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Formulae and equations for Edexcel A-level Chemistry

This page covers the following topics:

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.

Empirical formula

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.

Molecular formula

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.

Limiting reactants

1

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

Zn + 2HCl โ†’ ZnClโ‚‚ + Hโ‚‚. Calculate the number of moles for each reactant. Zn: n = m/Mแตฃ = 2.5/65 = 0.04. HCl: n = m/Mแตฃ = 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.

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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 CaCOโ‚ƒ.

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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 โ†’ Naโ‚‚S. 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.

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