1. Mathematical Chemistry- Stiochometry part2 From the Periodic Table we know the Atomic mass of Elements. E.g. the mass of Helium is f_ _r. Therefore the molecular mass of molecules can be found. E.g. H 2 O is 1_. When we want to react substances together exactly we cannot weigh out 1 particle. We want to weigh out so many grammes. There we use the ………….. Concept.

2. Whats the connection ? Mr. S Finn.

3. The mole

4. A chemical mole Contains 6.023 x 10 23 particles of a substance. It is the mass in grammes of the atomic/molecular mass. Contains 6.023 x 10 23 particles of a substance. It is the mass in grammes of the atomic/molecular mass. Atomic mass of carbon is 12. Atomic mass of carbon is 12. 1 mole of carbon has a mass of 12grammes and there are 6.023 x 10 23 particles (atoms) of carbon 1 mole of carbon has a mass of 12grammes and there are 6.023 x 10 23 particles (atoms) of carbon Mr. S Finn.

5. Way 1 to convert from moles to mass Orange question If 5 oranges cost 25dirham how much does 3 oranges cost? A fairly easy sum and the answer is 15dirham (Think of the mental steps required.) Orange question If 5 oranges cost 25dirham how much does 3 oranges cost? A fairly easy sum and the answer is 15dirham (Think of the mental steps required.) How many moles of Iron in 112g. How many moles of Iron in 112g. We know 56g is 1mole. We know 56g is 1mole. Therefore 1g is 1/56mole. Therefore 1g is 1/56mole. 112g is 1 x 112/56. 112g is 1 x 112/56. Answer 112g is 2moles. Answer 112g is 2moles. Mr. S Finn. RAK Academy

6. Way 2 to convert from moles to mass. Just cover up the thing you are trying to find. Just cover up the thing you are trying to find. Moles = mass/A R. Moles = mass/A R. Number of moles in 72g of Mg. Number of moles in 72g of Mg. Moles= 72/24. Moles= 72/24. Moles =3moles. Moles =3moles. With molecules just use M R instead of A r. With molecules just use M R instead of A r. mass molesArAr Mr. S Finn.

7. Consider Calcium carbonate. 1. How many moles are in 1g,10g,50g,100g and 150g. 2. What is the mass of 0.25moles 0.1mol and 0.05moles.

8. Calculating reacting masses. Consider the simple reaction between Iron and Sulfur. How much must we add of each, to get Iron Sulfide ONLY. Consider the simple reaction between Iron and Sulfur. How much must we add of each, to get Iron Sulfide ONLY. 1.Write down a balanced equation for the reaction. 1.Write down a balanced equation for the reaction. 2.Find the molecular masses of the reactants. 2.Find the molecular masses of the reactants. 3.Take into account the ratios in the equation. 3.Take into account the ratios in the equation.

9. Fe + S = FeS We can see that Iron and Sulfur react exactly in a 1:1 ratio. We can see that Iron and Sulfur react exactly in a 1:1 ratio. If we had 1 particle of each the Fe would be 56 and the Sulfur would be 32. If we had 1 particle of each the Fe would be 56 and the Sulfur would be 32. Instead of particles we have moles. Instead of particles we have moles. Therefore 56 grammes of Iron would react with 32grammes of Sulfur. Therefore 56 grammes of Iron would react with 32grammes of Sulfur. If we had 16 grammes of Sulfur. If we had 16 grammes of Sulfur. How many grammes of Iron would we need? How many grammes of Iron would we need?

10. The Limiting Reagent When 2 substances are reacted together and the quantaties are NOT correct. The substance that runs out 1 st is called the Limiting Reagent. When 2 substances are reacted together and the quantaties are NOT correct. The substance that runs out 1 st is called the Limiting Reagent. Consider 4g Sulfur reacting with 7g of Iron to produce Iron Sulfide. Which is the limiting reagent Iron or Sulfur. How much of this substance should have been added? Consider 4g Sulfur reacting with 7g of Iron to produce Iron Sulfide. Which is the limiting reagent Iron or Sulfur. How much of this substance should have been added?

11. Limiting Reagent 2 g Hydrogen are reacted with 17g of Oxygen. 2 g Hydrogen are reacted with 17g of Oxygen. How many grammes of water are produced ? How many grammes of water are produced ? What is the limiting reagent ? What is the limiting reagent ? How many grammes of the excess reagent are left over? How many grammes of the excess reagent are left over?

12. Worked Example 2H 2 + O 2 = 2H 2 O. 2H 2 + O 2 = 2H 2 O. 2 moles Hydrogen + 1mole oxygen. 2 moles Hydrogen + 1mole oxygen. 4g + 32g 4g + 32g We have 2g H 2 which is half of 4g so we should have half of 32g of Oxygen which is 16g. We have 2g H 2 which is half of 4g so we should have half of 32g of Oxygen which is 16g. Therefore we have 17 -16=1g in excess of Oxygen. Hydrogen is the limiting reagent. Therefore we have 17 -16=1g in excess of Oxygen. Hydrogen is the limiting reagent. 2moles of Hydrogen will produce 2moles of water. 2g of Hydrogen is 1mole which will produce 1mole of water =18g. 2moles of Hydrogen will produce 2moles of water. 2g of Hydrogen is 1mole which will produce 1mole of water =18g.

13. Working out Empirical formula ? A compound contains 60g of carbon and 20g of Hydrogen work out the formula. A compound contains 60g of carbon and 20g of Hydrogen work out the formula. Work out number of moles by dividing by M R. Work out number of moles by dividing by M R. Divide by the smallest to get the numbers of each element. Divide by the smallest to get the numbers of each element. Mr. S Finn. Gulf English School Salmiya Kuwait tel:94934378.

14. What’s the difference between empirical formula and molecular formula ? Consider this molecule. Consider this molecule. Its molecular formula is__________. Its molecular formula is__________. But its empirical formula is _________. But its empirical formula is _________.

15. Working out the molecular formula A compound has a molecular mass of 78 And an empirical formula CH. What is its molecular formula? A compound has a molecular mass of 78 And an empirical formula CH. What is its molecular formula? Molecular formula is a multiple ‘y’ of the empirical formula. Molecular formula is a multiple ‘y’ of the empirical formula. Simple maths C=12 H=1. Simple maths C=12 H=1. Therefore 13y =78; y=78/13=6. Therefore 13y =78; y=78/13=6. Molecular formula is C 6 H 6. Molecular formula is C 6 H 6.

16. What is the volume of one mole of particles ? By considering Kinetic Theory the volume of a gas will depend on PRESSURE,TEMPERATURE and Number of particles. By considering Kinetic Theory the volume of a gas will depend on PRESSURE,TEMPERATURE and Number of particles. The volume of 1 mole of any gas at 25 o C (298k) and 1 atmosphere pressure occupies 24dm 3. The volume of 1 mole of any gas at 25 o C (298k) and 1 atmosphere pressure occupies 24dm 3. Mr. S Finn. RAK Academy

17. Avagadro’s Law Equal volumes of all gases at the same temperature and pressure will always contain the same number of particles. Equal volumes of all gases at the same temperature and pressure will always contain the same number of particles. E.g How many moles are E.g How many moles are Present in 120cm 3 of Present in 120cm 3 of Oxygen gas. Oxygen gas. 24dm 3 at 25C or 22.4dm 3 at 0C. Volume Number of moles Mr. S Finn RAK Academy

18. Molar gas calculation From the triangle. From the triangle. Number of moles =volume/24 000cm 3. Number of moles =volume/24 000cm 3. Moles = 120/24000 Moles = 120/24000 Moles= 1/200 = 0.002moles. Moles= 1/200 = 0.002moles. Mr. S Finn. RAK Academy

20. Concentration How are moles measured in solutions ? Molar solution Molar solution A one molar solution(1M) contains 1 mole of solute in 1000cm 3 (dm 3 ) of solution (normally water). A one molar solution(1M) contains 1 mole of solute in 1000cm 3 (dm 3 ) of solution (normally water). E.G How many moles of HCl in 250cm 3 of a 2M E.G How many moles of HCl in 250cm 3 of a 2M Solution of Hydrochloric acid ? Solution of Hydrochloric acid ? In 1000ml there are 2M In 1000ml there are 2M In 1ml there are 2/1000 M In 1ml there are 2/1000 M In 250ml there are 2 x 250/1000 =0.5moles. In 250ml there are 2 x 250/1000 =0.5moles. Mr. S Finn. RAK Academy

21. Concentration in moles for solutions Units of concentration are mols/litre.(mol/l) Units of concentration are mols/litre.(mol/l) Therefore c=n/V Therefore c=n/V Where c is the concentration, n is the number of moles & V is the volume. Where c is the concentration, n is the number of moles & V is the volume. Worked example 1. If there are 2mols of HCl in 500ml of water. What is the concentration? Worked example 1. If there are 2mols of HCl in 500ml of water. What is the concentration? C=n/V = 2/0.5 = 4 mol/l. C=n/V = 2/0.5 = 4 mol/l. Worked example 2. If we have a concentration of 0.5mol/l and we have 0.5litres.How many moles do we have? Worked example 2. If we have a concentration of 0.5mol/l and we have 0.5litres.How many moles do we have?

22. C=n/V Therefore n=Cv, n=0.5x0.5 =0.25moles Therefore n=Cv, n=0.5x0.5 =0.25moles Worked example 3. If we have a concentration of 2mol/l.What volume of solution do we need if we want 0.4 mols in total? Worked example 3. If we have a concentration of 2mol/l.What volume of solution do we need if we want 0.4 mols in total? C=n/V ;V=n/c V= 0.4/2 =0.2litres C=n/V ;V=n/c V= 0.4/2 =0.2litres

23. Molar solution square Remember 1dm 3 =1litre Remember 1dm 3 =1litre Number of moles (n) Volume of Solution (V) (dm 3 ) concentration of solution (c) (mol/dm 3 ) Mr. S Finn. RAK Academy

24. Percentage Purity %Purity = Mass of desired product/Total mass x 100. %Purity = Mass of desired product/Total mass x 100. A limestone rock had a mass of 125kg. It was found to contain 100kg of Calcium carbonate. A limestone rock had a mass of 125kg. It was found to contain 100kg of Calcium carbonate. Calculate the %purity of the rock with respect to Calcium carbonate. Calculate the %purity of the rock with respect to Calcium carbonate.

25. Percentage Yield %Yield = Actual amount of product/Theoretical amount of product x 100. %Yield = Actual amount of product/Theoretical amount of product x 100. 125g of Copper carbonate was heated and 68.85g of Copper oxide was obtained. 125g of Copper carbonate was heated and 68.85g of Copper oxide was obtained. Calculate the %yield. Calculate the %yield.

26. Answer CuCO 3 = CuO + CO 2 CuCO 3 = CuO + CO 2 Work out the molecular masses. Work out the molecular masses. 125 = 81 + 44. 125 = 81 + 44. 125g =81g + 44g. 125g =81g + 44g. %Y = 68.85/81 X 100 %Y = 68.85/81 X 100 %Y = 90%. %Y = 90%.

27. Percentage Element in a compound %Element i.a.c = Ar /Mr x 100. %Element i.a.c = Ar /Mr x 100. Worked example % Oxygen in CaCO 3 Worked example % Oxygen in CaCO 3 % of Oxygen = 3x 16/3x 16 +12 +40 x 100 % of Oxygen = 3x 16/3x 16 +12 +40 x 100 ………………….= 48/100 x 100= 48% ………………….= 48/100 x 100= 48%

28. Water of crystalisation Anhydrous Copper sulfate is CuSO4 white Anhydrous Copper sulfate is CuSO4 white Hydrous Copper sulfate CuSO4.5H2O blue. Hydrous Copper sulfate CuSO4.5H2O blue. If 2.5g of hydrous copper sulfate is heated. How much CuSO4 and water do we get ? If 2.5g of hydrous copper sulfate is heated. How much CuSO4 and water do we get ?

29. Answer CuSO4.5H2O = CuSO4 + 5H2O. CuSO4.5H2O = CuSO4 + 5H2O. 2.5/250= 0.01MOLES will give 0.01moles CuSO4 MASS= 0.01 X 1.6 =1.6g. 2.5/250= 0.01MOLES will give 0.01moles CuSO4 MASS= 0.01 X 1.6 =1.6g. Therefore 2.5-1.6g of steam will be produced=0.9g or 0.05moles. Therefore 2.5-1.6g of steam will be produced=0.9g or 0.05moles.

30. Problem finding the water of crystalisation. 2.81g of hydrated NiSO 4 was heated to complete dryness.1.55g of anhydrous NiSO4 was obtained. What is the formula of the hydrated form ( i.e. how many waters of crstalisation does it have)? 2.81g of hydrated NiSO 4 was heated to complete dryness.1.55g of anhydrous NiSO4 was obtained. What is the formula of the hydrated form ( i.e. how many waters of crstalisation does it have)? Hint this an algebra problem Hint this an algebra problem

31. Answer Let the number of water of crystalisation be x. Let the number of water of crystalisation be x. NiSO 4.XH 2 O = NiSO 4 + XH 2 O NiSO 4.XH 2 O = NiSO 4 + XH 2 O 2.81g = 1.55g + 1.26g(by lcm) 2.81g = 1.55g + 1.26g(by lcm) = 1.55/155 + 1.26/18mols = 1.55/155 + 1.26/18mols 0.01 = 0.01 + 0.07 0.01 = 0.01 + 0.07 Divide by the smallest Divide by the smallest 1 = 1 + 7 1 = 1 + 7 Therefore the formula is NiSO 4.7H 2 O. Therefore the formula is NiSO 4.7H 2 O.

32. Calculating an unknown concentration by titration Example. Burette containing Sodium Hydroxide 1mol/l. 30ml were required neutralise the acid in the beaker Conical flask containing 25ml of Hydrochloric acid of unknown concentration

33. Proceedure Write down a balanced symbol equation. Write down a balanced symbol equation. Work out the number of moles for 1 substance.(Sodium Hydroxide). Work out the number of moles for 1 substance.(Sodium Hydroxide). By looking at the equation there must be the same or ½ or 2etc the number of moles. By looking at the equation there must be the same or ½ or 2etc the number of moles. Use c=n/V to calculate the concentration. Use c=n/V to calculate the concentration.

34. HCl + NaOH = NaCl +H 2 O For Sodium Hydroxide n= 30 x 10 -3 x 1 For Sodium Hydroxide n= 30 x 10 -3 x 1 = 30 x 10 -3 moles = 30 x 10 -3 moles From the equation NaOH must have reacted with the same number of moles of HCl. From the equation NaOH must have reacted with the same number of moles of HCl. Concentration of HCl must be:- Concentration of HCl must be:- C = 30 x 10 -3 /25 x 10- 3 C = 30 x 10 -3 /25 x 10- 3 C=1.25mol/l. C=1.25mol/l. To be continued………………… To be continued…………………

35. ‘I’ll be back’ on the IB and AS course next year