The document discusses key concepts related to moles, including: 1) A mole is defined as 6.02x1023 representative particles of a substance, which can be used to convert between the number of particles and moles. 2) The mass of one mole of a substance is its molar mass, which can be used to convert between mass and moles. 3) One mole of an ideal gas occupies a volume of 22.4L at STP, which can be used to convert between moles and volume. 4) Percent composition by mass and molar mass can be used to determine empirical and molecular formulas of compounds.

1. Chapter 10
Chemical Quantities
Section 10.1
The Mole: A measurement of Matter

2. What is a Mole?
A mole (mol) of a substance is 6.02 X 1023
representative particles of that substance.
Usually atoms or molecules.
6.02 X 1023 is called Avogadro’s number

3. Converting Number of Particles to Moles
NumbersAvogadro'
Molecules)or(AtomsParticlesofNumber
MolesofNumber 
e.g.
How many moles of magnesium in 1.25 X 10 23 atoms of magnesium
23
23
10x6.02
10x1.25
MolesofNumber  = 0.208 mol Mg
Page 291: #3 & 4
Particles
Moles
Av.
number
Molecules or atoms
Moles - Particles

4. Converting Moles to Number of Particles
23
10x6.02xMolesofNumberMolecules)or(AtomsParticlesofNumber 
e.g. How many atoms are in 2.12 mol of propane (C3H8)?
Number of atoms = 2.12 x 6.02 x 10 23 x 11
= 1.4 x 10 25 atoms
Page: 292 #5 & 6

5. The mass of a Mole of an Element
The mass of a mole of an element is its molar mass.
How can you calculate the molar mass of a compound?
To calculate the molar mass of a compound:
1. Find the number of grams of each element.
2. Add the masses of the elements in the compound.
e.g. what is the molar mass of calcium
oxide?
Chemical formula = CaO,
Ca = 40, O = 16 Molar mass = 40 + 16 = 56g
Page: 296, #7 & 8

6. Atoms
or
Molecules
or
Particles
Number of
Moles
Avogadro’s
Number
Moles – Atoms
Multiplies by number of atoms
If atoms

7. Section 10.2
Mole – Mass
&
Mole – Volume
Relationships

8. massMolar
substancetheofMass
MolesofNumber 
Converting Mass to Moles
e.g. How many moles of iron(III) oxide are
contained
in 92.2g of pure Fe2O3?
Mass = 92.2g
Molar mass = 2x55.8 + 3x16 = 159.6g/mol
Number of moles = 92.2/159.6 = 0.578 mol Fe2O3
Page: 299, # 18 & 19
Moles - Mass

9. Converting Moles to Mass
Mass = number of Moles x molar mass
e.g. What is the mass of 9.45 mol of aluminum
oxide?
Chemical Formula = Al2O3
Number of moles = 9.45 mol
Molar mass = 2x27 + 3x16 = 102g Al2O3
Mass = 9.45 x 102 =964g Al2O3
Page: 298, # 16 & 17

10. The volume of a Mole of an Element
The volume of a mole of an element is called molar volume
The molar volume of a gas at STP is 22.4 L
Standard Temperature and Pressure
To calculate the volume of a gas at STP:
Volume of gas = Number of Moles x 22.4
e.g. Determine the volume of 0.6 mol sulfur dioxide gas at
STP.
Volume of SO2 = 0.6 x 22.4 = 13.4 L SO2
Page: 301, #20 & 21Moles - Volume

11. Calculating Molar Mass from Density
Molar mass = density at STP x molar volume at STP
e.g. The density of a gaseous compound carbon and oxygen is
found
to be 1.964g/L at STP. What is the molar mass of the
compound? Molar mass = 1.964 x 22.4 = 44g/mol
Page: 302, #22 & 23
Moles - Density

12. Mass
Number of
Moles
Molar
Mass
Moles – Mass

13. Volume
Number of
Moles
22.4
Moles – Volume

14. Molar
mass
Density 22.4
Moles – Density

15. Section 10.3
Percent Composition and Chemical Formulas

16. The percent by mass of an element in a compound is the number of grams of the
element divided by the mass in grams of the compound, multiplied by 100
100
compoundofmass
elementofmass
elementofmass% 
e.g. When a 13.6 g sample of a compound containing only magnesium and oxygen.
5.4 g of oxygen is obtained. What is the percent composition of this compound?
The percent composition is the relative amounts of the elements in a compound.
Mass of compound = 13.6 g
Mass of oxygen = 5.4 g
Mass of magnesium = 13.6 – 5.4 = 8.2 g
%7.39100
6.13
4.5
100
compoundofmass
Oofmass
O% 
%3.60100
6.13
2.8
100
compoundofmass
Mgofmass
Mg% 
Page 306, # 32 & 33
Calculating Percent Composition from Mass Data

17. Calculating Percent Composition from the Chemical Formula
100
compoundofmassmolar
compoundmole1inelementofmass
elementofmass% 
e.g. Propane (C3H8) is one of the compounds obtained from petroleum.
Calculate the percent composition of propane.
%7.81100
08.44
0.36
100
propaneofmass
Cofmass
C% 
%3.18100
08.44
08.8
100
propaneofmass
Hofmass
H% 
Mass of C in C3H8 = 36.0 g
Mass of H in C3H8 = 8.08 g
Molar mass of C3H8 = 44.08 g/mol
Page: 307, # 34 & 35

18. Empirical Formula
The empirical formula of a compound shows the smallest whole-number
ratio of the atoms in the compound.
It shows the kinds and lowest relative count of atoms or moles of atoms
in molecules or a compound.
An empirical formula may or may not be the same as a molecular formula.
For example, the lowest ratio of hydrogen to oxygen in hydrogen peroxide
is 1:1
So the empirical formula of hydrogen peroxide is HO
The actual molecular formula of hydrogen peroxide has twice the number of
atoms as the empirical formula.
The molecular formula is (HO) x 2 or H2O2
Notice that the ratio of hydrogen to oxygen is still the same.
The molecular formula tells the actual number of each kind of atom present
in a molecule of the compound.
For carbon dioxide, the empirical and molecular formulas are the same – CO2

19. e.g. A compound is analyzed and found to contain 25.9% nitrogen and
74.1% oxygen. What is the empirical formula of the compound?
25.9% N 74.1% O
Step 1. Assume that you have 100 g of the compound,
25.9 g 74.1 g
Step 2. Divide each by its atomic mass.
25.9 74.1
14 16
Step 3. Divide each by the smallest number of moles.
The empirical formula is N2O5 Page: 310, # 36 & 37
1.85 mol 4.63 mol
1.85
1.85
4.63
1.85
1 2.5
• Multiply all by 2

20. Molecular Formula
is the same as the empirical formula or it is a simple whole-number
multiple of its empirical
Once you determined the empirical formula of a compound, you can
determine its molecular formula, but you must know the compound’s
molar mass and empirical formula mass.
Step 1. Calculate the empirical formula mass
e.g. Calculate the molecular formula of a compound whose molar mass
is 60 g/mol and empirical formula is CH4N
Empirical formula = CH4N
Molar mass = 60 g/mol
Molecular formula = C?H?N?
Empirical formula mass = 1x12 + 4x1.01 + 1x14 = 30.04 g/mol
Step 2. Divide the molar mass by the empirical formula mass
60 = 2
30.04
Molecular formula = 2 x CH4N = C2H8N2
Step 3. Multiply the formula subscripts by this value (factor)
Page: 312, # 38 & 39