Question: 1.When 4.626 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 14.51 grams of CO2 and 5.943 grams of H2O were produced. -Free Course Hero Question Answer.
Question Description: 1.When 4.626 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 14.51 grams of CO2 and 5.943 grams of H2O were produced.
In a separate experiment, the molar mass of the compound was found to be 70.13 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
- empirical formula =
- molecular formula =
2.When 2.991 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.858 grams of CO2 and 2.691 grams of H2O were produced.
In a separate experiment, the molar mass of the compound was found to be 40.06 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
- empirical formula =
- molecular formula =
Course Hero Answer & Explanation:
1)
Empirical formula = CH2
Molecular formula = C5H10
2)
Empirical formula = C3H4
Molecular formula = C3H4
Step-by-step explanation
1)
Moles CO2 = 14.51 g / 44 g/mol (molar mass) = 0.3298
moles H2O =5.943 g / 18 g/mol = 0.33 mol
From the balanced equation
CxHy + O2 —> CO2 + H2O
moles of carbon = moles of CO2
moles of carbon = 0.3298
moles of H = 2 moles H2O
moles of H = 2x 0.33 = 0.66
Mol ratio C:H = 0.32980.66 = 12
Empirical formula = CH2
molar mass of empirical formula = 2×1.01 + 12 = 14.02 g/mol
molecular factor = 14.0270.13 = 5
Molecular formula = 5(CH2)
Molecular formula = C5H10
2) moles C = CO2 = 9.858 g / 44g/mol = 0.224
moles of H = 2 x H2O = 2 x (2.691g/18g/mol) = 0.299
C:H ratio = 0.2240.299 =11.33
multiply by 3 to get integers
33.99≈34
Empirical formula = C3H4
empirical formula molar mass = 3×12.01 + 4×1.01 = 40.07 g/mol
molecular factor = 40.0740.06≈1
Molecular formula = C3H4