1. The yield in a reaction is how much product is formed in a chemical reaction. This is precisely determined by the amount of reactant(s) are used. This is one example of the law of conservation of mass. If there is 37.456g before a reaction then there will be 37.456g at the end of the reaction. Atoms cannot be created or destroyed – if they are there at the start of the reaction (as reactants) they’ll be there at the end as products. For example the reaction between iron (Fe) and sulphur (S) to make iron (II) sulphide (FeS). Fe+S  FeS If we write the Ar and Mr under the relevant formulae we get the following… Fe+S  FeS 56 32 88 Think of these 3 numbers:- 56, 32, 88. These represent the ratios (by mass) of the 3 substances. So we could say that exactly 56g of Fe reacts with exactly 32g of S to form 88g of FeS. YIELDS AND REACTING MASSES

2. It is now possible to “scale up” or scale down” the amounts in the equation. Whatever we do to one amount (e.g. double the mass), we must do the same to all the other amounts. Fe+S  FeS 56g 32g 88g 112g 64g 176g (double the line above) 11.2g 6.4g 17.6g (one tenth the line above) 22.4g 12.8g 35.2g (double the line above) Therefore, given the amount of one or more reactants, and given the balanced equation, we can calculate the yield. For example, if 22.4g of Fe are reacted with 12.8g of S the yield of FeS should be 35.2g. Start with… Twice as much 1/10 th as much 1/5 th as much

3. So far, all the calculations have involved numbers that bear a simple relationship to the masses in the equation. What if the numbers are more awkward? 1.Write down the balanced equation (this is usually done for you). 2.Write down the M r /A r under the appropriate chemical, and multiply by the balancing number. 3.Look at what mass you are given. 4.Scale the Mr/Ar of that substance to 1g, or 1kg, or 1 tonne. 5.Do the same scaling for all other substances in the equation. 6.Then multiply up the amount of substance to the quantity given, and multiply up all substances by the same factor. An example question:- 567g of nitrogen are reacted with excess H 2. What mass of ammonia is formed? 1.N 2 +3H 2  2NH 3 2. Write the M r 28 6 34 3.28g 6g 34g 4. Scale down to 1g 1g (28/28)g(6/28)g (34/28)g 0.21g 1.21g 5. Scale up to 567g 567g (0.21 x 567)g (1.21 x 567)g Work out the answer121.5g688.5g So the answer to the question is 688.5g of ammonia

4. So consider our decomposition reaction; CuCO 3  CuO+CO 2 124g 80g 44g As long as we know the amount of copper carbonate we begin with, we can predict how much copper oxide should be made. From this, we can then calculate percentage yield. To do this:- eg I used 3.67g of copper carbonate 1. Write eqn.CuCO 3  CuO+CO 2 2. Write the M r 1248044 3. Convert to grams124g 80g 44g 4. Scale down to 1g 1g (124/124)g(80/124)g (44/124)g 5. Scale up to 3.67g 3.67g (80x3.67/124)g (44x3.67/124)g Work out the answer2.37g1.30g If my yield was actually 1.96g of copper oxide the percentage yield obtained was % yield = actual yield x 100%OR1.96 x 100 = 82.7% theoretical yield 2.37 RFM of CuCO 3 Use your mass here and here Your yield here