1. Chapter 3: Formulas, Equations, and Moles
9/19/2018
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

2. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Balancing Chemical Equations
A balanced chemical equation shows that the law of conservation of mass is adhered to.
In a balanced chemical equation, the numbers and kinds of atoms on both sides of the reaction arrow are identical.
2NaCl(s)
2Na(s) + Cl2(g)
Remember, a chemical reaction can be thought of as a rearrangement of the atoms to form new compounds.
left side:
2 Na
2 Cl
right side:
2 Na
2 Cl
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

3. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Balancing Chemical Equations
HgI2(s) + 2KNO3(aq)
Hg(NO3)2(aq) + 2KI(aq)
left side:
1 Hg
2 N
6 O
2 K
2 I
right side:
1 Hg
2 I
2 K
2 N
6 O
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

4. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Balancing Chemical Equations
Write the unbalanced equation using the correct chemical formula for each reactant and product.
H2O(l)
H2(g) + O2(g)
Find suitable coefficients—the numbers placed before formulas to indicate how many formula units of each substance are required to balance the equation.
2H2O(l)
2H2(g) + O2(g)
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

5. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Balancing Chemical Equations
Reduce the coefficients to their smallest whole-number values, if necessary, by dividing them all by a common denominator.
4H2O(l)
4H2(g) + 2O2(g)
divide all by 2
Since balancing at this level is often trial-and-error, the coefficients need to be checked to make sure they are the smallest whole-number ones. An alternative is to use fractions but some students do not do well with fractional coefficients at this stage.
2H2O(l)
2H2(g) + O2(g)
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

6. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Balancing Chemical Equations
Check your answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation.
2H2O(l)
2H2(g) + O2(g)
left side:
4 H
2 O
right side:
4 H
2 O
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

7. Balancing Chemical Equations
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Balancing Chemical Equations
Do not change subscripts when you balance a chemical equation. You are only allowed to change the coefficients.
H2O(l)
H2(g) + O2(g)
unbalanced
2H2O(l)
2H2(g) + O2(g)
H2O2(l)
H2(g) + O2(g)
Balancing chemical equations only determines the proper numerical ratios between all of the substances (both reactants and products). It does not change any of the substances.
Balanced properly
Chemical equation changed!
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

8. Chemical Symbols on Different Levels
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Chemical Symbols on Different Levels
2H2O(l)
2H2(g) + O2(g)
microscopic:
2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water.
macroscopic:
0.56 kg of hydrogen gas react with 4.44 kg of oxygen gas to yield 5.00 kg of liquid water.
The mole relates microscopic to macroscopic.
How can we relate the two to each other?
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

9. Avogadro’s Number and the Mole
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Avogadro’s Number and the Mole
Molecular Mass: Sum of atomic masses of all atoms in a molecule.
Formula Mass: Sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.
HCl:
1.0 amu amu = 36.5 amu
C2H4:
2(12.0 amu) + 4(1.0 amu) = 28.0 amu
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

10. Avogadro’s Number and the Mole
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Avogadro’s Number and the Mole
Avogadro’s Number (NA): One mole of any substance contains x 1023 formula units.
One mole of any substance is equivalent to its molecular or formula mass.
HCl:
1 mole = 36.5 g
“Mole” is often abbreviated “mol”.
6.022 x 1023 molecules = 36.5 g
C2H4:
1 mole = 28.0 g
6.022 x 1023 molecules = 28.0 g
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

11. Avogadro’s Number and the Mole
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Avogadro’s Number and the Mole
Avogadro’s Number (NA): One mole of any substance contains x 1023 formula units.
One mole of any substance is equivalent to its molecular or formula mass.
HCl:
1 mole = 36.5 g
“Mole” is often abbreviated “mol”.
C2H4:
1 mole = 28.0 g
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

12. Stoichiometry: Chemical Arithmetic
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Stoichiometry: Chemical Arithmetic
Stoichiometry: The relative proportions in which elements form compounds or in which substances react.
aA + bB
cC + dD
Grams of A
Moles of A
Moles of B
Grams of B
Definition from the Oxford Dictionary of Chemistry, Oxford University Press, 2000.
Its origin is Greek for “element” “measure.”
“grams-to-moles-to-moles-to-grams”
You need the coefficients from the balanced chemical equation.
You can relate any of the substances in the chemical equation to each other.
Molar Mass of A
Mole Ratio Between A and B (Coefficients)
Molar Mass of B
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

13. Stoichiometry: Chemical Arithmetic
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Stoichiometry: Chemical Arithmetic
Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas:
2NaOH(aq) + Cl2(g)
NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react with 25.0 g Cl2?
Grams of
Cl2
Moles of
Cl2
Moles of
NaOH
Grams of
NaOH
Molar Mass
Mole Ratio
Molar Mass
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

14. Stoichiometry: Chemical Arithmetic
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Stoichiometry: Chemical Arithmetic
Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas:
2NaOH(aq) + Cl2(g)
NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react with 25.0 g Cl2?
25.0 g Cl2
Remind students about significant figures.
1 mol Cl2
2 mol NaOH
40.0 g NaOH x x x
70.9 g Cl2
1 mol Cl2
1 mol NaOH
= 28.2 g NaOH
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

15. Yields of Chemical Reactions
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Yields of Chemical Reactions
Actual Yield:
Theoretical Yield:
The amount actually formed in a reaction.
The amount predicted by calculations.
actual yield
Percent Yield =
x 100%
theoretical yield
Maximum percent yield is 100%.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

16. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant.
Excess Reactant: Any of the other reactants still present after determination of the limiting reactant.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

17. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers:
C2H4O(aq) H2O(l)
C2H6O2(l)
Because water is so cheap and abundant, it is used in excess when compared to ethylene oxide. This ensures that all of the relatively expensive ethylene oxide is entirely consumed.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

18. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers:
C2H4O(aq) H2O(l)
C2H6O2(l)
If 3 moles of ethylene oxide react with 5 moles of water, which reactant is limiting and which reactant is present in excess?
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

19. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers:
C2H4O(aq) H2O(l)
C2H6O2(l)
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

20. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation:
Li2O(s) + H2O(g)
2LiOH(s)
If 80.0 g of water are to be removed and 65.0 g of Li2O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced?
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

21. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
Li2O(s) + H2O(g)
2LiOH(s)
Which reactant is limiting?
Amount of H2O that will react with 65.0 g Li2O:
65.0 g Li2O
1 mol Li2O
1 mol H2O x x
= 2.17 moles H2O
29.9 g Li2O
1 mol Li2O
You only have enough Li2O to react with 2.17 mol H2O. Since you actually have more than that amount of H2O, you don’t have enough Li2O to react with all of the H2O.
Amount of H2O given:
80.0 g H2O
1 mol H2O x
= 4.44 moles H2O
18.0 g H2O
Li2O is limiting
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

22. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
Li2O(s) + H2O(g)
2LiOH(s)
How many grams of excess H2O remain?
2.17 mol H2O
18.0 g H2O x
= 39.1 g H2O (consumed)
1 mol H2O
80.0 g H2O g H2O = 40.9 g H2O
initial
consumed
remaining
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

23. Reactions with Limiting Amounts of Reactants
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Reactions with Limiting Amounts of Reactants
Li2O(s) + H2O(g)
2LiOH(s)
How many grams of LiOH are produced?
2.17 mol H2O
2 mol LiOH
23.9 g LiOH x x
= 104 g LiOH
1 mol H2O
1 mol LiOH
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

24. Concentrations of Reactants in Solution: Molarity
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Concentrations of Reactants in Solution: Molarity
Molarity: The number of moles of a substance dissolved in each liter of solution. In practice, a solution of known molarity is prepared by weighing an appropriate amount of solute, placing it in a container called a volumetric flask, and adding enough solvent until an accurately calibrated final volume is reached.
Solution: A homogeneous mixture.
Solute: The dissolved substance in a solution.
Solvent: The major component in a solution.
It’s the volume of solution that’s needed and not simply the volume of water (assuming water is the solvent). Either pictures or showing a volumetric flask might be helpful.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

25. Concentrations of Reactants in Solution: Molarity
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Concentrations of Reactants in Solution: Molarity
Molarity converts between mole of solute and liters of solution:
molarity =
moles of solute
liters of solution
1.00 mol of sodium chloride placed in enough water to make 1.00 L of solution would have a concentration equal to:
1.00 mol
mol
= 1.00
or 1.00 M
1.00 L L
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

26. Concentrations of Reactants in Solution: Molarity
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Concentrations of Reactants in Solution: Molarity
How many grams of solute would you use to prepare 1.50 L of M glucose, C6H12O6?
Molar mass C6H12O6 = g/mol
1.50 L
0.250 mol x
= mol
1 L
0.275 mol
180.0 g x
= 49.5 g
1 mol
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

27. Diluting Concentrated Solutions
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Diluting Concentrated Solutions
dilute solution
concentrated solution + solvent
initial
final
Mi Vi = Mf Vf
Since the number of moles of solute remains constant, all that changes is the volume of solution by adding more solvent.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

28. Diluting Concentrated Solutions
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Diluting Concentrated Solutions
Sulfuric acid is normally purchased at a concentration of 18.0 M. How would you prepare mL of M aqueous H2SO4?
Mi = 18.0 M Mf = M
Vi = ? mL Vf = mL
Add acid to water!
Mf Vf
0.500 M
250.0 mL
Vi = = x
= 6.94 mL Mi
18.0 M
Add 6.94 mL 18.0 M sulfuric acid to enough water to make mL of M solution.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

29. Solution Stoichiometry
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Solution Stoichiometry
aA + bB
cC + dD
Volume of
Solution of A
Moles of A
Moles of B
Volume of
Solution of B
Molarity of A
Mole Ratio Between A and B (Coefficients)
Molar Mass of B
These problems still come back to moles.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

30. Solution Stoichiometry
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Solution Stoichiometry
What volume of M H2SO4 is needed to react with 50.0 mL of M NaOH?
H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + 2H2O(l)
Volume of
Solution of H2SO4
Moles of
H2SO4
Moles of
NaOH
Volume of
Solution of NaOH
Molarity of H2SO4
Mole Ratio Between H2SO4 and NaOH
Molarity of NaOH
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

31. Solution Stoichiometry
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Solution Stoichiometry
H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + 2H2O(l)
Moles of NaOH available:
50.0 mL NaOH
0.100 mol
1 L x x
= mol NaOH
1 L
1000 mL
Volume of H2SO4 needed:
mol NaOH
1 mol H2SO4
1 L solution
1000 mL x x x
2 mol NaOH
0.250 mol H2SO4
1 L
10.0 mL solution (0.250 M H2SO4)
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

32. Chapter 3: Formulas, Equations, and Moles
9/19/2018
Titration
Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known.
HCl(aq) + NaOH(aq)
NaCl(aq) + 2H2O(l)
Once the reaction is complete you can calculate the concentration of the unknown solution.
How can you tell when the reaction is complete?
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

33. Chapter 3: Formulas, Equations, and Moles
9/19/2018
Titration
standard solution
(known concentration)
buret
Erlenmeyer
flask
Phenolphthalein is commonly used for HCl-NaOH titrations. It turns red in slightly basic solutions.
unknown concentration solution
An indicator is added which changes color once the reaction is complete
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

34. Chapter 3: Formulas, Equations, and Moles
9/19/2018
Titration
48.6 mL of a M NaOH solution is needed to react with 20.0 mL of an unknown HCl concentration. What is the concentration of the HCl solution?
HCl(aq) + NaOH(aq)
NaCl(aq) + 2H2O(l)
Volume of
Solution of NaOH
Moles of
NaOH
Moles of
HCl
Volume of
Solution of HCl
Molarity of NaOH
Mole Ratio Between NaOH and HCl
Molarity of HCl
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

35. Chapter 3: Formulas, Equations, and Moles
9/19/2018
Titration
HCl(aq) + NaOH(aq)
NaCl(aq) + 2H2O(l)
Moles of NaOH available:
48.6 mL NaOH
0.100 mol
1 L x x
= mol NaOH
1 L
1000 mL
Moles of HCl reacted:
mol NaOH
1 mol HCl x
= mol HCl
1 mol NaOH
Concentration of HCl solution:
mol HCl
1000 mL x
= M HCl
20.0 mL solution
1 L
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

36. Percent Composition and Empirical Formulas
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Percent Composition and Empirical Formulas
Percent Composition: Expressed by identifying the elements present and giving the mass percent of each.
Empirical Formula: It tells only the ratios of the atoms in a compound.
Molecular Formula: It tells the actual numbers of atoms in a compound. It can be either the empirical formula or a multiple of it.
molecular mass
empirical formula mass
multiple =
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

37. Percent Composition and Empirical Formulas
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Percent Composition and Empirical Formulas
A colorless liquid has a composition of 84.1 % carbon and 15.9 % hydrogen by mass. Determine the empirical formula. Also, assuming the molar mass of this compound is g/mol, determine the molecular formula of this compound.
Mass percents
Moles
Mole ratios
Subscripts
Molar masses
Relative mole ratios
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

38. Percent Composition and Empirical Formulas
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Percent Composition and Empirical Formulas
Assume g of the substance:
Mole of carbon:
84.1 g C
1 mol C x
= 7.01 mol C
12.0 g C
Mole of hydrogen:
15.9 g H
1 mol H x
= 15.9 mol H
1.0 g H
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

39. Percent Composition and Empirical Formulas
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Percent Composition and Empirical Formulas
Empirical formula:
C7.01H15.9
C7.01H15.9
= C1H2.27
7.01
7.01
smallest value for the ratio
C1H2.27
C1x4H2.27x4 = C4H9
need whole numbers
Need small, whole numbers (4x2.27=9.08 which is close enough to 9).
114.2 was given.
57.0 is the empirical formula mass for C4H9.
Molecular formula:
114.2
C4x2H9x2 = C8H18
multiple =
= 2
57.0
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

40. Determining Empirical Formulas: Elemental Analysis
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Determining Empirical Formulas: Elemental Analysis
Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO2 and H2O, which are separated and weighed by an automated instrument called a gas chromatograph.
hydrocarbon + O2(g)
xCO2(g) + yH2O(g)
If the unknown compound also contains oxygen, then the masses of carbon and hydrogen are determined first. The mass of oxygen is determined by subtracting those masses from that of the unknown compound.
carbon
hydrogen
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

41. Determining Molecular Masses: Mass Spectrometry
Chapter 3: Formulas, Equations, and Moles
9/19/2018
Determining Molecular Masses: Mass Spectrometry
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.