Compound containing C & H is combusted. From a 20.0 gram of compound 67.5 g of CO2 & 13.8 g of H2O are obtained. What is empirical formula?
回答 (1)
2018-05-20 1:11 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
Moles of C in compound = Moles of CO₂ = (67.5 g) / (44.0 g/mol) = 1.53 mol
Mass of H in compound = (Moles of H₂O) × 2 = [(13.8 g) / (18.0 g/mol)] × 2 = 1.53 mol
Mole ratio C : H = 15.3 : 15.3 = 1 : 1
Empirical formula = CH
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Moles of C in compound = Moles of CO₂ = (67.5 g) / (44.0 g/mol) = 1.53 mol
Mass of H in compound = (Moles of H₂O) × 2 = [(13.8 g) / (18.0 g/mol)] × 2 = 1.53 mol
Mole ratio C : H = 15.3 : 15.3 = 1 : 1
Empirical formula = CH
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