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, C_xH_y, 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, C_xH_y, 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 =

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