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Yield

Yield. Noadswood Science, 2012. Yield. To be able to calculate the yield from chemical reactions. Conservation Of Mass. What does conservation of mass mean? Mass is never lost or gained in chemical reactions

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Yield

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  1. Yield Noadswood Science, 2012

  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

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