Question

A hydrocarbon containing C and H has 75% (by weight) of carbon. 32g of this hydrocarbon was completely burnt in air. What is the weight (in kg) of $CO_2$ formed?

• A. 0.044
• B. 0.088
• C. 44
• D. 88

Hint:

{Conservation of Mass (Carbon):} All Carbon in the reactant goes to $CO_2$. Mass of Carbon in 32g Hydrocarbon = $75%$ of 32 = 24g. Moles of Carbon = $24/12 = 2$ moles. Moles of $CO_2$ = 2 moles. Mass = $2 \times 44 = 88$g. This bypasses formula determination!

Answer 1

The Correct Option is B

Step 1: Determine the Empirical Formula:
Let the mass of hydrocarbon be 100g.

• Carbon: 75g. Moles of C = $75 / 12 = 6.25$.
• Hydrogen: $100 - 75 = 25g$. Moles of H = $25 / 1 = 25$.

Mole Ratio C : H = $6.25 : 25 = 1 : 4$. Empirical Formula is $CH_4$ (Methane). Molar Mass of $CH_4 = 12 + 4(1) = 16$ g/mol.
Step 2: Calculate Moles of Hydrocarbon Used:
Given Mass = 32g. Moles of $CH_4 = \frac{32}{16} = 2$ moles.
Step 3: Combustion Stoichiometry:
Reaction: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. From stoichiometry, 1 mole of $CH_4$ produces 1 mole of $CO_2$. Therefore, 2 moles of $CH_4$ will produce 2 moles of $CO_2$.
Step 4: Calculate Mass of $CO_2$:
Molar Mass of $CO_2 = 12 + 2(16) = 44$ g/mol. Mass of $CO_2$ produced = $2 \text{ moles} \times 44 \text{ g/mol} = 88$ g.
Step 5: Convert to kg:
Question asks for weight in kg. $88 \text{ g} = \frac{88}{1000} \text{ kg} = 0.088 \text{ kg}$.