what is the emperical formula...?

Question

what is the emperical formula of a compound that contains 29% Na, 41% S, 30% O by mass?

Answer 1

divide each of the percentages by the atomic mass to get the molar ratios.
so Na = 29/23 = 1.26
S = 41/32 = 1.28
O = 30/16 = 1.875

then take each and divide by 1.26 which is the smallest
so Na = 1.26/1.26 = 1
S = 1.28/1.26 = 1.02
O = 1.875/1.26 = 1.49

so it's a 1 : 1 : 1.5 ratio. Multiplying by 2 everywhere to get integers makes it 2 : 2 : 3.

Na2S2O3

Answer 2

empirical formulas are formulas with the smallest whole number ratios of atoms in the compound.

Change your % to grams. The easiest way to do this is to assume you have 100 grams of the compound. Then 29% of that 100 grams would be Na (or 29 grams) and 41% of the 100 grams would be S( or 41 grams) and 30% of the 100 grams would be O (or 30 grams).

Now, change grams to moles by dividing by the atomic mass of each of the elements.("gpm" means grams per mole)

29gNa/23 gpm = 1.3 moles Na

41g S/32 gpm = 1.3 moles S

30 g O/16 gpm = 1.9 moles O

Next look at all your answers: divide all three by the smallest (1.3) to get a mole ratio.

1.3/1.3 = 1 Na

1.3/1.3 = 1 S

1.9/1.3 = 1.5 O

Since the oxygen is halfway between two numbers, multiply them all by 2. These become the subscripts in the empirical formula Na2S2O3