A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm$^3$ of CO$_2$ and 54g of water. The hydrocarbon should be?

1. Analyze the question

We are given the volume of $CO_2$ produced and the mass of $H_2O$ produced from the combustion of a hydrocarbon ($C_xH_y$). We need to determine the empirical formula (and subsequently the molecular formula) of the hydrocarbon.

2. Explain the concept

• Combustion reaction: $C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$
• Mole calculation: • At s.t.p, 1 mole of any gas occupies $22.4\text{ dm}^3$. • Moles of $CO_2 = \frac{\text{Volume}}{22.4} = \frac{89.6}{22.4} = 4\text{ moles}$. • Moles of $H_2O = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{54}{18} = 3\text{ moles}$.
• Relationship: The carbon atoms in the hydrocarbon end up in $CO_2$, and the hydrogen atoms end up in $H_2O$.

3. Show the solution

Step 1: Find the moles of Carbon ($C$) Since each mole of $CO_2$ contains 1 mole of $C$: Moles of $C = 4\text{ moles}$.

Step 2: Find the moles of Hydrogen ($H$) Since each mole of $H_2O$ contains 2 moles of $H$: Moles of $H = 3 \times 2 = 6\text{ moles}$.

Step 3: Determine the ratio The ratio of $C:H$ is $4:6$. This gives us the empirical formula $C_4H_6$.

Step 4: Verify against options

• A. $C_6H_6$ (Ratio 1:1)
• B. $C_4H_{10}$ (Ratio 4:10)
• C. $C_5H_{10}$ (Ratio 1:2)
• D. $C_4H_6$ (Ratio 4:6)

The calculated ratio matches option D.

4. Identify the answer

The correct option is D.

5. Tip

When solving combustion problems, always remember:

1. Carbon atoms come only from $CO_2$.
2. Hydrogen atoms come only from $H_2O$ (remember to multiply the moles of water by 2 because there are two hydrogen atoms in one water molecule!).
3. If you get a fraction, multiply to get the simplest whole-number ratio.