1:2 ratio, so moles of NH₃ is 1 * 2 = 2
Mass = Moles * M_r. So, mass of NH₃ is therefore 2 * 17 = 34
17 as a percentage of 34 is (17 ÷ 34) * 100 = 50%

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Question 2

[Total Marks: 1] Medium

10 g of calcium is combined with bromine to form CaBr₂ . What mass of CaBr₂ (M_r = 200) would be formed if the actual yield is 50 % compared to the theoretical yield?

Select an answer from the options

Correct, the answer is B!

The equation would be: Ca + Br₂ → CaBr₂
Moles = Mass ÷ M_r. So, moles of calcium is 10 ÷ 40 = 0.25
Calcium and CaBr₂ are in a 1:1 ratio, so moles of CaBr₂ is 0.25
Mass = Moles * M_r. So, mass of CaBr₂ is therefore 0.25 * 200 = 50
50 is the theoretical yield, so the actual yield would be 50% of this. So, 50 * 0.5 = 25

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Question 3

[Total Marks: 1] Easy

If 10 g of Al produced 1 g of H₂ as per the equation below, the yield of H₂ as a % would be

2Al + 3H₂SO₄ → 3H₂ + Al₂(SO₄)₃

Select an answer from the options

Correct, the answer is A!

Moles = Mass ÷ M_r. So, moles of Al is 10 ÷ 27 = 0.37
Al and H₂ are in a 2:3 ratio, so moles of H₂ is 0.37 ÷ 2 * 3 = 0.55
Mass = Moles * M_r. So, mass of H₂ is therefore 0.55 * 2 = 1.11
1 as a percentage of 1.11 is (1 ÷ 1.11) * 100 = 90%

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Question 4

[Total Marks: 1] Hard

When 200 g of FeS₂ are reacted with excess oxygen, 112.0 g of Fe₂O₃ are formed alongside sulfur dioxide. By writing a balanced equation, the percentage yield of Fe₂O₃ could be calculated as

Select an answer from the options

Correct, the answer is C!

The equation would be: 4FeS₂ + 7O₂ → 2Fe₂O₃ + 4SO₂
Moles = Mass ÷ M_r. So, moles of FeS₂ is 200 ÷ 120 = 1.66
FeS₂ and Fe₂O₃ are in a 4:2 ratio, so moles of Fe₂O₃ is 1.66 ÷ 4 * 2 = 0.833
Mass = Moles * M_r. So, mass of Fe₂O₃ is therefore 0.833 * 160 = 133.33
112 as a percentage of 133.33 is (112 ÷ 133.33) * 100 = 84%

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Question 5

[Total Marks: 1] Easy

When 65 g of Fe₂O₃ was added as per the following equation, 38.5 g of Fe was obtained. The percentage yield for this reaction would be

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Select an answer from the options

Correct, the answer is D!

Moles = Mass ÷ M_r. So, moles of Fe₂O₃ is 65 ÷ 160 = 0.406
Fe₂O₃ and Fe are in a 1:2 ratio, so moles of Fe is 0.406 * 2 = 0.8125
Mass = Moles * M_r. So, mass of Fe is therefore 0.8125 * 56 = 45.5
38.5 as a percentage of 45.5 is (38.5 ÷ 45.5) * 100 = 84.6%

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Question 6

[Total Marks: 1] Easy

If trying to form 28.4 g of P₄O₁₀, one would need how many moles of O₂ ?

Select an answer from the options

Correct, the answer is C!

The equation would be: 4P + 5O₂ → P₄O₁₀
Moles = Mass ÷ M_r. So, moles of P₄O₁₀ is 28.4 ÷ 284 = 0.1
P₄O₁₀ and O₂ are in a 1:5 ratio, so moles of O₂ is 0.1 * 5 = 0.5

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Question 7

[Total Marks: 1] Medium

When 10 g of impure MgCO₃ is heated as below, 0.075 moles of CO₂ are produced. What is the percentage purity of the MgCO₃ ? (Assume no impurities produce CO₂)

MgCO₃ _(s) → MgO_(s) + CO₂ _(g)

Select an answer from the options

Correct, the answer is D!

CO₂ and MgCO₃ are in a 1:1 ratio, so moles of pure MgCO₃ is 0.075
Mass = Moles * M_r. So, mass of pure MgCO₃ is therefore 0.075 * 84 = 6.3
6.3 as a percentage of 10 is (6.3 ÷ 10) * 100 = 63%

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Question 8

[Total Marks: 1] Easy

When 224 g of carbon monoxide react through the given equation, which of the following is true?

CO_(g) + ½O₂ _(g) → CO₂ _(g)

Select an answer from the options

Correct, the answer is A!

Moles = Mass ÷ M_r. So, moles of CO is 224 ÷ 28 = 8. This is also the moles of CO₂
CO and O₂ are in a 1:0.5 ratio, so moles of O₂ is 8 * 0.5 = 4

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Question 9

[Total Marks: 1] Hard

In the following reaction, with 350 g of each reactant, which one would be the limiting reagent?

Ca₃(PO₄)₂ + 3SiO₂ + 5C + 5O₂ + 3H₂O → 3CaSiO₃ + 5CO₂ + 2H₃PO₄

Select an answer from the options

Correct, the answer is B!

Limiting reagents are found by finding moles of each reactant and dividing by the coefficient to get a 1:1 ratio for all. The smallest is limiting. This can be done using a table, like below:

Species	Ca₃(PO₄)₂	SiO₂	C	O₂	H₂O
Mass	350	350	350	350	350
/\ ÷ M_r (Moles)	1.129	5.833	29.17	10.94	19.44
/\ ÷ Coefficient	1.129	1.944	5.834	5.47	6.48

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Question 10

[Total Marks: 1] Easy

The following reaction shows the formation of aspirin from salicylic acid and ethanoic anhydride:

2 C₇H₆O₃ + C₄H₆O₃ → 2C₉H₈O₄ + H₂O

The maximum mass of aspirin produced from 2.5 g of salicylic acid (C₇H₆O₃) and excess ethanoic anhydride would be

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_r. So, moles of C₇H₆O₃ is 2.5 ÷ 138 = 0.0181
C₇H₆O₃ and aspirin are in a 2:2 ratio, so moles of aspirin is 0.0181
Mass = Moles * M_r. So, mass of aspirin is therefore 0.0181 * 180 = 3.258

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Question 11

[Total Marks: 1] Easy

If 5.0 g of Ca are heated with 5.0 g of S, to completion, in the equation below, then the maximum mass _(g) of calcium sulphide that would form is

Ca_(s) + S_(s) → CaS_(s)

Select an answer from the options

Correct, the answer is D!

Moles = Mass ÷ M_r. So, moles of Ca is 5 ÷ 40 = 0.125 and moles of S is 5 ÷ 32 = 0.156
Ca is therefore limiting and so the answer must include 40 but not 32

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Question 12

[Total Marks: 1] Medium

If 10.0 g of phosphorus and 10.0 g of bromine produce 9.6 g of product, which reagent was limiting, and what is the percentage yield for this reaction as shown below?

2P_(s) + 3Br₂ _(g) → 2PBr₃ _(l)

Select an answer from the options

Correct, the answer is D!

Moles = Mass ÷ M_r. So, moles of P is 10 ÷ 31 = 0.322 and moles of Br is 10 ÷ 160 = 0.0625
Br is therefore limiting
Br and PBr₃ are found in a 3:2 ratio, so moles of PBr₃ is 0.0625 ÷ 3 * 2 = 0.04166
Mass = Moles * M_r. So, mass of PBr₃ is therefore 0. 04166 * 271 = 11.2916
9.6 as a percentage of 11.29 is (9.6 ÷ 11.29) * 100 = 85.02%

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Question 13

[Total Marks: 1] Easy

If 6.0 mol of C₂H₃Cl and 6.0 mol of O₂ are reacted as below, how many moles of CO₂ would be made?

C₂H₃Cl + 2.5O₂ → 2CO₂ + H₂O + HCl

Select an answer from the options

Correct, the answer is A!

Given C₂H₃Cl and O₂ are found in a 1:2.5 ratio, O₂ would be limiting if there was the same moles
O₂ and CO₂ are found in a 2.5:2 ratio, so moles of CO₂ is 6 ÷ 2.5 * 2 = 4.8

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Question 14

[Total Marks: 1] Hard

Equal masses of oxygen and hydrogen gas are reacted in a sealed reaction vessel to produce

Select an answer from the options

Correct, the answer is A!

The equation would be: O₂ + 2H₂ → 2H₂O. Let’s assume 1g of each is present
Moles = Mass ÷ M_r. So, moles of O₂ is 1 ÷ 32 = 0.0312 and moles of H₂ is 1 ÷ 2 = 0.5. So, after dividing by the coefficient, H₂ is 0.25. O₂ is therefore limiting
There would therefore be some H₂ left over in addition to the water produced, but no O₂

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Question 15

[Total Marks: 1] Medium

Completely burning 0.60 mol of a hydrocarbon produced CO₂ and H₂O of 1.80 and 2.40 mol respectively. The hydrocarbon must have been

Select an answer from the options

Correct, the answer is C!

When burning a hydrocarbon, the moles of CO₂ and H₂O produced give the moles of C and H respectively in the compound. So, moles of C is 1.8 and moles of H is 2.4
The ratio of C:H is therefore 1.8:2.4, which simplifies to 3:4
So, the compound would have been C₃H₄

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Question 16

[Total Marks: 1] Medium

The compound NH₄V₃O₈ is found in the following reactions:

2NH₃_(g) + V₂O₅_(s) + H₂O_(l) → 2NH₄VO₃ _(aq)

3NH₄VO₃ _(aq) + 2HCl_(aq) → NH₄V₃O₈_(aq) + 2NH₄Cl_(aq) + H₂O_(l)

If all reactants except ammonia are provided in excess, what is the maximum yield of NH₄V₃O₈ producible from ¼ moles of ammonia?

Select an answer from the options

Correct, the answer is D!

Ammonia and NH₄VO₃ are in a 2:2 ratio, so the moles of NH₄VO₃ would be ¼ in reaction 1
NH₄VO₃ and NH₄V₃O₈ are in a 3:1 ratio, so assuming all ¼ moles are used in reaction 2, the moles of NH₄V₃O₈ would be ¼ ÷ 3 = 1/12

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Question 17

[Total Marks: 3] Easy

A compound of mass 0.496 g consisted of only C, H and O. It was completely burned in oxygen, producing 1.56 g of CO₂ and 0.638 g of H₂O

(a) Calculate the number of moles of carbon dioxide formed [1]

(b) Calculate the number of moles of water formed [1]

(c) What is the empirical formula of the substance? [1]

Write out your answer in the box

(a) Moles = Mass ÷ M_r. So, moles of CO₂ is 1.56 ÷ 44 = 0.0355 mol. ✓

(b) Moles = Mass ÷ M_r. So, moles of H₂O is 0.638 ÷ 18 = 0.0354 mol. ✓

(c) The ratio of moles of CO₂ to H₂O is 1:1, so the empirical formula would be CH. ✓

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Question 1

[Total Marks: 1] Easy

6.0 dm³ of sulfur dioxide and 4.0 dm³ of oxygen react as shown below. The volume of sulfur trioxide, in dm³, formed would be

2SO₂ _(g) + O₂ _(g) → 2SO₃ _(g)

Select an answer from the options

Correct, the answer is C!

The ’true’ volume of sulfur dioxide would be 6 ÷ 2 = 3. The volume of O₂ would still be 4
Therefore, sulfur dioxide is limiting. (Don’t forget to divide by the coefficients for limiting reagents!)
Sulfur dioxide and sulfur trioxide are in 2:2 ratio. So, volume sulfur trioxide would be 6

Note: When all regents are in the gaseous state, you can use volumes just like you would moles, saving lots of work!

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Question 2

[Total Marks: 1] Easy

Choose the answer with the correct product from the reaction between 400 cm³ of H₂ and 300 cm³ of Cl₂? Assume constant temperature and pressure

Select an answer from the options

Correct, the answer is D!

The equation would be: H₂_(g) + Cl₂_(g) → 2HCl_(g)
Cl₂ is clearly limiting
Cl₂ and HCl are in a 1:2 ratio, so volume of HCl would be 300 * 2 = 600

Note: H₂ and Cl₂ are in a 1:1 ratio, hence the 100cm³ of H₂ left

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Question 3

[Total Marks: 1] Medium

According to the equation below, what volume of air is required to react with 5 dm³ of SO₂ ? Assume air contains 20% oxygen

2SO₂ _(g) + O₂ _(g) → 2SO₃ _(g)

Select an answer from the options

Correct, the answer is B!

SO₂ and O₂ are in a 2:1 ratio, so volume of O₂ required is 5 ÷ 2 = 2.5
If air contains 20% O₂ , the volume of air required would be 2.5 ÷ 0.2 = 12.5

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Question 4

[Total Marks: 1] Hard

Equal volumes of H₂ and O₂ reacted together will produce

Select an answer from the options

Correct, the answer is A!

The equation would be: O₂ + 2H₂ → 2H₂O
Let’s assume 1dm³ of each is present
‘True’ volume of O₂ is still 1 and ‘true’ volume of H₂ is 1 ÷ 2 = 0.5. (Don’t forget to divide by coefficients for limiting reagents!). So H₂ is limiting
There would therefore be some O₂ left over in addition to the water produced, but no H₂

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Question 5

[Total Marks: 1] Easy

For the following combustion of a hydrocarbon, if 300 cm³ of propane is burnt in excess oxygen, the volume of carbon dioxide produced is:

C₃H₈_(g) + 5O₂ _(g) → 3CO₂ _(g) + 4H₂O_(g)

Select an answer from the options

Correct, the answer is C!

Propane and CO₂ are in 1:3 ratio. The volume of CO₂ is therefore 300 * 3 = 900cm³

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Question 6

[Total Marks: 1] Medium

A 600 cm³ sample of Xe gas has a pressure of 2.335x10⁴ Pa at 150 K. What pressure will it exert if the volume and temperature are increased to 0.8 dm³ and 200 K?

Select an answer from the options

Correct, the answer is C!

The equation to use here is P1 * V1T1 = P₂ * V₂T2
So, 2.335x10⁴ * 0.6150 = P₂ * 0.8200. Then rearrange

Note: This equation generally requires T in Kelvin, P in Pa and V in dm³. If you want, you can use any units for V and P, but they must be the same on both sides

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Question 7

[Total Marks: 1] Medium

The temperature of 210 cm³ H₂ gas is lowered from 40.0 °C to 0 °C at constant pressure. What is the final volume?

Select an answer from the options

Correct, the answer is A!

The equation to use here is V1T1 = V₂T2
So, 210313 = V₂273. Then rearrange

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Question 8

[Total Marks: 1] Medium

4.00 mol of He, at 27 °C and 3.00 atm, would fit in a vessel of volume

Select an answer from the options

Correct, the answer is D!

The equation to use here is PV=nRT
So, 3x10⁵ * V = 4 * 8.31 * 300. Then rearrange to get V in m³ as 0.03324

Note: This equation requires P in Pa, V in m³ and T in Kelvin without exception!

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Question 9

[Total Marks: 1] Easy

0.673 g of a substance has a volume of 272 cm³ at 42 °C and 103.4 kPa. What is its molar mass?

Select an answer from the options

Correct, the answer is A!

The equation to use here is PV=nRT
So, 103.4x10³ * 2.72x10^-4 = n * 8.31 * 315. Then rearrange to get moles as 0.0107. M_r = Mass ÷ Moles
So, 0.673 ÷ 0.0107 = 62.64

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Question 10

[Total Marks: 1] Hard

A gas of density 2.48 g dm^-3 at 40 °C and 1.3 atm would have a molar mass closest to which answer?

Select an answer from the options

Correct, the answer is B!

The equations to use here are M_r = MassPV ÷ RT and Density = Mass ÷ Volume
So, M_r = Mass(1.3x10⁵ * V) ÷ (8.31 * 313)
This simplifies to M_r = Mass49.98 * V. So, 49.98 * M_r = MassV
Therefore, 49.98 * M_r = 2.48x10³. (Don’t forget to convert g/dm³ into g/m³!)
Then rearrange to get 49.62

Note: This question is tough!

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Question 11

[Total Marks: 3] Medium

Bromine can be prepared by the reaction of hydrogen bromide with manganese(IV) oxide as shown:

4HBr_(aq) + MnO₂ _(s) → Br₂ _(g) + MnBr₂ _(aq) + 2H₂O_(l)

If 17 g of HBr reacted completely with manganese(IV) oxide

(a) Calculate how many moles of hydrogen bromide reacted. [1]

(b) Calculate the volume of bromine gas produced (at STP). [2]

Write out your answer in the box

(a) Moles = Mass ÷ M_r. So, moles is 17 ÷ 79.9 = 0.2128. ✓

(b) HBr and Br₂ are in 4:1 ratio, so moles of Br₂ is 0.2128 ÷ 4 = 0.0532. ✓
Volume = Moles * 22.7. So, volume is 0.0532 * 22.7 = 1.207dm³ → 1.21dm³ (3sf). ✓

Note: STP uses 22.7 and RTP uses 24dm³!

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Question 12

[Total Marks: 2] Easy

If 42 g of C₃H₆ undergoes combustion as shown, calculate the volume of carbon dioxide produced at STP:

2C₃H₆_(g) + 9O₂ _(g) → 6CO₂ _(g) + 6H₂O_(l)

Write out your answer in the box

Moles = Mass ÷ M_r. So, moles of C₃H₆ is 42 ÷ 42 = 1
C₃H₆ and CO₂ are in 2:6 ratio, so moles of CO₂ is 1 ÷ 2 * 6 = 3. ✓
Volume = Moles * 22.7. So, volume is 3 * 22.7 = 68.1dm³. ✓

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Question 13

[Total Marks: 4] Medium

Excess HCl reacts with Be to produce 50 cm³ of hydrogen gas at 20 °C and 1.08x10⁴ Pa. The equation for this reaction is shown below:

Be + 2HCl → BeCl₂ + H₂

(a) Calculate the moles of hydrogen gas given off. [2]

(b) Calculate the mass of Be that was required for this reaction. [2]

Write out your answer in the box

(a) n = PVRT. ✓
So, moles of H₂ is (1.08x10⁴ * 5x10^-5)(8.31 * 293) = 2.22x10^-4. ✓

(b) H₂ and Be are in 1:1 ratio, so moles of Be is 2.22x10^-4. ✓
Mass = Moles * M_r. So, mass is 2.22x10^-4 * 9 = 0.001996 → 0.00200 (3sf). ✓

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Question 14

[Total Marks: 4] Medium

When manganese(IV) oxide (MnO₂) is heated it can decompose. Heating 5.00 g of MnO₂ would produce what volume of oxygen in dm³? (18 °C and 1.05x10⁵ Pa)

3MnO₂ → Mn3O₄ + O₂

Write out your answer in the box

Moles = Mass ÷ M_r. So, moles of MnO₂ is 5 ÷ 87 = 0.0574. ✓
MnO₂ and O₂ are found in 3:1 ratio. So, moles of O₂ is 0.0574 ÷ 3 = 0.01915. ✓
V = nRTP. So, volume of O₂ is (0.01915 * 8.31 * 291)1.05x10⁵ = 4.41x10^-4 m³ ✓
Therefore, V in dm³ is 4.41x10^-4 * 1000 = 0.441dm³ (3sf). ✓

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Question 1

[Total Marks: 1] Easy

1.01 g of KNO₃ is dissolved in water to make 0.500 dm³ of solution. What would be the concentration in mol dm^-3?

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_r. So, moles of KNO₃ is 1.01 ÷ 101 = 0.01
Concentration = Moles ÷ Volume. So, concentration is 0.01 ÷ 0.5 = 0.02

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Question 2

[Total Marks: 1] Easy

What volume in dm³ of 0.6 mol dm^-3 NaCl solution can be prepared from 0.12 mol of NaCl?

Select an answer from the options

Correct, the answer is B!

Volume = Moles ÷ Concentration. So, volume is 0.12 ÷ 0.6 = 0.2

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Question 3

[Total Marks: 1] Easy

How many moles of HBr are in 25 cm³ of 0.2 mol dm^-3 hydrogen bromide?

Select an answer from the options

Correct, the answer is A!

Moles = Volume * Concentration. So, moles is 0.025 * 0.2 = 0.005

Note: Volume must be in dm³ for any mole calculation!

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Question 4

[Total Marks: 1] Easy

The mass of ammonium nitrate required make a solution of volume 500 cm³ and concentration 0.40 mol dm^-3 is

Select an answer from the options

Correct, the answer is D!

Moles = Volume * Concentration. So, moles is 0.5 * 0.4 = 0.2
Mass = Moles * M_r. So, mass of ammonium nitrate is 0.2 * 80 = 16
Note: Ammonium nitrate is NH₄NO₃ – useful to learn

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Question 5

[Total Marks: 1] Easy

If a 150 cm³ sample of vinegar has a concentration of 0.25 mol dm^-3 of ethanoic acid, how many moles of ethanoic acid are present?

Select an answer from the options

Correct, the answer is B!

Moles = Volume * Concentration. But 150cm³ needs to be converted to dm³ by ÷ 1000

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Question 6

[Total Marks: 1] Hard

A sample of HNO₃ _(aq) of volume 20.0 cm³ and concentration 0.400 mol dm^-3 is made into NaNO₃ _(aq). What volume in cm³ of 0.200 mol dm^-3 NaOH_(aq) would be required?

Select an answer from the options

Correct, the answer is C!

The equation would be: HNO₃ _(aq) + NaOH_(aq) → NaNO₃ _(aq) + H₂O_(l)
The equation to use here is C₁ * V₁n₁ = C₂ * V₂n₂
So, 0.4 * 201 = 0.2 * V₂1. Then rearrange

Note: With the C₁ * V₁n₁ = C₂ * V₂n₂ equation the units can be anything, as long as they are the same on both sides. n stands for the coefficient in the equation relating to that species

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Question 7

[Total Marks: 1] Medium

How much 0.100 mol dm^-3 NaOH in cm³ is required for full titration with 10.0 cm³ of 0.050 mol dm^-3 H₂SO₄ ?

Select an answer from the options

Correct, the answer is C!

The equation would be: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
The equation to use here is C₁ * V₁n₁ = C₂ * V₂n₂
So, 0.1 * V12 = 0.05 * 101. Then rearrange

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Question 8

[Total Marks: 1] Medium

50.0 cm³ of sulphuric acid neutralises 36.2 cm³ of 0.225 mol dm^-3 NaOH. The concentration of the H₂SO₄ is thus

Select an answer from the options

Correct, the answer is C!

The equation would be: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
The equation to use here is (C₁ * V₁) ÷ n₁ = (C₂ * V₂) ÷ n₂
So, (0.225 * 36.2) ÷ 2 = (C₂ * 50) ÷ 1

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Question 9

[Total Marks: 1] Medium

In the below reaction, how much AgF would be precipitated if 45.0 cm³ of 0.300 M KF solution was added to excess silver nitrate?

AgNO₃ _(aq) + KF_(aq) → AgF_(s) + KNO₃ _(aq)

Select an answer from the options

Correct, the answer is B!

Moles = Concentration * Volume. So, moles of KF is 0.3 * 0.045 = 0.0135
KF and AgF are in 1:1 ratio, so moles of AgF = 0.0135
Mass = Moles * M_r. So, mass of AgF is 0.0135 * 127 = 1.71

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Question 10

[Total Marks: 1] Easy

The concentration of sulphate ions in 500.0 cm³ of 0.30 mol dm^-3 Fe₂(SO₄)₃ solution would be

Select an answer from the options

Correct, the answer is D!

There are three sulphate ions per molecule of Fe₂(SO₄)₃, so the concentration is 0.90 mol dm^-3

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Question 11

[Total Marks: 1] Medium

What volume of 3.0 M nitric acid is needed to react 2.83 g of zinc according to the equation shown?

3Zn_(s) + 8HNO₃ _(aq) → 3Zn(NO₃)₂_(aq) + 2NO_(g) + 4H₂O_(l)

Select an answer from the options

Correct, the answer is B!

Moles = Mass ÷ M_r. So, moles of zinc is 2.83 ÷ 65 = 0.0435
Zinc and nitric acids are in a 3:8 ratio. So, moles of nitric acid is 0.0435 ÷ 3 * 8 = 0.116
Volume = Moles ÷ Concentration. So, volume is 0.116 ÷ 3 = 0.0386

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Question 12

[Total Marks: 1] Medium

What is the concentration of Li⁺ ions resulting from 5 g of Li₂CO₃ •7H₂O being dissolved in water of 250 cm³?

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_r. So, moles of Li₂CO₃ •7H₂O is 5 ÷ 200
There are 2 Li atoms in each molecule, so (5 ÷ 200) * 2 gives the moles of Li⁺ atom

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Question 13

[Total Marks: 1] Medium

A student determined the concentration of HCl_(aq) by titration. When 90.0 cm³ of acid was titrated to end point, 18.0 cm³ of 0.18 M LiOH was required. What was the molarity of the HCl solution?

Select an answer from the options

Correct, the answer is C!

The equation would be: HCl + LiOH → LiCl + H₂O
The equation to use here is C₁ * V₁n₁ = C₂ * V₂n₂
So, 18 * 0.0181 = C₂ * 901. Then rearrange

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Question 14

[Total Marks: 1] Hard

Manganese(II) hydroxide is produced by reacting manganese(II) sulphate with potassium hydroxide as shown below. How much could be produced if 15 cm³ of 0.2 mol dm^-3 potassium hydroxide is added to 40 cm³ 0.15 mol dm^-3 manganese sulphate?

MnSO₄ _(aq) + 2KOH_(aq) → Mn(OH)₂_(s) + K₂SO₄ _(aq)

Select an answer from the options

Correct, the answer is A!

As we have information on both reactants, we need to establish which is limiting
Moles = Mass ÷ M_r. So, moles of KOH is 0.015 * 0.2. = 0.003. Moles of MnSO₄ is 0.040 * 0.15 = 0.006
The ‘true’ moles of KOH is 0.003 ÷ 2 = 0.0015, so KOH is limiting. (Coefficient of 2)
As KOH and Mn(OH)₂ are in 2:1 ratio, moles of Mn(OH)₂ is therefore 0.003 ÷ 2 = 0.0015
Mass = Moles * M_r. So, mass is 0.0015 * 89 = 0.1335

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Question 15

[Total Marks: 1] Hard

What mass of silver bromide can be produced from 25 cm³ of 0.17 mol dm^-3 sodium bromide and an excess of aqueous silver ions?

Select an answer from the options

Correct, the answer is B!

The equation would be: Ag⁺ + NaBr → AgBr + Na⁺
Moles = Volume * Concentration. So, moles of sodium bromide is 0.025 * 0.17
Sodium bromide and silver bromide are in 1:1 ratio, so moles of silver bromide is 0.025 * 0.17
Mass = Moles * M_r. So, mass is (0.025 * 0.17) * (107.87 + 79.9)

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Question 16

[Total Marks: 1] Medium

By writing an equation, calculate the concentration of NaOH if 70.00 cm³ successfully neutralises 50.00 cm³ of 0.3000 mol dm^-3 H₂SO₄

Select an answer from the options

Correct, the answer is B!

The equation would be: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
The equation to use here is C₁ * V₁n₁ = C₂ * V₂n₂
So, 0.3 * 501 = C₂ * 702. Then rearrange

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Question 17

[Total Marks: 1] Easy

Potassium hydroxide reacts with phosphoric(V) acid as shown below:

3KOH + H₃PO₄ → K₃PO₄ + 3H₂O

If 50.00 cm³ of 0.20 mol dm^-3 KOH reacts with 0.04 mol dm^-3 H₃PO₄ , the volume of H₃PO₄ , in cm³, necessary would be calculated by which of the following?

Select an answer from the options

Correct, the answer is D!

The equation to use here is C₁ * V₁n₁ = C₂ * V₂n₂
So, 0.2 * 503 = 0.04 * V₂1
Multiplying by 0.2 is the same as dividing by 5. Then rearrange

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Question 18

[Total Marks: 1] Easy

What volume of 0.500 M HF is required to react with 9.3 g of calcium carbonate according to the equation:

CaCO₃ _(s) + 2HF_(aq) → CaF₂_(aq) + H₂O_(l) + CO₂ _(g)

Select an answer from the options

Correct, the answer is C!

Moles = Mass ÷ M_r. So, moles of calcium carbonate is 9.3 ÷ 100 = 0.093
Calcium carbonate and HF are in 1:2 ratio. So, moles of HF is 0.093 * 2 = 0.186
Volume = Moles ÷ Concentration. So, volume is 0.186 ÷ 0.5 = 0.372dm³

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Question 19

[Total Marks: 1] Easy

60 cm³ of HCl of concentration 1 M is added to 40 cm³ of 1 M NaOH. The number of moles of HCl left unreacted would be given by which answer?

Select an answer from the options

Correct, the answer is C!

The equation would be: HCl + NaOH → NaCl + H₂O
Since they react in a 1:1 ratio, all NaOH would be used up and 20cm³ of HCl would be left
Moles = Concentration * Volume. So, moles is 0.02 * 1 = 0.02

Note: Don’t forget to convert cm³ to dm³!

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Question 20

[Total Marks: 1] Easy

What mass of PbCl₂ would be precipitated if excess Pb(NO₃)₂ solution were combined with 87.0 cm³ of 0.100 M KCl solution?

Select an answer from the options

Correct, the answer is B!

The equation would be: Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃
Moles = Concentration * Volume. So, moles of KCl is 0.1 * 0.087 = 0.0087
KCl and PbCl₂ are in 2:1 ratio, so moles of PbCl₂ is 0.0087 ÷ 2 = 0.00435
Mass = Moles * M_r. So, mas is 0.00435 * 278 = 1.209

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Question 1

[Total Marks: 1] Medium

50.00 cm³ of silver nitrate solution is reacted with 100.00 cm³ of calcium bromide solution. The concentrations of these solutions respectively could be:

2AgNO₃ + CaBr₂ → 2AgBr + Ca(NO₃)₂

Select an answer from the options

Correct, the answer is D!

Moles = Concentration * Volume. So, using the coefficients as moles, we can say for silver nitrate 2 = C1 * 50 and for calcium bromide 1 = C₂ * 100
Therefore, C1 = 0.04 and C₂ = 0.01. This is a ratio of 4:1, so, the answer would match this ratio

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Question 2

[Total Marks: 1] Hard

0.05 mol of the iodide of element A was dissolved in water to make a total volume of 1 dm³. 25 cm³ of this solution reacted with 25 cm³ of 0.2 mol dm^-3 silver nitrate solution. The formula of the iodide is

Select an answer from the options

Correct, the answer is D!

Since we don’t know element A, the equation would be: AIx + AgNO₃ → IxNO₃ + AgI where x denotes an unknown number
Moles = Concentration * Volume. So, moles of silver nitrate is 0.2 * 0.025 = 0.005. Assuming a 1:1 ratio, we can therefore say 0.005 moles of AIx was required
Since we used a 25cm³ sample, which is 1/40th of the total 1dm³, the total moles of AIx that would be required for this reaction is 0.005 * 40 = 0.2
This 0.2 mol is 4x the amount of moles of iodide A that were prepared, so the equation must have been 4x our assumed 1:1 ratio: AI₄ + AgNO₃ → I₄NO₃ + AgI

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Question 3

[Total Marks: 1] Easy

A 25 cm³ sample of a solution which is 0.05 M in HCl is combined with 10 cm³ of 0.05 M NaOH. The concentration of HCl remaining after reaction is given by

Select an answer from the options

Correct, the answer is A!

The equation would be: HCl + NaOH → NaCl + H₂O
Moles = Concentration * Volume. So, the moles of HCl left would be (0.05 * 25) - (0.05 * 10)

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Question 4

[Total Marks: 1] Medium

20 cm³ of barium fluoride solution, BaF₂_(aq), completely reacted with 40 cm³ of iron sulphate solution, Fe₂(SO₄)₃_(aq) to form 1.398 g of BaSO₄ precipitate. The concentration of the iron sulphate solution is

Select an answer from the options

Correct, the answer is B!

The equation would be: 3BaF₂ + Fe₂(SO₄)₃ → 3BaSO₄ + 2FeF₃
Moles = Mass ÷ M_r. So, moles of BaSO₄ is 1.398 ÷ 233 = 0.006
BaSO₄ and iron sulphate are in 3:1 ratio, so moles of iron sulphate is 0.006 ÷ 3 = 0.002
Concentration = Moles ÷ Volume. So, concentration is 0.002 ÷ 0.04 = 0.05

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Question 5

[Total Marks: 1] Medium

If 0.001 mole of a bromide required 40 cm³ of 0.05 M silver nitrate (AgNO₃) for reaction. The original bromide could be which of the following?

Select an answer from the options

Correct, the answer is B!

The equation would be: XBr + AgNO₃ → XNO₃ + AgBr
Moles = Concentration * Volume. So, moles of AgNO₃ is 0.05 * 0.04 = 0.002
This is 2x the moles of bromide, so the equation becomes XBr₂ + 2AgNO₃ → X(NO₃)₂ + 2AgBr

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Question 6

[Total Marks: 1] Easy

In a NaOH and HCl titration, 100 cm³ of 0.2 M NaOH is added to 50 cm³ of 0.1 M HCl. The molarity of NaOH remaining is thus

Select an answer from the options

Correct, the answer is B!

The equation would be: HCl + NaOH → NaCl + H₂O
Moles = Concentration * Volume. So, the moles of NaOH left would be (0.2 * 100) - (0.1 * 50)

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Question 7

[Total Marks: 1] Easy

How much AgCl is precipitated from 0.002 m³ of 0.1000 M AgNO₃ reacting with excess MgCl₂?

Select an answer from the options

Correct, the answer is A!

The equation would be: MgCl₂ + 2AgNO₃ → 2AgCl + Mg(NO₃)₂
Moles = Concentration * Volume. So, the moles of AgNO₃ is 0.1 * 2 = 0.2
AgNO₃ and AgCl are in 2:2 ratio, so moles of AgCl is 0.2
Mass = Moles * M_r. So, mass is 0.2 * 143.5 = 28.7

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Question 8

[Total Marks: 1] Easy

What mass of PbF₂ can be precipitated from 100 cm³ of 0.300 M Pb(NO₃)₂ and 100 cm³ of 0.200 M NaF?

Select an answer from the options

Correct, the answer is B!

The equation would be: Pb(NO₃)₂ + 2NaF → PbF₂ + 2NaNO₃
None of the answers use 0.3, so we assume NaF is limiting
Moles = Concentration * Volume. So, the moles of NaF is 0.2 * 0.1. Check the answers
Note: Remember its multiple choice. For speed, work out a partial answer and then scan the options!

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Question 9

[Total Marks: 5] Hard

To discover the percentage of calcium carbonate in 1.216g of clean, dry coral, an experiment was conducted. The coral was added to 10.0 cm³ of 5.00 mol dm^-3 HCl to dissolve. The solution created was made up to 25.0 cm³ with distilled water. This diluted solution required 28.0 cm³ of 1.00 mol dm^-3 NaOH for neutralisation. An equation for the initial reaction is shown below:

CaCO₃ _(s) + 2HCl_(aq) → CaCl₂_(aq) + CO₂ _(g) + H₂O_(l)

(a) Calculate the number of moles of NaOH added. [1]

(b) How many moles of HCl remained in the beaker after the coral was dissolved? [1]

(c) Using your answer to part b), how many moles of acid reacted with the coral? [1]

(d) What mass of CaCO₃ was present in the coral? [1]

(e) What was the % composition of CaCO₃ in the coral? [1]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 1 * 0.028 = 0.028 mol. ✓

(b) The equation would be: HCl + NaOH → NaCl + H₂O. So, if HCl and NaOH react in 1:1 ratio, the moles of HCl would be the same as NaOH, so = 0.028 mol. ✓

(c) Moles = Concentration * Volume. So, moles of HCl added is 5 * 0.01 = 0.05
If 0.028 remained, moles reacted is 0.05 - 0.028 = 0.022 mol. ✓

(d) HCl and CaCO₃ are in 2:1 ratio, so moles CaCO₃ is 0.022 ÷ 2 = 0.011
Mass = Moles * M_r. So, mass is 0.011 * 100 = 1.1 g. ✓

(e) 1.1 as a percentage of 1.216 is (1.1 ÷ 1.216) * 100 = 90.4605 → 90.5% (3sf). ✓

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Question 10

[Total Marks: 4] Medium

2.447 g of an ammonia-containing surface cleaner was diluted to 20.0 cm³. This solution required 28.51 cm³ of 0.4040 mol dm^-3 sulphuric acid for neutralisation as per the equation shown:

2NH₃_(aq) + H₂SO₄ _(aq) → (NH₄)₂SO₄ _(aq)

(a) Calculate the number of moles of H₂SO₄ required for this reaction. [1]

(b) Calculate the mass of ammonia present in the surface cleaner. [2]

(c) Show by calculation the percentage by mass of ammonia present in the household cleaner. [1]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 0.404 * 0.02851 = 0.0115 mol. ✓

(b) H₂SO₄ and ammonia are in 1:2 ratio, so moles ammonia is 0.0115 * 2 = 0.023
Mass = Moles * M_r. So, mass is 0.023 * 17 = 0.391 g. ✓

(c) 0.3916 as a percentage of 2.447 is (0.391 ÷ 2.447) * 100 = 15.979 → 16.0% (3sf). ✓

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Question 11

[Total Marks: 6] Hard

1.377 g of anhydrous potassium carbonate was made up to 500 cm³ using distilled water. 50.00 cm³ of this solution was titrated with CH₃COOH using methyl orange indicator. The titration required 45.3 cm³ of the acid for complete neutralisation.

(a) Write a balanced equation for this neutralisation reaction. [2]

(b) Determine the concentration of the potassium carbonate solution. [2]

(c) Calculate the concentration of the ethanoic acid solution. [2]

Write out your answer in the box

(a) K₂CO₃ + 2CH₃COOH → 2CH₃COOK + H₂CO₃ . ✓ ✓ (1 mark for balancing)

(b) Moles = Mass ÷ M_r. So, moles of potassium carbonate is 1.377 ÷ 138 = 0.009978. ✓
Concentration = Moles ÷ Volume. So, 0.009978 ÷ 0.5 = 0.0200 mol/dm³. ✓

(c) K₂CO₃ and ethanoic acid are in 1:2 ratio, so, moles of ethanoic acid is 0.009978 * 2 = 0.019956. ✓
Concentration = Moles ÷ Volume. So, 0.019956 ÷ 0.0453 = 0.4405 → 0.441 mol/dm³ (3sf). ✓

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Question 1

[Total Marks: 5] Medium

Consider the following reaction. 0.0487 mol of zinc(II) oxide was added to 39.0 cm³ of 2.42 mol dm^-3 nitric acid solution

ZnO + 2HNO₃ → Zn(NO₃)₂ + H₂O

(a) Calculate the amount (in mol) of nitric acid. [1]

(b) Which is the limiting reactant? [1]

(c) Determine the number of moles of zinc(II) nitrate formed. [1]

(d) The product of this reaction was collected as zinc(II) nitrate trihydrate (Zn(NO₃)₂•3H₂O). Calculate the mass of this product formed. [2]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 2.42 * 0.039 = 0.0944 mol. ✓

(b) We need to compare in 1:1 ratio, so ‘true’ moles of nitic acid is 0.09438 ÷ 2 = 0.04719
Moles of zinc oxide is 0.0487, so nitric acid is limiting. ✓

(c) Nitric acid and zinc nitrate are in 2:1 ratio, so, moles of zinc nitrate is 0.0944 ÷ 2 = 0.0472 mol. ✓

(d) Mass = Moles * M_r. So, mass is 0.0472 * 243 = 11.4696 → 11.5g (3sf). ✓ ✓ (1 mark for 243)

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Question 2

[Total Marks: 7] Hard

16.20x10^-2 dm³ of 0.304 mol dm^-3 aqueous AgNO₃ is added to 11.60x10^-2 dm³ of 0.4550 mol dm^-3 aqueous NaF as per the equation below:

AgNO₃ _(aq) + NaF_(aq) → AgF_(s) + NaNO₃ _(aq)

(a) Calculate the mass in grams of AgF obtained in this reaction. [4]

(b) By what percentage, more or less, of AgF would the other product produce? [3]

Write out your answer in the box

(a) We need to find out limiting reagent, so, Moles = Concentration * Volume
Moles of AgNO₃ is 0.304 * 16.20x10^-2 = 0.049248 ✓ and NaF is 0.4550 * 11.60x10^-2 = 0.05278. ✓
Therefore, AgNO₃ is limiting
AgNO₃ and AgF are in 1:1 ratio, so, moles of AgF = 0.049248. ✓
Mass = Moles * M_r. So, mass is 0.049248 * 127 = 6.2545 → 6.25g (3sf). ✓

(b) Using NaF, the moles of AgF = 0.05278. ✓
Mass = Moles * M_r. So, mass is 0.05278 * 127 = 6.7036. ✓
The percentage increase is therefore 6.7036 - 6.25446.2544 * 100 = 7.1735 → 7.17% increase (3sf). ✓

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Question 3

[Total Marks: 5] Hard

Aqueous XO₄ ^3- ions form a precipitate with aqueous rubidium ions, Rb⁺:

3Rb⁺_(aq) + XO₄ ^3-_(aq) → Rb₃ XO₄ _(s)

When 82.36 cm³ of 0.4080 mol dm^-3 aqueous rubidium ions is added to XO₄ ^3- ions, 4.336 g of precipitate is formed

(a) Calculate the amount (in moles) of Rb⁺ ions used. [1]

(b) Calculate the amount (in moles) of the precipitate formed. [1]

(c) Calculate the M_r of this precipitate. [1]

(d) Determine the A_r of X and identify the element this represents [2]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles is 0.408 * 0.08236 = 0.03360 mol. ✓

(b) Rb⁺ and the precipitate are in 3:1 ratio, so, moles of precipitate is 0.0336 ÷ 3 = 0.01120 mol. ✓

(c) M_r = Mass ÷ Moles. So, M_r is 4.336 ÷ 0.0112336 = 387.1 g/mol. ✓

(d) M_r of Rb₃ XO₄ is 387.1. So, X is 387.1 – (3 * 85.5) – ( 4 * 16) = 66.6. ✓
The closest element is Zn. ✓

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Question 4

[Total Marks: 8] Hard

A dye contains only the elements C, N, H and O. 2.036 g of the dye was oxidised. In doing so, 5.470 g of CO₂ and 0.697 g of H₂O were produced

(a) Calculate the percentage by mass of C and H in the dye. [4]

(b) The % by mass of N in the dye is 10.75 %. Find the empirical formula. [3]

(c) Knowing the M_r is 260 g mol^-1, find the molecular formula of the dye. [1]

Write out your answer in the box

(a) Moles = Mass ÷ M_r. So, moles of CO₂ = 5.470 ÷ 44 = 0.124318. This is the moles of C in the dye
For H₂O, moles is 0.697 ÷ 18 = 0.03872. The moles of H in the dye is thus 0.03872 * 2 = 0.07744. ✓
Mass = Moles * M_r. So, mass of C is 0.124318 * 12 = 1.4918g and H is 0.07744 * 1 = 0.07744g. ✓
1.4918 as a percentage of 2.036 is (1.4918÷ 2.036) * 100 = 73.271 → C = 73.3% (3sf) ✓
0.07744 as a percentage of 2.036 is (0.07744 ÷ 2.036) * 100 = 3.803 → H = 3.80% (3sf) ✓

(b) The % of O would be 100 – 73.3 – 3.80 – 10.75 = 12.15%. ✓
Then find the empirical formula as below: ✓

Element	C	N	H	O
Mass/Percentage	73.3	10.75	3.8	12.15
/\ ÷ M_r (Moles)	6.1083	0.7678	3.8	0.7593
/\ ÷ Smallest (Ratio)	8.04	1.011	5	1
Final ratio	8	1	5	1

The empirical formula is C₈NH₅O. ✓

(c) The empirical mass is 131. The M_r is twice this, so the formula is C₁₆ N₂H₁₀O₂ . ✓

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Question 5

[Total Marks: 9] Hard

98.98 g of hydrated lithium carbonate crystals, Li₂CO₃ •xH₂O , were made up to 0.5000 dm³. 12.500 cm³ of this solution was neutralized with 24.40 cm³ of 0.2000 mol dm^-3 HCl as shown:

Li₂CO₃ _(aq) + 2HCl_(aq) → 2LiCl_(aq) + CO₂ _(g) + H₂O_(l)

(a) Calculate the concentration of the lithium carbonate solution. [3]

(b) Find the mass of lithium carbonate present in the 0.500 dm³ solution. [2]

(c) Calculate the mass of H₂O in the hydrated crystals and hence the value of x. [4]

Write out your answer in the box

(a) Moles = Concentration * Volume. So, moles of HCl is 0.2 * 0.02440 = 0.00488. ✓
HCl and Li₂CO₃ are in 2:1 ratio, so moles of Li₂CO₃ is 0.00488 ÷ 2 = 0.00244. ✓
Concentration = Moles ÷ Volume. So, concentration is 0.00244 ÷ 0.0125 = 0.1952 mol/dm³. ✓

(b) 0.5 dm³ of solution is 4 * 12.5 cm³ so the moles of Li₂CO₃ is 0.1952 * 4 = 0.7808. ✓
Mass = Moles * M_r. So, mass is 0.7808 * 74 = 57.78g. ✓

(c) Mass of water is 98.98 – 57.78 = 41.2g. ✓
Moles = Mass ÷ M_r. So, moles is 41.2 ÷ 18 = 2.2889. ✓
So, moles of water ÷ Li₂CO₃ will give X. So, X is 2.2889 ÷ 0.7808 = 2.93 ✓
X is therefore 3. ✓

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Question 6

[Total Marks: 6] Medium

0.453 g of an alkali metal sulfate (A₂SO₄) was dissolved in water. Excess barium fluoride solution was then added to precipitate all the sulfate ions as barium sulfate, BaSO₄ . This precipitate was filtered and dried. It weighed 0.438 g.

(a) Calculate the amount of moles of BaSO₄ . [1]

(b) Determine the amount (in mol) of A₂SO₄ present. [2]

(c) Determine the molar mass of the alkali metal sulfate. [1]

(d) Deduce the identity of A, showing your workings. [2]

Write out your answer in the box

(a) Moles = Mass ÷ M_r. So, moles of BaSO₄ is 0.438 ÷ 233 = 0.00188 mol. ✓

(b) The equation is A₂SO₄ + BaF₂ → BaSO₄ + 2AF ✓
BaSO₄ and A₂SO₄ are in 1:1 ratio, so moles of A₂SO₄ = 0.00188 mol. ✓

(c) M_r = Mass ÷ Moles. So, mass of A₂SO₄ is 0.453 ÷ 0.00188 = 241 g/mol. ✓

(d) M_r of A₂ is 241– 32 – (4 * 16) = 145. ✓
A is therefore 145 ÷ 2 = 72.5 and the closest element is Germanium (Ge). ✓

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Question 7

[Total Marks: 11] Hard

An experiment was conducted to calculate x in Fe(NH₄)₂(SO₄)₂•xH₂O. A 3.54 g sample of this compound was dissolved in water and excess BaBr₂ _(aq) was added to precipitate ions. The precipitate containing BaSO₄ weighed 2.83 g when dried.

(a) Calculate the amount, in moles, of BaSO₄ in the precipitate. [1]

(b) Calculate the number of moles of sulfate in the 3.54 g of Fe(NH₄)₂(SO₄)₂•xH₂O. [2]

(c) State the quantity, in moles, of Fe in the sample. [1]

(d) Determine the mass of the following present in the sample of Fe(NH₄)₂(SO₄)₂•xH₂O. [3]
(i) Iron
(ii) Ammonium
(iii) Sulfate

(e) Using your previous answers, determine the number of moles of water present in the 3.54g. [2]

(f) Thus, calculate the value of x. [2]

Write out your answer in the box

(a) Moles = Mass ÷ M_r. So, moles of BaSO₄ is 2.83 ÷ 233 = 0.0121 mol. ✓

(b) The equation is Fe(NH₄)₂(SO₄)₂ + 2BaBr₂ → 2BaSO₄ + 2Fe(NH₄)Br₂ ✓
BaSO₄ and Fe(NH₄)₂(SO₄)₂ are in 2:1 ratio, so moles of Fe(NH₄)₂(SO₄)₂ is 0.012145 ÷ 2 = 0.00607.
So, the moles of sulfate is 0.00607 * 2 = 0.0121 mol. ✓

(c) There are twice as many moles of SO₄ as Fe, so moles of Fe is 0.0121 ÷ 2 = 0.00605 mol. ✓

(d) The moles of ammonium are the same as sulfate. Mass = Moles * M_r. So:
Mass of iron is 0.00607 * 56 = 0.339g. ✓
Mass of sulfate is 0.0121 * 96 = 1.16g. ✓
Mass of ammonium is 0.0121 * 19 = 0.230g. ✓

(e) Mass of water is 3.54 - 0.339 - 1.16 - 0.230 = 1.811g. ✓
Moles = Mass ÷ M_r. So, moles of water = 1.811 ÷ 18 = 0.101 mol. ✓

(f) Moles of water ÷ Fe(NH₄)₂(SO₄)₂ will give X. So, X is 0.101 ÷ 0.00605 = 16.7. ✓
X is therefore 17. ✓ (Allow 16.7)

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Question 8

[Total Marks: 5] Hard

To determine the percentage Cu in 3 samples of brass, each 0.748 g, a three-stage reaction pathway was utilised. This is shown below:

1) Cu_(s) + 2HNO₃ _(aq) + 2H⁺_(aq)   →   Cu²⁺_(aq) + 2NO₂ _(g) + 2H₂O_(l)

2) 4I^-_(aq) + 2Cu²⁺_(aq)   →   2CuI_(s) + I₂_(aq)

3) I₂_(aq) + 2S₂O₃ ^2-_(aq)   →   2I^-_(aq) + S₄O₆^2-_(aq)

	Titration Repeat
	1	2	3
Initial Volume of S₂O₃ ^2-	73.3	10.75	3.8
Final Volume of S₂O₃ ^2-	6.1083	0.7678	3.8
Volume of S₂O₃ ^2- added	8.04	1.011	5

(a) By using the data in the table and finding an appropriate average, calculate the average number of moles of 0.100 mol/dm³ S₂O₃ ^2- added in the final step. Assume all volumes are recorded in cm³. [2]

(b) Calculate the average percentage by mass of copper in the three samples of brass. [3]

Write out your answer in the box

(a) Average volume is 38.4 + 38.5 + 38.23 = 38.366. ✓
Moles = Concentration * Volume. So, average moles is 0.1 * 0.038366 = 0.00384 mol. ✓

(b) S₂O₃ ^2- and I₂ are in 2:1 ratio, so, the moles of I₂ = 0.00384 ÷ 2. However, I₂ and Cu²⁺ are in 1:2 ratio, so, moles of Cu²⁺ is (0.00384 ÷ 2) * 2, which = 0.00384.
Since Cu²⁺ and Cu are in 1:1 ratio, moles of Cu = 0.00384. ✓
Mass = Moles * M_r. So, mass is 0.00384 * 63.5 = 0.24384g ✓
0.24384 as a percentage of 0.748 is (0.24384 ÷ 0.748) * 100 = 32.5989 → 32.6% (3sf) ✓

Note: When working backwards through different reactions you assume all of the reactant comes from the product of the previous equation.

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