Which homologous series of organic compounds can be represented by the following general formula? (i) CnH2n+1 (ii) CnH2n OH

Here are the solutions to questions 6, 7, and 8.

6. (a) Which homologous series of organic compounds can be represented by the following general formula?

• (i) $\text{C}_n\text{H}_{2n+2}$
  • This is the general formula for Alkanes.
• (ii) $\text{C}_n\text{H}_{2n}$
  • This is the general formula for Alkenes.
• (iii) $\text{C}_n\text{H}_{2n+1}\text{OH}$
  • This is the general formula for Alkanols (or Alcohols).

6. (b) Give the name of the first compound in each series.

• (i) $\text{C}_n\text{H}_{2n+2}$ (Alkanes)
  • For $n=1$, the compound is $\text{CH}_4$. The first compound is Methane.
• (ii) $\text{C}_n\text{H}_{2n}$ (Alkenes)
  • For $n=2$, the compound is $\text{C}_2\text{H}_4$. The first compound is Ethene.
• (iii) $\text{C}_n\text{H}_{2n+1}\text{OH}$ (Alkanols)
  • For $n=1$, the compound is $\text{CH}_3\text{OH}$. The first compound is Methanol.

6. (c) (i) Describe a reaction by which a named compound of series in (a)(i) converted to a compound of series in (a)(i).

• Reaction: Cracking (thermal decomposition)
• Description: A larger alkane (e.g., decane, $\text{C}_{10}\text{H}_{22}$) can be converted into smaller alkanes (e.g., octane, $\text{C}_8\text{H}_{18}$) and alkenes (e.g., ethene, $\text{C}_2\text{H}_4$) by heating it to high temperatures (e.g., 500-800 °C) in the absence of air, often with a catalyst. This process breaks longer carbon chains into shorter ones, producing smaller alkanes which are still part of the same homologous series.
• Example: $\text{C}_{10}\text{H}_{22} \xrightarrow{\text{heat/catalyst}} \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4$

6. (c) (ii) How can a compound of series (a)(iii) be converted to a compound of series (a)(i)?

• (a)(iii) is Alkanol (Alcohol), (a)(i) is Alkane.
• Description: An alkanol (e.g., ethanol, $\text{CH}_3\text{CH}_2\text{OH}$) can first be dehydrated to an alkene (e.g., ethene, $\text{CH}_2=\text{CH}_2$) by heating it with concentrated sulfuric acid or passing its vapor over hot aluminium oxide. The resulting alkene can then be converted to an alkane (e.g., ethane, $\text{CH}_3\text{CH}_3$) through catalytic hydrogenation, where hydrogen gas is added across the double bond in the presence of a catalyst like nickel.
• Equations: $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \text{heat}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$ $\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni catalyst, heat}} \text{CH}_3\text{CH}_3$

7. (a) Differentiate empirical formula from molecular formula.

• The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound.
• The molecular formula represents the actual number of atoms of each element in a molecule of the compound.
• For example, the molecular formula of glucose is $\text{C}_6\text{H}_{12}\text{O}_6$, while its empirical formula is $\text{CH}_2\text{O}$.

7. (b) Calculate the percentage composition by mass of water in a hydrated magnesium chloride, $\text{MgCl}_2 \cdot 6\text{H}_2\text{O}$. Step 1: Calculate the molar mass of $\text{MgCl}_2 \cdot 6\text{H}_2\text{O}$. Relative atomic masses: Mg = 24.3, Cl = 35.5, H = 1.0, O = 16.0 $\text{Molar mass of } \text{MgCl}_2 \cdot 6\text{H}_2\text{O} = (1 \times 24.3) + (2 \times 35.5) + 6 \times ((2 \times 1.0) + 16.0)$ $= 24.3 + 71.0 + 6 \times (18.0)$ $= 24.3 + 71.0 + 108.0$ $= 203.3 \text{ g/mol}$

Step 2: Calculate the total mass of water in one mole of the compound. $\text{Mass of water} = 6 \times (2 \times 1.0 + 16.0) = 6 \times 18.0 = 108.0 \text{ g}$

Step 3: Calculate the percentage composition by mass of water. $\text{Percentage by mass of water} = \frac{\text{Mass of water}}{\text{Molar mass of } \text{MgCl}_2 \cdot 6\text{H}_2\text{O}} \times 100\%$ $= \frac{108.0}{203.3} \times 100\%$ $= 53.12\%$ The percentage composition by mass of water in $\text{MgCl}_2 \cdot 6\text{H}_2\text{O}$ is \boxed{\text{*53.12%*}}.

7. (c) Calculate the empirical formula for a compound with the following composition: lead 1.28 g, sulphur 1.28 g, oxygen 2.56 g (relative atomic wt of lead = 207, sulphur = 32, oxygen = 16). Step 1: Convert mass to moles for each element. $\text{Moles of Pb} = \frac{1.28 \text{ g}}{207 \text{ g/mol}} \approx 0.006183 \text{ mol}$ $\text{Moles of S} = \frac{1.28 \text{ g}}{32 \text{ g/mol}} = 0.04 \text{ mol}$ $\text{Moles of O} = \frac{2.56 \text{ g}}{16 \text{ g/mol}} = 0.16 \text{ mol}$

Step 2: Divide by the smallest number of moles to get the simplest ratio. The smallest number of moles is for Pb ($0.006183 \text{ mol}$). $\text{Ratio of Pb} = \frac{0.006183}{0.006183} = 1$ $\text{Ratio of S} = \frac{0.04}{0.006183} \approx 6.469 \approx 6.5$ $\text{Ratio of O} = \frac{0.16}{0.006183} \approx 25.877 \approx 26$

Step 3: Multiply the ratios by a small whole number to get whole-number ratios. Since the ratio for S is approximately 6.5, we multiply all ratios by 2. $\text{Pb}: 1 \times 2 = 2$ $\text{S}: 6.5 \times 2 = 13$ $\text{O}: 26 \times 2 = 52$ The empirical formula is $\boxed{\text{*Pb}_2\text{S}_{13}\text{O}_{52}*}$.

8. (a) Ammonia gas can be prepared by heating an ammonium salt with an alkali.

• (i) Name the most common pair of reagents suitable for this reaction.
  • The most common pair of reagents is Ammonium chloride ($\text{NH}_4\text{Cl}$) and Calcium hydroxide ($\text{Ca(OH)}_2$).
• (ii) Write the equation for the reaction. $2\text{NH}_4\text{Cl}_{(\text{s})} + \text{Ca(OH)}_{2(\text{s})} \xrightarrow{\text{heat}} \text{CaCl}_{2(\text{s})} + 2\text{H}_2\text{O}_{(\text{l})} + 2\text{NH}_{3(\text{g})}$

8. (b) Ammonia is very soluble in water and less dense than air. How does each of these properties determine the way in which ammonia is collected in a gas jar?

• Very soluble in water: Because ammonia is very soluble in water, it cannot be collected by displacement of water, as it would dissolve.
• Less dense than air: Because ammonia is less dense than air, it is collected by upward displacement of air (or downward delivery). The gas jar is placed upright, and ammonia fills it from the top, pushing the denser air out from the bottom.

8. (c) Give reasons for the following:

• (i) Solution of chlorine in water is acidic.
  • When chlorine gas dissolves in water, it reacts to form hydrochloric acid ($\text{HCl}$) and hypochlorous acid ($\text{HOCl}$). Both of these are acidic compounds, which makes the resulting solution acidic.
  • $\text{Cl}_{2(\text{g})} + \text{H}_2\text{O}_{(\text{l})} \rightleftharpoons \text{HCl}_{(\text{aq})} + \text{HOCl}_{(\text{aq})}$
• (ii) Yellow phosphorus is stored under water.
  • Yellow phosphorus is highly reactive and spontaneously ignites in air (oxidizes rapidly) at room temperature. Storing it under water prevents its contact with atmospheric oxygen, thereby preventing it from catching fire.