32.65%→32.65g of S

65.3%→65.3g of O

2.04%→2.04g of H 2) Next divide all the given masses by their molar mass.

32.65g of S/ 32gm-1 = 1.0203 moles of S

65.3g of O/ 16gm-1 = 4.08 moles of O

2.04g of H/ 1.008gm-1 = 2.024 moles of H 3) Then, pick the smallest answer in moles from the previous step and divide all the answers by that. Remember that if you calculate a number that is x0.9 round to the nearest whole number

1.0203 moles of S/ 1.0203 = 1

4.08 moles of O/1.0203 = 3.998 ≈ 4

2.024 moles of H/1.0203 = 1.984 ≈ 2 4) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula.

S = 1

O = 4

H = 2 H2SO4 Example#2: Given the mass of a reactant before a chemical reaction and the mass of a product after a reaction. When 0.273g of Mg is heated in a Nitrogen (N2) environment a chemical reaction occurs. The product of the reaction is 0.378g . Calculate the empirical formula. 1) In any empirical formula problem you must first find the mass % of the elements in the compound. Since the total mass of the final product was 0.378 we find that: 0.378g total-0.273g magnesium = 0.105g nitrogen 0.105g nitrogen/0.378g total (100) = 27.77% 0.273g magnesium/0.378g total (100) =72.23% 2) Then change the % to grams

27.77%→27.77g of N

72.23%→72.23g of Mg 3) Next, divide all the masses by their respective molar masses.

27.77g/14gm-1 = 1.98 moles N

72.23g/24.31gm-1 = 2.97 moles Mg 4) Pick the smallest answer of moles and divide all figures by that.

1.98 moles N/ 1.98 = 1

2.97 moles Mg/ 1.98 = 1.5 Since our answer for magnesium is not close enough to round to the nearest whole number we must choose a factor to multiply all the figures by that will yield us whole numbers

1*2= 2 N

1.5*2= 3 Mg 5) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula.

1.98 moles N/ 1.98 = 1

1.98 moles N/ 1.98 = 1

N = 2

1.98 moles N/ 1.98 = 1

Mg = 3 Mg3N2