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A 7.0 g sample of a hydrocarbon is subject to combustion analysis. The mass of CO2 collected is 22.0 g. What is the empirical formula of the compound? SHOW WORK
A 7.0 g sample of a hydrocarbon is subject to combustion analysis. The mass of CO2 collected is 22.0 g.?
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2017-11-16 1:46 am
Molar mass of H = 1.0 g/mol
Molar mass of C = 12.0 g/mol
Mass of 1 mole of CO₂ = (12.0 + 16.0×2) = 44.0 g
Mass of C in 1 mole of CO₂ = 12.0 g
Mass fraction of C in CO₂ = 12.0/44.0
Mass of C in 7.0 g of the hydrocarbon = Mass of C in CO₂ formed = (22.0 g) × (12.0/44.0) = 6.0 g
Mass of H in 7.0 g of the hydrocarbon = 1.0 g
In 7.0 g of the hydrocarbon :
No. of moles of C = (6.0 g) / (12.0 g/mol) = 0.5 mol
No. of moles of H = (1.0 g) / (1.0 g/mol) = 1 mol
Mole ratio C : H = 0.5 : 1 = 1 : 2
Empirical formula = CH₂
Molar mass of C = 12.0 g/mol
Mass of 1 mole of CO₂ = (12.0 + 16.0×2) = 44.0 g
Mass of C in 1 mole of CO₂ = 12.0 g
Mass fraction of C in CO₂ = 12.0/44.0
Mass of C in 7.0 g of the hydrocarbon = Mass of C in CO₂ formed = (22.0 g) × (12.0/44.0) = 6.0 g
Mass of H in 7.0 g of the hydrocarbon = 1.0 g
In 7.0 g of the hydrocarbon :
No. of moles of C = (6.0 g) / (12.0 g/mol) = 0.5 mol
No. of moles of H = (1.0 g) / (1.0 g/mol) = 1 mol
Mole ratio C : H = 0.5 : 1 = 1 : 2
Empirical formula = CH₂
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