Mod 1B: Moles Test
1)
%
A sample of element B was analysed in a mass spectrometer. The relative atomic
mass of element B can be calculated from the results shown in the table below.
Relative isotopic mass
Percentage composition
i)
/32
Isotope 1
58.0
68.2
Isotope 2
60.0
27.3
Isotope 3
62.0
4.5
Explain what is meant by the relative atomic mass of element B. [3]
the mass of 1 mole of atoms of an element (1)
compared with 12g (1) of carbon-12 (1)
or
the weighted mean mass of an atom of an element (1) compared with 1/12th of
the mass of an atom of carbon (1) – 12 (1)
ii) Using the information in the table, calculate the relative atomic mass of this sample of
B. Give your answer to 3 significant figures. [2]
(58x68.2/100) + (60x27.3/100) + (62x4.5/100) (1) = 58.7 (3sf) (1)
2)
a)
b)
Some dry – cleaning solvents include the chlorine compound perc.
Perc has the following percentage composition by mass: Cl = 85.6%
C = 14.4%
The relative molecular mass of perc is 166.
Calculate the molecular formula of perc. [3]
Cl 85.6/35.5 = 2.4 (1) 2:1 ratio Cl:C (1)  CCl2 (mass of 86) x 2 = 166
C
14.4/12
= 1.2
C2Cl4 (1)
Sodium chlorate, NaClO3, is a chlorine compound used as a weed killer.
When heated, NaClO3 releases oxygen gas.
2NaClO3(s)

2NaCl(s)
+
3O2(g)
Calculate the volume of O2 that can be formed at room temperature and pressure by
heating 4.26 g of NaClO3. [4]
1 mol of gas molecules occupies 24.0 dm3 at room temperature and pressure.
Mr(NaClO3) = 106.5 gmol-1 (1)
Moles NaClO3 = mass/Mr 4.26/106.5 = 0.04 moles (1)
Moles O2 = Ratio 2:3 NaClO3 : O2  0.06 moles (1)
Volume O2 = 0.06 x 24 = 1.44 (dm3) (1) no ratio = 0.96 (worth 3 marks)
3)
Farmers add ‘lime’, Ca(OH)2 to acid soils. A chemist neutralised 25.0cm3 0.200mol
dm-3 HCl with slaked lime.
Ca(OH)2(s)
a)
+
2HCl(aq)

CaCl2(aq)
What is the molar mass of Ca(OH)2? [1]
74.1 (g mol-1)
+
2H2O(l)
b)
How many moles of HCl were neutralised? [1]
0.005 moles
c)
Calculate the mass of Ca(OH)2 that neutralises this HCl. [2]
Moles of Ca(OH)2 is 2:1 ratio HCl:Ca(OH)2 = 0.5 x 0.005 = 0.0025 moles (1)
Mass = 0.0025 x 74.1 = 0.185g (1) 1 mark without if ratio missed out
d)
The chemist neutralised the same amount of HCl with NaOH. Explain why the chemist
would need to use more moles of NaOH than Ca(OH)2. [2]
Ca(OH)2 has more OH’s (1)
therefore twice the number of moles of NaOH needed (1) ORA
(could use reactions to explain using mole ratios (2)
4)
Industrially, chlorine, CI2(g), is prepared by passing an electric current through a
concentrated solution of sodium chloride, NaCI, known as brine.
In this question, assume brine has a concentration of NaCl of 4.00 mol dm -3.
a)
Calculate the mass of NaCl dissolved in 1.00 dm3 of brine. [2]
Mr NaCl = 58.5 g mol-1 (1)
Mass of NaCl dissolved = 58.5 x 4 = 234g (1)
b)
Calculate the volume of CI2(g) obtained from 1.00dm3 of brine. [2]
2NaCl(aq)
+
2H2O(I)

CI2(g)
+
2NaOH(aq)
+
H2(g)
Assume that 1 mole of CI2(g) occupies 24.0 dm3.
1dm3 of brine contains 4 moles of NaCl
Mole ratio 2:1 NaCl : Cl2
2 Moles of Cl2 made (1) or half answer of (i)
Volume = 2 x 24dm3 = 48dm3 of gas
5)
Magnesium reacts with oxygen to form magnesium oxide.
a) Write a balanced chemical equation showing the reaction, include state symbols
[2]
2Mg(s) +
O2(g) 
2MgO(s) balanced equation (1) / state symbols – (1)
b) Use oxidation numbers to show that magnesium has been oxidised and oxygen has
been reduced. [4]
Mg
0  +2 (1)
oxidation is loss of electrons (1)
O
0  -2 (1)
reduction is gain of electrons (1)
6)
The formula of a hydrated salt is ZnSO4.xH2O. A student heated the 8.985g of the
crystal in an oven for 24 hours. The final mass was 5.047g.
Find x in the formula and show all your working out. [4]
Works out mass of water = 3.938g (1)
Works out moles of water = 0.218 (1)
Works out moles of salt = 0.031 (1) Works out a value for x = 7 (1)