1. Chapter 2 The Mole: The Link Between the Macroscopic and the Atomic Worlds of Chemistry
In this chapter we explore the quantitative aspect of chemistry, also known as stoichiometry. The concept of the mole is a key part of stoichiometry.

2. The Mole as the Bridge Between the Macroscopic and Atomic Scales
Central concept
Connects microscopic and macroscopic.
Defined as the number of atoms in exactly 12 grams of 12C.
(For a lot of chemistry, it doesn’t matter what the numerical value of a mole equals!)

3. The Mole as the Bridge Between the Macroscopic and Atomic Scales
grams of 31P has the same number of atoms as grams of 12C.
Both quantities are referred to as one mole.

4. The Mole as a Collection of Atoms
The numerical value of the number of atoms in 12 grams of 12C is called Avogadro’s Number.

5. Converting Grams into Moles and Number of Atoms
Grams  Moles
Moles → Grams
Moles → Number of Atoms
Number of Atoms  Moles

6. The Mole as a Collection of Molecules
Need Avogadro’s number.
Need molar mass.
Mass corresponding to one mole.
For elements, same as atomic weight in grams unless the element is represented as a molecular element such as Cl2.

7. The Mole as a Collection of Molecules

8. The Mole as a Collection of Molecules
Also known as Molar Mass.
For compounds, use chemical formula.
Illustrates the importance of chemical formula:
Symbols
Subscripts

9. The Mole as a Collection of Molecules
Calculate the molecular mass of formaldehyde, H2CO.
Which atoms are present?
How many of each of these atoms?

10. The Mole as a Collection of Molecules
amu/H =
amu/C =
amu/O =
Molecular Mass of formaldehyde is grams/mole.

11. The Mole as a Collection of Molecules
With the molecular mass, many questions can be answered:
How many grams are in one mole of NH3?
How many moles are in 75 grams of H2?
What’s the average mass of one molecule of CO2?

12. The Mole as a Collection of Molecules
Correctly calculating molecular mass is an important skill in chemistry.

13. Percent Mass
Calculated with same data used to calculate molecular mass.
In the formaldehyde example, grams of the grams/mole come from carbon.

14. Determining the Formula of a Compound
From the % mass of the elements in a compound, we can determine the empirical formula of the compound.
If we know the molecular mass, we can determine the molecular formula from the empirical formula.

15. Two Views of Chemical Equations: Molecules Versus Moles
What does this equation mean?
2Mg(s) + O2(g) → 2MgO(s)
• Two atoms of Mg will react with one molecule of oxygen to produce 2 molecules of MgO.
• Also means 48 grams of Mg will react with 32 grams of oxygen to produce 80 grams of MgO.

16. Two Views of Chemical Equations: Molecules Versus Moles
First meaning is microscopic.
Second is macroscopic.

17. Mole Ratios and Chemical Equations
Chemical equations predict.
The numbers used to balance the equation are called stoichiometric coefficients.
These coefficients represent mole ratios.

18. Mole Ratios and Chemical Equations
2Mg(s) + O2(g) → 2MgO(s)
“Two moles of magnesium will react with 1 mole of oxygen to produce 2 moles of magnesium oxide.”

19. Mole Ratios and Chemical Equations
How many grams of oxygen are required to completely react with grams of magnesium?
2Mg(s) + O2(g) → 2MgO(s)

20. Mole Ratios and Chemical Equations
10.00 grams Mg = moles
1 mole of O2 will react with 2 moles of Mg.
moles of O2 will react with moles Mg.
1 mole of O2 = grams
moles O2 = grams

21. Mole Ratios and Chemical Equations
How many grams of magnesium oxide will be formed?

22. Mole Ratios and Chemical Equations
How many grams of magnesium oxide will be formed?
grams

23. Stoichiometry
The calculations just shown are an example of stoichiometry.
Stoichiometry is what makes chemistry a quantitative science.

24. Stoichiometry
Imagine that the previous problem had been stated as follows:
“How much magnesium oxide can be produced with 20 grams of magnesium and 10 grams of oxygen?”

25. Stoichiometry
Is the correct answer 30 grams?

26. Stoichiometry
Is the correct answer 30 grams?
No.
Why not?

27. Stoichiometry Is the correct answer 30 grams? No. Why not?
The mole ratio of Mg to O2 is not 2 to 1 in this problem.

28. The Stoichiometry of the Breathalyzer

29. The Nuts and Bolts of Limiting Reagents
The reaction between 20 grams of Mg and 10 grams of O2 is an example of a Limiting Reagent problem.
Reactions can be either stoichiometric or have a limiting reagent.

30. The Nuts and Bolts of Limiting Reagents
The mathematics of a limiting reagent problem extend beyond chemistry.
- Text example with nuts and bolts, section 2.11.
- Imagine other examples such as recipes and assembly lines.

31. The Nuts and Bolts of Limiting Reagents
In a limiting reagent problem, the most important task is to determine which reagent is the limiting reagent.
All subsequent calculations are based on this determination.

32. Density Mass of a sample divided by its volume.
Common unit is grams/cm3.
Applies to all forms of matter.
Most importantly, density is independent of the sample size.
Can be used to identify a substance.

33. Solute, Solvent, and Solution
A lot of chemistry is done with solutions.
Important terms associated with solutions:
Solvent
Solute
Solution

34. Solute, Solvent, and Solution
Solutions are homogeneous.
Solutions are mixtures.
The composition of a solution must be specified.

35. Solute, Solvent, and Solution
Compare these statements:
“I heated 50 mL of ethanol to 35 °C and held it at that temperature for 1 hour.”
“I heated 50 mL of an aqueous solution of sodium nitrate to 35 °C and held it at that temperature for 1 hour.”

36. Solute, Solvent, and Solution
Why is the second one ambiguous?

37. Solute, Solvent, and Solution
Why is the second one ambiguous?
The composition of the solution was not specified. This leads to the next topic:
Concentration.

38. Concentration Most useful with solutions.
Expressed as amount of solute per amount of solvent or amount of solute per amount of solution.
Important to know the difference between solvent and solution.

39. Molarity as a Way of Counting Particles in Solution
Most common concentration unit in chemistry is Molarity.
Equal to moles of solute/Liter of solution.
Symbol M used to express units mol/L.

40. Molarity as a Way of Counting Particles in Solution
Important Equation:
M × V = n
(Molarity of a solution) × (volume of solution expressed in Liters) = number of moles of solute in said volume

41. Dilution Calculations
Use M × V = n
Recall Molarity is moles solute/Liter solution.

42. Solution Stoichiometry
Balanced chemical reactions involve moles of substances, n.
Solution information must be converted to number of moles, n.
M × V = n
Important skill given the extent of solution work in chemistry.

43. Solution Stoichiometry
Net ionic equation sometimes used
Unreactive common ions (sometimes called spectator ions) are subtracted from balanced chemical equation.
H2C2O4(aq) +2Na+(aq) + 2OH-(aq) → 2Na+(aq) + C2O4-2(aq) + 2H2O(l)
H2C2O4(aq) + 2OH-(aq) → C2O4-2(aq) + 2H2O(l)