2. Avogadro’s Number
Avogadro’s number (symbol N) is the number of atoms in grams of carbon.
Its numerical value is 6.02 × 1023.
Therefore, a g sample of carbon contains 6.02 × 1023 carbon atoms.
Chapter 9

3. 1 mol = Avogadro’s number = 6.02 × 1023 units
The Mole
The mole (mol) is a unit of measure for an amount of a chemical substance.
A mole is Avogadro’s number of particles, which is 6.02 × 1023 particles.
1 mol = Avogadro’s number = 6.02 × 1023 units
We can use the mole relationship to convert between the number of particles and the mass of a substance.
Chapter 9

4. Analogies for Avogadro’s Number
The volume occupied by one mole of softballs would be about the size of the Earth.
One mole of Olympic shotput balls has about the same mass as the Earth.
Chapter 9

5. One Mole of Several Substances
C12H22O11
H2O
lead
mercury
K2Cr2O7
sulfur
copper
NaCl
Chapter 9

6. Mole Calculations We will be using the Unit Analysis Method again.
Recall:
First, we write down the unit requested.
Second, we write down the given value.
Third, we apply unit factor(s) to convert the given units to the desired units.
Chapter 9

7. Mole Calculations I How many sodium atoms are in 0.120 mol Na?
Step 1: we want atoms of Na
Step 2: we have mol Na
Step 3: 1 mole Na = 6.02 × 1023 atoms Na
= 7.22 × 1022 atoms Na
0.120 mol Na ×
1 mol Na
6.02 × 1023 atoms Na
Chapter 9

8. Mole Calculations I
How many moles of potassium are in 1.25 × 1021 atoms K?
Step 1: we want moles K
Step 2: we have 1.25 × 1021 atoms K
Step 3: 1 mole K = 6.02 × 1023 atoms K
= 2.08 × 10-3 mol K
1.25 × 1021 atoms K ×
1 mol K
6.02 × 1023 atoms K
Chapter 9

9. Molar Mass
The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.
The atomic mass of iron is amu.
Therefore, the molar mass of iron is g/mol.
Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is 2 times g or g/mol.
Chapter 9

10. Calculating Molar Mass
The molar mass of a substance is the sum of the molar masses of each element.
What is the molar mass of magnesium nitrate, Mg(NO3)2?
The sum of the atomic masses is:
( ) =
(62.01) = amu
The molar mass for Mg(NO3)2 is g/mol.
Chapter 9

11. 6.02 × 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 × 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.
Chapter 9

12. Mole-Mole Calculations
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 mole Ti ×
47.88 g Ti
1 mole Ti
Chapter 9

13. Mole Calculations III What is the mass of 2.55 × 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 ×
1 mol Pb
6.02×1023 atoms Pb
207.2 g Pb
1 mole Pb ×
= 87.9 g Pb
Chapter 9

14. Mole Calculations III
How many O2 molecules are present in g of oxygen gas?
We want molecules O2; we have grams O2.
Use Avogadro’s number and the molar mass of O2:
0.470 g O2 ×
1 mol O2
32.00 g O2
6.02×1023 molecules O2
1 mole O2 ×
8.84 × 1021 molecules O2
Chapter 9

15. Mass of a Single 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:
64.07 g SO2
1 mol SO2
6.02×1023 molecules SO2
1 mole SO2 ×
1.06 × g/molecule
Chapter 9

16. Molar Volume
At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.
The volume occupied by 1 mole of gas (22.4 L) is called the molar volume.
Standard temperature and pressure are 0C and 1 atm.
Chapter 9

17. Molar Volume of Gases We now have a new unit factor equation:
1 mole gas = 6.02 × 1023 molecules gas = 22.4 L gas
Chapter 9

18. One Mole of a Gas at STP
The box below has a volume of 22.4 L, the volume occupied by 1 mole of a gas at STP.
Chapter 9

19. Gas Density The density of gases is much less than that of liquids.
We can calculate the density of any gas at STP easily.
The formula for gas density at STP is:
= density, g/L
molar mass in grams
molar volume in liters
Chapter 9

20. Calculating Gas Density
What is the density of ammonia gas, NH3, at STP?
First we need the molar mass for ammonia:
(1.01) = g/mol
The molar volume NH3 at STP is 22.4 L/mol.
Density is mass/volume:
= g/L
17.04 g/mol
22.4 L/mol
Chapter 9

21. Molar Mass of a Gas
We can also use molar volume to calculate the molar mass of an unknown gas.
1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass?
We want g/mol; we have g/L.
1.96 g
1.00 L
22.4 L
1 mole ×
= g/mol
Chapter 9

22. Mole Unit Factors We now have three interpretations for the mole:
1 mol = 6.02 × 1023 particles
1 mol = molar mass
1 mol = 22.4 L at STP for a gas
This gives us 3 unit factors to use to convert between moles, particles, mass, and volume.
Chapter 9

23. Mole-Volume Calculation
A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?
We want moles; we have volume.
Use molar volume of a gas: 1 mol = 22.4 L.
4.50 L CH4 ×
= mol CH4
1 mol CH4
22.4 L CH4
Chapter 9

24. Mass-Volume Calculation
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 × ×
22.4 L O3
1 mol O3
48.00 g O3
= 7.20 g O3
Chapter 9

25. Molecule-Volume Calculation
How many molecules of hydrogen gas, H2, occupy L at STP?
We want molecules H2; we have L H2.
Convert volume to moles, and then moles to molecules:
0.500 L H2 ×
1 mol H2
22.4 L H2
6.02×1023 molecules H2
1 mole H2 ×
= 1.34 × 1022 molecules H2
Chapter 9

26. 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
All water contains % hydrogen and % oxygen by mass.
Chapter 9

27. 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 mole of the compound.
One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen.
2(1.01 g H) + 1(16.00 g O) = molar mass H2O
2.02 g H g O = g H2O
Chapter 9

28. 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:
2.02 g H
18.02 g H2O
× 100% = 11.2% H
16.00 g O
18.02 g H2O
× 100% = 88.79% O
Chapter 9

29. 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.
Chapter 9

30. Percent Composition of TNT
84.07 g C
g TNT
× 100% = 37.01% C
1.01 g H
g TNT
× 100% = 2.22% H
42.03 g N
g TNT
× 100% = 18.50% N
96.00 g O
g TNT
× 100% = 42.26% O
Chapter 9

31. Empirical Formulas
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.
Chapter 9

32. 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 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.
Chapter 9

33. Calculating Empirical Formulas
The molar mass of radium is g/mol, and the molar mass of oxygen is g/mol.
1 mol Ra
g Ra
1.640 g Ra ×
= mol Ra
1 mol O
16.00 g O
0.115 g O ×
= mol O
We get Ra O Simplify the mole ratio by dividing by the smallest number.
We get Ra1.01O1.00 = RaO is the empirical formula.
Chapter 9

34. Empirical Formulas from Percent 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 g carbon and 7.83 g hydrogen.
Chapter 9

35. Empirical Formula for Acetylene
Calculate the moles of each element:
1 mol C
12.01 g C
92.2 g C ×
= 7.68 mol C
1 mol H
1.01 g H
7.83 g H ×
= 7.75 mol H
The ratio of elements in acetylene is C7.68H7.75. Divide by the smallest number to get the formula:
7.68 C
= C1.00H1.01 = CH
7.75 H
Chapter 9

36. Molecular Formulas
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: = CH
(CH)n
26 g/mol
13 g/mol
n = 2 and the molecular formula is C2H2.
Chapter 9

37. Chapter Summary
Avogadro’s number is 6.02 × 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.
Chapter 9

38. Chapter Summary, continued
We can use the following flow chart for mole calculations:
Chapter 9

39. 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.
Chapter 9