# The Mole

```Quantitative Analysis
A.S. 2.1 (Chemistry 91161)
Year 12
4 Internal Credits
The Mole and Molar Mass
The amount (n)
of a substance
that contains
23
6.02 X 10
particles is
called a mole
The Mole
• A counting unit
• Similar to a dozen, except instead
of 12, it’s 602 billion trillion
602,000,000,000,000,000,000,000
• 1 mole = 6.02 X 1023 particles
• Enough soft drink cans to cover
the surface of the earth to a depth
of over 300 Km.
The Mole
• 1 dozen cars = 12 cars
• 1 mole of cars = 6.02 X 1023 cars
• 1 dozen Al atoms = 12 Al atoms
• 1 mole of Al atoms = 6.02 X 1023 atoms
Note that the NUMBER is always the same,
but the MASS is very different!
Mole can be abbreviated, but only to mol!!!
Learning Check
1. Number of atoms in 0.500 mole of Al
a) 500 Al atoms
b) 6.02 x 1023 Al atoms
c) 3.01 x 1023 Al atoms
2.Number of moles of S in 1.8 x 1024 S atoms
a) 1.0 mole S atoms
b) 3.0 mole S atoms
c) 1.1 x 1048 mole S atoms
Molecular Mass (M)
• The Mass of 1 mole of a substance (in grams)
• Equal to the atomic mass number
1 mole of C atoms
=
12.0 g
1 mole of Mg atoms
=
24.3 g
1 mole of Cu atoms
=
63.5 g
• Unit gmol-1
• 7gmol-1 means 1 mole of the substance would
weigh 7 g
Molecular Mass of Molecules and
Compounds
Mass (in grams) of 1 mole is equal to the sum
of the atomic masses e.g.
What is the Molar mass of CaCl2
1 mole Ca x 40.1 g/mol
+ 2 moles Cl x 35.5 g/mol
= 111.1 g/mol CaCl2
You try
1 mole of N2O4
= 92.0 g/mol
Learning Check!
A. Molar Mass of K2O
= 94
? Grams/mole
B. Molar Mass of
antacid Al(OH)3
? gmol-1
= 78
C. Prozac, C17H18F3NO,
is a widely used
antidepressant that
inhibits the uptake
of serotonin by the
brain. Find its
molar mass. 309 gmol-1
Balancing Equations
• 2Ca(s) + O2(g)
2CaO(s) + energy
• This means that 2 mols of Ca atoms react with every 1
mol of Oxygen molecules to produce 2 mols of CaO and
“n” mols of energy is released.
• If 4 mols of Ca was burnt how many mols of oxygen
would be required and how much energy is produced?
• How many mols of oxygen are reacted when 6 mols of
CaO are produced?
• If 40g of Ca are reacted what mass of CaO is produced?
• What is the mole ratio of ions formed when MgCl2
dissolves?
Use a balanced equation to
calculate quantities
• If 50g of ethane burns what is the mass of water
produced?
• 2C2H6 + 7O2
4CO2 + 6H2O
• M(C2H6) = 30
• n=m/M =50/30 = 1.67 moles ethane
• C2H6:3H2O Mole ratio STOICHIOMETRY
• 1.67 x 3 = 5 moles water
m
• M(H2O)= 18
• m = nxM = 5x18 = 90g water
n M
Converting Moles (n), Mass (g)
and Molecular Mass (M)
n = m/M
m
n M
Aluminum is often used for the structure of
light-weight bicycle frames. How many grams
of Al are in 3.00 moles of Al?
m=nxM
= 3 x 27
= 81g Al
m
Learning Check!
n M
The artificial sweetener aspartame
(Nutra-Sweet) formula C14H18N2O5 is
used to sweeten diet foods, coffee and
soft drinks. How many moles of
aspartame are present in 225 g of
aspartame?
0.77 moles
Learning Check!
How many atoms of K are present in 78.4 g of K?
2.01 x 6.02 x 1023 = 12.1 x 1023
What is the mass (in grams) of 1.20 X 1024
molecules of glucose (C6H12O6)?
n=1.20 X 1024 / 6.02 x 1023 =2 moles
M= 180g/mole so m=nxM =2x180 = 360g
How many Na+ in 25g of Na2CO3?
M(Na2CO3)= 106g/mol
N=m/M = 25/106 = 0.236moles
X 2 = 0.472 moles
Percent Composition
Gives the breakdown of mass between different atoms
e.g. Find the percentage of Carbon in Carbon monoxide
M(C) =12 M(O) =16
Percentage(C) = M(C) /M(CO) x 100 = 42.86%
What is the percent carbon in C5H8NO4 (the
glutamic acid used to make MSG monosodium
glutamate), a compound used to flavour foods and
tenderise
meats?
a) 8.2 %C
b) 27.8 %C
c) 41.1 %C
Types of Formulas
Empirical Formula
The formula of a compound that has the
smallest whole number ratio of the atoms
present.
e.g. CH3 means for every C there are 3H
Molecular Formula
The formula that gives the actual number
of each kind of atom found in the
molecule.
e.g. C2H6 means every molecule has 2C and
6H atoms
Chemical Formulas of Compounds
• Formulas give the relative numbers of atoms or
moles of each element in a formula - always a
whole number ratio.
NO2 2 atoms of O for every 1 atom of N
1 mole of NO2 : 2 moles of O atoms to every 1
mole of N atoms
• If we know or can determine the relative number
of moles of each element in a compound, we can
determine a formula for the compound.
Find Empirical Formula given the
composition
1. Determine the mass in grams of each
element present, if needed.
2. Calculate the number of moles of each
n
atom.
3. Divide each by the smallest number of
moles to obtain the simplest whole
number ratio.
4. Write empirical formula and determine its
molar mass.
m
M
Learning Check!
A sample of a brown gas, a major air pollutant, is
found to contain 2.34g N and 5.34g O. Determine
the empirical formula for this substance.
moles of N = 2.34g of N = 0.167 moles of N
14.01 g/mole
moles of O = 5.34g of O = 0.334 moles of O
16.00 g/mole
Therefore N0.167O0.334
Formula:
N 0.167 O 0.334  N O 2
0.167
0.167
Find Molecular Formula given the
molar mass
Calculate the molecular formula of a
compound with empirical formula
CH4 and a molar mass of 64g/mol
m
n M
1. Identify the empirical formula
and calculate its molar mass.
CH4
M(CH4) =16
2. Divide the given molar mass by
the empirical molar mass.
64/16 = 4
3. Multiply the empirical formula
by the answer from step 2.
C4H16
Learning Check!
A compound has an
empirical formula of
NO2. The colourless
liquid, used in rocket
engines has a mass
of 92.0 g. What is the
molecular formula of
this substance?
Molecular Formula from % Composition
A substance contains 60.80 % Na ; 28.60 % B ;
10.60 % H by mass. Its molar mass is 114g/mole.
What is the molecular formula of the substance?
HELP!
Consider a sample size of 100 grams
Determine the number of moles of each atom
Determine the simplest whole number ratio
Write empirical formula
Calculate molar mass of empirical
Divide molar mass by empirical molar mass
Na3B3H12
Concentration of Solution
n
c V
• n = amount of solute (moles)
• c = concentration of solution (moles/litre)
• V = volume of solution (litres)
Problem 1. Calculate concentration if
0.025 moles of HCl are present in
50ml of solution.
• c=n/V
• =0.025/0.050
• =0.5 molL-1
Problem 2. Calculate moles of Na2CO3
present in 21mL of 0.3molL-1 solution
• n = cV
• = 0.3 x 0.021
• = 0.0063 moles
n
c V
2.4g of sodium carbonate is dissolved in water to
make 250ml of solution. Calculate the concentration.
M(Na2CO3) = 106gmol-1
• n = m/M
• = 2.4/106
• = 0.023 moles
m
nM
• c = n/V
• = 0.023/0.250
• = 0.092molL-1
n
c V
Volumetric Analysis
• Acid Base Titrations
– Used to determine an
unknown concentration
ACID-BASE REACTIONS
Titrations
H2C2O4(aq) + 2 NaOH(aq) ---> Na2C2O4(aq) + 2 H2O(liq)
acid
base
Oxalic acid,
H2C2O4
Carry out this neutralisation reaction using a TITRATION.
Setup for titrating an acid with a base
Titration
1. Use pipette to measure your
(unknown concentration) solution
2. Add known solution from the buret
to find the average titre.
3. Indicator shows when exact
stoichiometric reaction has
occurred. (Acid = Base)
4. Calculate moles (n) of known
solution.
5. Use balanced equation to find the
unknown amount of moles.
6. Calculate the concentration of
unknown.
PROBLEM #1: Standardise a solution of
NaOH — i.e., accurately determine its
concentration.
35 mL of NaOH is pipetted
(by titration) with 25.2
? mL of
0.0998 M HCl. What is the
concentration of the NaOH?
2. Find average titre of HCl (L)
Titre 1
CONCORDANCE
3 consistent results
2
3
4
Av.
Vol. 25.0 25.4 29.4 25.2 25.2
ml
PROBLEM #1: Standardise a solution of
NaOH — i.e., accurately determine its
concentration.
4. Calculate moles (n) of HCl
n = cV
= 0.0998 x 0.0252
= 0.025 moles HCl used
n
c V
5. Use balanced equation to find the unknown amount of
moles NaOH
NaOH + HCl
NaCl + H20
1:1 ratio, therefore 0.025 moles NaOH
reacts with 0.025 moles HCl
PROBLEM #1: Standardise a solution of
NaOH — i.e., accurately determine its
concentration.
6. Calculate the concentration of NaOH
• c=n/V
•
= 0.025/0.035
•
= 0.71molL-1
n
c V
20 mL of Al(OH)3 is neutralised (by titration) with 18.6mL
of 0.2 M HCl. What is the concentration of the Al(OH)3?
Concentration of Al(OH)3 = 0.062molL-1
PROBLEM 3: You have 50.0 mL of 3.0
M NaOH and you want 0.50 M NaOH.
What do you do?
Add water to the 3.0 M solution to lower its
concentration to 0.50 M i.e. Dilute it!
But how
much
water
do we
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What
do you do?
c • V = Amount (n) of NaOH in original solution
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must also = 0.15
mol NaOH
n/C = Volume of final solution
(0.15 mol NaOH)/(0.50 mol/L) = 0.30 L
or
300 mL
n
c
V
PROBLEM: You have 50.0 mL of 3.0
M NaOH and you want 0.50 M
NaOH. What do you do?
Conclusion:
water to 50.0 mL
of 3.0 M NaOH to
make 300 mL of
0.50 M NaOH.
Diluting Solutions
No need to
convert ml to
L if both units
are the same
Finding the new concentration
X concentration = diluted
Total volume
concentration
e.g. 20ml of 0.5molL-1 is diluted to 100ml. What is
the final concentration?
0.02/0.1 X 0.5 =0.1molL-1
Find the original concentration of HCl when 20ml
giving a final concentration of 4x10-3molL-1
0.05molL-1
Test Yourself
• You have 50.0 mL of 3.0 M NaOH and you want
0.50 M NaOH. How much water do you need to
250ml
• What is the concentration of Cu2+ if 20ml of
0.64molL-1 of CuSO4 is diluted to 100ml?
0.128molL-1
• 25ml CuSO4 solution was diluted to 200ml. The
new concentration is 7.36x10-4 molL-1. What was
the original concentration?
5.89x10-3 molL-1
Density
• Convert L
g and g L
• E.g. Calculate the volume of 6
moles of ethanol given the
density of ethanol (CH3CH2OH)
is 0.789gml-1
• M = 46gmol-1
• m = Mn = 46 x 6 = 276g
• 276g/0.789gml-1 = 349.8ml or
0.350L
Apple Juice
• 20ml apple juice was titrated
against 0.1molL-1 NaOH and the
average was 10.36ml. Assuming all
apple juice is citric acid and that
1mole reacts with 3 moles of
NaOH. Calculate the concentration
of citric acid in gL-1. Apple juice
must contain between 0.3 and 0.8g
per 100ml. Is this considered apple
juice or apple drink?
• M(citric acid)=192gmol-1
Gravimetric analysis
Decomposing of solids
• Hydrated magnesium sulphate
(MgSO4.XH2O) is heated in a crucible
(total mass 37.95g) so the water is lost
from the crystals. When it is reweighed it
has a constant mass of 36.82g. The
crucible weighs 35.75g what is the mass
of the water and anhydrous MgSO4?
MgSO4 = 1.07g
H20 = 1.13g
Gravimetric analysis
• Calculate X the water of crystallisation in
MgSO4.XH2O
• M(MgSO4) = 120gmol-1 M(H2O) = 18gmol-1
• Calculate the number of moles of each
and the mole ratio.
• n(MgSO4) = 1.07/120 = 0.00892moles
• n(H2O) = 1.13/18 = 0.0628moles
m
• 0.0089:0.0628 = 1:7
n M
• X=7
Gravimetric analysis
• AgNO3 is added to 25ml of water containing
Cl- so that the ions are precipitated out.
• Ag+(aq) + Cl-(aq)
AgCl(s)
• They are found to weigh 0.32g. Calculate
the concentration of Cl- in solution.
• n= m/M = 0.32/143.5 = 0.00223 moles AgCl
• Cl- : AgCl so 0.00223 moles Cl• c = n/V = 0.00223/0.025 = 0.089molL-1 m
n M
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