Kirk Balk Academy
Y11 Revision Chemistry Topic 3 – Relative Formula Mass
Q1.Formulae and equations are used to describe chemical reactions.
(a)Aluminium reacts with sulfuric acid (H2SO4) to produce aluminium sulfate, Al2(SO4)3 and hydrogen (H2).
Complete and balance the equation for this reaction.
...... Al + ...... ...... +......
(2)
(b)Calcium carbonate reacts with nitric acid to produce calcium nitrate.
Calculate the relative formula mass (Mr) of calcium nitrate, Ca(NO3)2
Relative atomic masses (Ar): N = 14; O = 16; Ca = 40
......
......
Relative formula mass (Mr) = ......
(2)
(c)Zinc carbonate decomposes when heated.
A student heated 25 g zinc carbonate (ZnCO3).
The figure shows how he set up the apparatus.
The balanced chemical equation for the decomposition reaction is:
ZnCO3 (s) ZnO (s) + CO2 (g)
The student measured the mass of solid product after heating until there was no further change in mass.
The student did the experiment four times. The table below shows the results.
Experiment / 1 / 2 / 3 / 4Mass of solid product in g / 17.4 / 19.7 / 17.6 / 16.9
Calculate the mean mass of the solid product.
Do not use any anomalous results in your calculation.
......
......
Mean mass = ...... g
(2)
(Total 6 marks)
Q2.Iron is an essential part of the human diet. Iron(II) sulfate is sometimes added to white bread flour to provide some of the iron in a person’s diet.
(a) The formula of iron(II) sulfate is FeSO4
Calculate the relative formula mass (Mr) of FeSO4
Relative atomic masses: O = 16; S = 32; Fe = 56.
......
......
The relative formula mass (Mr) = ......
(2)
(b) What is the mass of one mole of iron(II) sulfate? Remember to give the unit.
......
(1)
(c) What mass of iron(II) sulfate would be needed to provide 28 grams of iron?
Remember to give the unit.
......
(1)
(Total 4 marks)
Q3. A chemist was asked to identify a nitrogen compound. The chemist carried out an experiment to find the relative formula mass (Mr) of the compound.
The Mr of the compound was 44.
Relative atomic masses: N = 14, O = 16
Draw a ring around the formula of the compound.
NONO2N2O4N2O
(1)
(Total 1 mark)
Q4. Calcium carbonate tablets are used to treat people with calcium deficiency.
(a) Calculate the relative formula mass (Mr) of calcium carbonate.
Relative atomic masses: C = 12; O = 16; Ca = 40......
......
Relative formula mass = ......
(2)
(Total 2 marks)
Q5. Iron ore contains iron oxide.
(i) Calculate the relative formulamass of iron oxide, Fe2O3.
Relative atomic masses: O = 16; Fe = 56.
......
Answer = ......
(2)
(Total 2 marks)
Q6.The main ingredient in Aqamed is a painkiller called paracetamol.
The figure below represents a molecule of paracetamol.
Give the molecular formula of paracetamol.
Calculate its relative formula mass (Mr).
Relative atomic masses (Ar): H = 1; C = 12; N = 14; O = 16
Molecular formula......
Relative formula mass......
......
Mr = ......
(2)
(d)Aspirin is a medicine for use by adults.
An aspirin tablet contains 300 mg of acetylsalicylic acid.
Calculate the number of moles of acetylsalicylic acid in one aspirin tablet.
Give your answer in standard form to three significant figures.
Relative formula mass (Mr) of aspirin = 180
......
......
......
......
Number of moles = ......
(4)
(Total 6 marks)
M1.(a) 2Al + 3H2SO4 Al2(SO4)3 + 3H2
formulae correct
1
balancing correct
1
(b) 40 + 2(14 + (3 × 16))
1
= 164
allow 164 with no working shown for 2 marks
1
(c) (17.4 + 17.6 + 16.9) / 3
1
= 17.3
allow 17.3 with no working shown for 2 marks
1
[6]
M2. (a) 152 correct answer with or without working = 2 marks
56 + 32 + (4 ×16) gains 1 mark
ignore any units
2
(b) 152g(rams)
ecf from the answer to (a) and g
must have unit g / gram / gramme / grams etc
accept g / mol or g per mole or g mole–1or g/mol or g per mol or g mol–1
do not accept g m
do not accept G
1
(c) 76(g)
ecf from their answer to (a) or (b) divided by 2
ignore units
1
[4]
M3. (a) N2O
1
[3]
M4. (a) 100
ignore units
40 + 12 + (3 × 16) for 1 mark
1
M5. (i) 160 ignore units
(2 × 56) + (3 × 16) for 1 mark
2
M6. (c) C8H9NO2
any order of elements
1
151
1
(d) mass of acetylsalicylic acid = 0.3 g
1
=
method mark – divide mass by Mr
1
= 0.00167 (mol)
allow 0.0016666(66)
1
1.67 × 10-3 (mol)
correct answer with or without working scores 4 marks
allow ecf from steps1, 2 and 3
1
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