(c) Another compound, $\boldsymbol{R}$, is from the same homologous series as isomer $\mathbf{1}$.
Compound $\boldsymbol{R}$ contains $62.1 \%$ carbon and $10.3 \%$ hydrogen by mass.
(i) Deduce the empirical formula of compound $\boldsymbol{R}$.
Show your working.
(ii) Deduce the molecular formula for compound $\mathbf{R}$.

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the moles of carbon and hydrogen**: Given the mass percentages of carbon (62.1%) and hydrogen (10.3%), we can assume a 100 g sample of compound $R$. Thus, the mass of carbon is 62.1 g and the mass of hydrogen is 10.3 g. The number of moles can be calculated using the molar masses: $ \text{Molar mass of C} = 12.01 \, \text{g/mol} $ and $ \text{Molar mass of H} = 1.008 \, \text{g/mol} $. The moles of carbon and hydrogen are calculated as follows: $$ \text{Moles of C} = \frac{62.1 \, \text{g}}{12.01 \, \text{g/mol}} \approx 5.17 \, \text{mol} $$ $$ \text{Moles of H} = \frac{10.3 \, \text{g}}{1.008 \, \text{g/mol}} \approx 10.21 \, \text{mol} $$ ‖ ‖ 2 ⋮ **Determine the simplest mole ratio**: To find the empirical formula, we divide the number of moles of each element by the smallest number of moles calculated. The smallest number of moles is approximately 5.17 (for carbon). Thus, we have: $$ \text{Ratio of C} = \frac{5.17}{5.17} = 1 $$ $$ \text{Ratio of H} = \frac{10.21}{5.17} \approx 1.97 \approx 2 $$ This gives us a ratio of 1:2 for carbon to hydrogen. Therefore, the empirical formula of compound $R$ is $ \text{CH}_2 $. ‖ ‖ 3 ⋮ **Calculate the molar mass of the empirical formula**: The molar mass of the empirical formula $ \text{CH}_2 $ is calculated as follows: $$ \text{Molar mass of CH}_2 = 12.01 \, \text{g/mol} + 2 \times 1.008 \, \text{g/mol} = 14.026 \, \text{g/mol} $$ ‖ ‖ 4 ⋮ **Determine the molecular formula**: To find the molecular formula, we need to know the molar mass of compound $R$. Since it is from the same homologous series as isomer 1, we can assume a molar mass of approximately 58 g/mol (as a common value for alkanes). We calculate the ratio of the molar mass of the compound to the empirical formula mass: $$ \text{Ratio} = \frac{58 \, \text{g/mol}}{14.026 \, \text{g/mol}} \approx 4.14 \approx 4 $$ Thus, the molecular formula is $ \text{C}_4\text{H}_8 $. ‖ ∻Answer∻ ⚹ Empirical formula: $ \text{CH}_2 $; Molecular formula: $ \text{C}_4\text{H}_8 $ ⚹ ∻Key Concept∻ ⚹ The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms. ⚹ ∻Explanation∻ ⚹ The empirical formula of compound $R$ is derived from the mass percentages of carbon and hydrogen, leading to $ \text{CH}_2 $. The molecular formula is determined based on the molar mass, resulting in $ \text{C}_4\text{H}_8 $. ⚹◊What is the relationship between the empirical formula and the molecular formula in terms of molecular weight?⍭ Generate me a similar question◊

Sia