1. Chapter 3 Equations, the Mole, and Chemical Formulas

2. Chemical Equation
Stoichiometry is the study of the quantitative relationships involving the substances in chemical reactions.
A chemical equation describes the identities and relative amounts of reactants and products in a chemical reaction.
Just like a chemical formula, a chemical equation expresses quantitative relations.

3. Chemical Reactions
A chemical equation is a shorthand notation to describe a chemical reaction Mg + O2 ® 2MgO magnesium react to form magnesium and oxygen oxide

4. Definitions Reactants are the substances consumed.
Products are the substances formed.
Coefficients are numbers before the formula of a substance in an equation.
A balanced equation has the same number of atoms of each element on both sides of the equation.

5. Writing Balanced Equations
Write the correct formula for each substance. H Cl2 ® HCl (unbalanced)
Add coefficients so the number of atoms of each element are the same on both sides of the equation. H Cl2 ® 2HCl (balanced)

6. Balancing Chemical Equations
Write the correct formula for each substance. C5H O2 ® CO2 + H2O
Assume one molecule of the most complicated substance. Adjust the coefficient of CO2 to balance C. C5H O2 ® 5CO2 + H2O
Adjust the coefficient of H2O to balance H. C5H O2 ® 5CO H2O
Adjust the coefficient of O2 to balance O. C5H O2 ® 5CO H2O
Check the balance by counting the number of atoms of each element.

7. Test Your Skill
Balance the equation C4H9OH + O2 ® CO2 + H2O

8. Test Your Skill Balance the equation C4H9OH + O2 ® CO2 + H2O
Answer: C4H9OH + 6O2 ® 4CO2 + 5H2O

9. Balancing Equations Sometimes fractional coefficients are obtained.
C5H O2 ® CO2 + H2O
C5H O2 ® 5CO2 + H2O
C5H O2 ® 5CO2 + 5H2O
C5H /2O2 ® 5CO2 + 5H2O
Multiply all coefficients by the denominator.
2C5H O2 ® 10CO H2O

10. Neutralization Reactions
Neutralization is the reaction of an acid with a base to form a salt and water.
An acid is a compound that dissolves in water to produce hydrogen ions.
A base dissolves in water to produce hydroxide ions. It is usually a soluble metal hydroxide.
A salt is an ionic compound consisting of the cation of a base and the anion of an acid.

11. Examples of Neutralization
HCl NaOH ® NaCl H2O
H2SO KOH ® K2SO H2O
2HClO4 + Ca(OH)2 ® Ca(ClO4)2 + 2H2O acid base ® salt water

12. Balancing Neutralization Reactions
The number of hydrogen ions provided by the acid must equal the number of hydroxide ions provided by the base.
H3PO4 + 3NaOH ® Na3PO4 + 3H2O
3H OH- ® H2O

13. Combustion Reactions A combustion reaction is the process of burning.
Most combustion reactions are the combination of a substance with oxygen.
When an organic compound burns in oxygen, the carbon is converted to CO2, and the hydrogen forms water, H2O.

14. Test Your Skill
Identify the type of the reaction, then balance it. (a) (C2H5)2O + O2 ® CO2 + H2O (b) H2SO4 + Ca(OH)2 ® CaSO4 + H2O

15. Oxidation and Reduction
Oxidation is the loss of electrons by a chemical process.
When sodium forms a compound, Na+ is formed. Sodium is oxidized.
Reduction is the gain of electrons by a chemical process.
When Cl- ions are formed from elemental chlorine, chlorine is reduced. 1

16. Oxidation-Reduction (“Redox”)
An oxidation-reduction reaction, or redox reaction, is one in which electrons are transferred from one species to another.
In every redox reaction, at least one species is oxidized and at least one species is reduced.
2Na(s) + Cl2(g) → 2NaCl(s) is a redox reaction because Na is oxidized and Cl is reduced. 1

17. Oxidizing and Reducing Agents
An oxidizing agent is the reactant that accepts electrons, causing an oxidation to occur.
The oxidizing agent is reduced.
A reducing agent is the reactant that supplies electrons, causing a reduction to occur.
The reducing agent is oxidized.
In the reaction of sodium with chlorine, Na is the reducing agent and Cl2 is the oxidizing agent. 1

18. Half-reactions
In a half-reaction, either the oxidation or reduction part of a redox reaction is given, showing the electrons explicitly.
Half-reactions emphasize the transfer of electrons in a redox reaction.
For 2Na(s) + Cl2(g) → 2NaCl(s) :
Na → Na+ + 1e- oxidation half-reaction
Cl2 + 2e- → 2Cl- reduction half-reaction 1

19. Oxidation States
The oxidation state is the charge on the monatomic ion, or the charge on an atom when the shared electrons are assigned to the more electronegative atom.
Electron pairs shared by atoms of the same element are divided equally.
In CaCl2, an ionic compound:
calcium has an oxidation state of +2.
chlorine has an oxidation state of -1. 1

20. Rules for Assigning Oxidation Numbers
Oxidation numbers for atoms in their elemental form are 0.
The oxidation number of a monatomic ion is equal to the charge on the ion.
In compounds, F is always -1. Other halogens are also -1 unless they are combined with a more electronegative element (O or a halogen above it in the periodic table). 1

21. Rules for Assigning Oxidation Numbers (cont’d)
4. In compounds, O is -2 except for peroxides (where it is -1) or when combined with F.
5. In compounds, H is +1 except in metal hydrides, where it is -1.
6. The sum of all the oxidation numbers of the atoms in a substance must sum to the charge on the substance. 1

22. Assigning Oxidation Numbers
Assign oxidation numbers for each atom in K2CrO4.
K = +1 by rule 2.
O = -2 by rule 5.
Cr = +6 by rule 6.
2(+1) (-2) = 0, the overall charge on the substance. 1

23. Test Your Skill
Assign oxidation numbers to each atom in the following substances.
(a) PF3 (b) CO (c) NH4Cl 1

24. Balancing Redox Equations
Determine oxidation numbers that are changing and write the skeleton half-reactions.
Balance each half-reaction separately.
Balance element being oxidized or reduced.
Balance all other elements except H and O.
Balance O by adding H2O as needed.
Balance H by adding H+ as needed.
Balance charges by adding e- as needed. 1

25. Balancing Redox Equations (cont’d)
Multiply one or both reactions by an integer so that the number of electrons in both half-reactions is the same.
Add the two half-reactions, canceling out the electrons and any other species that appears on both sides of the equation. 1

26. Balancing Redox Equations (cont’d)
Balance the following equation.
Cu + NO3- → Cu2+ + NO
Cu is oxidized from 0 to +2
Write a skeleton equation for Cu:
Cu → Cu2+
No other step necessary except to balance charges with electrons:
Cu → Cu2+ + 2e- 1

27. Balancing Redox Equations (cont’d)
N is reduced from +5 to +2.
Write a skeleton equation for N:
NO3- → NO
Balance O by adding water:
NO3- → NO + 2H2O
Balance H by adding H+:
4H+ + NO3- → NO + 2H2O
Balance charge by adding e-:
3e- + 4H+ + NO3- → NO + 2H2O 1

28. Balancing Redox Equations (cont’d)
Make number of electrons in both reactions the same by multiplying to least common multiple (6, in this case):
3×(Cu → Cu2+ + 2e-)
2×(3e- + 4H+ + NO3- → NO + 2H2O)
Combine the two reactions and cancel as appropriate:
3Cu + 8H+ + 2NO3- → 3Cu2+ + 2NO + 4H2O
Reaction is balanced. 1

29. Balancing Redox Reactions
In basic solutions, add OH- to each side of the reaction to eliminate any H+ by combining them to make H2O. 7

30. Test Your Skill Balance the following equation in acid solution:
Cr2O72- + C2H5OH → Cr3+ + CO2 7

31. Test Your Skill Balance the following equation in basic solution:
Zn + ClO- → Zn(OH)42- + Cl- 7

32. The Mole
One mole is the amount of substance that contains as many entities as the number of atoms in exactly 12 grams of the 12C isotope of carbon.
Avogadro’s number is the experimentally determined number of 12C atoms in 12 g, and is equal to x 1023.

33. One Mole of Several Elements
Argon in green balloons Hg Cu Na Fe Al

34. Converting Moles and Entities
One mole of anything contains x 1023 entities.
1 mol H = x 1023 atoms of H
1 mol H2 = x 1023 molecules of H2
1 mol CH4 = x 1023 molecules of CH4
1 mol CaCl2 = x 1023 formula units of CaCl2

35. Moles to Number of Entities
Avogadro’s number allows the interconversion
of moles and numbers of atoms or molecules.
Number of atoms or molecules
Moles of substance
Avogadro’s number

36. Example: Convert Moles and Entities
How many atoms are present in 0.35 mol of Na?
How many moles are present in 3.00 x 1021 molecules of C2H6?

37. Molar Mass
The molar mass (M) of any atom, molecule or compound is the mass (in grams) of one mole of that substance.
The molar mass in grams is numerically equal to the atomic mass or molecular mass expressed in u.
Molar mass converts from mass (in grams) to amount (in moles) or the reverse.

38. Molar and Atomic Masses

39. Converting Moles and Mass
Molar mass of substance
Moles of substance
Mass of substance

40. Molar Mass Conversion What is the mass of 0.25 moles of CH4?
Molar mass of CH4 = 16.0 g/mol.

41. Example: Molar Mass Conversions
How many moles of ethylene (C2H4, M = 28.0 g/mol) are present in 16 g of that compound?
What is the mass, in grams, of moles of Fe?

42. Test Your Skill
What mass of compound must be weighed out to have a mol sample of H2C2O4 (M = g/mol)?

43. Mass Percentage from Formula
Use the formula of the compound to calculate the mass of each element in the compound and use those numbers to calculate percentage composition.

44. Example: Percentage Calculation
What is the mass percentage composition of H2C2O4?

45. A Combustion Experiment
The sample burns in excess O2: the H2O
is trapped by the CaCl2 and CO2 is trapped by the NaOH.

46. Carbon and Hydrogen Content
The masses of C and H in the sample are calculated from the masses of H2O and CO2 formed in the combustion reaction.

47. Example: Mass C, H and O in Sample
A compound contains only C, H, and O. A g-sample burns completely in oxygen to form g of water and g of CO2. Calculate the mass of each element in this sample.

48. Calculate Empirical Formula

49. Calculate Empirical Formula
What is the empirical formula of a compound that contains g C and g H in a g sample?

50. Example: Empirical Formula
What is the empirical formula of a chromium oxide that is 68.4% Cr and 31.6% O?

51. Test Your Skill
What is the empirical formula of a compound that is 59.9% Ti and 40.1% O?

52. Molecular Formula
The molecular formula must be a whole number multiple of the empirical formula.
If the empirical formula is CH2, the molecular formula is (CH2)n where
The molecular mass must be measured experimentally.

53. Example: Molecular Formula
An empirical formula calculation based on a combustion analysis experiment shows a new compound has the empirical formula C4H8O. A mass spectrometry experiment determines that the molar mass of the compound is 216 g/mol. What is the molecular formula?

54. Mole Relationships in Equations

55. Guidelines for Reaction Stoichiometry
Write the balanced equation.
Calculate the number of moles of the species for which the mass is given.
Use the coefficients in the equation to convert the moles of the given substance into moles of the substance desired.
Calculate the mass of the desired species.

56. Reaction Stoichiometry

57. Example: Stoichiometry
What mass of SO3 forms from the reaction of 4.1 g of SO2 with an excess of O2?

58. Test Your Skill Given the equation 4FeS2 + 11O2 ® 2Fe2O3 + 8SO2
what mass of SO2 is produced from reaction of 3.8 g of FeS2 with excess oxygen?

59. Theoretical Yield
The previous calculation showed that given the equation 4FeS2 + 11O2 ® 2Fe2O3 + 8SO2 4.1 g SO2 is produced from reaction of 3.8 g of FeS2 with excess oxygen.
The 4.1 g SO2 is the theoretical yield – the maximum quantity of product that can be obtained from a chemical reaction, based on the amounts of starting materials.

60. Limiting Reactant
Limiting reactant: the reactant that is completely consumed when a chemical reaction occurs.

61. Limiting Reactant
In the below reaction of Cl2 (green) and Na (purple), Cl2 is the limiting reactant. Once the limiting reactant is consumed the reaction stops and no more product forms. The formed NaCl and the excess Na are present at the end of the reaction.

62. Example: Limiting Reactant
Calculate the mass of the NH3 product formed (theoretical yield) when 7.0 g of N2 reacts with 2.0 g of H2.

63. Strategy for Limiting Reactant
Mass of A (reactant)
Mass of B (reactant)
Molar mass of A
Molar mass of B
Moles of A
Moles of B
Coefficients in the equation
Choose smaller amount
Moles of Product
Moles of Product
Molar mass of product
Mass of product

64. Example: Limiting Reactant
Calculate the mass of the NH3 product formed (theoretical yield) when 7.0 g of N2 reacts with 2.0 g of H2.
N2 limiting, 8.5 g NH3

65. Actual Yield, Percent Yield
As shown in the picture, in many reactions not all of the product formed can be isolated.
Actual yield: mass of product isolated in a reaction.
Yields are generally reported as a percent:

66. Example: Calculating Percent Yield
Given the reaction,
PCl3 + Cl2 ® PCl5
what is the percent yield when 50.0 g of PCl3 reacts with 35.0 g Cl2 and a chemist isolated 61.3 g of PCl5.?