# Question #b369f

##### 1 Answer

#### Explanation:

The idea here is that the molar mass of the compound will always be a *multiple* of the molar mass of its **empirical formula**.

In other words, if you multiply the molar mass of the empirical formula by a **whole number**, let's say

#("BH"_ 3) xx color(blue)(n) = "B"_ color(blue)(n)"H"_ (3color(blue)(n))#

So, you already know the molar mass of the empirical formula, so use the molar mass of boron and the molar mass of hydrogen to calculate the molar mass of the empirical formula--you can find the two molar masses in the Periodic Table.

#"BH"_3: " " "10.811 g mol"^(-1) + 3 xx "1.008 g mol"^(-1) = "13.835 g mol"^(-1)#

You can now set up your equation like this

#"13.835 g mol"^(-1) xx color(blue)(n) = "27.67 g mol"^(-1)#

This will get you

#color(blue)(n) = (27.67 color(red)(cancel(color(black)("g mol"^(-1)))))/(13.835color(red)(cancel(color(black)("g mol"^(-1))))) = 1.998 ~~ 2#

You can thus say that the molecular formula of the compound is

#("BH"_ 3)_ color(blue)(2) => "B"_2"H"_6 -> # diborane