Question

A student used combustion analysis to determine that a volatile organic compound was 52.14% carbon, 13.13% hydrogen and 34.73% oxygen. She used the Dumas method using 0.185 g of the compound and collected data at 90.5 C, 764.5 torr, with a flask volume of 121.0 mL. Determine the molar mass and the molecular formula of this compound.

Answer 1

Let the compound be 100 g

The amount of carbon = 52.14 g

Number of moles of carbon = 52.14 / 12.01 =4.34 moles

hydrogen = 13.13 g

Number of moles of hydrogen = 13.13 / 1.008 = 13.026

oxygen = 34.73 g

Number of moles of oxygen = 34.73 / 15.999 = 2.171

Dividing these number of moles by a common factor 2.171 , we get the ratio C:H:O = 2:6:1

So the empirical formula of the compound will be : C₂H₆O

Empirical mass = ( 2 x 12.01) + ( 6 x 1.008) + 15.999 = 46.07 g

Now using the ideal gas law , pV = nRT , number of moles of the compound can be determined ,

90.5^oC = 363.65 K

764.5 torr = 1.0059211 atm

121.0 mL = 0.121 L

Now putting all the values in the formula , we get ,

1.0059211 x 0.121 = n x 0.0821 x 363.65

n = 0.004 moles

So 0.004 moles of the compound weighs 0.185 g

The molar mass of the compound will be 0.185 / 0.004 = 46.25 g / mol

Now dividing the molar mass by empirical mass , we get : 46.25 / 46.07 = 1.004

So multiplying this with empirical formula suggests that the molecular formula of the compound is same as the empirical formula , i.e. C₂H₆O