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Formulae and equations for Edexcel GCSE Chemistry

Formulae and equations

This page covers the following topics:

1. Empirical formula
2. Molecular formula
3. Balancing equations
4. 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

A balanced equation is a model of a chemical reaction that shows the formulae of the reactants and the products. In order for an equation to be balanced, the number of atoms must be equal on both sides of the equation.

Balancing equations

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.

2.5 g of zinc are reacted with 9.8 g of hydrogen chloride. Which 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 CaCOโ‚ƒ.

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

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.

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

4

The molecular formula of ethane is Cโ‚‚Hโ‚†. Give its empirical formula.

Both numbers in the formula can be divided by 2. The empirical formula is CHโ‚ƒ.

The molecular formula of ethane is Cโ‚‚Hโ‚†. Give its empirical formula.

5

An unknown compound has an empirical formula of SOโ‚ƒ and a Mแตฃ 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 Sโ‚„O(12).

An unknown compound has an empirical formula of SOโ‚ƒ and a Mแตฃ of 320. What is its molecular formula?

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