Chemistry experts pls solve?
2018-06-03 5:26 pm
Complete combustion of a sample of hydrocarbon X gave 0.66g of carbon dioxide and 0.36g of water. What is the empirical formula of X?
回答 (1)
2018-06-03 11:02 pm
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
n(C in X) = n(C in CO₂) = n(CO₂) = (0.66 g) / (44 g/mol) = 0.015 mol
n(H in X) = n(H in H₂O) = 2 × n(H₂O) = 2 × (0.36 g) / (18.0 g/mol) = 0.04 mol
Mole ratio C : H = 0.015 : 0.04 = 3 : 8
Empirical formula = C₃H₈
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
n(C in X) = n(C in CO₂) = n(CO₂) = (0.66 g) / (44 g/mol) = 0.015 mol
n(H in X) = n(H in H₂O) = 2 × n(H₂O) = 2 × (0.36 g) / (18.0 g/mol) = 0.04 mol
Mole ratio C : H = 0.015 : 0.04 = 3 : 8
Empirical formula = C₃H₈
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