更新1:
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What is the empirical formula of the compound?
2017-10-20 11:50 pm
Combustion analysis of 0.150 g of an unknown compound containing carbon, hydrogen, and oxygen produces 0.343 g CO2 and 8.76×10−2g H2O.
回答 (1)
2017-10-21 12:04 am
Molar masses in g/mol : C = 12.0, H = 1.0, O = 16.0
Mass of C in 0.150 g compound = Mass of C in CO₂ = (0.343 g) × [12.0/(12.0 + 16.0×2)] = 0.0935 g
Mass of H in 0.150 g compound = Mass of H in H₂O = (8.76 × 10⁻² g) × [1.0×2/(1.0×2 + 16.0)] = 0.00973 g
Mass of O in 0.150 g compound = (0.150 - 0.0935 - 0.00973) g = 0.0468 g
No. of moles of C in 0.150 g compound = (0.0935 g) / (12.0 g/mol) = 0.00779 mol
No. of moles of H in 0.150 g compound = (0.00973 g) / (1.0 g/mol) = 0.00973 mol
No. of moles of O in 0.150 g compound = (0.0468 g) / (16.0 g/mol) = 0.00293 mol
Mole ratio C : H : O
= 0.00779 : 0.00973 : 0.00293
= 0.00779/0.00293 : 0.00973/0.00293 : 0.00293/0.00293
= 2.66 : 3.32 : 1
= 2.66×3 : 3.32×3 : 1×3
≈ 8 : 10 : 3
Empirical formula = C₈H₁₀O₃
Mass of C in 0.150 g compound = Mass of C in CO₂ = (0.343 g) × [12.0/(12.0 + 16.0×2)] = 0.0935 g
Mass of H in 0.150 g compound = Mass of H in H₂O = (8.76 × 10⁻² g) × [1.0×2/(1.0×2 + 16.0)] = 0.00973 g
Mass of O in 0.150 g compound = (0.150 - 0.0935 - 0.00973) g = 0.0468 g
No. of moles of C in 0.150 g compound = (0.0935 g) / (12.0 g/mol) = 0.00779 mol
No. of moles of H in 0.150 g compound = (0.00973 g) / (1.0 g/mol) = 0.00973 mol
No. of moles of O in 0.150 g compound = (0.0468 g) / (16.0 g/mol) = 0.00293 mol
Mole ratio C : H : O
= 0.00779 : 0.00973 : 0.00293
= 0.00779/0.00293 : 0.00973/0.00293 : 0.00293/0.00293
= 2.66 : 3.32 : 1
= 2.66×3 : 3.32×3 : 1×3
≈ 8 : 10 : 3
Empirical formula = C₈H₁₀O₃
收錄日期: 2021-04-18 17:53:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171020155043AA2AKNM