Acombustion analysis of a 0.44g sample of an unknown compound yields 0.88g g co2 and 0.36g h2o. if the sample’s molar mass is 132 g/mol, what is the molecular formula of the sample?

The molecular formula of the compound is C₆H₁₂O₃

We'll begin by calculating the mass of Carbon, hydrogen and oxygen in the compound.

Mass of CO₂ = 0.88 g

Molar mass of CO₂ = 44 g/mol

Molar mass of C = 12 g/mol

Mass of C = 12/44 × 0.88

Mass of H₂O = 0.36 g

Molar mass of H₂O = 18 g/mol

Molar mass of H₂ = 2 g/mol

Mass of H = 2/18 × 0.36

For oxygen, O

Mass of C = 0.24 g

Mass of H = 0.04 g

Mass of compound = 0.44 g

Mass of O = 0.44 – (0.24 + 0.04)

Mass of O = 0.44 – 0.28

Mass of O = 0.16 g

Next, we shall determine the empirical formula of the compound.

Mass of C = 0.24 g

Mass of H = 0.04 g

Mass of O = 0.16 g

C = 0.24 / 12 = 0.02

H = 0.04 / 1 = 0.04

O = 0.16 / 16 = 0.01

C = 0.02 / 0.01 = 2

H = 0.04 / 0.01 = 4

O = 0.01 / 0.01 = 1

Thus, the empirical formula of the compound is C₂H₄O

Finally, we shall determine the molecular formula of the compound.

Empirical formula = C₂H₄O

Molar mass of compound = 132 g/mol

Molecular formula = [C₂H₄O]ₙ = molar mass

[C₂H₄O]ₙ = 132

[(12×2) + (4×1) + 16]n = 132

[24 + 4 + 16]n = 132

44n = 132

Divide both side by n

n = 132 / 44

Molecular formula = [C₂H₄O]₃

Molecular formula = C₆H₁₂O₃

Thus, the molecular formula of the compound is C₆H₁₂O₃

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The molecular formula is C₆H₁₂O₃.

First, calculate the mass of each element .

Mass of C =0.88 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.240 g C

Mass of H = 0.36 g H₂O × (2.016 g H/18.02 g H₂O) = 0.0403 g H

Mass of O = Mass of compound - Mass of C - Mass of H

= (0.44 – 0.240 – 0.0403) g = 0.160 g

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

Element   Mass/g    Moles       Ratio    Integers  

     C         0.240      0.0200       2.00          2

     H        0.0403    0.0400       4.01            4

     O        0.160       0.009 97     1                 1

The empirical formula is C₂H₄O.

Calculate the molecular formula

EF Mass = 44.05 g/mol

MF mass = 132 g/mol

MF = (EF)ₙ

n = MF Mass/EF Mass = 132 g/mol/44.05 g/mol = 2.997 ≈ 3

MF = (C₂H₄O)₃ = C₆H₁₂O₃