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1 of 29 © Boardworks Ltd 2009 2 of 29 © Boardworks Ltd 2009 Counting atoms and molecules When conducting a chemical reaction, it is often important to mix reactants in the correct proportions. This prevents contamination of the products by wasted reactants. However, atoms are very small and impossible to count out. In order to estimate the number of atoms in a sample of an element, it is necessary to find their mass. The mass of an atom is quantified in terms of relative atomic mass. 3 of 29 © Boardworks Ltd 2009 Relative atomic mass The relative atomic mass (Ar) of an element is the mass of one of its atoms relative to 1/12 the mass of one atom of carbon-12. relative atomic mass average mass of an atom × 12 = (Ar) mass of one atom of carbon-12 Most elements have more than one isotope. The Ar of the element is the average mass of the isotopes, taking into account the abundance of each isotope. This is why the Ar of an element is frequently not a whole number. 4 of 29 © Boardworks Ltd 2009 Relative molecular mass The relative molecular mass (Mr) of a covalent substance is the mass of one molecule relative to 1/12 the mass of one atom of carbon-12. Mr can be calculated by adding together the masses of each of the atoms in a molecule. Example: what is the Mr of H2SO4? 1. Count number of atoms (2 × H) + (1 × S) + (4 × O) 2. Substitute the Ar values (2 × 1.0) + (1 × 32.1) + (4 × 16.0) 3. Add the values together 2.0 + 32.1 + 64.0 = 98.1 5 of 29 © Boardworks Ltd 2009 Relative formula mass The equivalent of relative molecular mass for an ionic substance is the relative formula mass. This is the mass of a formula unit relative to 1/12 the mass of one atom of carbon-12. It is calculated in the same way as relative molecular mass, and is represented by the same symbol, Mr. Example: what is the Mr of CaCl2? 1. Count number of atoms (1 × Ca) + (2 × Cl) 2. Substitute the Ar values (1 × 40.1) + (2 × 35.5) 3. Add the values together 40.1 + 71.0 = 111.1 6 of 29 © Boardworks Ltd 2009 Calculating relative formula mass 7 of 29 © Boardworks Ltd 2009 8 of 29 © Boardworks Ltd 2009 Moles and Avogadro's number 9 of 29 © Boardworks Ltd 2009 Moles, mass and Ar / Mr 10 of 29 © Boardworks Ltd 2009 Moles, mass and Mr calculations 11 of 29 © Boardworks Ltd 2009 Avogadro’s law In 1811 the Italian scientist Amedeo Avogadro developed a theory about the volume of gases. Avogadro’s law: Equal volumes of different gases at the same pressure and temperature will contain equal numbers of particles. For example, if there are 2 moles of O2 in 50 cm3 of oxygen gas, then there will be 2 moles of N2 in 50 cm3 of nitrogen gas and 2 moles of CO2 in 50 cm3 of carbon dioxide gas at the same temperature and pressure. Using this principle, the volume that a gas occupies will depend on the number of moles of the gas. 12 of 29 © Boardworks Ltd 2009 Molar volumes of gases If the temperature and pressure are fixed at convenient standard values, the molar volume of a gas can be determined. Standard temperature is 273 K and pressure is 100 kPa. At standard temperature and pressure, 1 mole of any gas occupies a volume of 22.7 dm3. This is the molar volume. Example: what volume does 5 moles of CO2 occupy? volume occupied = no. moles × molar volume = 5 × 22.7 = 113.5 dm3 13 of 29 © Boardworks Ltd 2009 Ideal gas equation How is the number of moles in a gas at other temperatures and pressures calculated? The ideal gas equation relates pressure, volume, number of moles and temperature for a gas. pV = nRT p = pressure in Pa n = number of moles V = volume in m3 R = gas constant: 8.31JK-1 mol-1 T = temperature in Kelvin A gas that obeys this law under all conditions is called an ideal gas. 14 of 29 © Boardworks Ltd 2009 Ideal gas equation: converting units It is very important when using the ideal gas equation that the values are in the correct units. The units of pressure, volume or temperature often need to be converted before using the formula. Pressure to convert kPa to Pa: × 1000 Volume to convert dm3 to m3: to convert cm3 to m3: ÷ 1000 (103) ÷ 1 000 000 (106) Temperature to convert °C to Kelvin: + 273 15 of 29 © Boardworks Ltd 2009 Calculating the Mr of gases 16 of 29 © Boardworks Ltd 2009 Using the ideal gas equation 17 of 29 © Boardworks Ltd 2009 Ideal gas calculations 18 of 29 © Boardworks Ltd 2009 19 of 29 © Boardworks Ltd 2009 Types of formulae The empirical formula of a compound shows the relative numbers of atoms of each element present, using the smallest whole numbers of atoms. For example, the empirical formula of hydrogen peroxide is HO – the ratio of hydrogen to oxygen is 1:1. The molecular formula of a compound gives the actual numbers of atoms of each element in a molecule. The molecular formula of hydrogen peroxide is H2O2 – there are two atoms of hydrogen and two atoms of oxygen in each molecule. 20 of 29 © Boardworks Ltd 2009 Determining empirical formulae 21 of 29 © Boardworks Ltd 2009 Percentage by mass Elemental analysis is an analytical technique used to determine the percentage by mass of certain elements present in a compound. To work out the empirical formula, the total mass of the compound is assumed to be 100 g, and each percentage is turned into a mass in grams. If necessary, the mass of any elements not given by elemental analysis is calculated. The empirical formula of the compound can then be calculated as normal. 22 of 29 © Boardworks Ltd 2009 Calculating empirical formulae 23 of 29 © Boardworks Ltd 2009 Calculating molecular formulae The molecular formula can be found by dividing the Mr by the relative mass of the empirical formula. Example: What is the molecular formula of hydrogen peroxide given that its empirical formula is HO and the Mr is 34? 1. Determine relative mass of empirical formula: empirical formula mass = H + O = 1.0 + 16.0 = 17 2. Divide Mr by mass of empirical formula to get a multiple: relative molecular mass = 34 = 2 multiple = 17 mass of empirical formula 3. Multiply empirical formula by multiple: HO × 2 = H2O2 24 of 29 © Boardworks Ltd 2009 Formulae calculations 25 of 29 © Boardworks Ltd 2009 26 of 29 © Boardworks Ltd 2009 Glossary 27 of 29 © Boardworks Ltd 2009 What’s the keyword? 28 of 29 © Boardworks Ltd 2009 Multiple-choice quiz 29 of 29 © Boardworks Ltd 2009 30 of 35 © Boardworks Ltd 2009 31 of 35 © Boardworks Ltd 2009 Balancing equations An important principle in chemical reactions is that matter cannot be created or destroyed. It is important that symbol equations are balanced. A balanced equation has the same number of each type of atom on each side of the equation. Unbalanced: Na + Cl2 1 sodium 2 chlorine Balanced: 2Na + Cl2 2 sodium 2 chlorine NaCl 1 sodium 1 chlorine 2NaCl 2 sodium 2 chlorine This shows that two moles of sodium react with one mole of chlorine to make two moles of sodium chloride. 32 of 35 © Boardworks Ltd 2009 Balancing unfamiliar equations 33 of 35 © Boardworks Ltd 2009 Balancing ionic equations Equations containing ions should have the same overall charge on each side in order to be balanced. This can be achieved by balancing the equation in the normal way: Unbalanced: Ca2+ + Cl- 2 calcium 1 chloride +1 charge Balanced: Ca2+ + 2Cl- 2 calcium 2 chloride no charge 34 of 35 → CaCl2 2 calcium 2 chloride no charge → CaCl2 2 calcium 2 chloride no charge © Boardworks Ltd 2009 Balancing ionic equations problems 35 of 35 © Boardworks Ltd 2009 State symbols State symbols are letters that are added to a formula to indicate what state each reactant and product is in. The four state symbols are: s solid l liquid g gas aq aqueous These are added after the formula in brackets and subscript. For example: 2H2(g) 36 of 35 + O2(g) 2H2O(g) © Boardworks Ltd 2009 Adding state symbols 37 of 35 © Boardworks Ltd 2009 38 of 35 © Boardworks Ltd 2009 Reacting masses 39 of 35 © Boardworks Ltd 2009 Calculating reacting masses To calculate the mass of a product given the mass of a reactant, use the following steps: 1. Calculate no. moles of reactant: no. moles = mass / Mr 2. Determine mole ratio of reactant to product: ensure the equation is balanced 3. Calculate no. moles of product: use the mole ratio 4. Calculate mass of product: mass = moles × Mr 40 of 35 © Boardworks Ltd 2009 Reacting masses example What mass of sodium chloride is produced if 2.30 g of sodium is burnt in excess chlorine? 1. Calculate no. moles of Na: no. moles = mass / Mr = 2.30 / 23.0 = 0.100 2. Determine mole 2Na + Cl2 2NaCl ratio of Na to NaCl: ratio = 2:2 = 1:1 3. Calculate 0.100 moles Na = 0.100 moles NaCl no. moles of NaCl: 4. Calculate mass = moles × Mr = 0.100 × 58.5 mass of NaCl: = 5.85 g 41 of 35 © Boardworks Ltd 2009 Reacting masses calculations 42 of 35 © Boardworks Ltd 2009 More reacting masses calculations 43 of 35 © Boardworks Ltd 2009 44 of 35 © Boardworks Ltd 2009 What is concentration? The concentration of a solution is a measure of how much solute is dissolved per unit of solvent. concentration = amount of solute / volume of solvent amount of solute is measured in moles volume of solvent is measured in dm3 concentration is measured in mol dm-3. Volumes are often expressed in cm3, so a more useful equation includes a conversion from cm3 to dm3. concentration = (no. moles × 1000) / volume mol dm3 45 of 35 cm3 © Boardworks Ltd 2009 Concentration, moles and volume 46 of 35 © Boardworks Ltd 2009 Concentration calculations 47 of 35 © Boardworks Ltd 2009 Standard solutions A standard solution is a solution of known concentration. Standard solutions are made by dissolving an accurately weighed mass of solid in a known volume of solvent using a volumetric flask. The volumetric flask has a thin neck, which is marked with a line so it can be filled accurately to the correct capacity. The standard solution can then be used to find the concentration of a second solution with which it reacts. This is known as volumetric analysis or titration. 48 of 35 © Boardworks Ltd 2009 Preparing standard solutions 49 of 35 © Boardworks Ltd 2009 What is a titration? A titration is a procedure used to identify the concentration of a solution by reacting it with a solution of known concentration and measuring the volume required for a complete reaction. The number of moles in the standard solution is calculated. Using a balanced equation for the reaction, the number of moles in the solution of unknown concentration can also be calculated. Once the number of moles for the solution is known, the concentration can be easily calculated. 50 of 35 © Boardworks Ltd 2009 Carrying out a titration 51 of 35 © Boardworks Ltd 2009 Titration calculations examples What is the concentration of an NaOH solution if 25.0 cm3 is neutralized by 23.4 cm3 0.998 mol dm-3 HCl solution? 1. Calculate no. moles HCl: moles = (conc. × volume) / 1000 = (0.998 × 23.4) / 1000 = 0.0234 2. Determine ratio NaOH + HCl NaCl + H2O of NaOH to HCl: ratio NaOH:NaCl = 1:1 3. Calculate no. 0.0234 moles HCl = 0.0234 moles NaOH moles of NaOH: 4. Calculate conc. of NaOH: 52 of 35 conc. = (moles × 1000) / volume = (0.0234 × 1000) / 25.0 = 0.936 mol dm-3 © Boardworks Ltd 2009 Titration calculations 53 of 35 © Boardworks Ltd 2009 More titration calculations 54 of 35 © Boardworks Ltd 2009 55 of 35 © Boardworks Ltd 2009 What are the different types of yield? The percentage yield of a chemical reaction shows how much product was actually made compared with the amount of product that was expected. To calculate the percentage yield, the theoretical yield and the actual yield must be calculated. The theoretical yield is the maximum mass of product expected from the reaction, calculated using reacting masses. The actual yield is the mass of product that is actually obtained from the real chemical reaction. 56 of 35 © Boardworks Ltd 2009 Calculating yield The percentage yield of a reaction can be calculated using the following equation: percentage yield = (actual yield × 100) / theoretical yield Example: What is the percentage yield of a reaction where the theoretical yield was 75 kg but the actual yield was 68 kg? percentage yield = (actual yield × 100) / theoretical yield = (68 × 100) / 75 = 90.7% 57 of 35 © Boardworks Ltd 2009 What is atom economy? Atom economy is another measure of the efficiency of a chemical reaction. It is the mass of reactants that end up as the desired product – this is calculated as a percentage. This concept is useful to chemical industry, because it takes into account the atoms that end up in unwanted waste products as well as the yield of the reaction. This means a process that produces several worthless by-products could have a high yield but a low atom economy. Reactions with a high atom economy tend to be more environmentally friendly as they tend to produce less waste, use fewer raw materials and use less energy. 58 of 35 © Boardworks Ltd 2009 Calculating atom economy mass of desired products × 100 atom economy = total mass of reactants Example: What is the atom economy of a reaction where the actual yield was 25 000 tonnes but the mass of the reactants was 30 000 tonnes? mass of desired products × 100 atom economy = total mass of reactants 25 000 × 100 = 30 000 = 83.3% 59 of 35 © Boardworks Ltd 2009 Yield and atom economy calculations 60 of 35 © Boardworks Ltd 2009 61 of 35 © Boardworks Ltd 2009 Glossary 62 of 35 © Boardworks Ltd 2009 What’s the keyword? 63 of 35 © Boardworks Ltd 2009 Multiple-choice quiz 64 of 35 © Boardworks Ltd 2009