Chapter 6, Problem 59A

Interpretation Introduction

Interpretation: The empirical and molecular formula of a compound containing carbon, hydrogen, oxygen and sulfur needs to be determined.

Concept Introduction:The empirical formula of a compound gives the simplest ratio of the numbers of atoms of each element present in the compound. The molecular formula of the compound gives the actual numbers of atoms each element present in the compound and is always a positive whole number multiple of the empirical formula.

Answer to Problem 59A

The empirical and the molecular formulas of the compound are ${\text{C}}_4{\text{H}}_8{\text{O}}_4\text{S}$ .

Explanation of Solution

The compound contains the same number of carbon and oxygen atoms and twice as many hydrogen atoms as the number of carbon atoms. The compound also contains sulfur atom(s) and the number of sulfur atom(s) is equal to one-eight of the number of hydrogen atoms.

Since atoms are indivisible, hence, the compound must contain atleast one sulfur atom and therefore, the atom ratio of hydrogen:sulfur must be $8:1$ .

Since the compound contains equal numbers of carbon and oxygen atoms, hence, the atom ratio of carbon and oxygen must be $1:1$ .

Again, there are twice as many hydrogen atoms as the number of carbon atoms and therefore, the atom ratio of carbon and hydrogen must be $1:2$ .

Considering all the above facts, the atom ratio must be

  $\text{C:H:O:S = 4:8:4:1}$ .

Therefore, the empirical formula of the compound is ${\text{C}}_4{\text{H}}_8{\text{O}}_4\text{S}$ .

The atomic masses are

C: $12.011\text{ g}{\text{.mol}}^{-1}$

H: $\text{1}\text{.008 g}{\text{.mol}}^{-1}$

O: $\text{15}\text{.999 g}{\text{.mol}}^{-1}$

S: $\text{32}\text{.065 g}{\text{.mol}}^{-1}$

  $\begin{array}{c}\text{Emperical formula mass =}\\ \left(4\times 12.011+8\times 1.008+4\times 15.999+1\times 32.065\right)\text{ g}{\text{.mol}}^{-1}\\ =152.169\text{ g}{\text{.mol}}^{-1}\end{array}$

Let the molecular formula of the compound be ${\left({\text{C}}_4{\text{H}}_8{\text{O}}_4\text{S}\right)}_{\text{n}}$ where n is a whole number.

The molecular mass of the compound is $\text{152 g}{\text{.mol}}^{-1}$ .

As per the problem,

  $\begin{array}{c}152.169\times \text{n g}{\text{.mol}}^{-1}=152\text{ g}{\text{.mol}}^{-1}\\ \Rightarrow \text{n =}\frac{\left(152\text{ g}{\text{.mol}}^{-1}\right)}{\left(152.169\text{ g}{\text{.mol}}^{-1}\right)}\\ \Rightarrow \text{n = 0}\text{.998889}\\ \Rightarrow \text{n}\approx \text{1}\end{array}$

Therefore, the molecular formula of the compound is ${\text{C}}_4{\text{H}}_8{\text{O}}_4\text{S}$ .

Conclusion

The empirical and the molecular formulas of the compound are ${\text{C}}_4{\text{H}}_8{\text{O}}_4\text{S}$ .