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2. Yield • To be able to calculate the yield from chemical reactions

3. Conservation Of Mass • What does conservation of mass mean? • Mass is never lost or gained in chemical reactions • Mass is always conserved – the total mass of products at the end of the reaction is equal to the total mass of the reactants at the beginning • This fact allows you to work out the mass of one substance in a reaction if the masses of the other substances are known…

4. Practice • Often in chemical reactions is appears that mass has been lost / gained – why could this be? • In practice, it is not always possible to get all of the calculated amount of product from a reaction: - • Reversible reactions may not go to completion • Some product may be lost when it is removed from the reaction mixture • Some of the reactants may react in an unexpected way • The reactants could react with something which was not measured (i.e. oxygen within the air would add mass to the final product) • Some of the products might be hard to measure (i.e. a gas could be given off from the reaction)

5. Reacting Masses • To calculate the mass in reactions there are three steps: - • Write out the balanced equation • Work out the Mr (off the bits you want) • Apply the rule ‘divide to get one, multiply to get all (first to the substance given, then to the one with no information)’! • E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in the air?

6. Reacting Masses • E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in the air? 2Mg + O2→ 2MgO • Relative formula: - (2 x 24) → 2 x (24 + 16) 48 80 • Apply the rule: divide to get one, multiply to get all 48g Mg reacts to give 80g MgO 1g Mg reacts to give 1.67g MgO 60g Mg reacts to give 100g MgO ÷ 48 ÷ 48 x 60 x 60

7. Reacting Masses • To calculate the mass in reactions there are three steps: - • Write out the balanced equation • Work out the Mr (off the bits you want) • Apply the rule ‘divide to get one, multiply to get all (first to the substance given, then to the one with no information)’! • E.g. if we have 50g of CaCO3, how much CaO can we make?

8. Reacting Masses • E.g. if we have 50g of CaCO3, how much CaO can we make? CaCO3 → CaO + CO2 • Relative formula: - 40 + 12 (3 x 16) → 40 + 16 100 56 • Apply the rule: divide to get one, multiply to get all 100g CaCO3 reacts to give 56g CaO 1g CaCO3reacts to give 0.56g CaO 50g CaCO3reacts to give 28g CaO ÷ 100 ÷ 100 x 50 x 50

9. Reacting Masses • To calculate the mass in reactions there are three steps: - • Write out the balanced equation • Work out the Mr (off the bits you want) • Apply the rule ‘divide to get one, multiply to get all (first to the substance given, then to the one with no information)’! • E.g. what mass of carbon will react with hydrogen to produce 24.6g of propane?

10. Reacting Masses • E.g. what mass of carbon will react with hydrogen to produce 24.6g of propane? 3C + 4H2 → C3H8 • Relative formula: - 3 x 12 → (3 x 12)+ (8 x 1) 36 44 • Apply the rule: divide to get one, multiply to get all 36g C reacts to give 44g C3H8 0.82g C reacts to give 1g C3H8 20.1g C reacts to give 24.6g C3H8 ÷ 44 ÷ 44 x 24.6 x 24.6

11. Yield • The amount of product made is called the yield – in a chemical reaction no atoms are lost or gained but sometimes the yield is not what you would expect • Theoretical yield: maximum products that are made if reactants react • Actual yield: the amount of product which actually forms

12. Percentage Yield • The yield of a reaction is the actual mass of product obtained – the percentage yield can be calculated: - Percentage yield = mass product obtained x 100 theoretical mass • For example, the maximum theoretical mass of product in a certain reaction is 20g, but only 15g is actually obtained… Percentage yield = 15⁄20 × 100 = 75%

13. Experiment • Precipitation is the formation of an insoluble solid when two solutions are mixed - e.g. barium sulfate is produced by precipitation from barium nitrate and sodium sulfate solutions • Write a word and symbol equation for this reaction… barium nitrate + sodium sulfate  barium sulfate + sodium nitrate Ba(NO3)2 + Na2SO4 BaSO4 + 2NaNO3 • Using the standard procedure carry out the precipitation reaction…

14. Experiment • Pour 50cm3 water into a 100cm3 beaker • Weigh 2.6g barium nitrate • Combine the two and stir (until all barium nitrate is dissolved) • Pour this into the 250cm3 beaker • Measure out 75cm3 sodium sulfate into a 100cm3 beaker • Add the two solutions together • Stir well (notice the white precipitate) • Filter the mixture using a funnel and filter paper (make sure you weigh your filter paper) - wash the residue with a little water • Dry the precipitate - weigh to find our yield (- filter paper)!…

15. Calculating Yield • Calculate the maximum theoretical yield of barium sulfate and then work out your own percentage yield for the experiment… • Ar of barium = 137; sulfur = 32; nitrogen = 14; and oxygen = 16 barium nitrate + sodium sulfate  barium sulfate + sodium nitrate Ba(NO3)2 + Na2SO4 BaSO4 + 2NaNO3

16. Theoretical Yield Ba(NO3)2 + Na2SO4BaSO4 + 2NaNO3 • Relative formula (barium nitrate and barium sulfate): - 137 + (2 x 14) + (2 x (3 x 16)) → 137 + 32 + (4 x 16) 261 233 • Apply the rule: divide to get one, multiply to get all 261g Ba(NO3)2reacts to give 233g BaSO4 2.61g Ba(NO3)2reacts to give 2.33g BaSO4 Maximum theoretical yield for experiment = 2.33g BaSO4

17. Percentage Yield • The yield of a reaction is the actual mass of product obtained – the percentage yield can be calculated: - Percentage yield = mass product obtained x 100 theoretical mass • In this experiment the maximum theoretical yield is 2.33g – if you got, for example, 1.25g then Percentage yield = 1.25⁄2.33 × 100 = 53.6%

18. Empirical Formula • Empirical formula is a simple expression of the relative numbers of each type of atom in it… • The following steps are used to calculate empirical formula: - • List all the elements in the compound • Write underneath them their experimental masses or percentages • Divide each mass or percentage by the Ar for that particular element • Turn the numbers until you get a ratio by multiplying / dividing them by well chosen numbers • Get the ratio in its simplest form…

19. Empirical Formula – Example • Find the empirical formula of the iron oxide produced when 44.8g of iron react with 19.2g of oxygen (Ar iron = 56; oxygen = 16) Iron (Fe) Oxygen (O) 44.8g 19.2g 44.8/56 = 0.8 19.2/16 = 1.2 8 12 2 3 Simplest formula is 2 atoms of Fe to 3 atoms of O (Fe2O3) Experiment mass ÷ Ar for each element x 10 ÷ 4

20. Empirical Formula – Example • Find the empirical formula of sulfur oxide if 3.2g of sulfur reacts with oxygen to produce 6.4g sulfur oxide (Ar sulfur = 32; oxygen = 16) • Conservation of mass tells us that the mass of oxygen equals the mass of sulfur oxide minus the mass of sulfur – the mass of oxygen reacted = 6.4 - 3.2 = 3.2g Sulfur (S) Oxygen (O) 3.2g 3.2g 3.2/32 = 0.1 3.2/16 = 0.2 1 2 Simplest formula is 1 atoms of S to 2 atoms of O (SO2) Experiment mass ÷ Ar for each element x 10

21. Moles • A mole is a number – 6.023 x 1023 • When you get precisely this number of atoms of carbon-12 it weighs 12g • That number of atoms or molecules of any element or compound will weigh exactly the same number of grams as the relative atomic mass of the element or compound: - • Iron has an Ar of 56 – 1 mole of iron weighs 56g • Nitrogen has a Mr of 28 (2 x 14) – 1 mole of nitrogen weighs 28g • Carbon dioxide has a Mr of 44 (12 + 2 x 16) – 1 mole of carbon dioxide weighs 44g

22. Moles • To work out the number of moles in a given mass: - Number of moles = Mass (g) of element or compound Mr of element or compound • How many moles are there in 42g of carbon? 42/12 = 3.5 moles