(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.
(a)(i) Law of constant composition: a pure compound always contains the same elements combined in the same fixed proportion by mass.
(ii) Law of multiple proportions: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.
(b) For X: oxygen \(= 1.535 - 1.365 = 0.170\) g. For Y: oxygen \(= 1.450 - 1.160 = 0.290\) g.
(i) X: Cu \(= \dfrac{1.365}{63.5} = 0.0215\), O \(= \dfrac{0.170}{16} = 0.0106\); ratio Cu:O \(\approx 2:1\), so X = Cu2O.
Y: Cu \(= \dfrac{1.160}{63.5} = 0.0183\), O \(= \dfrac{0.290}{16} = 0.0181\); ratio Cu:O \(\approx 1:1\), so Y = CuO.
(ii) Mass of copper combining with 0.500 g oxygen:
I. In X: \(\dfrac{1.365}{0.170} \times 0.500 = \mathbf{4.01\ g}\).
II. In Y: \(\dfrac{1.160}{0.290} \times 0.500 = \mathbf{2.00\ g}\).
(iii) The result (4.01 g and 2.00 g of copper combining with the same 0.500 g of oxygen, a 2:1 ratio) illustrates the law of multiple proportions.
(c)(i) The fragrant odour is from the ester ethyl butanoate: CH3CH2CH2COOCH2CH3.
(ii) The white solid left is sodium propan-2-oxide (sodium isopropoxide): (CH3)2CHONa, i.e. CH3CH(ONa)CH3.
(d)(i) CH3CH2COOCH2CH3 is ethyl propanoate.
(ii) Structural formula: CH3CH2COOCH2CH3.
(iii) Reagents/conditions: propanoic acid and ethanol, warmed with a little concentrated H2SO4 as catalyst (esterification, under reflux).
(a)(i) Law of constant composition: a pure compound always contains the same elements combined in the same fixed proportion by mass.
(ii) Law of multiple proportions: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.
(b) For X: oxygen \(= 1.535 - 1.365 = 0.170\) g. For Y: oxygen \(= 1.450 - 1.160 = 0.290\) g.
(i) X: Cu \(= \dfrac{1.365}{63.5} = 0.0215\), O \(= \dfrac{0.170}{16} = 0.0106\); ratio Cu:O \(\approx 2:1\), so X = Cu2O.
Y: Cu \(= \dfrac{1.160}{63.5} = 0.0183\), O \(= \dfrac{0.290}{16} = 0.0181\); ratio Cu:O \(\approx 1:1\), so Y = CuO.
(ii) Mass of copper combining with 0.500 g oxygen:
I. In X: \(\dfrac{1.365}{0.170} \times 0.500 = \mathbf{4.01\ g}\).
II. In Y: \(\dfrac{1.160}{0.290} \times 0.500 = \mathbf{2.00\ g}\).
(iii) The result (4.01 g and 2.00 g of copper combining with the same 0.500 g of oxygen, a 2:1 ratio) illustrates the law of multiple proportions.
(c)(i) The fragrant odour is from the ester ethyl butanoate: CH3CH2CH2COOCH2CH3.
(ii) The white solid left is sodium propan-2-oxide (sodium isopropoxide): (CH3)2CHONa, i.e. CH3CH(ONa)CH3.
(d)(i) CH3CH2COOCH2CH3 is ethyl propanoate.
(ii) Structural formula: CH3CH2COOCH2CH3.
(iii) Reagents/conditions: propanoic acid and ethanol, warmed with a little concentrated H2SO4 as catalyst (esterification, under reflux).
