 1. Empirical formula
2. Molecular formula
3. Limiting reactants

The empirical formula gives the simplest ratio of atoms for each element within a chemical compound. To calculate the empirical formula, the percentage by mass of each component is divided by the relative atomic mass of the element. The molecular formula gives the total number of atoms of each element within a chemical compound. It can be calculated using the empirical formula and the relative molecular mass of the compound: the relative molecular mass is divided by the mass of the atoms in the empirical formula. The numbers in the empirical formula are then multiplied by the resulting number. The limiting reactant is the first reactant to be used up in a chemical reaction. When the limiting reactant is used up, the reaction stops and no more of the product can be formed. Therefore, the mass of the product is dependent on the mass of the limiting reactant. The left-over reactant is referred to as the reactant in excess. # 1

2.5 g of zinc are reacted with 9.8 g of hydrogen chloride. Which is the limiting reactant?

Zn + 2HCl → ZnCl₂ + H₂. Calculate the number of moles for each reactant. Zn: n = m/Mᵣ = 2.5/65 = 0.04. HCl: n = m/Mᵣ = 9.8/36 = 0.3. Consider the ratio of the reactants: 1 mol Zn : 2 mol HCl. 0.15 mol Zn : 0.3 mol HCl. As there are only 0.04 mol of Zn, Zn is the limiting reactant. # 2

A compound contains 40% of calcium, 12% of carbon, and 48% of oxygen. What is its empirical formula?

Moles of Ca = 40/40 = 1. Moles of C = 12/12 = 1. Moles of O = 48/16 = 3. Molar ratio is 1 : 1 : 3. Therefore, the empirical formula is CaCO₃. # 3

5.7 g of sodium are reacted with 6.1 g of sulfur. Which is the limiting reactant and which reactant is in excess?

Na is the limiting reactant and S is in excess

2Na + S → Na₂S. Moles of Na = 5.7/23 = 0.25. Moles of S = 6.1/32 = 0.19. Molar ratio = 2 : 1 = 0.38 : 0.19. There are only 0.25 moles of Na, so Na is the limiting reactant and S is in excess. # 4

The molecular formula of ethane is C₂H₆. Give its empirical formula.

Both numbers in the formula can be divided by 2. The empirical formula is CH₃. # 5

An unknown compound has an empirical formula of SO₃ and a Mᵣ of 320. What is its molecular formula?

The mass of the atoms in the empirical formula is 80. 320/80 = 4. Therefore, the numbers in the empirical formula are multiplied by 4, which gives S₄O(12). End of page