(a) State the following laws of chemical combination: (i) Law of constant composition (ii) Law of multiple’ proportion.
(b) Copper reacts with oxygen to form two oxides X and Y. On analysis, 1.535 g of X yielded 1.365 g copper and 1.450 g of Y yielded 1.160 g of cooper.
i) Determine the chemical formula of X and Y.
(ii) Calculate the mass of copper which can react with 0.500 g of oxygen to yield I. X II. Y.
(iii) Which of the laws of chemical combination is illustrated by the result in (b)(i) above. [ = 16, Cu = 63.51]
(c) Write the structure of the product responsible for the observation in each of the following reactions:
(i) A mixture of butanoic acid and ethanol warmed in the presence of concentrated H(_2)SO(_4) gives off a fragrant odour.
(ii) Sodium dissolves in propan-2-ol with effervescence to give a solution which on evaporation to dryness leaves a white precipitate.
(d) Consider the compound CH(_3)CH(_2)COOCH(_2)CH(_3).
(i) Name the compound (ii) Write the structural formula of the compound (iii) State the reagents and conditions for the formation of the compound.
Explanation
(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.