Science Quantitative chemistry You should be able to calculate the masses of reactants and products from balanced equations, and the percentage composition by mass of an element in a compound. Higher tier students should also be able to calculate the percentage yield of a reaction, and the empirical formula of a compound from information about reacting masses. Percentage composition Percentage composition is just a way to describe what proportions of the different elements there are in a compound. If you have the formula of a compound, you should be able to work out the percentage by mass of an element in it. Example The formula for sodium hydroxide is NaOH. It contains three different elements: Na, O and H. But the percentage by mass of each element is not simply 33.3 per cent, because each element has a different relative atomic mass. You need to use the A r values to work out the percentages. Here is how to do it: Question What is the percentage by mass of oxygen (O) in sodium hydroxide (NaOH)? First, work out the relative formula mass of the compound, using the A r values for each element. In the case of sodium hydroxide, these are Na = 23, O = 16, H = 1. (You will be given these numbers in the exam.) http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/atomic_structure/quantitative_chemistryrev_print.shtml 1/7
Next, divide the A r of oxygen by the M r of NaOH, and multiply by 100 to get a percentage. Answer M r of NaOH is 23 + 16 + 1 = 40 (16 40 ) 100 = 0.4 100 = 40% So the percentage by mass of oxygen in sodium hydroxide is 40%. Remember, if there is more than one atom of the element in the compound, you need to multiply your answer by the number of atoms. If your answer is more than 100 per cent, you have gone wrong! Conservation of mass Massmass: The amount of matter an object contains. Mass is measured in 'kg'. is never lost or gained in chemical reactions. We say that mass is conserved. In other words, the total mass of productsproduct: A product is a substance formed in a chemical reaction. at the end of the reaction is equal to the total mass of the reactants [reactant: Substances present at the start of a chemical reaction ] at the beginning. This fact allows you to work out the mass of one substance in a reaction if the masses of the other substances are known. For example: Percentage yield The principle of conservation of mass lets you calculate the theoretical mass of product expected in a chemical reaction. However, it is not always possible in practice to get the entire http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/atomic_structure/quantitative_chemistryrev_print.shtml 2/7
calculated amount of product. This is because: Reversible reactions may not go to completion Some product may be lost when it is removed from the reaction mixture Some of the reactants [reactant: Substances present at the start of a chemical reaction ] may react in an unexpected way Yield The yield of a reaction is the mass of product obtained: The theoretical yield is the maximum theoretical mass of product in a reaction (calculated using the idea of conservation of mass) The actual yield is the mass of product you get when you actually do the reaction The percentage yield is the ratio of actual mass of products obtained compared with the maximum theoretical mass. Percentage yield Higher tier The percentage yield of a reaction is calculated using this equation: percentage yield = (actual mass of product) (theoretical mass of product) 100 For example, the maximum theoretical mass of product in a certain reaction is 20 g, but only 15 g is actually obtained. Percentage yield = 15 20 100 = 75% Reversible reactions Many reactions, such as burning fuel, are irreversible they go to completion and cannot be reversed easily. Reversible reactions are different. In a reversible reaction, the http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/atomic_structure/quantitative_chemistryrev_print.shtml 3/7
productsproduct: A product is a substance formed in a chemical reaction. can react to produce the original reactants [reactant: Substances present at the start of a chemical reaction ] again. When writing chemical equations for reversible reactions, we do not use the usual one way arrow. Instead, we use two arrows, each with just half an arrowhead the top one pointing right, and the bottom one pointing left. For example: ammonium chloride ammonia + hydrogen chloride The equation shows that ammonium chloride (a white solid) can break down to form ammonia and hydrogen chloride. It also shows that ammonia and hydrogen chloride (colourless gases) can react to form ammonium chloride again. The animation below shows a reversible reaction involving white anhydrous copper(ii) sulfate and blue hydrated copper(ii) sulfate, the equation for which is: anhydrous copper(ii) sulfate + water hydrated copper(ii) sulfate The reaction between anhydrous copper(ii) sulfate and water is used as a test for water. The white solid turns blue in the presence of water. Now try a Test Bite. Reacting masses calculations Higher tier If you have a balanced equation for a reaction, you can calculate the masses of reactants and products. Sample question Look at this equation: CaCO 3 (s) CaO(s) + CO 2 (g) If we have 50g of CaCO 3, how much CaO can we make? http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/atomic_structure/quantitative_chemistryrev_print.shtml 4/7
First, work out the M r values for the two compounds: M r of CaCO 3 is 40 + 12 + 16 + 16 + 16 = 100 M r of CaO is 40 + 16 = 56 This means that 100 g of CaCO 3 would yield 56 g of CaO in this reaction. In the question we are told we have only half of that amount of CaCO 3, 50 g. So we will get half the amount of CaO, 28 g. So the mass of CaO we can make = 28 g Notice in this that 22 g of CO 2 would also be produced, as 50 28 = 22 Empirical formula Higher tier You can use information about reacting masses to calculate the formula of a compound. Here is an example: Question Suppose 3.2 g of sulfur reacts with oxygen to produce 6.4 g of sulfur oxide. What is the formula of the oxide? Use the fact that the A r of sulfur is 32 and the A r of oxygen is 16 Answer Find the mass of each element. Conservation of mass tells us that the mass of oxygen = the mass of sulfur oxide the mass of sulfur. The mass of oxygen reacted = 6.4 3.2 = 3.2 g So we have 3.2 g of sulfur and 3.2 g of oxygen. http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/atomic_structure/quantitative_chemistryrev_print.shtml 5/7
Now divide the mass of each element by its A r value. sulfur: 3.2 32 = 0.1 oxygen: 3.2 16 = 0.2 Finally, find the ratio of the elements. You can do this by dividing the results by the smallest of the numbers to give you the number of atoms of each element in the compound. In this case the smallest value is 0.1, so divide both results by that. S = 0.1 0.1 = 1 O = 0.2 0.1 = 2 (If one of the numbers ends in 0.5, multiply all the numbers by 2 this is because you cannot have half atoms in a compound.) So the ratio of sulfur to oxygen is 1:2 The number of atoms tells you that the formula for sulfur oxide is SO 2 Here is the calculation again in tabular form to help you remember the steps: Steps to calculation the formula of a compound Step Action S O 1 Find masses 3.2 3.2 2 Look up given A r values 32 16 3 Divide masses by A r 0.1 0.2 4 Find the ratio r 1 2 http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/atomic_structure/quantitative_chemistryrev_print.shtml 6/7
Result: the formula for the oxide = SO 2 Now try a Test Bite Higher tier. Back to Revision Bite http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/atomic_structure/quantitative_chemistryrev_print.shtml 7/7