Question

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

Answer 1

`(CFBrO)` `xx` `3` = ??

Explanation:

Molar mass of compound (`381.01` `g`/`mol`) has been provided for you.

Now you need to calculate the molar mass of `CFBrO`.
The molar mass of:
`C = 12.0` `g`/`mol`
`F = 19.0` `g`/`mol`
`Br = 79.9` `g`/`mol`
`O = 16.0` `g`/`mol`

The molar mass of empirical formula, therefore (`CFBrO`) is:
`12.0 + 19.0 + 79.9 + 16.0` = `126.9` `g`/`mol`.

In order to get the molecular formula of `CFBrO`,
divide molar mass of compound by molar mass of empirical formula, like this: `"molar mass of compound"/" molar mass of empirical formula"` = `"whole number"`

i.e. `381.01/126.9` = `3.00`
Rounded to nearest whole number

Now, multiply the whole number by the Empirical Formula.
Like this `(CFBrO)` `xx` `3` = `C_3F_3Br_3O_3`.

Therefore your molecular formula is `C_3F_3Br_3O_3`.