A compound containing nitrogen and oxygen was found in pop. What would the empirical formula of the compound be if there was 21g of nitrogen to every 16g of oxygen in the compound?

Question

What would the molecular formula of the compound be if it was found that the molar mass of the compound was 148 g/mol?

Answer 1

Convert g of each element to moles using its molar mass.

mol N = 21 g N x (1 mol N / 14 g N) = 1.5 mol N

mol O = 16 g O x (1 mol O / 16 g O) = 1.0 mol O

Divide the moles of each element by the lowest # of moles (1.0) to obtain the mole ratio of each element in the compound.

N = 1.5 mol / 1.0 mol = 1.5 mol N

O = 1.0 mol / 1.0 mol = 1.0 mol O

We need to multiply the moles of N and O by 2 since we need the mole ratios to equal a whole #.

N = 1.5 x 2 = 3 mol N

O = 1.0 x 2 = 2 mol O

The empirical formula is therefore, N₃O₂.

To obtain the molecular formula, first determine the empirical formula mass of N₃O₂.

N₃ = 3 x 14 = 42 g/mol

O₂ = 2 x 16 = 32 g/mol

EFM = 42 + 32 = 74 g/mol

Since the molar mass of the compound is given (148 g/mol), we can calculate a value called the "multiple", which will provide the exact # of moles of each element in the compound.

multiple = molar mass / empirical formula mass = 148 / 74 = 2

The molecular formula is therefore, 2 x (N₃O₂) = N₆O₄

Answer 2

Hi Megan

You address this problem using moles. For the ratios, 21 grams of Nitrogen x 1 mole/14 gr, yields 1.5 moles

Similarly 16 grams of Oxygen yields one mole. So the ratio of Nitrogen to Oxygen is 1.5 to 1. or 3 to 2.

The empirical formula is N3O2.

For the second question, consider that one EMPIRICAL mole would measure 3*14 + 2*16 = 74 grams.

But the molar mass of the compound is 148 grams. Since that is twice the empirical mass, The actual formula would need to be double the empirical formula . N6O4