Question

What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen?

Answer 1

Answer: $N_2O_5$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of N = 25.92 g

Mass of O = 74.07 g

Step 1 : convert given masses into moles.

Moles of N =$\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{25.92g}{14g/mole}=1.85moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{74.07g}{16g/mole}=4.63moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For N = $\frac{1.85}{1.85}=1$

For O =$\frac{4.63}{1.85}=2.5$

The ratio of Fe : O= 1:2.5

Converting them into whole numbers, the ratio will be 2: 5

Hence the empirical formula is $N_2O_5$

Answer 2

N2O5

Explanation! 

When given %, assume you have 100 g of the substance. Find moles, divide by lowest count. In this case you'll end up with

25.92 g N14.01 g N/mol N=1.850 mol N

74.07 g O16.00 g O/mol O=4.629 mol O

The ratio between these is 2.502 mol O/mol N, which corresponds closely with N2O5.