A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 81.7 mg produced 222 mg of CO2 and 45.4 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Empirical formula= C4H20

Molecular formula = C8H4O2

Explanation:

It is given that 81.7 mg produced 222 mg of CO2 and 45.4 mg of H2O.

Note:Molar mass of C = 12g/mol

Molar mass of CO2 = 44g/mol

Therefore mass of C in 222mg of CO2= 12/44*(0.222)

=0.0605g

Note: Molar mass of H=1g/mol

Molar mass of H2O=18g/mil

Therefore, mass of H in 45.4mg of H2O= 1/18*(0.0454)

=0.0025g

Therefore, Mass of O = 0.0817-0.0025-0.0605

=0.0187g

Divide through by the atomic mass:

C=0.0605/12=0.00504

H=0.002/1=0.002

O=0.0187/16=0.00117

Divide through by the smallest number

C=0.00504/0.00117 = 4.3 ; Approx 4

H=0.002/0.00117=1.7 ; Approx 2

O=0.00117/0.00117=1

The empirical formula= C4H20

To calculate Molecular formula, (C4H20)n = 162

((12*4)+(1*2)+16)n=162

66n=162

n=162/66=2.45 ; Approx 2

Molecular formula = C8H4O2

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