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Atomic mass

Atomic mass. RELATIVE ATOMIC MASS, A r (“Mass number”) = number of protons + number of neutrons. 4. He. SYMBOL. 2. PROTON NUMBER = number of protons (obviously). Relative formula mass, M r. Relative atomic mass of O = 16. Relative atomic mass of H = 1.

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Atomic mass

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

  2. Relative formula mass, Mr Relative atomic mass of O = 16 Relative atomic mass of H = 1 The relative formula mass of a compound is the relative atomic masses of all the elements in the compound added together. E.g. water H2O: 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

  3. More examples 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

  4. Calculating percentage mass Mass of element Ar x100% Percentage mass (%) = Relative formula mass Mr If you can work out Mr then this bit is easy… 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

  5. Empirical formulae 14/09/2014 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

  6. Example questions 14/09/2014 • 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.

  7. Calculating the mass of a product 14/09/2014 IGNORE the oxygen in step 2 – the question doesn’t ask for it Step 1: READ the equation: 2Mg + O2 2MgO E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air? 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

  8. 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? 14/09/2014 2) What mass of calcium oxide is produced when 10g of calcium burns? 2Ca + O2 2CaO 3) What mass of aluminium is produced from 100g of aluminium oxide? 2Al2O3 4Al + 3O2 • 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 • Mr: 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = 112 • 80g produces 112g so 10g produces (112/80) x 10 =14g of CaO Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 204g produces 108g so 100g produces (108/204) x 100 =52.9g of Al2O3

  9. Actual Yield 14/09/2014 • 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

  10. Percentage yield 14/09/2014 Actual yield made 35.6 x100% Percentage yield (%) = Percentage yield (%) = Maximum yield possible 52.9 E.g. What mass of aluminium is produced from 100g of aluminium oxide? 2Al2O3 4Al + 3O2 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. 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. x100% = 67.3%

  11. Atom economy 14/09/2014 Total Mr of useful products x100% Atom Economy (%) = Total Mr of reactants 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.

  12. Reversible Reactions 14/09/2014 A + B C + D e.g. Ammonium chloride Ammonia + hydrogen chloride NH4Cl NH3 + HCl Some chemical reactions are reversible. In other words, they can go in either direction: 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.

  13. Making Ammonia 14/09/2014 Nitrogen + hydrogen Ammonia Fritz Haber, 1868-1934 N2 + 3H2 2NH3 Nitrogen Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy. Hydrogen Recycled H2 and N2 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: • High pressure • 450O C • Iron catalyst

  14. Uses of Ammonia 14/09/2014 Nitrogen monoxide Ammonia gas Hot platinum catalyst Cooled Nitric acid Oxygen Water and oxygen Ammonia + nitric acid Ammonium nitrate NH3 + HNO3 NH4NO3 Ammonia is a very important chemical as it can be used to make plant fertilisers and nitric acid: Nitrogen monoxide 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.

  15. Haber Process Summary 14/09/2014 Nitrogen Mixture of NH3, H2 and N2. This is cooled causing NH3 to liquefy. To compromise all of these factors, these conditions are used to make a reasonable Yield of ammonia, quickly: Hydrogen • 200 atm pressure • 450O C • Iron catalyst Recycled H2 and N2 • 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

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