(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.
(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.
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Correct Answer: Option n
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.
(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.