C2.6 Quantitative Chemistry Foundation 1. Relative masses Use the periodic table to find the relative masses of the elements below. (Hint: The top number in each element box) Hydrogen Carbon Nitrogen Oxygen Sodium Magnesium Chlorine Potassium Calcium Zinc 2. Relative formula masses Use the relative masses you found out and combine them to make the relative formula masses of these substances. (Hint: The small numbers only affect the letter it is immediately behind) (Hint: Everything in a bracket is multiplied by the small number behind it) (Hint: Each new atom has a capital letter) Oxygen, O2 Water, H2O Ammonia, NH3 Carbon dioxide, CO2 Copper sulfate, CuSO4 Iron hydroxide, Fe(OH)3 Magnesium hydroxide, Mg(OH)2 Aluminium nitrate, Al(NO3)3
3. Empirical formula This helps you find out the ratio of atoms inside a compound. Worked example: Calcium chloride was made by reacting 10g of calcium and 17.8g of chlorine Symbol for element Ca Cl Mass in g 10 17.8 Relative atomic mass 40 35.5 Divide the mass of each 10 / 40 = 0.25 17.8 / 35.5 = 0.5 element by its relative atomic mass Divide the answers by 0.25 / 0.25 = 1 0.5 / 0.25 = 2 the smallest number to find the simplest ratio Empirical formula CaCl 2 Use the tables below to work out the empirical formulas for these compounds a) Sodium chloride, if a sample contains 4.6 g of sodium and 7.1 g of chlorine Symbol of element Na Cl Mass in g Relative atomic mass Divide the mass of each element by its relative atomic mass Divide the answers by the smallest number to find the simplest ratio Empirical formula b) Zinc chloride, if a sample contains 10.4 g of zinc and 11.36 g of chlorine Symbol of element Zn Cl Mass in g Relative atomic mass Divide the mass of each element by its relative atomic mass Divide the answers by the smallest number to find the simplest ratio Empirical formula
c) Copper sulfate, if a sample contains 19.05 g of copper, 9.6 g of sulfur and 19.2 g of oxygen Symbol of element Cu S O Mass in g Relative atomic mass Divide the mass of each element by its relative atomic mass Divide the answers by the smallest number to find the simplest ratio Empirical formula
4. Percentage by mass This tells you the percentage mass of each element in the compound. E.g. What is the percentage of oxygen in water? Worked example: What is the percentage of oxygen in potassium nitrate, KNO 3? Relative formula mass of potassium nitrate, KNO 3 K = 39, N = 14, O = 16 39 + 14 + (3x16) = 101 Number of atoms of oxygen in KNO 3 3 Relative atomic mass of oxygen 16 % of oxygen in KNO 3 3 x 16 x 100 = 47.5% 101 Use the tables below to complete the questions on percentage by mass a) Calculate the percentage by mass of oxygen in water, H 2 O Relative formula mass of compound Number of atoms of oxygen in the element you are looking for Relative atomic mass of the element you are looking for % of element in compound b) Calculate the percentage by mass of oxygen in carbon dioxide, CO 2 Relative formula mass of compound Number of atoms of oxygen in the element you are looking for Relative atomic mass of the element you are looking for % of element in compound c) Calculate the percentage by mass of oxygen in Calcium chloride, CaCl 2 Relative formula mass of compound Number of atoms of oxygen in the element you are looking for Relative atomic mass of the element you are looking for % of element in compound
5. Yield Theoretical yield is how much product you should make from your start chemicals. Actual yield is how much you actually make in a real reaction Percentage yield is how much actual yield you make compared to theoretical yield. The equation to work out percentage yield is: Percentage yield = actual yield x 100 Theoretical yield For example: In the reaction between hydrogen and oxygen the actual yield was 30g, but the theoretical yield was 36g. 30 / 36 x 100 = 83.3% Work out the percentage yields for the reactions below: Actual Yield in g Theoretical yield in g Percentage yield in % 75 100 20 80 8 64 0.32 1.00 Theoretical yield worked example: (Hint: In the equation, both sides of the arrow should add up to the same mass in g) What will be the yield of water when you react hydrogen and oxygen together? Balanced equation 2H 2 O 2 2H 2 O Calculate formula masses and add g Answer: H = 1 2 x (1+1) = 4g O = 16 2 x 16 = 32g H = 1, O = 16 (2 x 2) + (2 x 16) = 36g 36g of water will be made from 4g of hydrogen and 32g of oxygen
Complete the questions below: Balanced equation Calculate the formula masses and add g Answer H 2 + F 2 HF + Balanced equation Calculate the formula masses and add g Answer K + Br 2 KBr +
C2.6 Quantitative Chemistry 1. Relative formula masses Use the relative masses of elements in the periodic table to find the relative formula mass of each substance below. (Hint: The small numbers only affect the letter it is immediately behind) (Hint: Everything in a bracket is multiplied by the small number behind it) Oxygen, O2 Water, H2O Ammonia, NH3 Carbon dioxide, CO2 Copper sulfate, CuSO4 Iron hydroxide, Fe(OH)3 Magnesium hydroxide, Mg(OH)2 Aluminium nitrate, Al(NO3)3 2. Empirical formula This helps you find out the ratio of atoms inside a compound. See worked examples on p177 for help Use the tables below to work out the empirical formulas for these compounds a) Sodium chloride, if a sample contains 4.6 g of sodium and 7.1 g of chlorine Symbol of element Na Cl Mass in g Relative atomic mass Divide the mass of each element by its relative atomic mass Divide the answers by the smallest number to find the simplest ratio Empirical formula
b) Zinc chloride, if a sample contains 10.4 g of zinc and 11.36 g of chlorine Symbol of element Zn Cl Mass in g Relative atomic mass Divide the mass of each element by its relative atomic mass Divide the answers by the smallest number to find the simplest ratio Empirical formula c) Copper sulfate, if a sample contains 19.05 g of copper, 9.6 g of sulfur and 19.2 g of oxygen Symbol of element Cu S O Mass in g Relative atomic mass Divide the mass of each element by its relative atomic mass Divide the answers by the smallest number to find the simplest ratio Empirical formula
3. Percentage by mass This tells you the percentage mass of each element in the compound. E.g. What is the percentage of oxygen in water? Use p180 for worked examples Use the tables below to complete the questions on percentage by mass a) Calculate the percentage by mass of oxygen in water, H 2 O Relative formula mass of compound Number of atoms of oxygen in the element you are looking for Relative atomic mass of the element you are looking for % of element in compound b) Calculate the percentage by mass of oxygen in calcium chloride, CaCl 2 Relative formula mass of compound Number of atoms of oxygen in the element you are looking for Relative atomic mass of the element you are looking for % of element in compound c) Calculate the percentage by mass of oxygen in magnesium nitrate, Mg(NO 3 ) 2 Relative formula mass of compound Number of atoms of oxygen in the element you are looking for Relative atomic mass of the element you are looking for % of element in compound
4. Yield The equation to work out percentage yield is: Percentage yield = actual yield x 100 Theoretical yield Work out the percentage yields for the reactions below: Actual Yield in g Theoretical yield in g Percentage yield in % 75 100 20 80 8 64 0.32 1.00 Percentage Yield in % Theoretical yield in g Actual yield in g 25 50 80 40 Theoretical yield This tells you how much product you should be making from a reaction. Use p182 to find a worked example Complete the questions below using the tables: (Hint: You need to balance the equations before you work out the products) Balanced equation Calculate the formula masses and add g Answer Fe 2 O 3 + CO Fe + CO 2
Balanced equation Calculate the formula masses and add g Answer SO 2 + O 2 SO 3 Balanced equation Calculate the formula masses and add g Answer CaCO3 CaO + CO 2
C2.6 Quantitative Chemistry Higher 1. Empirical formula This helps you find out the ratio of atoms inside a compound. See worked examples on p177 for help Work out the empirical formulas of the compounds below: d) Sodium chloride, if a sample contains 4.6 g of sodium and 7.1 g of chlorine e) Zinc chloride, if a sample contains 10.4 g of zinc and 11.36 g of chlorine f) Copper sulfate, if a sample contains 19.05 g of copper, 9.6 g of sulfur and 19.2 g of oxygen 2. Percentage by mass This tells you the percentage mass of each element in the compound. E.g. What is the percentage of oxygen in water? Use p180 for worked examples Work out the percentage by mass of each atom in the formulas below: d) Calculate the percentage by mass of oxygen in water, H 2 O e) Calculate the percentage by mass of oxygen in calcium chloride, CaCl 2 f) Calculate the percentage by mass of oxygen in magnesium nitrate, Mg(NO 3 ) 2
3. Yield The equation to work out percentage yield is: Percentage yield = actual yield x 100 Theoretical yield Work out the percentage, theoretical or actual yields for the reactions below: a) The theoretical yield is 100 g but the actual yield was 75g. b) The theoretical yield is 80g whereas the actual yield was 20g c) The actual yield as 8 g whereas the expected yield was 64g d) The percentage yield was 25 g and the theoretical yield was 50g e) The percentage yield was 80g and the theoretical yield was 40g Theoretical yield This tells you how much product you should be making from a reaction. Use p182 to find a worked example (Hint: You need to balance the equations before you work out the products) a) Iron oxide (Fe 2 O 3 ) was reacted with Carbon monoxide (CO) to make Iron (Fe) and Carbon dioxide (CO 2 ) b) Sulfur dioxide (SO 2 ) was reacted with Oxygen (O 2 ) to produce Sulphur trioxide (SO 3 ) c) Calcium carbonate (CaCO 3 ) is thermally decomposed to produce Calcium oxide (CaO) and Carbon dioxide (CO 2 )
4. Calculating a reacting mass During a chemical reaction, no atoms are lost or made, they are just rearranged to make new substances. You can use relative masses and the balanced equation for a reaction to calculate the mass of a reactant or a product. Worked example In a firework, potassium nitrate (KNO 3 ) is decomposed to potassium nitrate (KNO 2 ) and oxygen (O 2 ). What mass of potassium nitrate is needed in a firework to make 1.6g of oxygen? Write the balanced equation 2KNO 3 2KNO 2 + O 2 Work out the relative masses of the substances needed in the calculation. Notice that we need to multiply the relative masses of KNO3 by 2 because there are two particles of KNO3 in the balanced equation Convert the relative masses into the units in the question Divide the answers by the smallest number to find the ratio Find the mass of potassium nitrate needed to make 1.6 g of oxygen Answer KNO3 = 39 (K) + 14 (N) + (3x16) (O) = 101 O2 = 2x16 = 32 2 x 101 32 202 32 202g 32g so 202/ 32 32/32 = 10.1 1 6.3 x 1.6 1 x 1.6 10.1 1.6 10.1g of potassium nitrate are needed to make 1.6g of oxygen Use the worked example above to complete the questions below. a) Copper oxide can be reduced by hydrogen. What mass of copper could be obtained from 79.5g of copper oxide? CuO + H 2 Cu + H 2 O b) Calculate the mass of carbon dioxide you could obtain by adding hydrochloric acid to 15g of calcium carbonate: CacO 2 + 2HCl CaCl 2 + H 2 O + CO 2
C2.6 Quantitative Chemistry Higher extension 1. Empirical formula This helps you find out the ratio of atoms inside a compound. See worked examples on p177 for help Work out the empirical formulas of the compounds below: g) Sodium chloride, if a sample contains 4.6 g of sodium and 7.1 g of chlorine h) Zinc chloride, if a sample contains 10.4 g of zinc and 11.36 g of chlorine i) Copper sulfate, if a sample contains 19.05 g of copper, 9.6 g of sulfur and 19.2 g of oxygen Extension Write a short paragraph to show how relative formula mass, empirical formula and molecular formula are connected.
2. Percentage by mass This tells you the percentage mass of each element in the compound. Use p180 for worked examples Work out the percentage by mass of each atom in the formulas below: a) Calculate the percentage by mass of oxygen in water, H 2 O b) Calculate the percentage by mass of oxygen in calcium chloride, CaCl 2 c) Calculate the percentage by mass of oxygen in magnesium nitrate, Mg(NO 3 ) 2 Extension Make notes to explain how to calculate the percentage composition by mass for a friend who has missed the lesson. Use one or more of the compounds in the following equation as an example: MgCO 3 + H 2 SO 4 MgSO 4 + H 2 O + CO 2
3. Yield The equation to work out percentage yield is: Percentage yield = actual yield x 100 Theoretical yield Work out the percentage, theoretical or actual yields for the reactions below: f) The theoretical yield is 100 g but the actual yield was 75g g) The theoretical yield is 80g whereas the actual yield was 20g h) The actual yield as 8 g whereas the expected yield was 64g i) The percentage yield was 25 g and the theoretical yield was 50g j) The percentage yield was 80g and the theoretical yield was 40g Theoretical yield This tells you how much product you should be making from a reaction. Use p182 to find a worked example (Hint: You need to balance the equations before you work out the products) d) Iron oxide (Fe 2 O 3 ) was reacted with Carbon monoxide (CO) to make Iron (Fe) and Carbon dioxide (CO 2 ) e) Sulfur dioxide (SO 2 ) was reacted with Oxygen (O 2 ) to produce Sulphur trioxide (SO 3 ) f) Calcium carbonate (CaCO 3 ) is thermally decomposed to produce Calcium oxide (CaO) and Carbon dioxide (CO 2 ) Extension Explain the reasons why the actual yield of a reaction may be less than the theoretical yield.
5. Calculating a reacting mass During a chemical reaction, no atoms are lost or made, they are just rearranged to make new substances. You can use relative masses and the balanced equation for a reaction to calculate the mass of a reactant or a product. Worked example In a firework, potassium nitrate (KNO 3 ) is decomposed to potassium nitrate (KNO 2 ) and oxygen (O 2 ). What mass of potassium nitrate is needed in a firework to make 1.6g of oxygen? Write the balanced equation 2KNO 3 2KNO 2 + O 2 Work out the relative masses of the substances needed in the calculation. Notice that we need to multiply the relative masses of KNO3 by 2 because there are two particles of KNO3 in the balanced equation Convert the relative masses into the units in the question Divide the answers by the smallest number to find the ratio Find the mass of potassium nitrate needed to make 1.6 g of oxygen Answer KNO3 = 39 (K) + 14 (N) + (3x16) (O) = 101 O2 = 2x16 = 32 2 x 101 32 202 32 202g 32g so 202/ 32 32/32 = 10.1 1 6.3 x 1.6 1 x 1.6 10.1 1.6 10.1g of potassium nitrate are needed to make 1.6g of oxygen Use the worked example above to complete the questions below. c) Copper oxide can be reduced by hydrogen. What mass of copper could be obtained from 79.5g of copper oxide? CuO + H 2 Cu + H 2 O d) Calculate the mass of carbon dioxide you could obtain by adding hydrochloric acid to 15g of calcium carbonate: CacO 2 + 2HCl CaCl 2 + H 2 O + CO 2