Answer:
[tex]\[C_3H_4\][/tex]
Explanation:
Since we are given the volume of the products at STP, we can use this to find the moles by dividing by 22.4 L since that is how much volume a gas occupies at a mole.
Solving:
[tex]}\subsection*{Moles:}At STP, 1 mole of any gas occupies 22.4 liters. Therefore:\[\text{Moles of } \text{CO}_2 = \frac{1.5 \text{ L}}{22.4 \text{ L/mol}} \approx \boxed{0.067} \text{ mol}\]\[\text{Moles of } \text{H}_2\text{O} = \frac{1.0 \text{ L}}{22.4 \text{ L/mol}} \approx \boxed{0.045} \text{ mol}\][/tex]
[tex]\hrulefill[/tex]
[tex]\subsection*{Mole Ratios:}The mole ratio of \(\text{CO}_2\) to \(\text{H}_2\text{O}\) is:\[\text{Ratio} = \frac{0.067 \text{ mol CO}_2}{0.045 \text{ mol H}_2\text{O}} \approx \boxed{1.5}\]The ratio is approximately \( \frac{3}{2} \).[/tex]
[tex]\hrulefill\subsection*{Balanced Reaction:}For each mole of \( C_xH_y \):\\- \(\text{CO}_2\) produced: \( x \) moles\\\\- \(\text{H}_2\text{O}\) produced: \( \frac{y}{2} \) moles\[C_xH_y + \text{O}_2 \rightarrow x~ \text{CO}_2 + \frac{y}{2} ~\text{H}_2\text{O}\][/tex]
[tex]\[\boxed{x = 3} \quad \text{and} \quad \frac{y}{2} = 2 \implies \boxed{y = 4}\][/tex]
[tex]\hrulefill[/tex]
Since the ratio was 3/2, we were able to find the empirical formula of the hydrocarbon to be: