# What is the molecular formula of a compound if its empirical formula is CFBrO and its molar mass is 381.01 g/mol?

May 24, 2016

$\left(C F B r O\right)$$\times$ $3$ = ??

#### Explanation:

Molar mass of compound ($381.01$ $g$/$m o l$) has been provided for you.

Now you need to calculate the molar mass of $C F B r O$.
The molar mass of:
$C = 12.0$ $g$/$m o l$
$F = 19.0$ $g$/$m o l$
$B r = 79.9$ $g$/$m o l$
$O = 16.0$ $g$/$m o l$

The molar mass of empirical formula, therefore ($C F B r O$) is:
$12.0 + 19.0 + 79.9 + 16.0$ = $126.9$ $g$/$m o l$.

In order to get the molecular formula of $C F B r O$,
divide molar mass of compound by molar mass of empirical formula, like this: $\text{molar mass of compound"/" molar mass of empirical formula}$ = $\text{whole number}$

i.e. $\frac{381.01}{126.9}$ = $3.00$
Rounded to nearest whole number

Now, multiply the whole number by the Empirical Formula.
Like this $\left(C F B r O\right)$$\times$ $3$ = ${C}_{3} {F}_{3} B {r}_{3} {O}_{3}$.

Therefore your molecular formula is ${C}_{3} {F}_{3} B {r}_{3} {O}_{3}$.