Final answer:
By performing a combustion analysis and using stoichiometry, the empirical formula of the unknown hydrocarbon is found to be C2H3. Without molar mass, only possible molecular formulas, which are multiples of the empirical formula, can be suggested.
Step-by-step explanation:
The combustion analysis of a hydrocarbon involves burning the compound and analyzing the products—namely, carbon dioxide (CO2) and water (H2O). For a complete combustion reaction of a hydrocarbon, CxHy, the general equation is:
CxHy + O2 → xCO2 + (y/2)H2O
From the problem, we can use the given masses of CO2 and H2O to determine the moles of carbon and hydrogen in the original hydrocarbon. The molecular formula of the hydrocarbon can be deduced using stoichiometry.
First, convert the mass of CO2 to moles to find the moles of carbon:
88.0 g CO2 × (1 mol CO2 / 44.0 g CO2) = 2.00 mol CO2
Since each mole of CO2 provides one mole of carbon, the sample contained 2.00 mol of carbon.
Then, convert the mass of H2O to moles to find the moles of hydrogen:
27.0 g H2O × (1 mol H2O / 18.0 g H2O) = 1.50 mol H2O
Each mole of H2O provides two moles of hydrogen, so the sample contained 3.00 mol of hydrogen.
Now, we can write the empirical formula of the hydrocarbon by dividing the moles of hydrogen and carbon by the smallest number:
C2.00H3.00 → C1H1.5
Since we cannot have half atoms, the empirical formula must be multiplied by two to get whole numbers:
C2H3
This is the empirical formula, but to identify the molecular formula, we need the molar mass of the compound or additional information. Since this is not provided, we can only suggest possible molecular formulas based on empirical formula, such as C2H6 (ethane), C4H6 (butadiene), etc., which must be multiples of the empirical formula.