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Presentation On Stoichiometry: The Mole And Chemical Equations

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Published in: Chemistry | Science
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This PPT is suitable for the IGCSE and O Level Chemistry students. It only provides theoretical background of the topic. The practice questions, worksheets and topical questions will be discussed during the tutoring sessions

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  1. The Mole and Chemical Equations Chemistry (IGCSE / O-Level)
  2. What will you learn in this lesson? • Calculating amounts of reactants and products • Finding limiting reactant • Percentage yield and percentage purity • Percentage composition by mass • Molar volume of gas
  3. Calculating Reacting Amounts — a chemical 'footbridge' • The balanced equation itself can be used as the numerical 'footbridge' between the two sides of equation use ratio moles of reactant equation moles Of product mass of reactant mass of product 5.13: A chernical 'footbridge'. Following the sequence 'up—across—down' helps to relate the mass Of product made to the mass of reactant used. The 'bridge' can also be used in the reverse direction.
  4. Calculating Reacting Amounts — a chemical 'footbridge' (contd.) the mass elummum orde produced W'hen alumlnlum reacts wrth oxygen 9.2 g w Ith ratio 4 mol : 2mOI 9.2g Of the 'footbridg (the •up' 9.2 9 of A' Into O. 34 the work o ut how many of 4 2 APS Step 3. 'don* Rage): Work out the m•gs of this alum'nlum oxldo (tho rulAtWe formula of is 102' : the 0.17 X 102 g 1'.34 • of Alps ptoduc•d 17." threa slgnlflunt ngLnes
  5. Calculating Reacting Amounts — a chemical 'footbridge' (contd.) What mass of oxygen combines with 32 g of methane when that gas is burnt completely in air? (Ar: C = 12; H = 1; O = 16) Remember to read question carefully which will help in understanding the compounds of our interest Also take the balancing number into account in doing your calculations (this is called the stoichiometry of the equation)
  6. Limiting Reactants It is the reactant that determines the reactant that is not in excess quantity of the product or the A2B Limiting reactant Excess reactant
  7. Limiting Reactants (contd.) A reaction stops when the limiting reactant is used up. For finding the limiting reactant, do the following calculations: 1. Workout the number of moles of each reactant involved 2. Then divide each by its balancing number in the balanced equation 3. The smallest number indicate the limiting reactant
  8. Limiting Reactants (contd.) l_lmmng reactants and reactants In excess 6.0 g of cobAt(11) carbonate was reacted with 0.08 mol of hydrochloric add to produce cdbal(u) chloride. Show' the cobalt(ll) carbonate was in excess. cocos(s) + 2HCt(aq) + H20(1) + C02(g) co = sg; C = 12; O = 16) M, of cobalt carbmate = + 12 + (3 x 16) = 119 g/mol nurür of c±ak carbonate = 6.0 I = 0.05 mol From the bAaræd ewatlcn: • I mol cobalt(ll) cubonate reacts with 2 mol hydrochloric aci d • 0.05 ma' cobalt(lt) carbonate would react with 0.10 hydrochlortc acid • the carbonate in —cess. OR coco* 0.05/1 = 0.05 HCI: 0.08/2 = 0.04 0.05 > 0.04 So, COCOA is in excess
  9. Limiting Reactants (contd.) 1 4gof magnesium is reacted with a hydrochloric acid solution that contains 5.48 g of the acid. NE(s) + 2HCl(aq) —4 mgC12(aq) + H2(g) Which of the readants is the limiting reagent? (AF mg = 24; a = 35.5; H = 1)
  10. Percentage Yield and Percentage Purity A reaction may not always yield the predicted amount of product. The lose may be due to several factors: 1. The reaction is not completed 2. Errors may be made in weighing the reactants or the products 3. Material may be lost in carrying out the reaction, or in transferring or separating the product
  11. Percentage Yield and Percentage Purity (contd.) Percentage Yield: A measure of the actual yield of the reaction when carried out experimentally compared to the theoretical yield actual yield % yield = x 100 predicted yield Percentage Purity: Measure of the purity of the product from reaction carried out experimentally. It is checked in crude product and the product formed with impurities mass of pure product % purity = x 100 mass of impure product
  12. Percentage Yield and Percentage Purity (contd.) CalculatÄg percentage yield Heating 12.4 g of copper(IT) carbonate in a crucible produced only 7.0 g of copper(n) oxide. What was the percentage yield of copper(lt) oxide? CuCOa 1 mol 64+ 12+ 48 —124 g Cuo + C02 1 mol 4- 1 mol 64+16 heating 12.4 g of coppergr) carbonate should have produced 8.0 g of copper(TI) oxide. So: expe&d yield = 8.0 g actual yield = 7.0 g and percentage = x 100 = 87.5%
  13. Percentage Yield and Percentage Purity (contd.) 1 AStudentre•cts4SgofSlumidiurn powderwith chlorine gas, 17.8 gofaluminiumchlofideare producedlx the percentage yield of this reactiony + 3CJ2{g) 27; 3585)
  14. Percentage Yield and Percentage Purity (contd.) A sample of 10.15 g of the crude copper is analyzed by various methods and shown to contain 9.95 g of pure copper, with the remaining mass being made up of other metals. What will be the purity of copper? % purity of the copper sample of cowr in sample X 100 of impure 9.95 x 100 10.15 = 98.03% (answer to four significant figures)
  15. Percentage Composition by Mass The percentage by mass of each element in a compound What percentage of the mass of the compound Is nttrogen? The %rmula of ammonium nhte is NH4N03 (it contains the tons NH4 + and N03 Using the Ar values for N, H and O we get: 80 ma— of nitrogen In form& z 28 mass of as a fradionof the total mass of of total mass X 100
  16. Percentage Composition by Mass (contd.) Carry out a similar calcUlatian for yourself to work out the percentage by mass of nitrogen in another fertiliser, ammonium surate, (NH4hS04. Similar calculations can be used to workout the percentage by mass of water of crystallization in crystals of a hydrated salt What Is the percentage mass of water In crystals of hydrated magnesium sulfate? formula of hydrated magnesium sulfate is MgS04•7H20. Using the Ar values for Mg, S, O and H we M, = 24 + 32 (4 x 16) + (7 X 18) = 246 mass of water fornuda = 126 mass of water as A fraction & the total x 100 = 51.2% mass of water In the =
  17. Percentage Composition by Mass (contd.) Now carry out a similar calculation to work out the percentage mass of water in crystals of hydrated sodium carbonate, Na2C03• 10H20. 4 Calculate the percentage by mass of nitrogen in the following fertilisers and nitrogen-containing compounds: a ammonium phosphate, (NH4)3P04 b glycine, CH2(NH2)COOH (an amino acid) (Ar: H = 1, C = 12, N = 14, O — 16, P = 31, S = 32)
  18. Molar Volume of Gas Weighing solids or liquids is relatively straightforward as compared to gases. It is much easier to calculate the volume of gas bstance hydrogen (H2) oxygen (02) carbon dioxide (C02) ethane (C2Hs) Molar mass, Molar gas volume, g/ mol 2 32 30 dm3/ mol 24 24 24 Number of particles 6.02 x 1023 hydrogen molecules 5.02 x 1023 oxygen molecules 6.02 x 1023 carbon dioxide molecules 6.02 x 1023 ethane molecules
  19. Molar Volume of Gas (contd.) Avogadro's Law: Equal volumes of any gas, under the same conditions of temperature and pressure, contain the same number of molecules. This leads to simple rule about the relationship of the volume of 1 mole of gas. • One mole of any gas occupies the volume of 24 dm3 (or 24 liters) at room temperature and pressure (r.t.p.) • The molar gas volume of any gas therefore has the value of 24 dm3 at r.t.p. • 1 dm3 (l) = 1000 cm3 (ml)
  20. Molar Volume of Gas (contd.) number of moles = volume *to. of molar volum 24 dm3 at r. t. p.
  21. Molar Volume of Gas (contd.) If 8 g of sulfu- are bunt, what volume of S02 is produced? conAder of s•tur burnlng In oxygen. sulhE 4- dioxide 32 g 24 soa(g) I mol 24 dm3 rumber Of Of sufir burnt — 32g,•• = 0.25 mol From the etpatlon: 1 md sulfur 1 mol of soa 0.2S mol of suhlr Oi2S mol of S02 so. from the above ruk•. nunber af moles — vdwne of sulfur dioxi& 0.25 x 24 dmg = 6 dm3
  22. Molar Volume of Gas (contd.) Now, ushg the relati.msHp between the volume of a gas and the number of moles it contains, work out the : a the valume of gas present in 22g of carbon dioxide at r.t.p. b the mass of a sample of arrogen gas with a volume of 35 dm3 at r.t,p.

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