1. Calculations from Equations Chapter 9
Larry Emme
Chemeketa Community College

2. A Short Review

3. Avogadro’s Number of Particles
6.02 x 1023 Particles
1 MOLE
Mole Weight

4. The equation is balanced.
For calculations of mole-mass-volume relationships.
The chemical equation must be balanced.
The number 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 mole
1 mole

5. Warm-Up Problem!
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When this compound is combusted with oxygen in the lab, carbon dioxide is produced. Find/write/calculate the following:
The molecular formula of the hydrocarbon.
The balanced equation.

6. Warm-Up Problem!
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When this compound is combusted with oxygen in the lab, carbon dioxide is produced. Find/write/calculate the following:
The molecular formula of the hydrocarbon.
C7H14

7. Warm-Up Problem! C7H14 (g) + O2 (g) → CO2 (g) + H2O (l)
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When this compound is combusted with oxygen in the lab, carbon dioxide is produced. Find/write/calculate the following:
2. The balanced equation.
C7H14 (g) O2 (g) → CO2 (g) H2O (l)
C7H14 (g) O2 (g) → 7 CO2 (g) H2O (l)
2 C7H14 (g) O2 (g) → 14 CO2 (g) H2O (l)

8. Introduction to Stoichiometry: The Mole-Ratio Method

9. Examples of questions asked:
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.

10. 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.

11. Examples

12. N2 + 3H2  2NH3
1 mole
2 mole
3 mole

13. N2 + 3H2  2NH3
1 mole
2 mole
3 mole

14. 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.

15. 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)
Mole Ratio

16. Mole-Mole Calculations

17. 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

18. 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 mole
5 mole
3 mole
1 mole
5 mole
Step 1 Moles starting substance: 10 mole H2SO4
Step 2 The conversion needed is moles H2SO4  moles H3PO4
Mole Ratio

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

20. Mole-Mass Calculations

21. 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.

22. 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.

23. Examples

24. 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
grams H3PO4  moles H3PO4  moles H2SO4
The conversion needed is
Mole Ratio

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

26. Mass-Mass Calculations

27. 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.

28. Diagram to Successful Calculations
mole wt
of A
moles of A
Chemical
equation
moles of B
grams of A
mole wt
of B
grams of B

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

30. Limiting-Reactant and Yield Calculations

31. Limiting Reagent (Limiting Reactant) (Limiting Ingredient)

32. The limiting reagent is one of the reactants in a chemical reaction.
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.

33. How many tricycles 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.

34. How many tricycles can be assembled from the parts shown?
The limiting part is the number of pedal assemblies.

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

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

37. First Balance Your Assembly
2 Wheels + 1 Frame + 1 Pedal→ 1 Bike

38. Limiting Reactant  + + 
Determine the number of that can be made given these quantities of reactants and the reaction equation:  + + 

39. Limiting Reactant  + + 
Determine the number of that can be made given these quantities of reactants and the reaction equation:  + + 

40. Steps Used to Determine the Limiting Reagent

41. 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 reagent; the other reactant is in excess.
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.

42. Examples

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

44. MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
How many grams 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 → mole reactant → mole AgBr → g AgBr

45. 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 reagent.
The limiting reactant is MgBr2 because it forms less AgBr.

46. 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 50.0 g of MgBr2.
The conversion needed is
g MgBr2 → mole MgBr2 → mole AgNO3 → g AgNO3
The amount of AgNO3 that remains is
100.0 g AgNO3 -
92.3 g AgNO3 =
7.7 g AgNO3

47. Learning Check True/False:
You can compare the quantities of reactant when you work a limiting reactant problem. The reactant you have the least of is the limiting reactant.
True
False
Answer in Lecture!

48. Learning Check
Which is the limiting reactant when 3.00 moles of copper are reacted with 3.00 moles of silver nitrate in the following equation?
  Cu + 2AgNO3  Cu(NO3)2 + 2Ag Cu
AgNO3
Cu(NO3)2 Ag
Answer in Lecture!

49. Percent Yield

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

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

52. 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.

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

54. 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 → mole MgBr2 → mole AgBr → g AgBr

55. 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

56. Percent Purity

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

58. g CO2 → mole CO2 → mole CaCO3 → g CaCO3
A 5.00 gram sample of impure CaCO3 (limestone) produces 1.89 grams of CO2 when heated. Calculate the percent purity of CaCO3 according to:
CaCO3 (s)  CaO (s) + CO2
Step 1 Determine the actual grams (theoretical ) of CaCO3 in the original sample.
The conversion needed is
g CO2 → mole CO2 → mole CaCO3 → g CaCO3

59. A 5. 00 gram sample of CaCO3 produces 1. 89 grams of CO2 when heated
A 5.00 gram sample of CaCO3 produces 1.89 grams of CO2 when heated. Calculate the percent purity according to:
CaCO3 (s)  CaO (s) + CO2
Step 2 Calculate the percent purity.

60. 3 Cu(s) + 8 HNO3(aq)  3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
484 gram of copper ore is oxidized by HNO3 to yield grams of Cu(NO3)2. Find the percent purity of copper in the ore according to:
3 Cu(s) + 8 HNO3(aq)  3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
The conversion needed is
g Cu(NO3)2 → mole Cu(NO3)2 → mole Cu → g Cu

61. Extra Problem!
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When 2.60 g of this compound is combusted with 8.70 g of oxygen in the lab, 5.90 g of carbon dioxide is produced. Find/write/calculate the following:
The molecular formula of the hydrocarbon.
The balanced equation.
The limiting reactant.
The theoretical weight of carbon dioxide.
The percent yield.

62. Extra Problem!
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When 2.60 g of this compound is combusted with 8.70 g of oxygen in the lab, 5.90 g of carbon dioxide is produced. Find/write/calculate the following:
The molecular formula of the hydrocarbon.
C7H14

63. Extra Problem! C7H14 (g) + O2 (g) → CO2 (g) + H2O (l)
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When 2.60 g of this compound is combusted with 8.70 g of oxygen in the lab, 5.90 g of carbon dioxide is produced. Find/write/calculate the following:
2. The balanced equation.
C7H14 (g) O2 (g) → CO2 (g) H2O (l)
C7H14 (g) O2 (g) → 7 CO2 (g) H2O (l)
2 C7H14 (g) O2 (g) → 14 CO2 (g) H2O (l)

64. Extra Problem!
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When 2.60 g of this compound is combusted with 8.70 g of oxygen in the lab, 5.90 g of carbon dioxide is produced. Find/write/calculate the following:
3. The limiting reactant.
4. Theoretical weight.
The limiting
reactant.
Theoretical
weight.

65. Extra Problem!
A hydrocarbon with a mole weight of 98.2, is made of 85.6% carbon and 14.4% hydrogen. When 2.60 g of this compound is combusted with 8.70 g of oxygen in the lab, 5.90 g of carbon dioxide is produced. Find/write/calculate the following:
5. The percent yield.

66. The End