What is the empirical formula for 28gN and 64gO?

Answer 1

28g Nitrogen and 64g Oxygen

Find their atomic masses N= 14 and O= 16

Then divide 28/14 and 64/16, this will give you 2 and 4

Then divide the 2 and 4 by the lowest answer which is 2. This will give you the ratio.

2/2 = 1 and 4/2 = 2.

so you will have NO2

That's all there is to it. Hope I helped :)

Answer 2

The molar mass of N is 14 g/mol and the molar mass of O is 16 g/mol. To find the moles of each element, divide the given mass by the respective molar mass. For N: 28 g / 14 g/mol = 2 mol. For O: 64 g / 16 g/mol = 4 mol. This gives an empirical formula of N2O4.