U1M1 Mole Concept
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Mole Concept
a) define mole, Avogadro’s Law and molar mass
b) perform calculations involving Avogadro’s Law and molar mass
c) construct balanced ionic equations from full equations
d) state molar volumes of gases at stp and rtp
e) define molar concentration and mass concentration
f) perform calculations involving molar conc., mass conc., and volumes
of gases
g) calculate empirical and molecular formulae given appropriate data
h) perform calculations involving titrimetric analyses
Mole – one mole of a substance contains as many elementary particles as there
are carbon atoms in 0.012 kg of 12C
Avogadro’s Law – equal volumes of gases under the same conditions of
temperature and pressure contain the same number of molecules
Molar masses of elements are equivalent to the relative atomic mass of the
element with the units of grams.
Avogadro’s Law allows one to determine the ratio in which GASES are reacted
or produced without using stp or rtp considerations.
Example CH4 (g) +2O2 (g)  CO2 (g) + 2H2O (l)
If 40 cm3 of methane was reacted with excess oxygen, what volume of carbon
dioxide would be produced at rtp? Using the molar ratio in the balanced
equation of methane to carbon dioxide (1:1) it can be easily determined. The
volume of CO2 (g) would also be 40 cm3. However since water is a liquid,
Avogadro’s Law would NOT apply to this substance.
Ionic Equations
Step 1: Write the equation and balance it if necessary
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
Step 2: Split the ions. (Only compounds that are aqueous are split into
ions.)
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) → AgCl(s) + Na+(aq) + NO3Step 3: Cancel out spectator ions. (Spectator ions are ions that remain
the same in their original states before and after a chemical reaction.)
Step 4: Write a balanced ionic equation
Ag+(aq) + Cl-(aq) → AgCl(s)
U1M1 Mole Concept
Molar volumes of gases
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1 mole of any gas occupies 22.4 dm3 at stp (standard temperature and pressure)
1 mole of any gas occupies 24 dm3 at rtp (room temperature and pressure)
Volume of gas in dm3 = # of moles of gas x 22.4 at stp (OR 24 at rtp)
Empirical and Molecular Formula
Molecular formula (M.F) is the actual ratio of atoms in a compound e.g. C4H8
Empirical formula (E.F) is the simplest whole number ratio of atoms in a
compound. e.g. CH2
Note M.F C4H8 and the simplest ratio is 1C and 2H hence the E.F is CH 2
Example #1
1.15 g of Na react with 0.8 g of S. Calculate the empirical formula of the
compound
Na
S
Comments
1.15
0.80
Can also be
mass (g)
percentages as well
23
32
Ar
1.15/23
0.8/32
First step to find # of
mass/Ar
= 0.05
= 0.025
mol of each element
= # of mol
0.05 / 0.025
0.025/0.025
To find ratio divide
old ratio of mol
=2
=1
each # by the smallest
# present
x1
x1
If # in old ratio ends in
Conversion
.0 then conversion
factor
factor is x 1 , if ratio
end in .5 then
conversion factor is x
2, if # ends in .3 then
conversion factor is x
3, if # in ration ends in
.25 then conversion
factor is x 4,
If # in ratio ends in .9
round up to the
nearest whole number
1
new ratio of mol 2
Na2S
E.F
U1M1 Mole Concept
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Example #2 An oxide of aluminium contains 1.35 g of Al and 1.2g of O.
Calculate the empirical formula of the oxide
Al
1.35
O
1.2
Ar
mass/Ar
= # of mol
old ratio of mol
27
1.35/27
= 0.05
0.05 / 0.05
=1
16
1.2/16
= 0.075
0.075/0.05
= 1.5
Conversion
factor
x2
x2
new ratio of mol
E.F
2
mass (g)
3
Al2O3
Comments
Can also be
percentages as well
First step to find # of
mol of each element
To find ratio divide
each # by the smallest
# present
If # in ratio ends in .0
then conversion factor
is x 1 , if ratio end in
.5 then conversion
factor is x 2, if # ends
in .3 then conversion
factor is x 3, if # in
ration ends in .25 then
conversion factor is x
4,
If # in ratio ends in .9
round up to the
nearest whole number
U1M1 Mole Concept
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Example #3 A compound contains 24.24% carbon, 4.04% hydrogen and
71.72% chlorine. The molar mass of the compound is 99 g. Calculate the
empirical and molecular formula of the compound
%
C
24.24
H
4.04
Cl
71.72
Ar
%/Ar
= # of mol
old ratio of
mol
12
24.24/12
= 2.02
2.02/2.02
=1
1
4.04/1
= 4.04
4.04/2.02
=2
35.5
71.72/35.5
= 2.02
2.02/2.02
=1
Conversion
factor
x1
x1
x1
new ratio of
mol
E.F
1
2
1
Comments
Can also be
percentages as well
First step to find # of
mol of each element
To find ratio divide
each # by the smallest
# present
If # in ratio ends in .0
then conversion
factor is x 1 , if ratio
end in .5 then
conversion factor is x
2, if # ends in .3 then
conversion factor is x
3, if # in ration ends
in .25 then conversion
factor is x 4,
If # in ratio ends in .9
round up to the
nearest whole number
CH2Cl
Molecular formula (M.F) = Empirical formula (E.F) x a
a=
a=
molar mass
formula mass using E.F
99
45
=2
 M.F = CH2Cl x 2 = C2H4Cl2
U1M1 Mole Concept
Worksheet
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U1M1 Mole Concept
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Molar concentration
The term molar concentration means # of moles of a substance in 1000 cm3 or
1 dm3 of solution. Units are either mol dm-3 or M. However mol dm-3 is
preferred.
Mass concentration
The term mass concentration means mass of a substance in grams in 1000 cm3
or 1 dm3 of solution. Units are g dm-3.
Formulae to remember
mass concentration = molar concentration x molar mass of compound
# of moles = volume of solution in dm3 x molar concentration
Checkpoint B
Complete the table below related to molar and mass concentration
C = 12, H = 1, O = 16, Mg = 24, N = 14, Na = 23, S = 32
Substance
Molar
Volume
# of
Molar
Mass
mass
of
moles concentration
concentration
solution
mol dm-3
g dm-3
3
(cm )
NaOH
200
1
H2SO4
98
500
0.5
HNO3
1000
20
CH3COOH
2000
0.25
HCl
36.5
500
36.5
NB: known substances are those whose molar concentration is KNOWN.
Unknown substances are those whose molar concentrations are NOT
KNOWN.
Calculations that are done in a titration are:1) determine the # of moles of the known substance used
2) calculate the # of moles of the unknown substance that has reacted by using
the ratio of reacting substances which would be obtained by the BALANCED
EQUATION
3) calculate the molar and possibly the mass concentration of the unknown
substance.
U1M1 Mole Concept
Note: mL = cm3 M = mol dm-3
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Sample calculation
If 14.7 mL of 0.012 M NaOH is required to completely neutralise 25 mL of HCl,
what is the molar concentration of the acid? NaOH + HCl  NaCl + H2O
a) # of mol of NaOH = 0.0147 x 0.012 = 1.764 x 10-4
b) ratio alkali: acid = 1: 1
# mol of HCl that has reacted = 1.764 x 10 -4
c) molar conc of acid = # of mol / vol in dm3 = 1.764 x 10-4 / 0.025
= 0.007056 mol dm-3
Worksheet
1. If 8.6 mL of 0.05 mol dm-3 HNO3 is required to neutralise 25 cm3 of
Mg(OH)2, what is the molar and mass concentration of the Mg(OH) 2?
2HNO3 + Mg(OH)2  Mg(NO3)2 + 2H2O
2. If the molar concentration of Ba(OH)2 is 0.0249 mol dm-3 and 25 mL was
used, what volume of HCl of molar concentration 0.11 mol dm-3 would be
required for complete neutralisation? 2HCl + Ba(OH)2  MgCl2 + 2H2O
U1M1 Mole Concept
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3. Calculate the molarity of an acetic acid solution if 34.57 mL of this solution
are needed to neutralize 25.19 mL of 0.1025 M sodium hydroxide.
CH3COOH (aq) + NaOH (aq)
Na+(aq) + CH3COOH-(aq) + H2O (l)