An 8.4g sample of hydrogen containing carbon and hydrogen atoms only contains 7.2g of carbon (C). (Relative atomic of hydrogen (H) = 1 and carbon (C) = 12). a. What is the empirical formula of this compound? b. If this compound has a molar mass of 84, what is its molecular formula?

Question

An 8.4g sample of hydrogen containing carbon and hydrogen atoms only contains 7.2g of carbon (C). (Relative atomic of hydrogen (H) = 1 and carbon (C) = 12). a. What is the empirical formula of this compound? b. If this compound has a molar mass of 84, what is its molecular formula?

Asked on March 26, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-26T12:36:56.689Z

Q11. An 8.4g sample of hydrogen (H, containing carbon and hydrogen atoms only) contains 7.2g of carbon (C). (Relative atomic mass of hydrogen (H) = 1 and carbon (C) = 12) a. What is the empirical formula of this compound? Step 1: Calculate the mass of hydrogen (H). The total mass of the compound is 8.4 g, and the mass of carbon is 7.2 g. $$ \text{Mass of H} = \text{Total mass of compound} - \text{Mass of C} \\ \text{Mass of H} = 8.4 \text{ g} - 7.2 \text{ g} \\ \text{Mass of H} = 1.2 \text{ g} $$ Step 2: Convert the mass of each element to moles. $$ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Relative atomic mass of C}} = \frac{7.2 \text{ g}}{12 \text{ g/mol}} = 0.6 \text{ mol} \\ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Relative atomic mass of H}} = \frac{1.2 \text{ g}}{1 \text{ g/mol}} = 1.2 \text{ mol} $$ Step 3: Find the simplest whole number ratio of moles. Divide the number of moles of each element by the smallest number of moles (which is 0.6 mol). $$ \text{Ratio of C} = \frac{0.6 \text{ mol}}{0.6 \text{ mol}} = 1 \\ \text{Ratio of H} = \frac{1.2 \text{ mol}}{0.6 \text{ mol}} = 2 $$ The simplest whole number ratio of C:H is 1:2. Step 4: Write the empirical formula. The empirical formula is $\boxed{\text{CH}_2}$. b. If this compound has a molar mass of 84, what is its molecular formula? Step 1: Calculate the empirical formula mass. Empirical formula = $\text{CH}_2$ $$ \text{Empirical formula mass} = (1 \times \text{Relative atomic mass of C}) + (2 \times \text{Relative atomic mass of H}) \\ \text{Empirical formula mass} = (1 \times 12) + (2 \times 1) \\ \text{Empirical formula mass} = 12 + 2 = 14 \text{ g/mol} $$ Step 2: Determine the ratio (n) between the molar mass and the empirical formula mass. Given molar mass = 84 g/mol. $$ n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} \\ n = \frac{84 \text{ g/mol}}{14 \text{ g/mol}} \\ n = 6 $$ Step 3: Multiply the subscripts in the empirical formula by n to get the molecular formula. Molecular formula = $(\text{Empirical formula})_n = (\text{CH}_2)_6$ $$ \text{Molecular formula} = \boxed{\text{C}_6\text{H}_{12}} $$

Accepted Answer · 2026-03-26T12:36:56.689Z

Q11. An 8.4g sample of hydrogen (H, containing carbon and hydrogen atoms only) contains 7.2g of carbon (C). (Relative atomic mass of hydrogen (H) = 1 and carbon (C) = 12) a. What is the empirical formula of this compound? Step 1: Calculate the mass of hydrogen (H). The total mass of the compound is 8.4 g, and the mass of carbon is 7.2 g. $$ \text{Mass of H} = \text{Total mass of compound} - \text{Mass of C} \\ \text{Mass of H} = 8.4 \text{ g} - 7.2 \text{ g} \\ \text{Mass of H} = 1.2 \text{ g} $$ Step 2: Convert the mass of each element to moles. $$ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Relative atomic mass of C}} = \frac{7.2 \text{ g}}{12 \text{ g/mol}} = 0.6 \text{ mol} \\ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Relative atomic mass of H}} = \frac{1.2 \text{ g}}{1 \text{ g/mol}} = 1.2 \text{ mol} $$ Step 3: Find the simplest whole number ratio of moles. Divide the number of moles of each element by the smallest number of moles (which is 0.6 mol). $$ \text{Ratio of C} = \frac{0.6 \text{ mol}}{0.6 \text{ mol}} = 1 \\ \text{Ratio of H} = \frac{1.2 \text{ mol}}{0.6 \text{ mol}} = 2 $$ The simplest whole number ratio of C:H is 1:2. Step 4: Write the empirical formula. The empirical formula is $\boxed{\text{CH}_2}$. b. If this compound has a molar mass of 84, what is its molecular formula? Step 1: Calculate the empirical formula mass. Empirical formula = $\text{CH}_2$ $$ \text{Empirical formula mass} = (1 \times \text{Relative atomic mass of C}) + (2 \times \text{Relative atomic mass of H}) \\ \text{Empirical formula mass} = (1 \times 12) + (2 \times 1) \\ \text{Empirical formula mass} = 12 + 2 = 14 \text{ g/mol} $$ Step 2: Determine the ratio (n) between the molar mass and the empirical formula mass. Given molar mass = 84 g/mol. $$ n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} \\ n = \frac{84 \text{ g/mol}}{14 \text{ g/mol}} \\ n = 6 $$ Step 3: Multiply the subscripts in the empirical formula by n to get the molecular formula. Molecular formula = $(\text{Empirical formula})_n = (\text{CH}_2)_6$ $$ \text{Molecular formula} = \boxed{\text{C}_6\text{H}_{12}} $$