THE MOLEHow many particles are there really?

6.02 x 1023

"Happy Mole Day to You" Chemistry Song

The Rule

n =mM

Amount of substance (mol)

Mass of a given amount of substance (g)

Molar Mass (g mol -1)

Worked ExampleCalculate the mass of 0.35 mol of magnesium

nitrate (Mg(NO3)2.

What is the formula we use?????

How do we work out M

Solutionm = n x M(Mg(NO3)2 = 24.3 + 2(14.0 + 3 x 16.0)

= 24.3 + 2 x 62.0= 148.3 g mol-1

So:m(Mg(NO3)2 = n(Mg(NO3)2 x

M(Mg(NO3)2

= 0.35 mol x 148.3 g mol-1

= 52 gWhy only two significant figures?

Counting by WeighingIf 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??

Worked Examplea) 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.

Solutiona) Since

M(CO2) = 12.0 + 2 x 16.0 = 44.0 g mol-1

n =m

M

n(CO2) =m(CO2)M(CO2)

n(CO2) =22 g

44.0 g mol-1

= 0.50 mol

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

Percentage CompositionThe 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.

The RulePercentage Composition

% by massof the element=

Mass of element in 1 mole of compoundMass of 1 mole of compound

x 100

Worked ExampleCalculate 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

SolutionStep 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

= 54102 x 100

= 52.9%

The company management, therefore, knows that aluminium comprises about 53% by mass of any sample of alumina.

Empirical formulasThe 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.

Empirical FormulasCompound 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

Empirical FormulasEmpirical 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

Empirical Formulas – the processStep 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

Worked exampleA 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.

SolutionCarbon Oxygen

Step 1: Record the mass (m) of each element. If masses are given as a percentage assume that the sample weighs 100gStep 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 2Step 4: Obtain the simplest whole number ratio

27.312.0 = 2.27 = 4.54

72.7

16.0

2.272.27 = 1 = 2

4.54

2.27

The empirical formula of this compound is therefore CO2

27.3 72.7

1 2

Molecular FormulaWhile 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

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.

Worked exampleA 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?

SolutionThe 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

= 6The molecular formula of the compound is, therefore, 6 x CHIe C6H6

End of Outcome 1http

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