1. THE MOLE How many particles are there really? 6.02 x 1023
"Happy Mole Day to You" Chemistry Song

2. The Rule m n = M Mass of a given amount of substance (g)
Amount of substance (mol)
Mass of a given
amount of substance (g)
Molar Mass
(g mol -1)
n = m M

3. Worked Example
Calculate the mass of 0.35 mol of magnesium nitrate (Mg(NO3)2.
What is the formula we use?????
How do we work out M

4. Solution m = n x M (Mg(NO3)2 = 24.3 + 2(14.0 + 3 x 16.0)
= g mol-1
So:
m(Mg(NO3)2 = n(Mg(NO3)2 x M(Mg(NO3)2
= 0.35 mol x g mol-1
= 52 g
Why only two significant figures?

5. Counting by Weighing
If we have a known mass of a substance can we possibly know the number of particles present?
We can by combining both the equations we have learnt so far.
m = n x M and N = n x NA
How do we do this??

6. Worked Example
a) Calculate the amount (in mol) of CO2 molecules present in 22 g of carbon dioxide.
b) What is the number of molecules present in this mass of CO2.

7. Solution Since M(CO2) = 12.0 + 2 x 16.0 = 44.0 g mol-1 m n = M m(CO2)
n(CO2) =
m(CO2)
M(CO2)
n(CO2) =
22 g
44.0 g mol-1
= 0.50 mol

8. Solution Part b)
b) Since N = n x NA, where N is the number of specified particles: N(CO2) = n(CO2) x NA = 0.50 x 6.02 x 1023 = 3.0 x 1023 So, 22 g of carbon dioxide contains 3.0 x 1023 ?? Atoms, ions or molecules? Molecules

9. Percentage Composition
The values for molar masses of elements in compounds can be used to calculate the percentage composition of a compound once its formula is known.
This type of calculation is important in chemistry.
If for example, a company is producing aluminium from alumina (Al2O3), the management will want to know the mass of aluminium that can be extracted given a quantity of alumina.

10. The Rule Percentage Composition Mass of element in 1 mole of compound
% by mass
of the element =
Mass of element in 1 mole of compound
Mass of 1 mole of compound
x 100

11. Worked Example
Calculate the percentage of aluminium in alumina (Al2O3).
Solution:
Step 1. Find the molar mass of Al2O3
M(Al2O3) = (2 x 27) + (3 x 16)
= 102 g mol-1

12. Solution Step 2: Find the percentage of aluminium in the alumina.
Since 1 mol Al2O3 contains 2 mol Al atoms:
Mass of aluminium in 1 mol (102 g) of Al2O3 = 2 x 27 g
= 54 g
% of Al in Al2O3 =
Mass of Al in 1 mole of Al2O3
Mass of 1 mole of Al2O3
x 100 = 54
102
x 100
= 52.9%
The company management, therefore, knows that aluminium
comprises about 53% by mass of any sample of alumina.

13. Empirical formulas
The empirical formula of a compound is the formula that gives the simplest whole number ratio, by number of moles, of each element in the compound.
You may have wondered how chemists have determined the formulas of compounds such as water, carbon dioxide, nitrogen dioxide and other compounds.

14. Whole number ratio of elements in compound
Empirical Formulas
Compound
Empirical Formula
Whole number ratio of elements in compound
Water
H2O
H : O
2 : 1
Carbon dioxide
CO2
C : O
1 : 2
Calcium Carbonate
CaCO3
Ca : C : O
1 : 1 : 3

15. Empirical Formulas
Empirical formulas are determined experimentally, usually by determining the mass of each element present in a given mass of compound.
To determine the empirical formula of a compound, therefore, an experimentally determined ratio by mass must be converted to a ratio by numbers of atoms.
This is done by calculating the amount (in mol) of each element

16. Empirical Formulas – the process
Step 1: Measure the mass (m) of each
element in the compound
Step 2: Calculate the amount in mole (n)
of each element in the compound
Step 3: Calculate the simplest whole number ratio of moles
of each element in the compound
Step 4: Determine the empirical formula of the compound

17. Worked example
A compound of carbon and oxygen is found to contain 27.3% carbon and 72.7% oxygen by mass. Calculate the empirical formula of the compound.

18. Solution The empirical formula of this compound is therefore CO2
Carbon
Oxygen
Step 1: Record the mass (m) of each element. If masses are given as a percentage assume that the sample weighs 100g
Step 2: Calculate the amount in moles (n) of each element
Step 3: Divide the number of moles of each element by the smallest value from step 2
Step 4: Obtain the simplest whole number ratio
27.3
72.7
27.3
12.0
= 2.27
= 4.54
72.7
16.0
2.27
= 1
= 2
4.54 1 2
The empirical formula of this compound is therefore CO2

19. Molecular Formula
While an empirical formula gives the simplest whole number ratio of each element in a compound, a molecular formula gives the actual number of atoms in one molecule of the compound.
Note that the empirical formula for a compound can be the same or different from its molecular formula

20. Molecular Formula
The molecular formula is always a whole number multiple of the
empirical formula. A molecular formula can be obtained from the
empirical formula if the molar mass of a compound is known.

21. Worked example
A compound has the empirical formula CH. The molar mass of this compound is 78 g mol-1. What is the molecular formula of this compound?
Solution
The molecule must contain a whole number of (CH) units.
The molar mass of a (CH) unit is 13 g mol-1.
If the compound has a molar mass of 78 g mol-1, then:
The number of CH units in a molecule =
molar mass of compound
molar mass of one unit =
78 g mol-1
13 g mol-1
= 6
The molecular formula of the compound is, therefore, 6 x CH
Ie C6H6

22. End of Outcome 1