Chapter 6, Problem 64A

Interpretation Introduction

Interpretation: The empirical formula of the compound must be found out for which C, H, N and O are 0.2990 g, 0.058499 g, 0.2318 g and 0.1328 g respectively.

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 64A

The empirical formula of the compound is C₃H₇N₂O.

Explanation of Solution

0.7221 g sample contains 0.2990 g, 0.058499 g, 0.2318 g and 0.1328 g C, H, N and O respectively Moles of C $=\frac{0.2990}{12}=\mathrm{0.025.}$

Moles of H $=\frac{0.058499}{1}=0.0585$

Moles of N $=\frac{0.2318}{14}=0.0165$

Moles of O $=\frac{0.1328}{16}=0.0083$

Smallest mole ratio C: H: N: O $\begin{array}{l}=\frac{0.025}{0.0083}:\frac{0.0585}{0.0083}:\frac{0.0165}{0.0083}:\frac{0.0083}{0.0083}\\ =3:7:2:1\end{array}$

Thus empirical formula is C₃H₇N₂O.

Conclusion

Empirical formula of the compound is found to be C₃H₇N₂O.