# 1. 2 A Short Review 3 The mole weight of an element is its atomic mass in grams. It contains 6.02 x 10 23 atoms (Avogadro’s number) of the element.

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2 A Short Review

3 The mole weight of an element is its atomic mass in grams. It contains 6.02 x 10 23 atoms (Avogadro’s number) of the element.

4 The mole weight of an element or compound is the sum of the atomic masses of all its atoms.

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6 Avogadro’s Number of Particles 6.02 x 10 23 Particles Mole Weight 1 MOLE

7 1 mole = 6.02 x 10 23 molecule 1 mole = 6.02 x 10 23 formula units1 mole = 6.02 x 10 23 atoms1 mole = 6.02 x 10 23 ions

8 2 2 Al + Fe 2 O 3  Fe + Al 2 O 3  For calculations of mole-mass-volume relationships. –The chemical equation must be balanced. 2 mol 1 mol The equation is balanced. –The number in front of a formula represents the number of moles of the reactant or product.

9 Introduction to Stoichiometry: The Mole-Ratio Method

10 Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products. Examples of questions asked: How much product from a specific amount of reactant. How much reactant do you start with to get so much product. What if the reaction does not go to completion? What is the percent yield. What if the sample is impure? What is the percent purity.

11 Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction. –The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

12ExamplesExamples

13 N 2 + 3H 2  2NH 3 1 mol2 mol3 mol

14 1 mol2 mol3 mol N 2 + 3H 2  2NH 3

15 The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem. The mole ratio is used in the solution of every type of stoichiometry problem.

16 The Mole Ratio Method 1.Convert the quantity of starting substance to moles (if it is not already moles) 2.Convert the moles of starting substance to moles of desired substance. 3.Convert the moles of desired substance to the units specified in the problem.

17 Identify the starting substance from the data given in the problem statement. Convert the quantity of the starting substance to moles, if it is not already done. Step 1 Determine the number of moles of starting substance.

18 How many moles of NaCl are present in 292.215 grams of NaCl? The mole weight of NaCl =58.443 g.

19 The number of moles of each substance in the balanced equation is indicated by the coefficient in front of each substance. Use these coefficients to set up the mole ratio. Step 2 Determine the mole ratio of the desired substance to the starting substance.

20 Step 2 Determine the mole ratio of the desired substance to the starting substance. Multiply the number of moles of starting substance (from Step 1) by the mole ratio to obtain the number of moles of desired substance.

21 In the following reaction how many moles of PbCl 2 are formed if 5.000 moles of NaCl react? 2NaCl(aq) + Pb(NO 3 ) 2 (aq)  PbCl 2 (s) + 2NaNO 3 (aq)

22 Step 3. Calculate the desired substance in the units specified in the problem. If the answer is to be in moles, the calculation is complete If units other than moles are wanted, multiply the moles of the desired substance (from Step 2) by the appropriate factor to convert moles to the units required.

23 Step 3. Calculate the desired substance in the units specified in the problem.

24 Step 3. Calculate the desired substance in the units specified in the problem.

25 Step 3. Calculate the desired substance in the units specified in the problem.

26 Mole-Mole Calculations

27 Phosphoric Acid Phosphoric acid (H 3 PO 4 ) is one of the most widely produced industrial chemicals in the world. Most of the world’s phosphoric acid is produced by the wet process which involves the reaction of phosphate rock, Ca 5 (PO 4 ) 3, F with sulfuric acid (H 2 SO 4 ). Ca 5 (PO 4 ) 3 F(s) + 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4

28 Mole Ratio Calculate the number of moles of phosphoric acid (H 3 PO 4 ) formed by the reaction of 10 moles of sulfuric acid (H 2 SO 4 ). Ca 5 (PO 4 ) 3 F + 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 Step 1 Moles starting substance: 10.0 mol H 2 SO 4 Step 2 The conversion needed is moles H 2 SO 4  moles H 3 PO 4 1 mol5 mol3 mol1 mol5 mol

29 Step 2 The conversion needed is moles Ca 5 (PO 4 ) 3 F  moles H 2 SO 4 Calculate the number of moles of sulfuric acid (H 2 SO 4 ) that react when 10 moles of Ca 5 (PO 4 ) 3 F react. Ca 5 (PO 4 ) 3 F + 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 Mole Ratio Step 1 The starting substance is 10.0 mol Ca 5 (PO 4 ) 3 F 1 mol5 mol3 mol1 mol5 mol

30 Mole-Mass Calculations

31 1.The object of this type of problem is to calculate the mass of one substance that reacts with or is produced from a given number of moles of another substance in a chemical reaction. 2.If the mass of the starting substance is given, we need to convert it to moles.

32 3.We use the mole ratio to convert moles of starting substance to moles of desired substance. 4.We can then change moles of desired substance to mass of desired substance if called for by the problem.

33ExamplesExamples

34 Mole Ratio Calculate the number of moles of H 2 SO 4 necessary to yield 784 g of H 3 PO 4 Ca 5 (PO 4 ) 3 F+ 5H 2 SO 4  3H 3 PO 4 + HF + 5CaSO 4 grams H 3 PO 4  moles H 3 PO 4  moles H 2 SO 4 The conversion needed is

35 Mole Ratio moles NH 3  moles H 2  grams H 2 Calculate the number of grams of H 2 required to form 12.0 moles of NH 3. N 2 + 3H 2  2NH 3 The conversion needed is

36 Mass-Mass Calculations

37 Solving mass-mass stoichiometry problems requires all the steps of the mole-ratio method. 1.The mass of starting substance is converted to moles. 2.The mole ratio is then used to determine moles of desired substance. 3.The moles of desired substance are converted to mass of desired substance.

38 Diagram to Successful Calculations grams of A mole wt of A moles of Amoles of B grams of B mole wt of B Chemical equation

39 Calculate the number of grams of NH 3 formed by the reaction of 112 grams of H 2. N 2 + 3H 2  2NH 3 grams H 2  moles H 2  moles NH 3  grams NH 3

40 Limiting-Reactant and Yield Calculations

41 Limiting Reagent

42 It is called the limiting reagent because the amount of it present is insufficient to react with the amounts of other reactants that are present. The limiting reagent limits the amount of product that can be formed. The limiting reagent is one of the reactants in a chemical reaction.

43 How many bicycles can be assembled from the parts shown? From eight wheels four bikes can be constructed. From four frames four bikes can be constructed. From three pedal assemblies three bikes can be constructed. The limiting part is the number of pedal assemblies.

44 H 2 + Cl 2  2HCl  + 7 molecules H 2 can form 14 molecules HCl 4 molecules Cl 2 can form 8 molecules HCl 3 molecules of H 2 remain H 2 is in excess Cl 2 is the limiting reagent

45 Steps Used to Determine the Limiting Reagent

46 1.Calculate the amount of product (moles or grams, as needed) formed from each reactant. 2.Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reagent; the other reactant is in excess. 3.Calculate the amount of the other reactant required to react with the limiting reagent, then subtract this amount from the starting quantity of the reactant. This gives the amount of the substance that remains unreacted.

47ExamplesExamples

48 How many moles of HCl can be produced by reacting 4.0 mol H 2 and 3.5 mol Cl 2 ? Which compound is the limiting reagent? Step 1 Calculate the moles of HCl that can form from each reactant. H 2 + Cl 2 → 2HCl Step 2 Determine the limiting reagent. The limiting reagent is Cl 2 because it produces less HCl than H 2.

49 How many grams of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr 2 and 100.0 g of AgNO 3 are mixed together? How many grams of the excess reactant remain unreacted? Step 1 Calculate the grams of AgBr that can form from each reactant. MgBr 2 (aq) + 2AgNO 3 (aq) → 2AgBr(s) + Mg(NO 3 ) 2 (aq) The conversion needed is g reactant → mol reactant → mol AgBr → g AgBr

50 How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr 2 and 100.0 g of AgNO 3 are mixed together? How many grams of the excess reactant remain unreacted? Step 2 Determine the limiting reagent. MgBr 2 (aq) + 2AgNO 3 (aq) → 2AgBr(s) + Mg(NO 3 ) 2 (aq) The limiting reactant is MgBr 2 because it forms less AgBr.

51 How many grams of the excess reactant (AgNO 3 ) remain unreacted? Step 3 Calculate the grams of unreacted AgNO 3. First calculate the number of grams of AgNO 3 that will react with 50 g of MgBr 2. MgBr 2 (aq) + 2AgNO 3 (aq) → 2AgBr(s) + Mg(NO 3 ) 2 (aq) The conversion needed is g MgBr 2 → mol MgBr 2 → mol AgNO 3 → g AgNO 3 The amount of MgBr 2 that remains is 100.0 g AgNO 3 -92.3 g AgNO 3 =7.7 g AgNO 3

52 Percent Yield

53 The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

54 Many reactions fail to give a 100% yield of product. This occurs because of side reactions and the fact that many reactions are reversible.

55 The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant. The actual yield is the amount of product finally obtained from a given amount of reactant.

56 The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.

57 Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr 2 (aq) + 2AgNO 3 (aq) → 2AgBr(s) + Mg(NO 3 ) 2 (aq) Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed. The conversion needed is g MgBr 2 → mol MgBr 2 → mol AgBr → g AgBr

58 Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction: MgBr 2 (aq) + 2AgNO 3 (aq) → 2AgBr(s) + Mg(NO 3 ) 2 (aq) Step 2 Calculate the percent yield. must have same units

59 Percent Purity

60 The percent purity of a reaction is the ratio of the theoretical yield (calculated) to the sample weight multiplied by 100.

61 A 5.00 gram sample of impure CaCO 3 (limestone) produces 1.89 grams of CO 2 when heated. Calculate the percent purity of CaCO 3 according to: CaCO 3 (s)  CaO (s) + CO 2  Step 1 Determine the actual grams (theoretical ) of CaCO 3 in the original sample. The conversion needed is g CO 2 → mol CO 2 → mol CaCO 3 → g CaCO 3

62 A 5.00 gram sample of CaCO 3 produces 1.89 grams of CO 2 when heated. Calculate the percent purity according to: CaCO 3 (s)  CaO (s) + CO 2  Step 2 Calculate the percent purity.

63 484 gram of copper ore is oxidized by HNO 3 to yield 1000. grams of Cu(NO 3 ) 2. Find the percent purity of copper in the ore according to: 3 Cu (s) + 8 HNO 3(aq)  3 Cu(NO 3 ) 2(aq) + 2 NO (g) + 4 H 2 O (l) The conversion needed is g Cu(NO 3 ) 2 → mol Cu(NO 3 ) 2 → mol Cu → g Cu

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