What's the empirical formula of 36.48% Na 25.41% S 38.11% O

A compound is found to contain 36.48% Na, 25.41% S, and 38.11% O. find its empirical formula. % in such problems are by weight, & can be changed to grams out of a 100 grams total using molar mass, find moles:

36.48g Na @ 23 g/mol = 1.6 moles

Na 25.41g S, @ 32 g/mole = 0.8 mol

S 38.11g O. @ 16 g/mol = 2.4 mol

O ratio the moles, by dividing all by the smallest: 1.6 mol Na / 0.8 = 2 mol Na 0.8 mol S / 0.8 = 1 mole

S 2.4 mol O / 0.8 = 3 mol O

your empirical formula os Na2SO3

It was hard when I took chemistry too! I passed with a B