1. Stoichiometry Hein and Arena Version 1.1
Eugene Passer Chemistry Department Bronx Community College
© John Wiley and Sons, Inc.
Version 1.1

2. Chapter Outline 9.1 A Short Review 9.4 Mole-Mass Calculations
9.2 Introduction to Stoichiometry: The Mole-Ratio Method
9.5 Mass-Mass Calculations
9.6 Limiting-Reactant and Yield Calculations
9.3 Mole-Mole Calculations

3. A Short Review

4. The molar mass of an element is its atomic mass in grams.
It contains x 1023 atoms (Avogadro’s number) of the element.

5. The molar mass of an element or compound is the sum of the atomic masses of all its atoms.

7. Avogadro’s Number of Particles
6.022 x 1023 Particles
1 MOLE
Molar Mass

8. 1 mole = x 1023 ions
1 mole = x 1023 atoms
1 mole = x 1023 molecules

9. The equation is balanced.
For calculations of mole-mass-volume relationships.
The chemical equation must be balanced.
The coefficient in front of a formula represents the number of moles of the reactant or product.
The equation is balanced.
Al + Fe2O3  Fe + Al2O3  2
2 mol
1 mol

10. Introduction to Stoichiometry: The Mole-Ratio Method

11. Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products.
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.

12. Examples

13. N2 + 3H2  2NH3
1 mol
2 mol
3 mol

14. N2 + 3H2  2NH3
1 mol
2 mol
3 mol

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
Convert the quantity of starting substance to moles (if it is not already moles)
Convert the moles of starting substance to moles of desired substance.
Convert the moles of desired substance to the units specified in the problem.

17. Step 1 Determine the number of moles of starting substance.
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 in moles.

18. How many moles of NaCl are present in 292. 215 grams of NaCl
How many moles of NaCl are present in grams of NaCl? The molar mass of NaCl = g.

19. How many moles of NaCl are present in 292. 215 grams of NaCl
How many moles of NaCl are present in grams of NaCl? The molar mass of NaCl = g.

20. Step 2 Determine the mole ratio of the desired substance to the starting substance.
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.

21. 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.

22. In the following reaction how many moles of PbCl2 are formed if 5
In the following reaction how many moles of PbCl2 are formed if moles of NaCl react?
2NaCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2NaNO3(aq)

23. 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.

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

27. Mole-Mole Calculations

28. Phosphoric Acid
Phosphoric acid (H3PO4) 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, Ca5(PO4)3F, with sulfuric acid (H2SO4).
Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO4

29. Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4).
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 Moles starting substance: mol H2SO4
Step 2 The conversion needed is moles H2SO4  moles H3PO4
Mole Ratio

30. Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of sulfuric acid (H2SO4) that react when 10 moles of Ca5(PO4)3 react.
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 The starting substance is 10.0 mol Ca5(PO4)3F
Step 2 The conversion needed is moles Ca5(PO4)3F  moles H2SO4
Mole Ratio

31. Mole-Mass Calculations

32. 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.
If the mass of the starting substance is given, we need to convert it to moles.

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

34. Examples

35. Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 1 Step by Step
Step 1 The starting substance is 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
Step 3 Convert moles of H3PO4 to moles of H2SO4 by the mole-ratio method.
Mole Ratio

36. Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 2 Continuous
grams H3PO4  moles H3PO4  moles H2SO4
The conversion needed is
Mole Ratio

37. Calculate the number of grams of H2 required to form 12.0 moles of NH3.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 1 The starting substance is 12.0 moles of NH3
Step 2 Calculate moles of H2 by the mole-ratio method.
Mole Ratio
Step 3 Convert moles of H2 to grams of H2.

38. The conversion needed is
Calculate the number of grams of H2 required to form 12.0 moles of NH3.
N2 + 3H2  2NH3
Method 2 Continuous
The conversion needed is
moles NH3  moles H2  grams H2
Mole Ratio

39. Mass-Mass Calculations

40. Solving mass-mass stoichiometry problems requires all the steps of the mole-ratio method.
The mass of starting substance is converted to moles.
The mole ratio is then used to determine moles of desired substance.
The moles of desired substance are converted to mass of desired substance.
Baby Aufort is Boy!

41. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 1 The starting substance is 112 grams of H2. Convert 112 g of H2 to moles.
grams  moles
Step 2 Calculate the moles of NH3 by the mole ratio method.

42. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 3 Convert moles NH3 to grams NH3.
moles  grams

43. Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 2 Continuous
grams H2  moles H2  moles NH3  grams NH3

44. Limiting-Reactant and Yield Calculations

45. Limiting Reactant

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

47. How many bicycles can be assembled from the parts shown?
From eight wheels four bikes can be constructed.
From three pedal assemblies three bikes can be constructed.
From four frames four bikes can be constructed.
The limiting part is the number of pedal assemblies.
9.2

48. H2 + Cl2  2HCl  + Cl2 is the limiting reactant
4 molecules Cl2 can form 8 molecules HCl
Cl2 is the limiting reactant
3 molecules of H2 remain H2 is in excess
7 molecules H2 can form 14 molecules HCl
9.3

49. Steps Used to Determine the Limiting Reactant

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

51. Examples

52. How many moles of HCl can be produced by reacting 4. 0 mol H2 and 3
How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reactant?
H2 + Cl2 → 2HCl
Step 1 Calculate the moles of HCl that can form from each reactant.
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it produces less HCl than H2.

53. MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Calculate the grams of AgBr that can form from each reactant.
The conversion needed is
g reactant → mol reactant → mol AgBr → g AgBr

54. MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Determine the limiting reactant.
The limiting reactant is MgBr2 because it forms less Ag Br.

55. MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
How many grams of the excess reactant (AgNO3) remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
The amount of AgNO3 that remains is
100.0 g AgNO3 -
92.3 g AgNO3 =
7.7 g AgNO3

56. Reaction Yield

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

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

59. 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.

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

61. MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgBr → g AgBr

62. MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
must have same units

63. The End