Q3.35P - Chemistry: Molecular Nature Of Matter And Change

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

Step-by-Step Solution

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 1: 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,

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

Step 2: To calculate Molecular formula of compound having empirical formula CH 2 ( M w = 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 compound data-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 is data-custom-editor="chemistry" ${\mathrm{C}}_3{\mathrm{H}}_6$

Step 3: To calculate molecular formula of compound having empirical formula NH 2 ( M w = 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 ${\mathrm{C}}_2{\mathrm{H}}_4$

Step 4: To calculate Molecular formula of compound having empirical formula NO 2 ( M w = 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 compound data-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 is data-custom-editor="chemistry" ${\mathrm{N}}_2{\mathrm{O}}_4$

Step 5: To calculate Molecular formula of compound having empirical formula CHN ( M w = 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 compound data-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 is data-custom-editor="chemistry" ${\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{N}}_5$