Page 58 - Spotlight A+ Form 4 & 5 Chemistry KSSM

Chemistry Answer
Paper 2 Element Carbon, C Hydrogen, H Oxygen, O
Section A
1. (a) To remove the layer of the magnesium oxide on its Mass (g) 54.55 9.09 36.36
surface.
(b) To allow the oxygen to enter the crucible for the Number of moles 54.55 9.09 36.36
complete combustion of magnesium. of atoms 12 1 16
(c) (i) = 4.546 = 9.09 = 2.273
Element Magnesium, Mg Oxygen, O
22.30 – 20.50 23.50 – 22.30 4.546 9.09 2.273
Mass (g) Simplest ratio 2.273 2.273 2.273
=1.80 = 1.2
= 2 = 4 = 1
Number of
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moles of 1.8 = 0.075 1.2 = 0.075 Thus, the empirical formula of compound Y is
atoms 24 16 C H O.
2
4
(ii) Assume that the molecular formula of Y
Simplest ratio 1 1 = (C H O) n
2
4
Thus the empirical formula of magnesium oxide is Given that the molecular mass of (C H O) = 88
2
n
4
MgO. n[2(12) + 4(1) + 16] = 88
(ii) 2Mg(s) + O (g) → 2MgO(s) 44n = 88
2
(d) (i) n = 88
44
Glass tube
Metal oxide W n = 2
Rubber Therefore, the molecular formula of compound Y
tubing
Ethanol, Glass tube = (C H O)
Air hole Spirit 2 4 2
C H OH = C H O
lamp 2 5 4 8 2
Glass tube
Hydrochloric Section B
Water acid 4. (a) (i) The molecular formula of a compound shows the
Zinc, Zn actual number of atoms of each element present in
a molecule of the compound.
granules (ii) The relative molecular mass of glucose
ANSWER FORM 4 2. (a) The empirical formula of a compound is the chemical (b) (i) Consider 100 g of compound Q. 2
Wooden block
= 6(12) + 12(1) + 6(16)

(ii) The hydrogen gas is allowed to flow through the
= 180
apparatus for 10 to 15 seconds to remove all the
A molecule of glucose consists of 6 carbon atoms,

air.
12 hydrogen atoms and 6 oxygen atoms.

The empirical formula of glucose is CH O.
formula that shows the simplest ratio of the number of
atoms of each element in the compound.
Element
hydrogen.
14.29%

85.71%
(ii) The heating, cooling and weighing steps are
(b) (i) Method I. Magnesium is more reactive than Percentage Carbon, C Hydrogen, H
repeated until a constant mass is obtained. Mass (g) 85.71 14.29
(c) (i) Mass of copper
= 61.71 – 55.31 = 6.4 g Number of 85 14.29
(ii) Number of moles of copper moles of atoms 12 1
= 6.4 = 0.1 mol = 7.1425 = 14.29
64
(iii) Mass of oxygen 7.1425 14.29
= 63.31 – 61.71 = 1.6 g Simplest ratio 7.1425 7.1425
(iv) Number of moles of oxygen = 1 = 2
= 1.6 = 0.1 mol Thus, the empirical formula Q is CH . 2
16
(v) 0.1 mol of copper atom combines with 0.1 mol (b) (ii) Assume that the empirical formula of Q = (CH )
2 n
of oxygen atom. Thus, 1 mol of copper atom = 42
combines with 1 mol of oxygen atom. Empirical Given that the relative molecular mass of
formula of copper oxide is CuO. (CH ) = 42
2 n
3. (a) The molecular formula is the chemical formula that n[12+ 2(1)] = 42
shows the actual number of atoms of each element 14n = 42
found in a molecule of a compound. n = 42
(b) C H 12 14
n = 3
6
(c) 6(12) + 12(1) = 84 Therefore, the molecular formula of compound Q

(d) CH 2
(e) (i) Consider 100 g of compound Y. = (CH )
2 3
= C H
3 6
546 547