# What is the empirical formula of a barium salt that contains 17.14*g of the metal, and 8.86*g of chlorine?

Mar 26, 2017

You mean the $\text{empirical formula} \ldots \ldots \ldots . .$ and here it is $B a C {l}_{2}$.

#### Explanation:

You mean the $\text{empirical formula} \ldots \ldots \ldots . .$

The $\text{empirical formula}$ is the simplest whole number defining constituent atoms in a species. And it is calculated by taking percentage mass of each element, and dividing thru by atomic mass, and then modifying these percentages to give simple whole numbers.

And thus %Ba $=$ (17.14*g)/(17.14*g+8.86*g)xx100%=65.9%

And thus %Cl $=$ (8.86*g)/(17.14*g+8.86*g)xx100%=34.1%

You will note of course that the percentages sum to 100%. Why $\text{of course?}$

And then we assume $100 \cdot g$ of compound and divide thru by the atomic masses of each constituent:

$\text{Moles of barium}$ $=$ $\frac{65.9 \cdot g}{137.3 \cdot g \cdot m o {l}^{-} 1} = 0.480 \cdot m o l .$

$\text{Moles of chlorine}$ $=$ $\frac{34.10 \cdot g}{35.5 \cdot g \cdot m o {l}^{-} 1} = 0.960 \cdot m o l .$

If we divide thru by the smallest molar quantity, CLEARLY, we get an empirical formula of $B a C {l}_{2}$...........