(a)(i) Law of constant composition: This law states that all pure samples of the same chemical compound contain the same elements combined in the same proportion by mass.
(ii) Law of multiple proportion: This law states that if two elements form more than one compound, then the different masses of one which combines with a fixed mass of the other are in the ratio of small whole numbers.

(b)(i) To determine the chemical formula of X
Mass of the oxide = 1.535g
Mass of copper = 1.365g
Mass of oxygen = 1.535 - 1.365 = 0.170
Number of moles of oxygen = \(\frac{0.170}{16}\) = 0.011
Number of moles of copper = \(\frac{1.365}{63.5}\) = 0.0211
Mole ratio of Cu : O = \(\frac{0.021}{ 0.011}\) : \(\frac{ 0.011}{0.011}\)
2 : 1
I. Chemica: formula of X = Cu\(_2\)O.
To determine the chemical formula of Y
Mass of the oxide = 1.450
Mass of copper = 1.160
Mass of oxygen = 1.450 - 1.160 = 0.290
Number of moles of Cu = \(\frac{1.160}{63.5}\) = 0.018
Number of moles of oxygen = \(\frac{0.290}{16}\) = 0.018
Mole ratio of Cu : O = \(\frac{0.018}{0.018}\) : \(\frac{0.018}{0.018}\)
1 : 1
II. Formila of Y = CuO

(ii) To calculate the mass of copper that can react with 0 500g of oxygen.
I. Mass of oxygen in X = 1.535 - 1.3
C5 = 0.170g.
If 0.170g of oxygen reacted with 1,365g
Cu 0.500g of oxygen will react with \(\frac{1.365 \times 0.500}{ 0.170}\) = 4.0g

II. Mass of oxygen in Y = 1.450 - 1.160 = 0.290
0.500g of oxygen = \(\frac{1.160 \times 0.500}{0.0290}\)
= 2.0g
(iii) Law of multiple proportion.

(c)(i) CH\(_3\)CH\(_2\)CH\(_2\)COOCH\(_2\)CH\(_3\)
(ii) CH\(_3\)(CH)CH\(_3\) OR CH\(_3\)CH(ONa)CH\(_3\)

(d)(i) Ethyl propanoate
(ii)


(iii) Propanoic acid and ethanol in the presence of concentated H\(_2\)SO\(_4\)/HCL and heat.