# What is the empirical formula for the compound which is comprised of 2.677 g Ba and 3.115 g Br?

Feb 27, 2016

$B a B {r}_{2}$

#### Explanation:

Masses:
$B a = 2.677 g$
$B r = 3.115 g$

Atomic Masses:

$B a = 137.327 \frac{g}{m o l}$
$B r = 79.904 \frac{g}{m o l}$

Looking for the values in mol:
$B a = 2.677 \cancel{g B a}$ $\times$ (1molBa)/(137.327cancel(gBa) = $0.01949 m o l B a$
$B r = 3.115 \cancel{g B r}$ $\times$ (1molBr)/(79.904cancel(gBr)=$0.038983 m o l B r$

Therefore:
$B a$ = $\frac{0.01949 m o l}{0.01949 m o l}$=1
$B r$=$\frac{0.03898 m o l}{0.01949 m o l}$=2