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Rest of the question. A liquid with a mass of 0.0184 g is vaporized to a final volume of 5.58 mL at 45 ºC and standard pressure. Given that the compound’s empirical formula is CH2, determine the molar mass of the compound.
Given that the compound’s empirical formula is CH2, what is the molar mass of the compound?
回答 (1)
2016-11-04 2:39 am
For the vapor of the compound :
Mass, m = 0.0184 g
Volume, V = 5.58 mL = 0.00558 L
Absolute temperature, T = (273 + 45) K = 318 K
Pressure, P = 1 atm
Gas constant, R = 0.08206 L atm / (mol K)
Molar mass, M = ? g/mol
PV = nRT and n = m/M
Hence, PV = (m/M)RT
and thus M = mRT/(PV)
Molar mass, M = 0.0184 × 0.08206 × 318 / (1 × 0.00558) = 86.0 g/mol
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Additionally, let (CH₂)n be the molecular formula of the compound.
Molar mass of C = 12.0 g/mol
Molar mass of H = 1.0 g/mol
Molar mass of the compound (in g/mol) :
n (12.0 + 1.0×2) = 86.0
n ≈ 6
The molecular formula = C₆H₁₂
Mass, m = 0.0184 g
Volume, V = 5.58 mL = 0.00558 L
Absolute temperature, T = (273 + 45) K = 318 K
Pressure, P = 1 atm
Gas constant, R = 0.08206 L atm / (mol K)
Molar mass, M = ? g/mol
PV = nRT and n = m/M
Hence, PV = (m/M)RT
and thus M = mRT/(PV)
Molar mass, M = 0.0184 × 0.08206 × 318 / (1 × 0.00558) = 86.0 g/mol
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Additionally, let (CH₂)n be the molecular formula of the compound.
Molar mass of C = 12.0 g/mol
Molar mass of H = 1.0 g/mol
Molar mass of the compound (in g/mol) :
n (12.0 + 1.0×2) = 86.0
n ≈ 6
The molecular formula = C₆H₁₂
收錄日期: 2021-04-18 15:45:31
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https://hk.answers.yahoo.com/question/index?qid=20161103182311AAqF0yg