# Question #fb2b9

Aug 2, 2017

The empirical formula is ${N}_{2} {O}_{5}$

#### Explanation:

The empirical formula is the smallest integer ratio of the atoms in the compound. It may or may not be the same as the molecular formula.

Pretend that we have 100 g of the unknown compound. The % amount of nitrogen and oxygen adds up to 100%, so there are no other elements present in the compound.

First, take the ratio of masses of each element:

$N : O$

$25.9 : 74.1$

Now convert this to moles by dividing each mass by the elements' respective molar mass, because that tells us the ratio of how many particles of each atom we have:

$\left(\frac{25.9}{14}\right) : \left(\frac{74.1}{16}\right) \Rightarrow 1.85 : 4.63$

Next, we need to simplify this ratio to whole numbers. To do this, divide both numbers by the smallest number:

$\frac{1.85}{1.85} : \frac{4.63}{1.85} \Rightarrow 1 : 2.50$

Finally, these numbers must be whole numbers (because we can't have a fraction of an atom) so multiply everything in the ratio by the lowest possible integer to make everything a whole number:

$2 \cdot \left(1 : 2.50\right) = 2 : 5$

So, the empirical formula is ${N}_{2} {O}_{5}$