# A substance consisting only of Na, B and H is 60.80% Na and 28.60% B. What is the empirical formula of the substance?

Oct 7, 2016

$N a B {H}_{4}$

#### Explanation:

As is standard with these sorts of problems, we assume a $100 \cdot g$ mass of compound to calculate the empirical formula:

$N a : \frac{60.80 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.65 \cdot m o l$

$B : \frac{28.60 \cdot g}{10.81 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.65 \cdot m o l$

$H : \frac{10.60 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $=$ $10.52 \cdot m o l$

And then we divide thru each mass by the ATOMIC mass of each constituent to give a molar ratio.

How did I know that there were $10.6 \cdot g$ hydrogen? The question did not give me this percentage composition.

And now we divide thru by the smallest molar quantity, i.e. $2.65 \cdot m o l$, to give an empirical formula of:

$N a B {H}_{4}$

This is sodium borohydride, a very common laboratory reagent.