1. Yield Noadswood Science, 2012

2. Yield Monday, January 25, 2016  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 M r (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 + O 2 → 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 x 60

7. Reacting Masses  To calculate the mass in reactions there are three steps: -  Write out the balanced equation  Work out the M r (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 CaCO 3, how much CaO can we make?

8. Reacting Masses  E.g. if we have 50g of CaCO 3, how much CaO can we make? CaCO 3 → CaO + CO 2  Relative formula: - 40 + 12 (3 x 16) → 40 + 16 100 56  Apply the rule: divide to get one, multiply to get all 100g CaCO 3 reacts to give 56g CaO 1g CaCO 3 reacts to give 0.56g CaO 50g CaCO 3 reacts to give 28g CaO ÷ 100 x 50

9. Reacting Masses  To calculate the mass in reactions there are three steps: -  Write out the balanced equation  Work out the M r (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 + 4H 2 → C 3 H 8  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 C 3 H 8 0.82g C reacts to give 1g C 3 H 8 20.1g C reacts to give 24.6g C 3 H 8 ÷ 44 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(NO 3 ) 2 + Na 2 SO 4  BaSO 4 + 2NaNO 3  Using the standard procedure carry out the precipitation reaction…

14. Experiment  Pour 50cm 3 water into a 100cm 3 beaker  Weigh 2.6g barium nitrate  Combine the two and stir (until all barium nitrate is dissolved)  Pour this into the 250cm 3 beaker  Measure out 75cm 3 sodium sulfate into a 100cm 3 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…  A r of barium = 137; sulfur = 32; nitrogen = 14; and oxygen = 16 barium nitrate + sodium sulfate  barium sulfate + sodium nitrate Ba(NO 3 ) 2 + Na 2 SO 4  BaSO 4 + 2NaNO 3

16. Theoretical Yield Ba(NO 3 ) 2 + Na 2 SO 4  BaSO 4 + 2NaNO 3  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(NO 3 ) 2 reacts to give 233g BaSO 4 2.61g Ba(NO 3 ) 2 reacts to give 2.33g BaSO 4 Maximum theoretical yield for experiment = 2.33g BaSO 4

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 A r 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 (A r 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 (Fe 2 O 3 ) Experiment mass ÷ A r 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 (A r 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 (SO 2 ) Experiment mass ÷ A r for each element x 10

21. Moles  A mole is a number – 6.023 x 10 23  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 A r of 56 – 1 mole of iron weighs 56g  Nitrogen has a M r of 28 (2 x 14) – 1 mole of nitrogen weighs 28g  Carbon dioxide has a M r 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 M r of element or compound  How many moles are there in 42g of carbon? 42/12 = 3.5 moles