Read GCSE Chemistry C2 Calculations Revision text version

Useful atomic masses: C = 12, H = 1, O = 16, N = 14, Cu = 64, S = 32, Cl = 35.5, Mg = 24 GCSE Chemistry C2 Calculations Revision Formula mass or Mr ­ This is the mass of a molecule calculated by adding the atomic masses of each of the atoms in the molecule. Eg C2H6 ­ Ethane (Atomic masses C = 12, H = 1). IN a molecule of ethane there are 2 atoms of carbon and 6 atoms of hydrogen. Find the formula mass of the following. Formula mass = 2 x 12 = 24 1. CO2 2. CuSO4 3. NH4(SO4) 6x1 = 6 4. MgO 5. C6H12O6 Total mass = 30 Mr = 30 Percentage of an element in a compound. ­ This is used to calculate how much of an element is in a compound. Eg What is he percentage of carbon in ethane. Step 1. Work out the formula mass of the compound Step 2. Work out the mass of the element you are interested in in the compound Step 3. Divide the mass of the element you are interested in by the total mass of the compound and multiply by 100. Step 1 and 2 .Formula mass = 2 x 12 = 24 (this is the mass of carbon in the compound) 6x1 = 6 Total mass = 30 Mr = 30 Step 3 = 24 x 100 = 80% of ethane is carbon. Work out the % of carbon in compounds 1 and 5. 30 Empirical formula ­ Find the formula of a compound from the mass of the elements in the compound. Eg find the empirical formula of magnesium oxide. 120g of magnesium reacts in air to form 200g of magnesium oxide. Step 1. Find the mass of each element in a compound (sometimes you have to calculate this). Step 2. Divide the mass of each element by its atomic mass. Step 3. Divide your answers by the lowest answer to get a ratio. Find the empirical formula for the following: Step 4. Write the formula. Hydrocarbon 80% carbon & 20% Step 1 and 2 = Mg O Hydrogen. 120g 200g ­ 120g = 80g Sulphur oxide ­ 40% sulphur. 120g/24 80g/16 Lead chloride ­ 82.8g of lead made = 5 5 111.2g of lead chloride. Step 3. = 5/5 = 1 5/5 = 1 Calcium compound with 0.8g of calcium, 0.64g of oxygen and 0.04g of hydrogen. Formula = MgO Atom economy - Atom economy is a measure of how efficient a reaction is. A high atom economy (<50%) means you have less waste, and the process supports sustainable development. Step 1. Calculate the molecular mass of the useful product. Step 2. Calculate the molecular mass of all the products added together. Step 3. Divide the mass of the useful product by the total mass of all products and multiply by 100. The following reaction is used to produce ethene which is used in other chemical processes calculate the atom economy. C12H26 C2H4 + C10H22 Work out the atom economy for the Step 1. C2H4 = (2 x 12) + (4 x 1) = 28 following reactions. The second product Step 2. Total formula mass of all products C2H4 and C10H22. will be the useful product. = 28 + (10 x 12) + (22 x 1) HCl + KOH KCl + H2O = 28 + 142 4CuO + CH4 CO2 + 2H2O + Cu = 170 CaCO3 + 2HCl CaCl2 + H2O + CO2 Step 3. = 28 / 170 = 0.164 = 0.164 x 100 = 16.4%

Reacting amounts This is used to calculate how much product can be produced or reactant is needed in a reaction. Step 1. Write the mass of the substance you are given above that substance and put a `?' above the substance you are trying to find the mass of. Step 2. Write the formula mass of each of the two substances you are interested in underneath each one. (don't forget to multiply by the number in front of the formula) Step 3. Make an equation and solve. Eg Find the mass of ethene needed to produce 10 tonnes of ethanol. C2H4 + Step 1. H2O C2H5OH

10 T ?T C2H4 + H2O C2H5OH Step 2. 28 46 Step 2. Formula mass C2H4 = (12 x 2) + (4 x 1) = 28 C2H5OH = (2 x 12) + (6 x 1) + (1 x 16) = 46 Step 3. 10T = ?T = 10T x 46 = 16.4 Tonnes 28 46 28

How much CO2 is released from 20g of CaCO3? CaO + CO2 CaCO3 How much carbon would be needed to make 8.8g of CO2? C+O CO2 How much AlO3 is needed to make 4.5kg of Al? 2AlO3 4Al + 3O2

Percentage Yield - This is used to calculate the percentage of a product actually produced compared to the amount you would expect to be produced based on the amount of reactants you started with. The first part of the calculation is the same as "reacting amounts". Step 1. Write the mass of the reactant you are given above that substance and put a `?' above the substance you are trying to find the yield of. Step 2. Write the formula mass of each of the two substances you are interested in underneath each one (don't forget to multiply by the number in front of the formula). Step 3. Make an equation and solve. This gives you your theoretical yield ­ the maximum yield you could obtain. Step 4. Divide the yield (mass) of the product you are given by the theoretical yield and multiply by 100. Eg 12tonnes of ethanol is produced from 30 tonnes of ethane work out the percentage yield. C2H4 + Step 1. H2O C2H5OH

30T ?T C2H4 + H2O C2H5OH Step 2. 28 46 Step 2. Formula mass C2H4 = (12 x 2) + (4 x 1) = 28 C2H5OH = (2 x 12) + (6 x 1) + (1 x 16) = 46 Step 3. 30T = ?T = 30T x 46 = 49.3Tonnes 28 46 28 Step 4. 12T x 100 = 24.3% 49.3

Find the % yield if 10g CO2 is released from 20g of CaCO3? CaCO3 CaO + CO2

Find the % yield if 5g of CO2 is produced from 5g of C? C+O CO2

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