The structure of the atom ParticleRelative MassRelative Charge Proton11 Neutron10 Electron0 MASS NUMBER = number of protons + number of neutrons SYMBOL PROTON NUMBER = number of protons (obviously)

25/08/2015 Atomic mass SYMBOL PROTON NUMBER = number of protons (obviously) RELATIVE ATOMIC MASS, A r (“Mass number”) = number of protons + number of neutrons

Isotopes An isotope is an atom with a different number of neutrons: Each isotope has 8 protons – if it didn’t then it just wouldn’t be oxygen any more. Notice that the mass number is different. How many neutrons does each isotope have? H.T – The A r of an element compares the mass of atoms with the C 12 isotope. It is an average value of the isotopes of the element.

25/08/2015 Relative formula mass, M r The relative formula mass of a compound is blatantly the relative atomic masses of all the elements in the compound added together. E.g. water H 2 O: Therefore M r for water = 16 + (2x1) = 18 Work out M r for the following compounds: 1)HCl 2)NaOH 3)MgCl 2 4)H 2 SO 4 5)K 2 CO 3 H=1, Cl=35 so M r = 36 Na=23, O=16, H=1 so M r = 40 Mg=24, Cl=35 so M r = 24+(2x35) = 94 H=1, S=32, O=16 so M r = (2x1)+32+(4x16) = 98 K=39, C=12, O=16 so M r = (2x39)+12+(3x16) = 138 Relative atomic mass of O = 16Relative atomic mass of H = 1

25/08/2015 More examples CaCO x16100 HNO x16 2MgO2 x ( )80 3H 2 O3 x ((2x1) + 16) 4NH 3 2KMnO 4 3C 2 H 5 OH 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 H 2 O = 1 mole of H 2 O

25/08/2015 Calculating percentage mass If you can work out M r then this bit is easy… Calculate the percentage mass of magnesium in magnesium oxide, MgO: A r for magnesium = 24Ar for oxygen = 16 M r for magnesium oxide = = 40 Therefore percentage mass = 24/40 x 100% = 60% Percentage mass (%) = Mass of element A r Relative formula mass M r x100% Calculate the percentage mass of the following: 1)Hydrogen in hydrochloric acid, HCl 2)Potassium in potassium chloride, KCl 3)Calcium in calcium chloride, CaCl 2 4)Oxygen in water, H 2 O

25/08/2015 Empirical formulae Empirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO 3, H 2 0 and KMnO 4 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 = 0.04For 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 Fe 2 O 3 H.T

25/08/2015 Example questions 1)Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g of oxygen. 2)Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen. 3)Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen. H.T

25/08/2015 Calculating the mass of a product E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air? Step 1: READ the equation: 2Mg + O 2 2MgO IGNORE the oxygen in step 2 – the question doesn’t ask for it Step 3: LEARN and APPLY the following 3 points: 1)48g of Mg makes 80g of MgO 2)1g of Mg makes 80/48 = 1.66g of MgO 3)60g of Mg makes 1.66 x 60 = 100g of MgO Step 2: WORK OUT the relative formula masses (M r ): 2Mg = 2 x 24 = 48 2MgO = 2 x (24+16) = 80 H.T

25/08/2015 Work out M r : 2H 2 O = 2 x ((2x1)+16) = 36 2H 2 = 2x2 = g of water produces 4g of hydrogen 2.So 1g of water produces 4/36 = 0.11g of hydrogen 3.6g of water will produce (4/36) x 6 = 0.66g of hydrogen M r : 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = g produces 112g so 10g produces (112/80) x 10 = 14g of CaO M r : 2Al 2 O 3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al 2 O 3 1)When water is electrolysed it breaks down into hydrogen and oxygen: 2H 2 O 2H 2 + O 2 What mass of hydrogen is produced by the electrolysis of 6g of water? 3) What mass of aluminium is produced from 100g of aluminium oxide? 2Al 2 O 3 4Al + 3O 2 2) What mass of calcium oxide is produced when 10g of calcium burns? 2Ca + O 2 2CaO H.T

25/08/2015 Actual Yield 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

25/08/2015 Percentage yield 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% H.T E.g. What mass of aluminium is produced from 100g of aluminium oxide? 2Al 2 O 3 4Al + 3O 2 M r : 2Al 2 O 3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = g 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 (%) = x100% = 67.3%

25/08/2015 Atom economy This is simply a measure of the amount of starting materials that end up as useful products. It is important for sustainable development and economical reasons that industrial reactions have High Atom Economy. Atom Economy (%) = Total M r of useful products Total M r of reactants x100% H.T

25/08/2015 Reversible Reactions Some chemical reactions are reversible. In other words, they can go in either direction: A+BC+D NH 4 ClNH 3 + HCl e.g. Ammonium chlorideAmmonia + hydrogen chloride 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. H.T

25/08/2015 Making Ammonia Nitrogen + hydrogen Ammonia N 2 + 3H 2 2NH 3 High pressure 450 O C Iron catalyst Recycled H 2 and N 2 Nitrogen Hydrogen Mixture of NH 3, H 2 and N 2. This is cooled causing NH 3 to liquefy. Fritz Haber, 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. To produce ammonia from nitrogen and hydrogen you have to use three conditions:

25/08/2015 Uses of Ammonia Nitrogen monoxide Hot platinum catalyst Ammonia gas Oxygen Cooled Water and oxygen Nitrogen monoxide Nitric acid Ammonia + nitric acid Ammonium nitrate NH 3 + HNO 3 NH 4 NO 3 Ammonia is a very important chemical as it can be used to make plant fertilisers and nitric acid: More ammonia can then be used to neutralise the nitric acid to produce AMMONIUM NITRATE (a fertiliser rich in nitrogen). The trouble with nitrogen based fertilisers is that they can also create problems – they could contaminate our drinking water.

25/08/2015 Haber Process Summary 200 atm pressure 450 O C Iron catalyst Recycled H 2 and N 2 Nitrogen Hydrogen Mixture of NH 3, H 2 and N 2. This is cooled causing NH 3 to liquefy. To compromise all of these factors, these conditions are used to make a reasonable yield of ammonia, quickly: 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