Question

A sample of pure compound is found to have Na = 0.0887 mole, O = 0.132 mole, C = 2.65×$10^{22}$ atoms. What is the empirical formula of the compound?

Answer 1

The “empirical formula” is formulated by calculating the nearest whole number ratio of the atoms or moles present in a compound. In order to formulate the empirical formula, one to follow the following steps,

1. The first step involves the conversion of mass to a number of moles in the element.

2. The next step involves the division of the number of moles present in each element by the "smallest value" and then rounding that value to the nearest "whole number".

3. The rounded value represents the ratio of the moles present in the element.

4. This value is then represented in subscript along with the element's chemical symbol.

In the above case,  

No. of moles in each element:

Na = 0.0887 mol

O = 0.132 mol

$C\quad =\quad 2.65\quad \times \quad { 10 }^{ 22 }\quad =\quad 0.0441\quad mol$

Division and rounding of steps:

$Na:\quad \frac { 0.0887 }{ 0.0441 } \quad =\quad 2.01\quad \approx \quad 2O:\quad \frac { 0.132 }{ 0.0441 } \quad =\quad 2.99\quad \approx \quad 3C:\quad \frac { 0.0441 }{ 0.0441 } \quad =\quad 1$

Adding the numbers to the elements we get:

${ Na }_{ 2 }{ CO }_{ 3 }$

Empirical Formula: ${ Na }_{ 2 }{ CO }_{ 3 }$