1. Concept Chapter 8 The Mole Lecture Presentation John Singer
Jackson College

2. A 12.01 g sample of carbon contains 6.02 x 1023 carbon atoms.
Avogadro’s Number
Avogadro’s number (symbol N) is the number of atoms in grams of carbon.
A g sample of carbon contains x 1023 carbon atoms.
Avogadro’s number = 6.02 x 1023

3. 1 mol = Avogadro’s number = 6.02 x 1023 units
The Mole
The mole (mol) is a unit of measure for an amount of a chemical substance.
A mole is 6.02 x 1023 particles.
1 mol = Avogadro’s number = 6.02 x 1023 units

4. Mole Calculations

5. Mole Calculations I How many sodium atoms are in 0.240 mol Na?
We want atoms of Na.
We have mol Na.
1 mole Na = 6.02 x 1023 atoms Na.
= 1.44 x 1023 atoms Na
0.240 mol Na x
1 mol Na
6.02 x 1023 atoms Na

6. How many moles of aluminum are in 3.42 x 1021 atoms Al?
Mole Calculations I
How many moles of aluminum are in x 1021 atoms Al?
We want moles Al.
We have 3.42 x 1021 atoms Al.
1 mol Al = 6.02 x 1023 atoms Al.
= 5.68 x 10–3 mol Al
3.42 x 1021 atoms Al x
1 mol Al
6.02 x 1023 atoms Al

7. Molar Mass
The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.
The atomic mass of carbon is amu.
Therefore, the molar mass of carbon is g/mol.

8. Molar Mass
Since nitrogen occurs naturally as a diatomic, N2, the molar mass of nitrogen gas is:
2 x g = g/mol

9. Molar Mass

10. Calculating Molar Mass
The molar mass of a substance is the sum of the atomic masses of each element.

11. Calculating Molar Mass
What is the molar mass of copper(II) nitrite, Cu(NO2)2?
The sum of the atomic masses is as follows:
( ) =
(46.01) = amu
The molar mass for Cu(NO2)2 is g/mol.

12. 6.02 x 1023 particles = 1 mol = molar mass
Mole Calculations II
Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance.
6.02 x 1023 particles = 1 mol = molar mass
If we want to convert particles to mass, we must first convert particles to moles, and then we can convert moles to mass.

13. Mole–Mass Calculation
What is the mass of 1.33 moles of titanium, Ti?
We want grams.
We have 1.33 moles of titanium.
Use the molar mass of Ti: 1 mol Ti = g Ti
= 63.7 g Ti
1.33 mol Ti x
47.88 g Ti
1 mol Ti

14. Atoms–Mass Calculation
What is the mass of 2.55 x 1023 atoms of lead?
We want grams.
We have atoms of lead.
Use Avogadro’s number and the molar mass of Pb.
2.55 × 1023 atoms Pb x
1 mol Pb
6.02×1023 atoms Pb
207.2 g Pb x
= 87.8 g Pb

15. Mass–Molecule Calculation
How many F2 molecules are present in 2.25 g of fluorine gas?
We want molecules F2.
We have grams F2.
Use Avogadro’s number and the molar mass of F2.
2.25 g F2 x
1 mol F2
38.00 g F2
6.02 x 1023 molecules F2 x
3.56 x 1022 molecules O2

16. Mass of an Atom or Molecule
What is the mass of a single molecule of sulfur dioxide? The molar mass of SO2 is g/mol.
We want mass/molecule SO2, we have the molar mass of sulfur dioxide.
Use Avogadro’s number and the molar mass of SO2 as follows:
64.07 g SO2
1 mol SO2
6.02 x 1023 molecules SO2 x
1.06 x 10–22 g/molecule

17. Molar Volume
At standard temperature and pressure, 1 mol of any gas occupies 22.4 L.
The volume occupied by 1 mol of gas (22.4 L) is called the molar volume.
Standard temperature and pressure are 0 C and 1 atm.

18. One Mole of a Gas at STP
The box below has a volume of 22.4 L, which is the volume occupied by 1 mol of a gas at STP.

19. Molar Volume of Gases We now have a new unit factor equation:
1 mol gas = 6.02 x 1023 molecules gas = 22.4 L gas

20. Molar Volume of Gases

21. Gas Density
The density of gases is much less than that of liquids.
We can easily calculate the density of any gas at STP.
The formula for gas density at STP is as follows:

22. Calculating Gas Density
What is the density of methane gas, CH4, at STP?
First we need the molar mass for methane.
(1.01) = g/mol
The molar volume for CH4 at STP is 22.4 L/mol.
Density is mass/volume.
= g/L
16.05 g/mol
22.4 L/mol

23. Molar Mass of a Gas 4.29 g 22.4 L = 64.1 g/mol 1.50 L 1 mol
An unknown gas has 4.29 g and occupies 1.50 L at STP. What is the molar mass?
We want g/mol; we have g/L.
4.29 g
1.50 L
22.4 L
1 mol x
= 64.1 g/mol

24. Mole Calculations III We now have three interpretations for the mole:
1 mol = 6.02 x 1023 particles
1 mol = molar mass
1 mol = 22.4 L at STP
This gives us three unit factors to use to convert among moles, particles, mass, and volume.

25. Calculating Molar Volume
A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?
Use molar volume of a gas: 1 mol = 22.4 L.
4.50 L CH4 x
= mol CH4
1 mol CH4
22.4 L CH4

26. Calculating Mass Volume
What is the mass of 3.36 L of ozone gas, O3, at STP?
We want mass O3; we have 3.36 L O3.
Convert volume to moles, then moles to mass.
3.36 L O3 x x
22.4 L O3
1 mol O3
48.00 g O3
= 7.20 g O3

27. Calculating Molecule Volume
How many molecules of argon gas, Ar, occupy L at STP?
We want molecules Ar; we have L Ar.
Convert volume to moles, and then moles to molecules.
0.430 L Ar x
1 mol Ar
22.4 L Ar
6.02 x 1023 molecules Ar x
= 1.16 x 1022 molecules Ar

28. Percent Composition
The percent composition of a compound lists the mass percent of each element.
For example, the percent composition of water, H2O, is 11% hydrogen and 89% oxygen.

29. Calculating Percent Composition
There are a few steps to calculating the percent composition of a compound. Let’s practice using H2O.
Assume you have 1 mol of the compound.
One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen. Therefore,
2(1.01 g H) + 1(16.00 g O) = molar mass H2O
2.02 g H g O = g H2O

30. More Calculating Percent Composition
Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water.

31. Percent Composition Problem
TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3.
7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N g O) = g C7H5(NO2)3
84.07 g C g H g N g O = g C7H5(NO2)3

32. Percent Composition of TNT

33. Empirical Formula
The empirical formula of a compound is the simplest whole-number ratio of ions in a formula unit or atoms of each element in a molecule.
The molecular formula of benzene is C6H6.
The empirical formula of benzene is CH.
The molecular formula of octane is C8H18.
The empirical formula of octane is C4H9.

34. Calculating Empirical Formulas
We can calculate the empirical formula of a compound from its composition data.
We can determine the mole ratio of each element from the mass to determine the empirical formula of radium oxide, Ra?O?.
– A g sample of radium metal was heated to produce g of radium oxide. What is the empirical formula?
– We have g Ra and – = g O.

35. Calculating Empirical Formulas
– The molar mass of radium is g/mol, and the molar mass of oxygen is g/mol.
We get Ra O Simplify the mole ratio by dividing by the smallest number.
We get Ra1.01O1.00 = RaO is the empirical formula.

36. Empirical Formulas from Mass Composition
We can also use percent composition data to calculate empirical formulas.
Assume that you have 100 grams of sample.
Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula?
– If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen.

37. Empirical Formula for Acetylene
Calculate the moles of each element.
The ratio of elements in acetylene is C7.68H7.75. Divide by the smallest number to get the following formula:
=CH

38. n = 2 and the molecular formula is C2H2.
The empirical formula for acetylene is CH. This represents the ratio of C to H atoms on acetylene.
The actual molecular formula is some multiple of the empirical formula, (CH)n.
Acetylene has a molar mass of 26 g/mol. Find n to find the molecular formula:
n = 2 and the molecular formula is C2H2.

39. Chapter Summary
Avogadro’s number is 6.02 x 1023, and is 1 mole of any substance.
The molar mass of a substance is the sum of the atomic masses of each element in the formula.
At STP, 1 mole of any gas occupies 22.4 L.

40. Chapter Summary, Continued
We can convert between the number of particles and moles of a substance using Avogadro’s number (1 mol = 6.02 x 1023 particles).
We can convert between mass of a substance and moles of a substance using the molar mass.
We can convert between the volume of a gas at STP and moles of a gas using molar volume at STP (1 mol = 22.4 L).

41. Chapter Summary, Continued
The percent composition of a substance is the mass percent of each element in that substance.
The empirical formula of a substance is the simplest whole-number ratio of the elements in the formula.
The molecular formula is a multiple of the empirical formula.