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Additional Chemistry Calculations

Additional Chemistry Calculations. Relative atomic and Formula Masses

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Additional Chemistry Calculations

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  1. 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. Relative formula mass can be found by adding up the relative atomic masses of each element in a compound. E.g. Carbon Dioxide (CO2) 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 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 44g. Simple! The relative atomic mass can be found by looking at the periodic table, It is always the larger of the two numbers. x 2 because it’s “O2”

  2. 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? • Work out the mass of MgO • Convert to grams • Work out the percentage 24 + 16 = 40 40g 24 X 100% = 60% is Magnesium, so 40% must be Oxygen! Work out the formula mass of the compound Convert this into grams Work out the percentage by using this equation: 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 H2O. Hydrogen peroxide's empirical formula is HO, but it’s molecular formula is H2O2. E.g. If 9g of Aluminium react with 35.5g of Chlorine, what is the empirical formula of the compound formed? Aluminium 9 27 Chlorine 35.5 35.5 • Mass of element • Total mass of compound x 100% Divide the mass of each element by its relative atomic mass to find out the number of moles reacted Create a ratio and simplify if necessary Write a formula based on the ratio = 1/3 moles of Aluminium atoms = 1 mole of Chlorine atoms Al : Cl 1/3 : 1 1 : 3 AlCl3

  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. 2NaOH + Cl2 NaOCl + NaCl + H2O How much Chlorine gas should we bubble through 100g of Sodium Hydroxide to make Bleach? 1. NaOH 23 + 16 + 1 = 40g is one mole of NaOH 2. We have 100g in our reaction so… 100 = 2.5 moles 40 3. The chemical equation tells us that we need 2 moles of Sodium Hydroxide (2NaOH) for every mole of Chlorine (Cl2). So we need: 2.5 = 1.25 moles of Chlorine 2 4. 35.5 x 2 = 71g is one mole of Cl2 So we need 1.25 x 71 = 88.75g of Chlorine to react with 100g of Sodium Hydroxide. Bleach Salt Water Sodium Hydroxide Chlorine Work out the mass of one mole of Sodium Hydroxide Calculate how many moles you have in your reaction Work out how many moles of Chlorine you need Convert this into a mass for Chlorine

  4. 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 2.5g of C. However, when they tried it out, they only made 1.5. What is the percentage yield of this reaction? 1.5 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! Amount of product produced Maximum amount of product possible x 100% x 100% = 60% percentage yield

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