Question

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

Answer 1

`BaBr_2`

Explanation:

Masses:
`Ba=2.677g`
`Br=3.115g`

Atomic Masses:

`Ba=137.327g/(mol)`
`Br=79.904g/(mol)`

Looking for the values in mol:
`Ba=2.677cancel(gBa)` `xx` `(1molBa)/(137.327cancel(gBa)` = `0.01949molBa`
`Br=3.115cancel(gBr)` `xx` `(1molBr)/(79.904cancel(gBr)`=`0.038983molBr`

Therefore:
`Ba` = `(0.01949mol)/(0.01949mol)`=1
`Br`=`(0.03898mol)/(0.01949mol)`=2

Empirical Formula`=BaBr_2`