Stoichiometric Calculations Introduction Dr Seemal Jelani

PROTON NUMBER = number of protons (obviously) Atomic mass RELATIVE ATOMIC MASS, Ar (“Mass number”) = number of protons + number of neutrons He 2 4 SYMBOL PROTON NUMBER = number of protons (obviously) Dr Seemal Jelani

Relative formula mass, Mr The relative formula mass of a compound is the relative atomic masses of all the elements in the compound added together. Relative atomic mass of O = 16 E.g. water H2O: Relative atomic mass of H = 1 Therefore Mr for water = 16 + (2x1) = 18 Work out Mr for the following compounds: HCl NaOH MgCl2 H2SO4 K2CO3 H=1, Cl=35 so Mr = 36 Na=23, O=16, H=1 so Mr = 40 Mg=24, Cl=35 so Mr = 24+(2x35) = 94 H=1, S=32, O=16 so Mr = (2x1)+32+(4x16) = 98 K=39, C=12, O=16 so Mr = (2x39)+12+(3x16) = 138 Dr Seemal Jelani

More examples CaCO3 40 + 12 + 3x16 100 HNO3 1 + 14 + 3x16 2MgO 2 x (24 + 16) 80 3H2O 3 x ((2x1) + 16) 4NH3 2KMnO4 3C2H5OH 4Ca(OH)2 Moles – The relative formula mass of a substance, in grams, is known as 1 mole of that substance. E.g. 18g of H2O = 1 mole of H2O Dr Seemal Jelani

Calculating percentage mass If you can work out Mr then this bit is easy… Percentage mass (%) = Mass of element Ar Relative formula mass Mr x100% Calculate the percentage mass of magnesium in magnesium oxide, MgO: Ar for magnesium = 24 Ar for oxygen = 16 Mr for magnesium oxide = 24 + 16 = 40 Therefore percentage mass = 24/40 x 100% = 60% Calculate the percentage mass of the following: Hydrogen in hydrochloric acid, HCl Potassium in potassium chloride, KCl Calcium in calcium chloride, CaCl2 Oxygen in water, H2O Dr Seemal Jelani

A classic exam question: Empirical formulae 27/02/2019 Empirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO3, H20 and KMnO4 are all empirical formulae. Here’s how to work them out: A classic exam question: Find the simplest formula of 2.24g of iron reacting with 0.96g of oxygen. Step 1: Divide both masses by the relative atomic mass: For iron 2.24/56 = 0.04 For oxygen 0.96/16 = 0.06 Step 2: Write this as a ratio and simplify: 0.04:0.06 is equivalent to 2:3 Step 3: Write the formula: 2 iron atoms for 3 oxygen atoms means the formula is Fe2O3 Dr Seemal Jelani

Example questions 27/02/2019 Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g of oxygen. Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen. Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen. Dr Seemal Jelani

Applications of balancing Dr Seemal Jelani

Calculating the mass of a product 27/02/2019 Calculating the mass of a product E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air? IGNORE the oxygen in step 2 – the question doesn’t ask for it Step 1: READ the equation: 2Mg + O2 2MgO Step 2: WORK OUT the relative formula masses (Mr): 2Mg = 2 x 24 = 48 2MgO = 2 x (24+16) = 80 Step 3: LEARN and APPLY the following 3 points: 48g of Mg makes 80g of MgO 1g of Mg makes 80/48 = 1.66g of MgO 60g of Mg makes 1.66 x 60 = 100g of MgO Dr Seemal Jelani

Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 When water is electrolysed it breaks down into hydrogen and oxygen: 2H2O 2H2 + O2 What mass of hydrogen is produced by the electrolysis of 6g of water? 27/02/2019 Work out Mr: 2H2O = 2 x ((2x1)+16) = 36 2H2 = 2x2 = 4 36g of water produces 4g of hydrogen So 1g of water produces 4/36 = 0.11g of hydrogen 6g of water will produce (4/36) x 6 = 0.66g of hydrogen 2) What mass of calcium oxide is produced when 10g of calcium burns? 2Ca + O2 2CaO Mr: 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = 112 80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO 3) What mass of aluminium is produced from 100g of aluminium oxide? 2Al2O3 4Al + 3O2 Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3 Dr Seemal Jelani

Actual Yield 27/02/2019 Even though no atoms are ever gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of product. Because: The reaction may not totally finish – it may be reversible Some of the product may be lost when it is separated from the reaction mixture – filtered Some of the reactants may react in different ways to the expected reaction Dr Seemal Jelani

Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 27/02/2019 Percentage yield The amount of product obtained is known as the yield. When compared to the maximum theoretical (calculated) amount as a percentage, it is called percentage yield. Percentage yield (%) = Actual yield made Maximum yield possible x100% E.g. What mass of aluminium is produced from 100g of aluminium oxide? 2Al2O3 4Al + 3O2 Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al However, only 35.6g of Al was actually obtained during the experiment. What is the percentage yield. Percentage yield (%) = 35.6 52.9 x100% Dr Seemal Jelani = 67.3%

Reversible Reactions 27/02/2019 Some chemical reactions are reversible. In other words, they can go in either direction: A + B C D e.g. Ammonium chloride Ammonia + hydrogen chloride NH4Cl NH3 + HCl When a reversible reaction occurs in a closed system (Where nothing can escape), equilibrium is reached when both reactions occur at exactly the same rate in each direction. The relative amounts of all the reacting substances at equilibrium depend on the conditions of the reaction. Dr Seemal Jelani

Making Ammonia Nitrogen + hydrogen Ammonia N2 + 3H2 2NH3 27/02/2019 Fritz Haber, 1868-1934 Guten Tag. My name is Fritz Haber and I won the Nobel Prize for chemistry. I am going to tell you how to use a reversible reaction to produce ammonia, a very important chemical. This is called the Haber Process. Nitrogen + hydrogen Ammonia N2 + 3H2 2NH3 To produce ammonia from nitrogen and hydrogen you have to use three conditions: High pressure 450O C Iron catalyst Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy. Nitrogen Hydrogen Recycled H2 and N2 Dr Seemal Jelani

Ammonia + nitric acid Ammonium nitrate Uses of Ammonia 27/02/2019 Ammonia is a very important chemical as it can be used to make plant fertilisers and nitric acid: Water and oxygen Nitrogen monoxide Ammonia gas Oxygen Hot platinum catalyst Nitrogen monoxide Nitric acid Cooled More ammonia can then be used to neutralise the nitric acid to produce AMMONIUM NITRATE (a fertiliser rich in nitrogen). Ammonia + nitric acid Ammonium nitrate NH3 + HNO3 NH4NO3 The trouble with nitrogen based fertilisers is that they can also create problems – they could contaminate our drinking water. Dr Seemal Jelani

Haber Process Summary 27/02/2019 A low temperature increases the yield of ammonia but is too slow A high temperature improves the rate of reaction but decreases the yield too much A high pressure increases the yield of ammonia but costs a lot of money To compromise all of these factors, these conditions are used to make a reasonable Yield of ammonia, quickly: 200 atm pressure 450O C Iron catalyst Recycled H2 and N2 Nitrogen Hydrogen Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy. Dr Seemal Jelani