Calculations in Chemistry You need to know how

Calculations in Chemistry • You need to know how to carry out several calculations in Additional and Triple Chemistry • This booklet gives you a step by step guide to carrying out these equations and examples of each one. Follow the same rules for the questions in your exam and chemistry calculations are actually quite easy, honest! • These are: – Additional: • Relative formula mass • Mole masses • Percentage of element in a compound • Formula of a compound from its percentage composition • Masses of reactants and products • Percentage yield • Atom economy – Triple: • Titration calculations • Bond energy calculations

Additional Chemistry Calculations Relative atomic and Formula Masses The mass of an atom is too small to deal with in real terms, so we use ‘relative’ masses – Carbon is given a mass of 12, and everything else is compared with it and given a mass, e. g. Oxygen is ‘heavier’, so its relative mass is 16. The relative atomic mass can be found by looking at the periodic table, It is always the larger of the two numbers. Relative formula mass can be found by adding up the relative atomic masses of each element in a compound. E. g. Carbon Dioxide (CO 2) Carbon has a relative atomic mass of 12 Oxygen has a relative atomic mass of 16 The relative formula mass of Carbon Dioxide is therefore: 12 + (16 x 2) = 44 x 2 because it’s “O 2” Moles Because saying ‘relative formula (or atomic!) mass in grams’ is a bit clumsy, we simply say ‘moles’ instead. This means that 1 mole of Carbon Dioxide is 44 grams, or 44 g. Simple!

Percentage of an element in a compound We can use the relative atomic mass (Ar) of elements and the relative formula mass (Mr) of compounds to find out the percentage composition of different elements. E. g. What percentage mass of white Magnesium Oxide is actually Magnesium, and how much is Oxygen? 1. 2. 3. Work out the mass of Mg. O 24 + 16 = 40 Convert to grams 40 g 1. Work out the formula mass of the compound 2. Convert this into grams 3. Work out the percentage by using this equation: Mass of element x 100% Total mass of compound Work out the percentage 24 X 100% = 60% is Magnesium, so 40% must be Oxygen! 40 Formula of a compound from its percentage composition We can also do this backwards! If we know the percentage composition of a compound we can work out the ratio of atoms. This is known as the Empirical Formula. Sometimes this is the same as the molecular formula, but not always (e. g. water has an empirical and molecular formula of H 2 O. Hydrogen peroxide's empirical formula is HO, but it’s molecular formula is H 2 O 2. E. g. If 9 g of Aluminium react with 35. 5 g of Chlorine, what is the empirical formula of the compound formed? Aluminium 9 = 1/3 moles of Aluminium atoms 27 Chlorine 35. 5 = 1 mole of Chlorine atoms 35. 5 1. Divide the mass of each element by its relative atomic mass to find out the number of moles reacted 2. Create a ratio and simplify if necessary 3. Write a formula based on the ratio Al : Cl 1/3 : 1 1 : 3 Al. Cl 3

Masses of reactants and products This is an important calculation when we want to know how much of each reactant to react together. For example, sodium hydroxide reacts with chlorine gas to make bleach. If we have too much Chlorine, some will be wasted. Too little and not all of the sodium hydroxide will react. 2 Na. OH + Cl 2 Na. OCl + Na. Cl + H 2 O Sodium Chlorine Bleach Hydroxide Salt Water How much Chlorine gas should we bubble through 100 g of Sodium Hydroxide to make Bleach? 1. Na. OH 23 + 16 + 1 = 40 g is one mole of Na. OH 2. We have 100 g in our reaction so… 100 = 2. 5 moles 40 1. Work out the mass of one mole of Sodium Hydroxide 2. Calculate how many moles you have in your reaction 3. Work out how many moles of Chlorine you need 4. Convert this into a mass for Chlorine 3. The chemical equation tells us that we need 2 moles of Sodium Hydroxide (2 Na. OH) for every mole of Chlorine (Cl 2). So we need: 2. 5 = 1. 25 moles of Chlorine 2 4. 35. 5 x 2 = 71 g is one mole of Cl 2 So we need 1. 25 x 71 = 88. 75 g of Chlorine to react with 100 g of Sodium Hydroxide.

Triple Chemistry Calculations Titrations are used to find out the volumes of solutions that react exactly. If you know the concentration of one of the solutions, and the volumes that react together are known as well, you can use this to work out the concentration of the other solution. It is important that you know that mol/dm 3 is a term for concentration of a solution. It means that X moles of substance are dissolved in 1000 cm 3 (1 dm 3) of a solvent. 1 mol/dm 3 of Sodium Hydroxide means that 40 g (the mass of one mole) is dissolved in 1000 cm 3 of the solvent. 20 g dissolved in the same volume would be 0. 5 mol/dm 3. E. g. a student put 25. 0 cm 3 of sodium hydroxide solution with an unknown concentration into a conical flask using a pipette. The sodium hydroxide reacted with exactly 20. 0 cm 3 of 0. 50 mol/dm 3 hydrochloric acid added from a burette. What was the concentration of the sodium hydroxide solution? 1. The equation for this reaction is: Calculate the number of moles reacted in the known substance: (Concentration/1000) x volume used Na. OH + HCl Na. Cl + H 2 O 0. 50 moles of Hydrochloric acid are dissolved in 1000 cm 3 of acid. We used 20 cm 3 in this reaction. So… 0. 50 1000 X 20 = 0. 10 moles of HCl in 20 cm 3 of acid.

Titration Calculations continued… We know from the equation that 1 mole of HCl reacts with 1 mole of Na. OH. So, if there is 0. 10 moles of HCl in the reaction, there must be 0. 10 moles of Na. OH as well. Now we can work out the concentration of Na. OH. If 0. 10 moles of Na. OH are in 25 cm 3 of solution, we can use the equation in the box on the right to work out how many are in 1000 cm 3, therefore the concentration! 2. Use the volume of the second unknown solution to work out the concentration in 1 dm 3 (number of moles/volume of solution) x 1000 0. 10 X 1000 = 0. 40 moles of Na. OH in 1000 cm 3 of solution = 0. 4 mol/dm 3 25 This is the concentration!

Percentage Yield Rather than talk about the yield of a chemical reaction in terms of mass (grams, tonnes etc. ) we can talk about the percentage yield. This gives us an idea of the amount of product that the reaction really makes, compared to what it could possibly make under perfect conditions. There are many reasons why we don’t make 100% every time, such as: – – The reaction may be reversible Some product could be left behind in the apparatus The reactants may not be pure It may be difficult to separate the products if more than one are made. Using this reaction “A + B C”, it was found that in perfect conditions, scientists could make 2. 5 g of C. However, when they tried it out, they only made 1. 5. What is the percentage yield of this reaction? Amount of product produced x 100% Maximum amount of product possible 1. 5 x 100% = 60% percentage yield 2. 5 The higher the percentage yield and atom economy, the better the reactions are for the Earth’s resources, as there’s less waste! Percentage Atom Economy Sometimes reactions will make more than one product. Working out the atom economy is a way of finding out how much of the reactant ends up in the useful product. Relative formula mass of useful product x 100% Relative formula mass of all products Ethanol (C 2 H 5 OH) can be converted into ethene (C 2 H 4) to make polyethene. However, the reaction also produces water. C 2 H 5 OH C 2 H 4 + H 2 O Mass of useful product = (12 x 2) + (1 x 2) = 28 Mass of all products = 28 + [(1 x 4) + 16] = 28 + 20 = 48 28 48 x 100% = 58% atom economy

And Finally… Bond Energy Calculations The energy needed to break a bond between two atoms is called the bond energy. It is measured in k. J/mol – kilojoules per mole. We can use this to work out the energy change (ΔH) for chemical reactions. However, we need a list of the bond energies to do this: Bond energy (k. J/mol) C-C 347 H-Cl 432 C-O 358 H-O 464 C-H 413 H-N 391 C-N 286 H-H 436 C-Cl 346 O=O 498 Cl-Cl 243 NΞN 945 You don’t need to remember this! To calculate bond energy reactions we need to know two things: – – How much energy is needed to break the chemical bonds in the reactants How much energy is released when the new bonds are formed in the products e. g. Hydrogen and Nitrogen are used in the Haber process to make ammonia. What is the overall energy change in this reaction? 3 H 2 + N 2 = 2 NH 3 2. Energy needed to break one mole of nitrogen, and three moles of hydrogen 945 + (3 x 436) = 2253 k. J/mol 3. Energy released when 3 N-H bonds are formed in one mole of ammonia = 3 x -391 = 1173 Energy released in two moles of ammonia = 2 x -1173 = -2346 k. J Overall energy change = +2253 – 2346 = -93 k. J 1. Draw out the structure of each atom so you can work out what bonds are involved 2. Work out the energy needed to break the reactants into atoms. 3. Work out the energy released when the bonds in the products are formed 4. Work out the energy change The bond energies are: NΞN = 945 k. J/mol H-H = 391 k. J/mol