Hydrocarbon Chemistry?
2017-12-18 11:25 am
An unknown hydrocarbon gave 0.07484 g of CO2 and 0.03063 g of H2O on combustion. What is the empirical formula of this unknown hydrocarbon?
回答 (1)
2017-12-18 4:35 pm
Mass fraction of C in CO₂ = (12.0 g/mol) / [(12.0 + 16.0×2) g/mol] = 12.0/44.0
Mass fraction of H in H₂O = (1.0×2 g/mol) / [(1.0×2 + 16.0) g/mol] = 2.0/18.0
Mass of C in the compound = Mass of C in CO₂ formed = (0.07484 g) × (12.0/44.0) = 0.02041 g
Mass of H in the compound = Mass of H in H₂O formed = (0.03053 g) × (2.0/18.0) = 0.003392 g
Mole ratio C : H = (0.02041/12.0) : (0.003392/1.0) = 0.001701 : 0.003392 = 1 : 2
Hence, empirical formula = CH₂
Mass fraction of H in H₂O = (1.0×2 g/mol) / [(1.0×2 + 16.0) g/mol] = 2.0/18.0
Mass of C in the compound = Mass of C in CO₂ formed = (0.07484 g) × (12.0/44.0) = 0.02041 g
Mass of H in the compound = Mass of H in H₂O formed = (0.03053 g) × (2.0/18.0) = 0.003392 g
Mole ratio C : H = (0.02041/12.0) : (0.003392/1.0) = 0.001701 : 0.003392 = 1 : 2
Hence, empirical formula = CH₂
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https://hk.answers.yahoo.com/question/index?qid=20171218032538AAQ04uc