1. STANDARD GRADE CHEMISTRY CALCULATIONS Calculations involving the mole. 1 mole of a solid substance is the formula mass of the substance in grams. This is known as the gram formula mass (gfm). mass n gfm The triangle shown below can be used to give the following relationships:- 1.Mass = number of moles x gram formula mass. 2. Number of moles = mass = mass of substance. n = number of moles gfm = gram formula mass mass gram formula mass You must learn this and be able to apply it in calculations.

2. mass n gfm Worked example 1. Calculate the mass of 0.25 moles of butane (C 4 H 10 ). Calculations for you to try. 1. Calculate the mass present in 2.5 moles of calcium carbonate (CaCO 3 ). 2. Calculate the mass of ammonium sulphate, (NH 4 ) 2 SO 4, present in 0.1 mol of ammonium sulphate? gfm = 100g mass = 2.5 x 100 = 250g gfm = 132g mass = 0.1 x 132 = 13.2g Step 3:- Using the triangle we have mass = number of moles x gram formula mass mass = 0.25 x 58 mass = 14.5 g Step 2:- Calculate the gram formula mass (4 x 12) + (10 x 1) = 58g Step 1 :-Write the formula for butane C 4 H 10 Standard Grade Chemistry

3. mass n gfm Worked example 2. Calculate the number of moles in 5.05 g of potassium nitrate, (KNO 3 ). Step 1 :-Write the formula for potassium nitrate KNO 3 Step 2:- Calculate the gram formula mass (1 x 39) + (1 x 14) + (3 x 16) = 101g Step 3:- Using the triangle we have number of moles = = = 0.05 mole mass gram formula mass 5.05 101 Calculations for you to try. 1. Calculate the number of moles in 132 g of carbon dioxide, CO 2. 2. Calculate the number of moles in 4g of bromine, Br 2. gfm = 44g Number of moles = 132 / 44 = 3 moles gfm = 160 Number of moles = 4 / 160 = 0.025 moles

4. n C V (l) Mole calculations involving solutions. The concentration of a solution is measured in moles per litre (mol/l) The triangle shown below can be used to give the following relationships:- n = number of moles. C = concentration. V(l) = volume in litres 1. number of moles = concentration x volume (in litres). 2. concentration = number of moles volume (in litres) Standard Grade Chemistry You must learn this and be able to apply it in calculations. Remember this is in litres

5. Worked example 1. Calculate the number of moles in 200cm 3 of 0.5 mol/l sodium hydroxide solution. Standard Grade Chemistry Step 2 :-Using the triangle gives number of moles = concentration x volume (in litres). = 0.5 x 0.2 = 0.1 moles Step 1 :-Change the volume into litres.0.2 litres Calculations for you to try. 1. Calculate the number of moles in 50 cm 3 of 0.1 mol/l zinc sulphate solution. 2. Calculate the number of moles in 0.2 litres of 2 mol/l sodium hydroxide solution n C V (l) Volume = 50 / 1000 = 0.05 litres Number of moles = 0.1 x 0.05 = 0.0005 moles Volume = 0.2 litres Number of moles = 0.2 x 2 = 0.4 moles

6. Calculations involving, concentration, moles and mass. In this type of calculation both triangles are used. Standard Grade Chemistry Worked example 1. Calculate the mass required to prepare 200cm 3 of 0.1 mol/l sodium hydroxide, (NaOH), solution. Step 1 :-Calculate the number of moles of sodium hydroxide in 200cm 3 of 0.1 mol/l of solution. Step 2 :-Calculate the mass of NaOH in 0.02 moles. Calculation for you to try. Calculate the mass of zinc sulphate in 500 cm 3 of 0.2 mol/l ZnSO 4 (aq) n C V (l) n = c x V(l) = 0.1 x 0.2 = 0.02 moles Volume = 500 / 1000 = 0.5 litres No. of moles = 0.2 x 0.5 = 0.1 Mass = 0.1 x 161.5 = 16.15 g gfm of NaOH = (1 x 23) + (1 x 16) + (1 x 1) = 40 mass n gfm mass = n x gfm = 0.02 x 40 = 0.8 g

7. Worked example 2. Calculate the concentration of a solution that contains 7.45 g of potassium chloride (KCl) in 250cm 3 of solution. mass gfm 7.45 74.5 Step 1 :-Calculate the number of moles of potassium chloride in 7.45g Number of moles = = = 0.1 mass n gfm Step 2 :-Calculate the concentration of the solution. Concentration = = = 0.4 mol/l No. of moles Volume in litres 0.1 0.25 n C V (l) Calculation for you to try. Calculate the concentration of a solution that contains 5.85 g of sodium Chloride, NaCl, in 200cm 3 of solution. Volume = 200 / 1000 = 0.2 litres No. of moles in 5.85 g of NaCl = 5.85 / 58.5 = 0.1 Concentration = 0.1 / 0.2 = 0.5 mol/l Standard Grade Chemistry