Question

What is the molecular formula for each of the following compounds?

a) Empirical formula${\mathbf{CH}}_{\mathbf{2}}\left(M_w=42.08g/\mathrm{mol}\right)$

b) Empirical formula${\mathbf{NH}}_{\mathbf{2}}\left(M_w=32.05g/\mathrm{mol}\right)$

c) Empirical formula${\mathbf{NO}}_{\mathbf{2}}\left(M_w=92.02g/\mathrm{mol}\right)$

d) Empirical formula$\mathbf{CHN}\left(M_w=135.14g/\mathrm{mol}\right)$

Short answer

1. Molecular formula of ${\mathbf{CH}}_{\mathbf{2}}$is${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{6}}$
2. Molecular formula of ${\mathbf{NH}}_{\mathbf{2}}$is ${\mathbf{N}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}$
3. Molecular formula of ${\mathbf{NO}}_{\mathbf{2}}$is${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}$
4. Molecular formula of $\mathbf{CHN}$is${\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{N}}_{\mathbf{5}}$

Step-by-step solution

Defining Empirical formula and Molecular formula

Empirical formula is given by the simplest number ratio of the number of different atoms present in the compound.

Molecular formula gives the actual number of different atoms present in the same molecule.

The whole number multiple given as the ratio of molar mass to the empirical formula mass of each element deduces molecular formula.

The whole number multiple is given as,

$\mathrm{whole}-\mathrm{number}\mathrm{multiple}=\frac{\mathrm{molar}\mathrm{mass}(\mathrm{g}/\mathrm{mol})}{\mathrm{empirical}\mathrm{formula}\mathrm{mass}(\mathrm{g}/\mathrm{mol})}$

To calculate Molecular formula of compound having empirical formulaCH2(Mw=42.08 g/mol)

The empirical formula (EF) mass is given as,

$\begin{array}{c}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}{\mathrm{CH}}_2={\mathrm{M}}_{\mathrm{w}}\left(\mathrm{C}\right)+2\times \left({\mathrm{M}}_{\mathrm{w}}\right(\mathrm{H}\left)\right)\\ =12.01\mathrm{g}/\mathrm{mol}+2(1.01\mathrm{g}/\mathrm{mol})\\ =14.03\mathrm{g}/\mathrm{mol}\end{array}$

The whole-number multiple for compounddata-custom-editor="chemistry" ${\mathrm{CH}}_2$

$\mathrm{whole}-\mathrm{number}\mathrm{multiple}=\frac{42.03(\mathrm{g}/\mathrm{mol})}{14.03(\mathrm{g}/\mathrm{mol})}\approx 3$

Multiplying subscripts of data-custom-editor="chemistry" ${\mathrm{C}}_1{\mathrm{H}}_2$by 3, the molecular formula is given by

$={\mathrm{C}}_{(1\times 3)}\times {\mathrm{H}}_{\text{(}2\times 3)}$

Thus, the molecular formula isdata-custom-editor="chemistry" ${\mathrm{C}}_3{\mathrm{H}}_6$

To calculate molecular formula of compound having empirical formula NH2(Mw=32.05 g/mol)

The empirical formula (EF) mass is given as,

$\begin{array}{c}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}{\mathrm{NH}}_2={\mathrm{M}}_{\mathrm{w}}\left(\mathrm{N}\right)+2\times \left({\mathrm{M}}_{\mathrm{w}}\right(\mathrm{H}\left)\right)\\ =14.01\mathrm{g}/\mathrm{mol}+2(1.01\mathrm{g}/\mathrm{mol})\\ =16.03\mathrm{g}/\mathrm{mol}\end{array}$

The whole- number multiple for compound${\mathrm{NH}}_2$

$\mathrm{whole}-\mathrm{number}\mathrm{multiple}=\frac{32.05(\mathrm{g}/\mathrm{mol})}{16.03(\mathrm{g}/\mathrm{mol})}\approx 2$

Multiplying subscripts of ${\mathrm{N}}_1{\mathrm{H}}_2$by 2, the molecular formula is given by

$={\mathrm{N}}_{(1\times 2)}\times {\mathrm{H}}_{\text{(}2\times 2)}$

Thus, the molecular formula is${\mathrm{C}}_2{\mathrm{H}}_4$

To calculate Molecular formula of compound having empirical formulaNO2(Mw=92.02 g/mol)

The empirical formula (EF) mass is given as,

$\begin{array}{c}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}{\mathrm{NO}}_2={\mathrm{M}}_{\mathrm{w}}\left(\mathrm{N}\right)+2\times \left({\mathrm{M}}_{\mathrm{w}}\right(\mathrm{O}\left)\right)\\ =14.01\mathrm{g}/\mathrm{mol}+2(16.00\mathrm{g}/\mathrm{mol})\\ =46.01\mathrm{g}/\mathrm{mol}\end{array}$

The whole- number multiple for compounddata-custom-editor="chemistry" ${\mathrm{NO}}_2$

$\mathrm{whole}-\mathrm{number}\mathrm{multiple}=\frac{92.02(\mathrm{g}/\mathrm{mol})}{46.01(\mathrm{g}/\mathrm{mol})}\approx 2$

Multiplying subscripts of data-custom-editor="chemistry" ${\mathrm{N}}_1{\mathrm{O}}_2$by 2, the molecular formula is given by

$={\mathrm{N}}_{(1\times 2)}\times {\mathrm{O}}_{\text{(}2\times 2)}$

Thus, the molecular formula isdata-custom-editor="chemistry" ${\mathrm{N}}_2{\mathrm{O}}_4$

To calculate Molecular formula of compound having empirical formulaCHN(Mw=135.14 g/mol)

The empirical formula (EF) mass is given as,

$\begin{array}{c}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{CHN}={\mathrm{M}}_{\mathrm{w}}\left(\mathrm{C}\right)+{\mathrm{M}}_{\mathrm{w}}\left(\mathrm{H}\right)+{\mathrm{M}}_{\mathrm{w}}\left(\mathrm{N}\right)\\ =12.01\mathrm{g}/\mathrm{mol}+1.01\mathrm{g}/\mathrm{mol}+14.01\mathrm{g}/\mathrm{mol}\\ =27.03\mathrm{g}/\mathrm{mol}\end{array}$

The whole-number multiple for compounddata-custom-editor="chemistry" $\mathrm{CHN}$

$\mathrm{whole}-\mathrm{number}\mathrm{multiple}=\frac{135.14(\mathrm{g}/\mathrm{mol})}{27.03(\mathrm{g}/\mathrm{mol})}\approx 5$

Multiplying subscripts of data-custom-editor="chemistry" ${\mathrm{C}}_1{\mathrm{H}}_1{\mathrm{N}}_1$by 5, the molecular formula is given by

$={\mathrm{C}}_{(1\times 5)}\times {\mathrm{H}}_{(1\times 5)}\times {\mathrm{N}}_{(1\times 5)}$

Thus, the molecular formula isdata-custom-editor="chemistry" ${\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{N}}_5$