# What is the empirical formula of a species that is 7.19% by mass with respect to phosphorus, and 92.81% by mass with respect to bromine? What is the molecular formula is the molecular mass is 431*g*mol^-1?

Jul 23, 2017

I think you have quoted a question from this site.

#### Explanation:

And this gives (i) microanalysis of P(7.19%), and Br(92.81%). You are lucky that I bothered to search for the question, as I am no computer buff, and these data should have been included with the question.......

For these data we determine the empirical formula......by assuming an $100 \cdot g$ mass of compound.

$\text{Moles of phosphorus}$ $=$ $\frac{7.19 \cdot g}{31 \cdot g \cdot m o {l}^{-} 1} = 0.232 \cdot m o l$.

$\text{Moles of bromine}$ $=$ $\frac{92.81 \cdot g}{79.9 \cdot g \cdot m o {l}^{-} 1} = 1.161 \cdot m o l$.

In each case I divided the elemental mass thru by the molar mass of the ELEMENT.

We divide thru by the lowest molar quantity (that of $P$), to get an empirical formula of $P : \frac{0.232}{0.232} = 1$and $B r : \frac{1.161}{0.232} = 5$.. to give an empirical formula, the simplest whole number ratio of constituent elements in a species, of $P B {r}_{5}$.

Now, we know that.........

{"empirical formula"}xxn="molecular formula", and from (ii) the molecular mass, we determine that.....

$\left\{31.00 + 5 \times 79.9\right\} \cdot g \cdot m o {l}^{-} 1 \times n = 431 \cdot g \cdot m o {l}^{-} 1$

Clearly, $n = 1$, and so.........

$\text{empirical formula" = "molecular formula} = P B {r}_{5}$