Chapter 6, Problem 51A

Interpretation Introduction

Interpretation: Empirical and molecular formula of the hydrocarbon must be found for which the % of C is 81.71 and molar mass is 44.1 g/mol.

Concept Introduction :

Empirical formula of a molecule is the smallest unit which is repeated in the molecule.

Molecular formula of a compound is the representation of the molecule with the number of different atoms present in the molecule.

Molecular formula $\text{=empirical formula×n}$ where n is any positive integer like 1, 2, 3 and so on.

Answer to Problem 51A

The empirical formula is C₃H₈and molecular formula is also C₃H₈.

Explanation of Solution

The molecular formula can be found from the empirical formula using the following relationship.

Molecular formula $\text{=empirical formula×n}$ where n is any positive integer like 1, 2, 3 and so on.

% of C $=81.71$ and % of H is $(100-81.71)=\mathrm{18.29.}$

Moles of C $=\frac{81.71}{12}=6.8$ and moles of H is $=\frac{18.29}{1}=18.29$ .

Smallest mole ratio C: H $=\frac{6.8}{6.8}:\frac{18.29}{6.8}=1:2.68\approx 3:8$ .

Thus empirical formula is C₃H₈.

Now molar mass 44.1= (C₃H₈) _n or, n $=\frac{44.1}{(3\times 12+8\times 1)}=\frac{44.1}{44}\approx 1.$

Thus molecular formula is C₃H₈.

Conclusion

Molecular formula as well as empirical formula of the compound is C₃H₈.