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GCSE Revision (Higher) Formulae The chemical formula of a compound shows how many of each type of atom join together to make the units which make up the compound. For example, in iron sulfide every iron atom is joined to one sulfur atom, so we show its formula as FeS. In sodium oxide, there are two sodium atoms for every oxygen atom, so we show its formula as Na2O. Notice that the '2' is written as a subscript, so Na2O would be wrong. This diagram shows that one carbon atom and two oxygen atoms combine to make up the units of carbon dioxide. Its chemical formula is written as CO2. Carbon dioxide units contain one carbon atom and two oxygen atoms Sometimes you see more complex formulae such as Na2SO4 and Fe(OH)3 : a unit of Na2SO4 contains two sodium atoms, one sulfur atom and four oxygen atoms joined together a unit of Fe(OH)3 contains one iron atom, three oxygen atoms and three hydrogen atoms - the brackets show that the 3 applies to O and H Chemical equations Word equations are useful to show which chemicals react together (the reactants) and which chemicals are produced (the products). Symbol equations allow chemists to work out the masses that will react or be produced. Copper and oxygen reaction - getting a balanced equation Balanced symbol equations show what happens to the different atoms in reactions. For example, copper and oxygen react together to make copper oxide. Take a look at this word equation for the reaction: copper + oxygen → copper oxide copper and oxygen are the reactants because they are on the left of the arrow copper oxide is the product because it is on the right of the arrow If we just replace the words shown above by the correct chemical formulas, we will get an unbalanced equation, as shown here: Cu + O2 → CuO Notice that there are unequal numbers of each type of atom on the left-hand side compared with the right-hand side. To make things equal, you need to adjust the number of units of some of the substances until you get equal numbers of each type of atom on both sides. Here is the balanced symbol equation: 2Cu + O2 → 2CuO You can see that now there are two copper atoms and two oxygen atoms on each side. This matches what happens in the reaction. Two atoms of copper react with two atoms of oxygen to form two molecules of copper oxide Remember: never change a formula to balance an equation State symbols Sometimes it is useful to know whether the reactants and products in a chemical reaction are solids, gases, liquids or dissolved in water. We can add state symbols to a symbol equation to show this. State symbol and meaning Symbol Meaning (s) Solid (l) Liquid (g) Gas (aq) Aqueous (dissolved in water) For example, for the reaction between sodium and water, this is the symbol equation with state symbols: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Conservation of mass Mass is never lost or gained in chemical reactions. We say that mass is always conserved. In other words, the total mass of products at the end of the reaction is equal to the total mass of the reactants at the beginning. This is because no atoms are created or destroyed during chemical reactions. The principle of conservation of mass allows you to work out the mass of one substance in a reaction if the masses of the other substances are known. For example: Percentage yield For a given mass of reactants, it is possible to calculate the maximum mass that could be made of a certain product. This mass is called the predicted yield. The actual yield is the mass of product made when the reaction is actually carried out. Calculating percentage yield The percentage yield is calculated using this formula: Equation: Percentage yield = Actual yield obtained/(divided) by the Theoretical or predicted yield x 100 For example, if the predicted yield is 20 g but the actual yield is 15 g: Percentage yield = (15 ÷ 20) × 100 = 75% Losing product A 100 per cent yield means that no product has been lost, while a 0 per cent yield means that no product has been made. There are several reasons why the percentage yield of a product might be less than 100 per cent. These include: loss when filtering loss when evaporating loss when transferring liquids not all reactants reacting to make product Percentage purity Percentage purity of a substance can be calculated by dividing the mass of the pure chemical by the total mass of the sample, and then multiplying this number by 100. Calcium carbonate ? Calcium oxide 28g + Carbon dioxide 22g Calcium carbonate is made up of 28 grams of calcium oxide and 22 grams of carbon dioxide In practice, it is not always possible to get the entire calculated amount of product from a reaction. This is because: reversible reactions may not be completed some product may be lost when it is removed from the reaction mixture some of the reactants may react in an unexpected way Empirical formulae The empirical formula of a compound is the simplest whole number ratio of atoms of each element in the compound. It is determined using data from experiments and therefore empirical. For example, the molecular formula of glucose is C6H12O6 but the empirical formula is CH2O. This is because we can divide each number in C6H12O6 by 6 to make a simpler whole number ratio. On the other hand, a compound which has the empirical formula of CH2 could have a molecular formula of C2H4, C3H6, C4H8 or even C13H26. You can use information about reacting masses to calculate the formula of a compound. Here is an example: 3.2 g of sulphur reacts with oxygen to produce 6.4 g of sulphur oxide. What is the formula of the oxide? You need to use the fact that the Ar (relative atom mass) of sulphur is 32 and the Ar of oxygen is 16. Step Action Result 1 Write the element symbols S O 2 Write the masses 6.4 g – 3.2 g = 3.2 g 3.2 g 3 Write the Ar values 32 16 4 Divide masses by Ar 3.2 ÷ 32 = 0.1 3.2 ÷ 16 = 0.2 5 Divide by the smallest number 0.1 ÷ 0.1 = 1 0.2 ÷ 0.1 = 2 6 Write the formula SO2 The action at Step 5 usually gives you the simplest whole number ratio straightaway. Sometimes it does not, so you might get 1 and 1.5. In this example, you would multiply both numbers by 2, giving 2 and 3 (instead of rounding 1.5 up to 2). Converting the empirical formula to a molecular formula From the empirical formula, you can work out the molecular formula if you know the relative formula mass (Mr) of the compound. Add up the atomic masses of the atoms in the empirical formula. For example, the empirical formula of a hydrocarbon is CH2 and its Mr is 42. the mass of the atoms in the empirical formula is 14 42 ÷ 14 = 3 so you need to multiply the numbers in the empirical formula by 3 The molecular formula of the hydrocarbon is therefore C3H6. Empirical formulae experiments Finding the empirical formula of a metal oxide The empirical formula of magnesium oxide can be calculated using the following experiment, which finds the mass of the magnesium and oxygen atoms in a sample of the compound. 1. Weigh a crucible (with its lid). 2. Put a sample of clean magnesium ribbon into the crucible and weigh it with the lid. Calculate the mass of magnesium by subtracting the mass of the empty crucible. 3. Strongly heat the crucible over a Bunsen burner for several minutes. 4. Carefully lift the lid from time to time to allow sufficient air into the crucible for the magnesium to fully oxidise without letting any magnesium oxide escape. 5. Continue heating until the mass of the crucible reaches a constant (maximum) mass, indicating that the reaction is complete. 6. Measure the mass of the crucible and contents again. Calculate the mass of the magnesium oxide by subtracting the mass of the empty crucible. 7. To work out the empirical formula, you need the mass of the magnesium (from step 2) and the mass of the oxygen atoms as well. To find the mass of the oxygen atoms, subtract the mass of magnesium used from the mass of the magnesium oxide (from step 6). 8. Now divide each of the two masses by the relative atomic masses of the elements and simplify the ratio. Finding the molecular formula - crystallisation Some ionic compounds have a fixed number of water molecules trapped inside their crystal lattice. The formula of a hydrated ionic compound can be calculated by dehydrating it: 1. Record the mass of an evaporating basin. 2. Add a known mass of hydrated copper sulfate. 3. Heat over a Bunsen burner, gently stirring, until the blue copper sulfate has turned completely white and the mass has reached a constant (minimum) mass, indicating that all the water has been lost. 4. Record the mass of the evaporating basin and contents and then use this to calculate the mass of the white anhydrous copper sulfate remaining, and also the mass of the water lost. 5. Divide the mass of the white copper sulfate by the Mr (relative formula mass) of CuSO4 (159.5) and divide the mass of water by the Mr of water (18). Simplify the ratio of waters to copper sulfates to find the whole number represented here by ‘x’: CuSO4.xH2O Reacting mass calculations If you know the mass of a reactant and a product, you can use simple ratios to calculate reacting masses and product masses. Example 1 – reactant mass Question: When 12 g of carbon is burned in air, 44 g of carbon dioxide is produced. What mass of carbon is needed to produce 11 g of carbon dioxide? Answer: 12 g of carbon makes 44 g of carbon dioxide. 12 ÷ 44 g of carbon will make 1 g of carbon dioxide. You will need 11 × (12 ÷ 44 g) = 3 g of carbon to make 11 g of carbon dioxide. Example 2 – product mass Question: When 5.0 g of calcium carbonate is decomposed by heating, it produces 2.2 g of carbon dioxide. What mass of calcium carbonate is needed to produce 8.8 g of carbon dioxide? Answer: 5 g of calcium carbonate makes 2.2 g of carbon dioxide. 5 ÷ 2.2 g of calcium carbonate will make 1 g of carbon dioxide. You will need 8.8 × (5 ÷ 2.2 g) = 20 g of carbon to make 8.8 g of carbon dioxide. More reacting mass calculations The mass of a product or reactant can be calculated using the balanced equation. Follow these steps: 1. Write out the balanced symbol equation. Underline the two substances you are interested in. 2. Write the given mass of a substance under its formula. 3. Work out the total relative formula mass (Mr) for each substance (the one you know and the one you are trying to find out). Write these under their formulae. 4. Calculate: unknown mass = known mass ÷ total Mr of known substance × total Mr of unknown substance Worked example Question: Calculate the mass of iron (Fe) that can be extracted from 8 g of Fe2O3 in the reaction with carbon. Answer: Step 4Fe + 3CO2 1 2Fe2O3 + 3C → ?g 2 8g 4 × 56 = 224 3 2 × 160 = 320 4 mass of iron formed = (8 ÷ 320) × 224 = 5.6 g Mole calculations in solutions Concentration calculations You can calculate the amount of a substance (in moles) in a solution if you know the volume and concentration. You can also work out the concentration of an acid reacting with an alkali, or vice versa. Concentration, amount of solute and volume of solution are linked by this equation: Concentration in mol/dm3 = amount in mol ÷ volume in dm3 This equation can be rearranged to find the amount of solute or volume of solution. Divide a volume in cm3 by 1000 to convert it to a volume in dm3. Calculating concentration Question: 0.5 mol of solute is dissolved in 250 cm3 of solution. Calculate its concentration. Answer: 250 cm3 = 250 ÷ 1000 = 0.25 dm3 Concentration = 0.5 ÷ 0.25 = 2.0 mol/dm3 Calculating amount of solute Rearranging the equation: amount = concentration × volume Question: Calculate the amount of solute dissolved in 2 dm3 of a 0.1 mol/dm3 solution. Answer: Amount = 0.1 × 2 = 0.2 mol/dm3 Calculating volume Rearranging the equation: volume = amount ÷ concentration Question: Calculate the volume of a 2 mol/dm3 solution that contains 0.5 mol of solute. Answer: Volume = 0.5 ÷ 2 = 0.25 dm3 Note that 0.25 dm3 is the same as 250 cm3 (0.25 × 1000). Concentrations can also be expressed in grams per decimetre cubed (g/dm3). To convert the two different units of concentration, use the relative formula mass (Mr). Percentage yield For a given mass of reactants, it is possible to calculate the maximum mass that could be made of a certain product. This mass is called the predicted yield. The actual yield is the mass of product made when the reaction is actually carried out. Chemical formulae and equations 1 What is the correct formula for carbon dioxide? C2 O CO2 CO2 2 The formula for sulfuric acid is H2SO4. How many of each type of atom are there? Two hydrogens, one sulfur and four oxygens One hydrogen, two sulfurs and four oxygens Two hydrogens, two sulfurs and four oxygens 3 The formula for ammonium sulfate is (NH4)2SO4. How many hydrogen atoms are there in this formula? 4 2 8 4 What is used in the middle of a chemical equation? = → Δ In the reaction CH4 + O2 → CO2 + H2O, what is the formula of a reactant? 5 CH4 CO2 H2O 6 When hydrogen (H2) reacts with oxygen (O2), water is produced (H2O). Which of these shows a balanced symbol equation for this reaction? H2 + O2 → H2O 2H2 + O2 → 2H2O 2H2 + 2O2 → 2H2O 7 What is the state symbol for solid? (s) (l) (g) 8 What is the meaning of the state symbol (aq)? Liquid Dissolved in water Gaseous 9 What is the empirical formula of C4H8O2? CH2O C8H16O4 C2H4O 10 4.0 g of magnesium oxide was found to contain 2.4 g of magnesium. The rest was oxygen. What is the empirical formula of magnesium oxide? MgO2 Mg2O MgO Chemical calculations 1 If 24 g of magnesium reacts with 16 g of oxygen to produce 40 g of magnesium oxide, what mass of magnesium would you need to start with to produce 10 g of magnesium oxide? 6 g of magnesium 12 g of magnesium 4 g of magnesium 2 What is the relative formula mass of calcium carbonate, CaCO3? 50 100 68 3 The relative formula mass of CaO is 56 and the relative formula mass of CO2 is 44. Work out the mass of CaO that can be obtained from 200 g of CaCO3. CaCO3 → CaO + CO2. 56 g 224 g 112 g 4 What mass of CO2 will be produced when 50 g of CaCO3 decomposes? 22 g 44 g 11 g 5 What is 20 cm3 in dm3? 0.002 dm3 0.02 dm3 0.2 dm3 6 What is the concentration of a solution if 0.25 mol is dissolved in 500 cm3 of solution? 1 mol/dm3 2 mol/dm3 0.5 mol/dm3 7 How many moles of solute are there in 25 cm3 of a solution that has a concentration of 0.4 mol/dm3? 0.01 moles 10 moles 0.0625 moles 8 If a solution of NaOH is 0.1 mol/dm3, what is its concentration in g/dm3? 40 g/dm3 4 g/dm3 0.4 g/dm3 9 In a reaction, the predicted yield was 80 g, but the actual yield was 60 g. What was the percentage yield? 75% 133% 50% 10 What is the percentage yield if the actual yield was 2.4 g and the predicted yield was 3.0 g? 75% 80% 125% Answers 1 What is the correct formula for carbon dioxide? CO2 2 The formula for sulfuric acid is H2SO4. How many of each type of atom are there? Two hydrogens, one sulfur and four oxygens 3 The formula for ammonium sulfate is (NH4)2SO4. How many hydrogen atoms are there in this formula? 8 4 What is used in the middle of a chemical equation? 5 → In the reaction CH4 + O2 → CO2 + H2O, what is the formula of a reactant? CH4 6 When hydrogen (H2) reacts with oxygen (O2), water is produced (H2O). Which of these shows a balanced symbol equation for this reaction? 2H2 + O2 → 2H2O 7 What is the state symbol for solid? (s) 8 What is the meaning of the state symbol (aq)? Dissolved in water 9 What is the empirical formula of C4H8O2? C2H4O 10 4.0 g of magnesium oxide was found to contain 2.4 g of magnesium. The rest was oxygen. What is the empirical formula of magnesium oxide? MgO 1 If 24 g of magnesium reacts with 16 g of oxygen to produce 40 g of magnesium oxide, what mass of magnesium would you need to start with to produce 10 g of magnesium oxide? 6g of magnesium If 24 g of Mg produces 40 of MgO, and you want 10 g, you need to divide all the reacting quantities by 4. So 24 g of Mg divided by 4 equals 6 g. You divided the mass of Mg by 10. 2 What is the relative formula mass of calcium carbonate, CaCO3? 100 The relative formula mass is calculated by adding together the masses of the atoms in the compound: 40 + 12 + 16 + 16 + 16 = 100. 3 The relative formula mass of CaO is 56 and the relative formula mass of CO2 is 44. Work out the mass of CaO that can be obtained from 200 g of CaCO3. CaCO3 → CaO + CO2. 112 g The ratio of RFMs for this reaction is 100 → 56 + 44. But starting with 200 g of CaCO3 means we will end up with twice as much CaO: 112 g. 4 What mass of CO2 will be produced when 50 g of CaCO3 decomposes? 22 g Working from the ratio of RFMs for this reaction, we see that 100 g of CaCO3 will produce 44 g of CO2. So 50 g of CaCO3 will make 22 g of CO2. 5 What is 20 cm3 in dm3? 0.02 dm3 To convert cm3 to dm3, you divide by 1000. So 20 ÷ 1000 = 0.02 dm3. 6 What is the concentration of a solution if 0.25 mol is dissolved in 500 cm3 of solution? 0.5 mol/dm3 Convert the volume to dm3 (0.5) and then divide the moles by the volume: 0.25 ÷ 0.5 = 0.5 mol/dm3. 7 How many moles of solute are there in 25 cm3 of a solution that has a concentration of 0.4 mol/dm3? 0.01 moles Convert the volume to dm3 (0.025) and then multiply this by the concentration: 0.025 x 0.4 = 0.01 moles. 8 If a solution of NaOH is 0.1 mol/dm3, what is its concentration in g/dm3? 4 g/dm3 To convert mol/dm3 to g/dm3, multiply by the RFM. So 0.1 x 40 = 4 g/dm3. 9 In a reaction, the predicted yield was 80 g, but the actual yield was 60 g. What was the percentage yield? 75% Percentage yield is calculated by dividing the actual yield by the predicted yield and then multiplying by 100. So (60 ÷ 80) x 100 = 75%. 10 What is the percentage yield if the actual yield was 2.4 g and the predicted yield was 3.0 g? 80% Percentage yield is calculated by dividing the actual yield by the predicted yield and then multiplying by 100. So (2.4 ÷ 3.0) x 100 = 80%.