Page 13 - Q & A STPM 2022 Chemistry

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Solving, x = 75.0%
Percent abundance of X = 75.0%
35
Percent abundance of X = 25.0%
37
3
m
m
3
9
(iii) — 70: X X + — 70: — × — = —–
35
35
e
e
4
4
16
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m
1
m
6
3
— 72: X X + — 72: 2 × — × — = —–
37
35
e
e
4
16
4
1
m
1
m
1
— 74: X X + — 74: — × — = —–
37
37
e
e
16
4
4
Therefore, the mass spectrum of molecule X is as below:
10
Term
8
1 Relative 6
abundance 4 9
2 6
1
70 72 74 m/e
Section C Essay Questions
Question 10
(a) The four naturally occurring stable isotopes of iron are given in the table
below.
Relative atomic mass 53.94 55.93 56.94 57.93
Relative abundance/% 5.82 91.80 2.10 0.28
Determine the relative atomic mass of iron.
(b) Due to the differing proportions of isotopes, the relative atomic mass of
iron depends on its source. Two samples of iron were analysed. The mass
spectrum of the first sample was taken and found to be the same as in the
above table. When 1.12 g of the second sample was converted to iron(III)
chloride, the mass of the product is 3.25 g. [Relative atomic mass: Cl = 35.5]
(i) Write an equation for the conversion of iron to iron(III) chloride.
(ii) How many moles of chlorine was used in the reaction?
(iii) Determine the relative atomic mass of iron from this experiment.
(iv) Compare the relative atomic mass of iron with the value obtained in (a).
Decide whether the two samples have identical isotopic compositions.
Answer:
(a) Relative atomic mass of Fe
53.94(5.82) + 55.93(91.80) + 56.94(2.10) + 57.93(0.28)
= ————————–—————————————–
100
= 55.84
8
01 STPM Q&A Chem T1 layout.indd 8 4/19/19 9:28 AM