1. Chemical Stoichiometry
Chapter 3
Chemical Stoichiometry

2. 3.1: Atomic Mass/Weight The average mass of a sulfur atom is 32.06 amu
The average mass of a sodium atom is amu Na

3. 3.2: Molecular Mass/Weight Formula Mass/Weight
The formula of an Acetylsalicylic Acid (Aspirin) molecule is C9H8O4
Acetylsalicylic Acid (Aspirin) has molecular mass of amu

4. 3.2: Molecular Mass/Weight Formula Mass/Weight
Aluminum sulfate is an ionic compound with the formula Al2(SO4)3
Aluminum sulfate has a formula mass of amu.

5. 3.3: Isotopes
Isotopes are atoms of an element that differ only in the number of neutrons in the nucleus of the atom.
Chlorine has two isotopes:
chlorine – Cl
chlorine – Cl

6. 3.3: Isotopes
In any random sample of chlorine about 1 in 4 atoms is chlorine – 37.
The atomic mass of chlorine is therefore about 35.5 amu.

7. 3.3: Isotopes
How do atoms of chlorine – 35 and chlorine – 37 differ?

8. 3.4: Moles of Atoms and Avogadro’s Number
What is a mole of atoms?

9. 3.4: Moles of Atoms and Avogadro’s Number
What is a mole of atoms?
One mole of magnesium atoms contains
6.02 x 1023 Mg atoms.

10. 3.4: Moles of Atoms and Avogadro’s Number
What is the mass of one mole of Mg atoms?
The atomic mass of Mg is amu.
The molar mass of Mg is grams.

11. 3.5: Moles of Molecules
What is a mole of molecules?

12. 3.5: Moles of Molecules What is a mole of molecules?
One mole of CHCl3 (chloroform) contains 6.02 x 1023 molecules of CHCl3.

13. 3.5: Moles of Molecules What is a mole of molecules?
One mole of CHCl3 (chloroform) contains 6.02 x 1023 molecules of CHCl3.
What else does it contain?

14. 3.5: Moles of Molecules
How many H atoms are in one mole of CHCl3?

15. 3.5: Moles of Molecules How many H atoms are in one mole of CHCl3?
6.02 x 1023 H atoms

16. 3.5: Moles of Molecules
How many Cl atoms are in one mole of CHCl3?

17. 3.5: Moles of Molecules How many Cl atoms are in one mole of CHCl3?
3(6.02 x 1023) Cl atoms

18. 3.5: Moles of Molecules
What is the molar mass of CHCl3?

19. 3.5: Moles of Molecules
What is the molar mass of CHCl3? g

20. Key Concept
You have one mole of ascorbic acid (Vitamin C) C6H8O6. What else do you have?
6.02 x 1023 molecules of C6H8O6
grams of C6H8O6
6 moles C or 6(6.02 x 1023) C atoms
8 moles H or 8(6.02 x 1023) H atoms
6 moles O or 6(6.02 x 1023) O atoms

21. Key Concept
If you have a certain number of moles of any compound you can always find the moles of each element present?
How many mol C are in 0.80 mol of CO2?
How many mol O are in 0.80 mol of CO2?
How many mol H are in 1.2 mol of C12H22O11?
How many mol S are in 3.8 mol of Cr(SO4)3?
How many mol O are in 3.8 mol of Cr(SO4)3?

22. The Law of Definite Proportions
Compounds are pure substances that contain two or more elements combined together in FIXED (or definite) proportions.

23. 3.6: Percentage Composition or Percent by Mass
What is the percent by mass of hydrogen in water?
What is the percent by mass of oxygen in water?
How does this illustrate the law of definite proportions?
11%
89%

24. 3.6: Percentage Composition or Percent by Mass
A glass of water contains 126g of water. How many grams of hydrogen are present?
14.1g H

25. NO Two unknown compounds are tested.
Compound 1 contains 15.0g of hydrogen
and 120.0g oxygen. Compound 2
contains 2.0g of hydrogen and 32.0g
oxygen. Are the compounds the same? NO

26. Compound 1:
%H = [15.0 / ( )] x 100%
= 11.1%
%O = [120.0 / ( )] x 100%
= 88.9%
Compound 2:
%H = [2.0 / ( )] x 100%
= 5.9%
%O = [32.0 / ( )] x 100%
= 94.1%
NOT THE SAME COMPOUNDS

27. How do we find the formula of compound 1? Compound 2:
%H = [15.0 / ( )] x 100%
= 11.1%
%O = [120.0 / ( )] x 100%
= 88.9%
How do we find the formula of compound 1?
Compound 2:
%H = [2.0 / ( )] x 100%
= 5.9%
%O = [32.0 / ( )] x 100%
= 94.1%
How do we find the formula of compound 2?

28. 3.7: Derivation of Formulas
Key Concept
To calculate a formula of a substance we need a mole ratio.
Do you understand why?

29. 3.7: Derivation of Formulas
To calculate a formula of a substance we need a mole ratio.
Analysis of a gas shows that it is composed of mol carbon and 0.36 mol hydrogen. What is the empirical formula gas?
CH4

30. 3.7: Derivation of Formulas
What is the empirical formula of a compound that is 27.29% carbon and 72.71% oxygen.
Now see if you can answer our original question.
CO2

31. What is the formula of compound 1? What is the formula of compound 2?
%H = [15.0 / ( )] x 100%
= 11.1%
%O = [120.0 / ( )] x 100%
= 88.9%
What is the formula of compound 1?
Compound 2:
%H = [2.0 / ( )] x 100%
= 5.9%
%O = [32.0 / ( )] x 100%
= 94.1%
What is the formula of compound 2?

32. Any guess as to what the molecular formula of compound 2 is?
%H = [2.0 / ( )] x 100%
= 5.9%
%O = [32.0 / ( )] x 100%
= 94.1%
Any guess as to what the molecular formula of compound 2 is?

33. 3.7: Derivation of Formulas
A sample of hematite contains 34.97g of iron and 15.03g of oxygen. What is the empirical formula of hematite?
Fe2O3

34. Determine the empirical and molecular formulas for a compound with the following elemental composition: % C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol.

35. Determine the empirical and molecular formulas for a compound with the following elemental composition: % C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol.

36. Determine the empirical and molecular formulas for a compound with the following elemental composition: % C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol.
CH2O

37. Determine the empirical and molecular formulas for a compound with the following elemental composition: % C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol.
CH2O
(1.008) + 1 (16.00) = 30.03
180./30.03 ≈ 6
6(CH2O) = C6H12O6

38. 3.7: Derivation of Formulas
3.22g of a compound decomposes when heated into 1.96g of KCl and oxygen. What is the empirical formula of the compound?
KClO3

39. 3.8: Solutions
Solute: The substance that dissolves (the minor component of a solution).
KMnO4

40. 3.8: Solutions
Solvent: The substance in which the solute dissolves (the major component of a solution).

41. Solution: A homogeneous mixture of the solute and solvent.
KMnO4 solution

42. 3.8: Solutions
Dilute Concentrated

43. Concentration of a solution
Molarity(M) is the moles of solute per liter of solution.
moles of solute
Molarity(M) =
Liters of solution

45. Preparation of Solutions
28.3 grams of nickel(II) chloride
Add water to make the desired volume (500.0 ml)

46. Molarity
Calculate the molarity of an aqueous nickel (II) chloride solution containing 28.3 grams of nickel (II) chloride in mL of solution.
0.437M NiCl2

47. Molarity
The maximum solubility of lead (II) chromate, PbCrO4, is 4.3 x 10-5 g/L. What is the molarity of a saturated solution of PbCrO4?
1.3 x 10-7M PbCrO4

48. Molarity
How many moles of sulfuric acid, H2SO4, are contained in 0.80L of a 0.050M solution of sulfuric acid?
How many grams of H2SO4 would be needed to make this solution?
0.040mol H2SO4
3.9g H2SO4

49. Dilution
If we take a more concentrated solution we can dilute it to a lower concentration by adding water to it.
We determine the concentration (molarity) of the diluted solution by using the following formula.
M1V1 = M2V2

50. Dilution
Calculate the concentration of the resulting solution when enough water is added to 250.0mL of 0.60M NaOH to make 300.0mL of solution?
0.50M NaOH

51. Dilution
How much water would need to be added to 125mL of a 1.50M solution of HCl to dilute the solution to a concentration of 0.570M?
204 ml

52. Dilution
250. ml of 0.50M Cu(NO3)2 solution is mixed with 400. ml of 0.75M CoCl2. What is the concentration of the Cu2+ in the final solution?
0.75M CoCl2
0.50M Cu(NO3)2

53. Dilution
250. ml of 0.50M Cu(NO3)2 solution is mixed with 400. ml of 0.75M CoCl2. What is the concentration of the Cu2+ in the final solution?
0.19M Cu2+

54. Dilution
250. ml of 0.50M Cu(NO3)2 solution is mixed with 400. ml of 0.75M CoCl2. What is the concentration of the Cl- in the final solution?
0.75M CoCl2
0.50M Cu(NO3)2

55. Dilution
250. ml of 0.50M Cu(NO3)2 solution is mixed with 400. ml of 0.75M CoCl2. What is the concentration of the Cl- in the final solution?
0.92M Cl-

56. Stoichiometry: Calculations based on balanced equations.
We use the balanced equation to determine a mole ratio.
Then we use “Moles to Move”.

57. Stoichiometry: Calculations based on balanced equations.
How many moles of Al2I6 are produced from 0.40 mol Al? → +
Al + I2 → Al2I6
2Al + 3I2 → Al2I6

58. Calculate the moles of oxygen produced by the thermal decomposition of 100.0g of potassium chlorate.
2KClO3 → 2KCl + 3O2

59. Calculate the moles of oxygen produced by the thermal decomposition of 100.0g of potassium chlorate.
2KClO3 → 2KCl + 3O2
1.224 mol O2

60. The antacid milk of magnesia contains Mg(OH)2
The antacid milk of magnesia contains Mg(OH)2. What mass of NaOH would be needed to produce 16g of Mg(OH)2? MgCl2 + 2NaOH → Mg(OH)2 + 2NaCl .

61. The antacid milk of magnesia contains Mg(OH)2
The antacid milk of magnesia contains Mg(OH)2. What mass of NaOH would be needed to produce 16g of Mg(OH)2?
MgCl2 + 2NaOH → Mg(OH)2 + 2NaCl
22g NaOH

62. What mass of oxygen is required to burn 702g of octane?

63. What mass of oxygen is required to burn 702g of octane?
Write the balanced equation for this reaction.

64. 2460g O2 What mass of oxygen is required to burn 702g of octane?
2C8H O2 → 16CO2 + 18H2O
2460g O2

65. 0.571L HCl What volume of 0.750M HCl can be made from 25.0g of NaCl?
NaCl + H2SO4 → HCl + NaHSO4
0.571L HCl

66. What volume of 0. 2089M KI solution reacts with 43. 88mL of 0
What volume of M KI solution reacts with 43.88mL of M Cu(NO3)2?
2Cu(NO3)2 + 4KI → 2CuI + I2 + 4KNO3
0.1614L Kl

67. Titration
Titration is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of a reactant.
Because volume measurements play a key role in titration, it is also known as volumetric analysis.
A reagent, called the titrant, is placed in a calibrated buret and reacted with another solution.
The titrant is normally a solution of known concentration (a standard solution). It is used to react with an unknown or the analyte (the substance that you are attempting to analyze).
An indicator is used to identify the endpoint of the reaction. The point at which the reaction is complete. The endpoint is often indicated by a permanent change in color due to the indicator.

68. Titration
Before endpoint
Endpoint
Overshoot

69. Titration – Setting Up and Performing a Titration

70. 5105g of a monoprotic acid with a molecular mass of 72
0.5105g of a monoprotic acid with a molecular mass of 72.54g is dissolved in water and titrated to an endpoint with mL of NaOH. What is the concentration of NaOH?
0.3818M NaOH

71. 15.62ml of the standardized NaOH from the previous problem is used to titrate 0.614g of an unknown monoprotic acid to an endpoint. What is the molar mass of the unknown acid?
103 g/mol

72. 5H2C2O4 + 2KMnO4 + 3H2SO4 → 2MnSO4 + K2SO4 + 10CO2 + 8H2O
A solution of 20.00mL of oxalic acid was titrated to an endpoint with 23.23mL of M potassium permanganate. What is the concentration of the oxalic acid solution.
5H2C2O4 + 2KMnO4 + 3H2SO4 → 2MnSO4 + K2SO4 + 10CO2 + 8H2O
0.2647M H2C2O4

73. Homework (Do on three separate sheets of paper).
Pre-Lab Assignment for the Lab: “Acid – Base Titration & Volumetric Analysis – Inquiry”.
Lab Summary.
Organize your own data table (See Data and Observations).

74. Percentage Yield actual yield Percent Yield = X 100 theoretical yield
Percentage yield can be a part of stoichiometry problems

75. Percentage Yield
1.274g of CuSO4 produces 0.392g of Cu. What is the percentage yield?
CuSO4 + Zn → Cu + ZnSO4 →

76. Percentage Yield
1.274g of CuSO4 produces 0.392g of Cu. What is the percentage yield?
CuSO4 + Zn → Cu + ZnSO4
77.3%

77. Limiting Reagents
A limiting reagent (reactant) is the reactant that is used up in a chemical reaction.
All other reactants are said to be in excess.

78. Baking Soda + Vinegar: An introduction to Limiting Reagents (≈1:00)

79. NaHCO3 + HC2H3O2 → NaC2H3O2 + H2CO3
Limiting Reagent
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2CO3
195g NaHCO3 and 152ml 3.0M HC2H3O2

80. NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
Limiting Reagent
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
195g NaHCO3 and 152ml 3.0M HC2H3O2

81. NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
Limiting Reagent
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
195g NaHCO3 and 152ml 3.0M HC2H3O2
Which reactant is limiting?
HC2H3O2

82. NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
Limiting Reagent
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
195g NaHCO3 and 152ml 3.0M HC2H3O2
Which reactant is limiting?
How many grams of CO2 are produced?
20.1g CO2

83. NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
Limiting Reagent
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
195g NaHCO3 and 152ml 3.0M HC2H3O2
Which reactant is limiting?
How many grams of CO2 are produced?
How many grams of NaHCO3 remain after the reaction?

84. NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
Limiting Reagent
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
195g NaHCO3 and 152ml 3.0M HC2H3O2
How many grams of NaHCO3 remain after the reaction?
157g NaHCO3 remain

85. A mixture of 5. 0g H2 and 10. 0g of O2 is ignited forming water
A mixture of 5.0g H2 and 10.0g of O2 is ignited forming water. How much water will the reaction produce?
0.625 mol or 11.3g H2O