Question 1 Report
(a)(i) What is 2 homologous series?
(ii) Give two homologous series present in petroleum
(iii) Give one example of a compound belonging to each of the homologous series in (a)(ii).
(iv) Name two fractions obtained from the fractional distillation of petroleum
(v) Why is there a gradual change in the physical properties of petroleum fractions?
b) Write a two-step balanced chemical equation for the reaction of (i) ethanol with excess concentration traoxosulphate (VI) acid at high temperature
(ii) excess ethanol with concentrated tetraoxosulphate (VI) acid at lower temperature.
(c) An organic compound of relative molecular mass 46, on analysis vas ound to contain 52.0% carbon, 13.3% hydrogen and 34.7% oxygen.
(i) Determine its I. empirical formula, II. molecular formula.
(ii) Draw two possible structures of the compound and name one of them [O = 16; C = 12; H = 1 ]
(a)(i) A homologous series is a family of organic compounds having the same general formula and similar chemical properties, in which successive members differ by a \(\mathrm{CH_2}\) unit.
(ii) Two homologous series present in petroleum are:
(iii) Examples are:
(iv) Two fractions obtained from the fractional distillation of petroleum are petrol (gasoline) and kerosene.
(v) There is a gradual change in physical properties because the fractions contain hydrocarbons with gradually increasing molecular masses and carbon-chain lengths. Consequently, the intermolecular forces and boiling points increase from one fraction to the next.
(b)(i) Ethanol with excess concentrated tetraoxosulphate(VI) acid at high temperature:
At about \(170^\circ\mathrm{C}\), ethanol is dehydrated to ethene in two steps:
(ii) Excess ethanol with concentrated tetraoxosulphate(VI) acid at lower temperature:
At about \(140^\circ\mathrm{C}\), diethyl ether is formed in two steps:
(c)(i)(I) Empirical formula
| Element | C | H | O |
|---|---|---|---|
| Percentage composition | 52.0 | 13.3 | 34.7 |
| Divide by atomic mass | \(52.0/12=4.33\) | \(13.3/1=13.3\) | \(34.7/16=2.17\) |
| Divide by 2.17 | 2 | 6 | 1 |
Empirical formula = \(\boxed{\mathrm{C_2H_6O}}\).
(c)(i)(II) Molecular formula
Empirical formula mass = \((2\times12)+(6\times1)+16=46\).
\[n=\frac{\text{relative molecular mass}}{\text{empirical formula mass}}=\frac{46}{46}=1\]
Molecular formula = \((\mathrm{C_2H_6O})_1=\boxed{\mathrm{C_2H_6O}}\).
(c)(ii) Two possible structural formulae of \(\mathrm{C_2H_6O}\)
The first structure is ethanol. The second structure is methoxymethane, also called dimethyl ether.
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