Chapter 6, Problem 50A

Interpretation Introduction

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

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

Answer to Problem 50A

The empirical and the molecular formula of the compound are ${\text{C}}_2{\text{H}}_2\text{ON}$ and ${\text{C}}_6{\text{H}}_6{\text{O}}_3{\text{N}}_3$ .

Explanation of Solution

The mass percentages of carbon, hydrogen, oxygen and nitrogen in the compound are given as

C = $42.87\%$

H = $3.598\%$

O = $28.55\%$

N = $25.00\%$

Therefore, $100.00\text{ g}$ of the compound contains $\text{42}\text{.87 g}$ carbon, $\text{3}\text{.598 g}$ hydrogen, $\text{28}\text{.55 g}$ oxygen and $\text{25}\text{.00 g}$ nitrogen.

The atomic masses are

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

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

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

N: $14.007\text{ g}{\text{.mol}}^{-1}$

  $\begin{array}{c}\text{Moles carbon =}\frac{\left(42.87\text{ g}\right)}{\left(12.011\text{ g}{\text{.mol}}^{-1}\right)}\\ =3.569\text{ mol}\end{array}$

  $\begin{array}{c}\text{Moles hydrogen =}\frac{\left(\text{3}\text{.598 g}\right)}{\left(1.008\text{ g}{\text{.mol}}^{-1}\right)}\\ =3.569\text{ mol}\end{array}$

  $\begin{array}{c}\text{Moles oxygen =}\frac{\left(\text{28}\text{.57 g}\right)}{\left(15.999\text{ g}{\text{.mol}}^{-1}\right)}\\ =1.786\text{ mol}\end{array}$

  $\begin{array}{c}\text{Moles nitrogen =}\frac{\left(\text{25}\text{.00 g}\right)}{\left(14.007\text{ g}{\text{.mol}}^{-1}\right)}\\ =1.785\text{ mol}\end{array}$

Simplest ratio of the numbers of moles (equal to the numbers of atoms) of each element in one mole

  $\begin{array}{l}\left(3.569\text{ mol}\right):\left(3.569\text{ mol}\right):\left(1.786\text{ mol}\right):\left(1.785\text{ mol}\right)\\ =\left(\frac{3.569\text{ mol}}{1.785\text{ mol}}\right):\left(\frac{3.569\text{ mol}}{1.785\text{ mol}}\right):\left(\frac{\text{1}\text{.786 mol}}{1.785\text{ mol}}\right):\left(\frac{\text{1}\text{.786 mol}}{1.785\text{ mol}}\right)\\ =1.999:1.999:1.000:1.000\\ \approx 2:2:1:1\end{array}$

Therefore, the empirical formula of the compound contains 2 moles of carbon and hydrogen each and 1 mole each of oxygen and nitrogen. Therefore, the empirical formula of the compound is ${\text{C}}_2{\text{H}}_2\text{ON}$ .

  $\begin{array}{c}\text{Emperical formula mass =}\\ \left(2\times 12.011+2\times 1.008+1\times 15.999+1\times 14.007\right)\text{ g}{\text{.mol}}^{-1}\\ =56.044\text{ g}{\text{.mol}}^{-1}\end{array}$

Let the molecular formula of the compound be ${\left({\text{C}}_2{\text{H}}_2\text{ON}\right)}_{\text{n}}$ where n is a whole number.

The molecular mass of the compound lies between 165 and 170 grams. It is quite evident from the empirical formula of the compound that only when n is equal to 3, the empirical formula mass will lie between 165-170 grams.

Therefore, the molecular formula of the compound is ${\text{C}}_6{\text{H}}_6{\text{O}}_3{\text{N}}_3$ .

Conclusion

The empirical and the molecular formula of the compound are ${\text{C}}_2{\text{H}}_2\text{ON}$ and ${\text{C}}_6{\text{H}}_6{\text{O}}_3{\text{N}}_3$ .