Applied Science
28. Mole Concept III
Lecture Notes
PFP Applied Science Semester 1
Copyright © 2023
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Learning Outcomes
•
Apply the concept of solution concentration (in mol/L)
•
Apply the concept of solution concentration (in g/L)
•
Process the results of volumetric experiments due to differing solvent volume
•
Solve simple problems involving volumetric experiments (Appropriate guidance
will be provided where unfamiliar reactions such as redox are involved.
Calculations on percentage yield and percentage purity are not required.)
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Making an Aqueous Solution
An aqueous solution is often made by dissolving a solute (i.e. dissolved
substance) in a solvent (i.e. substance which dissolves a solute).
Concentration is a measurement of the amount of a solute present in
every unit volume of the solution.
For example,
Solid
Mass, g or number of
moles, mol are
accurate measurements
of amount of solute
+
Final Solution
Distilled Water
Volume in L
is an accurate
measurement of
amount of solvent
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Making an Aqueous Solution
4
1
2
Try the following:
1. Select a “Solute”
4b
3
2. Select “solid” or “solution”
3. Drag the meter to the
liquid
4. Add the solid/aqueous
solute, and observe:
a) Colour of the solution
b) Concentration value
Source:
https://phet.colorado.edu/sims/html/conce
ntration/latest/concentration_en.html
What do you observe?
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Concentration of an Aqueous Solution
Concentration =
Mass (g)
Volume (L)
OR Concentration =
Number of moles (mol)
Volume (L)
Concentration in “mol/L” is also called
Molarity which is represented by the unit “M”
E.g. 5 M = 5 mol/L
For example,
Solid
Mass, g or number of
moles, mol are
accurate measurements
of amount of solute
+
Final Solution
Distilled Water
Volume in L
is an accurate
measurement of
amount of solvent
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Mind-Mapping
No. of
particles
 molar mass
÷ (6.02  1023)
MOLE
 (6.02  1023)
MASS
÷ molar mass
÷ Volume
of solution
÷ Volume
of solution
 Volume of
solution
Concentration
(of aq. solution)
mol/L
 molar mass
g/L
÷ molar mass
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Example
Solution A is prepared by dissolving 171 g of sugar in water.
Solution A has a resulting volume of 5 L.
a) Calculate the concentration (in g/L) of Solution A.
Concentration =
171 g
= 34.2 g/L
5L
This means 34.2 g of sugar is
dissolved in every 1 L of this solution
b) How much sugar can be found in 250 mL of Solution A?
Alt mtd:
1000 mL → 34.2 g
250 mL → 250/1000 × 34.2 = 8.55 g of sugar Mass = Vol × Conc
c) Sugar has the chemical formula C12H22O11.
Calculate the concentration (in M) of Solution A.
= 250/1000 × 34.2
Molar mass of C12H22O11 = (12 × 12) + (22 × 1) + (11 × 16) = 342 g/mol
No. of moles = 171 ÷ 342 = 0.5 mol
0.5 mol
Molarity =
= 0.1 M
5L
Alt mtd (mol/L  g/L):
34.2 g/L ÷ 342 = 0.1 M
This means 0.1 mol of sugar is
dissolved in every 1 L of this solution
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Discussion 1
a) Calculate the molarity (i.e. concentration in M) of 0.206 g
hydrogen fluoride in 200 mL of solution.
No. of moles = ?
Concentration =
??
??
b) Calculate the mass of sodium chloride required to prepare
0.5 L of 0.2 M aqueous sodium chloride?
0.2 M means ?
1000 mL → ? mol
500 mL → ? mol of NaCl is required
Molar mass = ?
Mass = ?
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Discussion 2 (Self-read)
Calculate the volume (in mL) of a 23.8 g/L solution which will contain
0.287 mol of ammonia (NH3)?
Mass = 0.287  (14 + 1  3) = 4.879 g
Since 23.8 g is found in 1 L
4.879 g →
4.879 𝑔
= 0.205 L
23.8 𝑔/𝐿
= 205 mL
Alternative 1:
Concentration =
→ Volume =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
4.879 𝑔
= 0.205 L = 205 mL
23.8 𝑔/𝐿
Alternative 2:
Concentration (mol/L  g/L)
= 23.8 g/L ÷ (14 + 1  3)
= 1.4 M
Volume =
0.287 𝑚𝑜𝑙
= 0.205 L
1.4 𝑀
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Example
The concept of concentration can also be applied to scenarios involving
chemical reactions. As always, determine the number of moles present
before applying the concept of mole ratio in the balanced equations.
80 mL of 2.0 M of aq. NaCl is reacted in the following reaction.
What is the mass of AgCl formed?
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
80 mL, 2.0 M
= 0.08 × 2
= 0.16 mol
AgNO2 : AgCl
1 : 1
0.16 : ?
?g
→ 0.16 mol AgCl produced
→ 0.16  (108 + 35.5) = 22.96 g
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Discussion 3
Aq. lead(II) nitrate is reacted with 50 mL of 0.811 M aq. potassium
iodide. Calculate the mass of lead(II) iodide formed.
Pb(NO3)2 (aq) + 2 KI (aq) → 2 KNO3 (aq) + PbI2 (s)
? mL
?M
?g
No. of moles of KI = ?
Mole ratio of
Mass of PbI2 = ?
KI: PbI2
?:?
?:?
→ ? mol PbI2
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Discussion 4
25.0 mL of aq. sodium hydroxide is completely reacted with
20.0 mL of 3.0 M aq. sulfuric acid. Calculate the concentration (g/L)
of the sodium hydroxide used in this reaction.
H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O (l)
? mL
?M
? mL
? g/L
No. of moles of H2SO4 = ?
Mole ratio of
H2SO4 : NaOH
?:?
?:?
→ ? mol of NaOH
Concentration of NaOH (in M) =
??
??
Concentration of NaOH (in g/L) = ?
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Discussion 5 (Self-read)
25.0 mL of 2.0 M aq. sodium hydroxide is reacted with 10.0 mL of
3.0 M aq. sulfuric acid. Calculate the mass of water formed.
2 NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2 H2O (l)
25 mL, 2 M
= 0.025 × 2.0
= 0.05 mol
10 mL, 3 M
= 0.01 × 3.0
= 0.03 mol
NaOH : H2SO4
2 : 1
0.05 : ?
→ 0.025 mol H2SO4 needed
→ 0.03 mol H2SO4 available
→ NaOH is limiting reactant
?g
NaOH : H2O
2:2
0.05 : ?
→ 0.05 mol H2O
→ 0.05 × (1+1+16)
= 0.9 g H2O produced
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What changes solution concentration?
Try the following:
1. Select “Solution
Values” to be shown
2. Select a “Solute”
3
3b
3. Change the amount of
solute and/or solution
volume, and observe:
a) Colour of the
solution
1
2
Source:
https://phet.colorado.edu/sims/html/molari
ty/latest/molarity_en.html
b) Concentration
value
What do you observe?
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Dilution – Volume and Molarity
Dilution is the process of adding a solvent (usually water) to a solution.
Solution of a desired concentration can be prepared by adding water to
a relatively more concentrated solution.
For example,
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Dilution Factor
Dilution factor is the magnitude through which a solution is diluted by.
Consider the initial and final volume of this solution, how many
times is the original solution diluted by? Ans. 10 times!
Dilution factor
=
𝐅𝐢𝐧𝐚𝐥 𝐕𝐨𝐥𝐮𝐦𝐞
𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐕𝐨𝐥𝐮𝐦𝐞
or
Dilution factor
=
𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧
𝐅𝐢𝐧𝐚𝐥 𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧
If this is diluted by 10×
• Molarity ↓ by 10×
• Volume ↑ by 10×
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Dilution – Solute Amount
Since Moles of Solute = Molarity × Volume
Moles of Solute before dilution = Moles of Solution after dilution
Mb4Vb4 = MaftVaft
The number of particles in
the solution did not
change after dilution
There were five
solute particles in the
original solution
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Example
25 mL of 12.0 M hydrochloric acid (HCl) is topped up with
water in a measuring cylinder till the 500 mL mark.
Calculate the resulting molarity of the solution.
Dilution factor =
Final volume
500 mL
=
= 20
Initial volume
25 mL
Solution is diluted by 20× → final molarity will ↓ by 20×
=
12 M
= 0.6 M
20
Alternatively:
Mb4Vb4 = MaftVaft
(12 M)(0.025 L) = Maft(0.5 L)
Maft = 0.6 M
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Discussion 6
a) 600 mL of water is added to 200 mL of 3 M sugar solution.
What is the final molarity of this solution?
Dilution factor =
??
??
Solution is diluted by ?× → final molarity will <↓/↑> by ?×
=?
b) HCl of concentration 4.0 M is to be prepared from 25 mL of
12.0 M HCl. How many mililitres of water should be added?
Dilution factor =
??
??
Solution is diluted by ?× → final molarity will <↓/↑> by ?×
=?
Since 25 mL is present, volume of water to add = ?
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Empirical and Molecular Formula
Empirical formula
• Shows the simplest whole number ratio of different atoms in
a compound
• Usually experimentally determined
Molecular formula
• Shows the actual number and kind of atoms present in a
compound
Example:
• C2H4 and C3H6 are molecular formulae
• Both chemical compounds have the empirical formula CH2
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Determining the Empirical Formula
Step 1: Find the masses of different elements in a sample of the compound.
Step 2: Convert the masses into moles of atoms of the different elements.
Step 3: Express the moles of atoms as the smallest possible ratio of integers.
Step 4: Write the empirical formula using the number for each atom in the
integer ratio as the subscript in the formula.
A 100 g sample of ethylene contains 85.7% carbon and 14.3 % hydrogen by
mass. Find the empirical formula of ethylene.
Element
C
H
Step 1
Mass (g)
85.7
14.3
Step 2
Moles (mol)
85.7/12
14.3/1
(mass ÷ molar mass)
= 7.142
= 14.3
7.142/7.142
=1
14.3/7.142
=2
Step 3
Mole ratio
(÷ smallest mol)
Step 4
Empirical Formula
CH2
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Discussion 7 (Self-read)
The mass of a piece of iron is 1.62 g. If the iron is exposed to
oxygen under conditions in which oxygen combines with all of the
iron to form a pure oxide of iron, the final mass increases to 2.31 g.
Find the empirical formula of the compound.
Mole ratios:
Element
Fe
O
Mass (g)
1.62
2.31 – 1.62 = 0.69
Moles (mol)
1.62/56
0.69/16
(mass ÷ molar mass)
= 0.02893
= 0.04313
Mole ratio
(÷ smallest mol)
0.02893/0.02893 0.04313/0.02893
=1
= 1.5
Empirical Formula
Mole ratio
Fe : O
3
(state as fraction) 1 : 2
(both sides × 2) 2 : 3
Fe2O3
• Avoid rounding off
to nearest whole
numbers when the
ratios are around
1
̶ 0.25:1 (i.e. 4 : 1)
1
3
̶ 0.33:1 (i.e. : 1)
1
̶ 0.5:1 (i.e. 2:1),
• In such cases, we
multiply by a
number to make
both sides of the
ratios become
whole numbers
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Discussion 8a
Fructose is the sugar found in honey and fruits. It contains 40.0%
carbon, 6.71% hydrogen by mass and the remaining x % is oxygen.
Assuming the Fructose has a mass of 100 g:
a) Determine the empirical formula of the compound.
Element
Mass (g)
Moles (mol)
(mass ÷ molar mass)
Mole ratio
(÷ smallest mol)
Empirical Formula
C
H
O
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Determining the Molecular Formula
Step 1: Determine the empirical formula of the compound.
Step 2: Calculate the molar mass of the empirical formula unit.
Step 3: Determine the molar mass of the compound (usually given).
Step 4: Divide the molar mass of the compound by the molar mass of the
empirical formula unit to get n, the number of empirical formula units
per molecule.
Step 5: Write the molecular formula.
A compound with the empirical formula C2H5 has a molar mass of 58 g/mol.
Find the molecular formula.
Step 1
Step 4
Empirical formula was given as C2H5
Molar mass of empirical unit = 2 × 12 + 5 × 1 = 29 g/mol
Molar mass of compound = 58 g/mol
n = 58 ÷ 29 = 2
Step 5
(C2H5)2 = C4H10
Step 2
Step 3
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Discussion 8b (Cont. from 8a)
Fructose is the sugar found in honey and fruits. It contains 40.0%
carbon, 6.71% hydrogen by mass and the remaining x % is oxygen.
a) Determine the empirical formula of the compound. Assume we
have 100 g of Fructose.
b) The molar mass of fructose is 180 g/mol. Determine the molecular
formula.
Element
Mass (g)
C
H
40
6.71
O
100 – 40 – 6.71 = 53.29
Moles (mol)
40/12
6.71/1
53.29/16
(mass ÷ molar mass)
= 3.33
= 6.71
= 3.33
Mole ratio
3.33/3.33
=1
6.71/3.33
=2
3.33/3.33
=1
(÷ smallest mol)
Empirical Formula
Molar mass of empirical unit = ??
CH2O
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Learning Outcomes
•
Apply the concept of solution concentration (in mol/L)
•
Apply the concept of solution concentration (in g/L)
•
Process the results of volumetric experiments due to differing solvent volume
•
Solve simple problems involving volumetric experiments (Appropriate guidance
will be provided where unfamiliar reactions such as redox are involved.
Calculations on percentage yield and percentage purity are not required.)