Ahydrocarbon contains 92.3 grams of carbon and the rest is hydrogen. calculate its empirical formula and molecular formula. (mr = 78)

To find the Empirical Formula and Empirical Formula:

First note down the percentages of the elements in the given compound.

Here, as the compound is a hydrocarbon, we know it involves Carbon and Hydrogen bonded to each other.

Carbon's atomic mass = 12

Hydrogen's atomic mass = 1.008

The compound contains 92.3g of Carbon. It implies it has;

mass of hydrogen = 100 - 92.3 = 7.7g

To find the no.of gram atoms of an element we divide the percentage mass by atomic mass of the said element.

no. of gram atoms of Carbon =  92.3/12 = 7.6916 gram atoms of Carbon

Similarly for Hydrogen,

no. of gram atoms of hydrogen = 7.7/1.008 = 7.63 gram atoms of Hydrogen

Divide all the value of gram atoms by the smallest value among them.

Thus, to determine ratios:

7.69/7.63 = 1.008

Carbon and Hydrogen are in 1:1 ratio.

The ratio between number of atoms gives us the empirical formula. Hence, empirical formula of given compound is: CH.

Molecular mass of a compound = n(empirical formula mass)

Empirical Formula mass = (12 * 1 ) + (1.008 * 1) = 13.008

Molecular mass is given in the question. Therefore, the equation becomes:

78 = n(13.008)

n = 78/13.008

n = 5.99 ≅  6

Therefore, the molecular formula of the given hydrocarbon is: C₆ H₆

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