Q3.33P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the empirical formula and empirical formula mass for each of the following compounds?

a) ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}$

b) ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}$

c) ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}$

d) ${\mathbf{Ba}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}$

e) ${\mathbf{Te}}_{\mathbf{4}}{\mathbf{I}}_{\mathbf{16}}$

Step-by-Step Solution

Answer

(a) Empirical formula is${\mathrm{CH}}_2$

          Empirical formula mass is $14.03\mathrm{g}/\mathrm{mol}$

(b) Empirical formula is ${\mathrm{CH}}_3\mathrm{O}$

     Empirical formula mass is $31.04\mathrm{g}/\mathrm{mol}$

(c) Empirical formula is ${\mathrm{N}}_2{\mathrm{O}}_5$

       Empirical formula mass is $108.02\mathrm{g}/\mathrm{mol}$

(d) Empirical formula is ${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$

       Empirical formula mass is $601.93\mathrm{g}/\mathrm{mol}$

(e) Empirical formula is ${\mathrm{TeI}}_4$

             Empirical formula mass is $635.20\mathrm{g}/\mathrm{mol}$

Step 1: Defining Empirical formula and Empirical formula mass

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

Empirical formula mass is calculated by multiplying each of the constituents’ subscript by its own atomic weight on the periodic table and then adding them together.

Step 2: To calculate Empirical formula and Empirical formula mass of C 2 H 4

The ratio of carbon atoms to hydrogen atoms is given as 2 : 4.

This empirical formula is ${\mathrm{CH}}_2$

The empirical formula (EF) mass is given as,

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

Step 3: To calculate Empirical formula and Empirical formula mass of C 2 H 6 O 2

The ratio of carbon atoms to hydrogen atoms to oxygen atoms is given as 2 : 6 : 2

This ratio can be reduced to 1 : 3 : 1

This empirical formula is ${\mathrm{CH}}_3\mathrm{O}$

The empirical formula (EF) mass is given as,

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}{\mathrm{CH}}_3\mathrm{O}& =& {\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{C}+3\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{H}\right)+{\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{O}\\ & =& 12.01\mathrm{g}/\mathrm{mol}+3\times \left(1.01\mathrm{g}/\mathrm{mol}\right)+16\mathrm{g}/\mathrm{mol}\\ & =& 31.04\mathrm{g}/\mathrm{mol}\end{array}$

Step 4: To calculate Empirical formula and Empirical formula mass of N 2 O 5

The ratio of nitrogen atoms to oxygen atoms is given as 2 : 5 and this cannot be further reduced.

This empirical formula is ${\mathrm{N}}_2{\mathrm{O}}_5$

The empirical formula (EF) mass is given as,

 $\begin{array}{rcl}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}{\mathrm{N}}_2{\mathrm{O}}_5& =& 2\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{N}\right)+5\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{O}\right)\\ & =& 2\times \left(14.01\mathrm{g}/\mathrm{mol}\right)+5\times \left(16.00\mathrm{g}/\mathrm{mol}\right)\\ & =& 108.02\mathrm{g}/\mathrm{mol}\end{array}$

Step 5: To calculate Empirical formula and Empirical formula mass of Ba 3 ( P O 4 ) 2

The ratio of Ba atoms to P atoms to O atoms is given as 3 : 8 : 2 and this cannot be further reduced.

This empirical formula is${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$

The empirical formula (EF) mass is given as,

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}{\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2& =& 3\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{Ba}\right)+2\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{P}\right)+8\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{O}\right)\\ & =& 3\times \left(137.33\mathrm{g}/\mathrm{mol}\right)+2\times \left(30.97\mathrm{g}/\mathrm{mol}\right)+8\times \left(16\mathrm{g}/\mathrm{mol}\right)\\ & =& 601.93\mathrm{g}/\mathrm{mol}\end{array}$

Step 6: To calculate Empirical formula and Empirical formula mass of Te 4 I 16

The ratio of Te atoms to I atoms 4 : 16 and this can be further reduced to 1 : 4

This empirical formula is${\mathrm{TeI}}_4$

The empirical formula (EF) mass is given as,

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}{\mathrm{TeI}}_4& =& {\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{Te}+4\times \left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{I}\right)\\ & =& 127.60\mathrm{g}/\mathrm{mol}+4\times \left(126.90\mathrm{g}/\mathrm{mol}\right)\\ & =& 635.20\mathrm{g}/\mathrm{mol}\end{array}$