1. Chapter 8 The Mole Concept by Christopher G. Hamaker
Illinois State University
© 2014 Pearson Education, Inc. 1

2. Avogadro’s Number
In a lab, we cannot work with individual atoms or molecules. They are too small.
Avogadro’s number (symbol N) is the number of atoms in grams of carbon.
Its numerical value is 6.02 x It is called 1 mole
Therefore, a g sample of carbon contains 6.02 x 1023 carbon atoms and it is 1 mole.
Similarly, g sample of water contains
6.02 x 1023 H2O molecules (1 mole water).

3. Analogies for Avogadro’s Number
The volume occupied by one mole of softballs would be about the size of Earth.
One mole of Olympic shot put balls has about the same mass as that of Earth.
One mole of hydrogen atoms laid side by side would circle Earth about 1 million times?.

4. 1 mol = Avogadro’s number = 6.02 x 1023 units
Mole Calculations I
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 x 1023 particles (any particles- atoms, molecules, ions, protons, neutrons, electrons, etc)
1 mol = Avogadro’s number = 6.02 x 1023 units
We can use the mole relationship to convert between the number of particles and the mass of a substance.

5. Mole Calculations I, Continued
How many sodium atoms are in 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. Mole Calculations I, Continued
How many moles of aluminum are in 3.42 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. Mole Calculations II We will be using the unit analysis method again.
Recall the following steps:

8. 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.
Since nitrogen occurs naturally as a diatomic, N2, the molar mass of nitrogen gas is two times g or g/mol.
Similarly, the molar mass of H2O is g/mol (sum of masses of 2H and one O)
1 amu = e-24 g

9. One Mole of Several Substances
C12H22O11
H2O
mercury
sulfur
NaCl
copper
lead
K2Cr2O7
Pb =207.20, Cu = S = 32.06, Hg = C = 12.01, K = 39.10, H = 1.01, Cr = 51.99, O = 16.00, N = 14.01

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 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.

11. 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.

12. 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 mole Ti x
47.88 g Ti
1 mole Ti

13. 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 to convert atoms to moles and then molar mass of Pb.
2.55 × 1023 atoms Pb x
1 mol Pb
6.02×1023 atoms Pb
207.2 g Pb
1 mole Pb x
= 87.9 g Pb

14. CONCEPT CHECK!
Which of the following g samples contains the greatest number of atoms?
Magnesium
Zinc
Silver
The correct answer is “a”. Magnesium has the smallest average atomic mass (24.31 u, Zn = 65.38, Ag = 107.9).
Since magnesium is lighter than zinc and silver, it will have more moles in a g substance than Zn and Silver.
(Mg = u, Zn = 65.38, Ag = 107.9)

15. Rank the following according to number of atoms (greatest to least):
EXERCISE!
Rank the following according to number of atoms (greatest to least):
107.9 g of silver
70.0 g of zinc
21.0 g of magnesium
b, a, c; The greater the number of moles, the greater the number of atoms present. Zinc contains 1.07 mol, Ag contains 1.00 mol, and Mg contains mol.
(Mg = u, Zn = 65.38, Ag = 107.9)

16. Mass–Molecule Calculation
How many F2 molecules are present in 2.25 g of fluorine gas?
We want molecules F2.
We have grams F2. Convert g to moles using molar mass F2
Use Avogadro’s number for converting moles to molecules
2.25 g F2 x
1 mol F2
38.00 g F2
6.02 x 1023 molecules F2 x
3.56 x 1022 molecules F2

17. 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 of a molecule of 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

18. 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.
STP = 0 0C, 1 atm, NTP = 20 0C, 1 atm

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. 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.

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:
= density, g/L
molar mass in grams
molar volume in liters

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 CH4 at STP is 22.4 L/mol.
Density is mass/volume.
What is the density of CO2
= g/L
16.05 g/mol
22.4 L/mol

23. Molar Mass of a Gas
We can also use molar volume to calculate the molar mass of an unknown gas.
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
= 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?
We want moles; we have volume.
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.
All water contains % hydrogen and % oxygen by mass.

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.
2.02 g H
18.02 g H2O
x 100% = 11.2% H
16.00 g O
18.02 g H2O
x 100% = 88.79% O

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
84.07 g C
g TNT
x 100% = 37.01% C
5.05 g H
g TNT
x 100% = 2.22% H
42.03 g N
g TNT
x 100% = 18.50% N
96.00 g O
g TNT
x 100% = 42.26% O

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.
Na2O AlF3 Ca2O2

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, Continued
– 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 x
= mol Ra
1 mol O
16.00 g O
0.115 g O x
= 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.

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.
1 mol C
12.01 g C
92.2 g C x
= 7.68 mol C
1 mol H
1.01 g H
7.83 g H x
= 7.75 mol H
The ratio of elements in acetylene is C7.68H7.75. Divide by the smallest number to get the following formula:
7.68 C
= C1.00H1.01 = CH
7.75 H

38. Molecular Formula
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.

39. What is the empirical formula? What is the molecular formula?
EXERCISE!
The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol.
What is the empirical formula?
What is the molecular formula?
Assume 100.0g g of carbon is mol C (49.3/12.01) g of hydrogen is mol H (6.9/1.008) g of oxygen is mol O (43.8/16.00). The ratio of C:H:O is 1.5:2.5:1 (C: 4.105/ = 1.5; H: 6.845/ = 2.5). Multiplying each by 2 to get a whole number becomes a ratio of 3:5:2. The empirical formula is therefore C3H5O2. The molar mass of the empirical formula is g/mol, which goes into the molar mass of the molecular formula 2 times (146/73.07). The molecular formula is therefore C6H10O4.

40. From % composition get the empirical formula: mol of C = 49. 3g/12
From % composition get the empirical formula: mol of C = 49.3g/12.01g mol−1 = 4.11 mol mol of H = 6.9g/1.008g mol−1 = 6.85 mol mol of O = 43.8g/16.00g mol−1 = 2.74 mol C: 4.11/2.74 = 1.5 H: 6.85/2.74 = 2.5 O: 1 (x 2) No fractions, simplest whole number is EF = C3H5O2 Its (EF) molar mass is 73g/mol. Adipic acid molar mass is 146g/mol, twice that of EF. Therefore molecular formula is 2 x C3H5O2 = C6H10O4

41. Critical Thinking: Avogadro’s Number
In 1911, Ernest Rutherford estimated the value of Avogadro’s number as 6.11 x 1023 using alpha particles from radium.
The most recent measurements of Avogadro’s number were made using X-ray diffraction of silicon crystals.
Currently, the most accurate value for Avogadro’s number is x 1023.

42. 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.

43. 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 (g/mol).
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).

44. 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.