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Quantitative aspects of chemical change
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Grade 10 Physical Science
CAPS
2016
The mole concept
The mole concept
• Atoms are small – chemists know this.
• But somewhere along the line – they have to count them.
• The mole is the SI Unit for the measurement of the amount
of matter.
• A mole is a large group of atoms or molecules, which can
be described as a specific amount of matter, or the
number of particles.
The mole concept
• The particles of an element is called “atoms.”
• The particles of a covalent compound is called
“molecules.”
• The particles of an ionic compound is called “ions.”
“Particles”
NaNa
Na
Na
H
H
OH
HO𝐻2𝑂
𝑁𝑎𝐶𝑙
NaCl
Na
Cl
The mole concept
• A mole is that quantity of matter that has the
• same number of particles as there are in 12g of carbon
12.
• How much is this “number of particles?”
• Avogardo’s number
Who is Avogardo?
Some clever Italian scientist….
What is this number?
6.02 × 1023
The mole concept
•How big is that?
• It’s 602 000 000 000 000 000 000 000
•That’s nice….how BIG IS THAT?!
Go and read on page 196
The mole concept
• This means that…
• 1 mole of a substance contains 6.02 × 1023 units
• Example:
• 1 mole of sodium means 6.02 × 1023 sodium atoms
• 1 mole of 𝐻2𝑂 means 6.02 × 1023 𝐻2𝑂 molecules.
The mole concept
• So just like:
• A dozen = 12
• A pair = 2
• A case = 24
• SO a Mole = 6.02 × 1023
The mole concept
• So if I could put all of these
particles onto a scale – what
would it’s mass be?
• The mass of 1 mole of a substance is equal to it’s atomic
mass in grams.
• 1 mole Na = 23g
• 1 mole Mg = g
• 1 mole 𝑂2 = g2432
The mole concept
• So if I could put all of these
particles onto a scale – what
would it’s mass be?
• The mass of 1 mole of a substance is equal to it’s atomic
mass in grams.
• 1 mole 𝑀𝑔𝑂 = g
• 1 mole 𝑁𝑎𝑂𝐻 = g
• 1 mole 𝐻2𝑆𝑂4 = g
244098
The mole concept
Molecular Mass
• Since particles are so small, it’s difficult to use g or kg to
calculate their masses.
• Hence we used the atomic mass unit (u)
The mole concept
There’s a difference
• Symbol: M
• Unit: 𝑔.𝑚𝑜𝑙−1
• Used for an
element• Eg.𝑀 𝐶𝑢 = 63.53𝑔.𝑚𝑜𝑙−1
• 𝑀 𝑀𝑔 =• M 𝐶𝑎 =
Atomic mass Molar mass Formula mass
• Symbol: 𝑀• Unit: 𝑔.𝑚𝑜𝑙−1
• Used for an ionic
compound.• Eg.𝑀 𝑁𝑎𝐶𝑙 = 58.5𝑔.𝑚𝑜𝑙−1
• 𝐴𝑟 𝑀𝑔𝑆𝑂4 =• 𝐴𝑟 𝐾𝑂𝐻 =
• Symbol: M
• Unit: 𝑔.𝑚𝑜𝑙−1
• Used for an
compound• Eg.𝑀 𝐻2𝑂 = 18𝑔.𝑚𝑜𝑙−1
• 𝑀 𝐻𝐶𝑙 =• 𝑀 𝐶𝑂2 =
24𝑔.𝑚𝑜𝑙−1
40𝑔.𝑚𝑜𝑙−136.5𝑔.𝑚𝑜𝑙−1
44𝑔.𝑚𝑜𝑙−1
120𝑔.𝑚𝑜𝑙−1
56𝑔.𝑚𝑜𝑙−1
The mole concept
There’s a difference
• Symbol: 𝐴𝑟• No unit
• Used for an
element• Eg.𝐴𝑟 𝐶𝑢 = 63.53• 𝐴𝑟 𝑀𝑔 =• 𝐴𝑟 𝐶𝑎 =
Relative Atomic mass
Relative Molar mass
Relative Formula mass
• Symbol:𝑀𝑟
• No unit
• Used for an ionic
compound.• Eg.𝑀𝑟 𝑁𝑎𝐶𝑙 = 58.5• 𝑀𝑟 𝑀𝑔𝑆𝑂4 =• 𝑀𝑟 𝐾𝑂𝐻 =
• Symbol: 𝑀𝑟
• No unit
• Used for a
compound.• Eg.𝑀𝑟 𝐻2𝑂 = 18• 𝑀𝑟 𝐻𝐶𝑙 =• 𝑀𝑟 𝐶𝑂2 =
2440
36.544
12056
The mole concept
Let’s practice…
Substance Number of moles
Mass in grams
Particles Number of particles
Carbon 1 12 Atoms 6.02 × 1023
1
1
1
Cobalt Atoms 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑59
Carbon dioxide
molecules
𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑44
Sodium Chloride
ions 𝟐 × 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑58.5
homeworkExercise 20 pg. 198-199
The relationship between:
mole mass molar mass
n m M( ) (g) (g.mol-1)
𝑛 =𝑚
𝑀
mol
mass
Molar mass
( )
(𝒈)
(𝒈.𝒎𝒐𝒍−𝟏)
TB. PG 229
The equation #1
Relationship between mole, mass and molar mass
𝑛 =𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠
𝑁𝐴
mol
Avogardo’s number:
𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑
( )
TB. PG 229
The equation #2
Relationship between mole, mass and molar mass
worksheet
Relationship between mole, mass and molar mass
Empirical formulaThis is the simplest ratio in which the elements of the
compound bonds with each other.
𝐻2𝑂
This formula says that 1 water molecule has:
• 2 hydrogen atoms and 1 oxygen atom.
• 1 mole 𝐻2𝑂 contains 6.02 × 1023 water molecules
• So that means there are 2 × 6.02 × 1023hydrogen atoms and 1 × 6.02 × 1023 oxygen atoms.
OH
H
worksheet
Empirical formula
HomeworkExercise 21 pg. 202-203
Percentage Composition
Used to determine the composition of a substance according
to a percentage division.
Example 1
Determine the percentage composition of 𝑁𝑎𝐻𝐶𝑂3.
Step 1
Find the formula mass of 𝑁𝑎𝐻𝐶𝑂3. 84𝑔.𝑚𝑜𝑙−1
Step 2Find the atomic mass of each element in the
compound.
𝑀 𝑁𝑎= 23𝑔.𝑚𝑜𝑙−1
𝑀 𝐻= 1𝑔.𝑚𝑜𝑙−1
𝑀 𝐶= 12𝑔.𝑚𝑜𝑙−1
𝑀 𝑂= 16𝑔.𝑚𝑜
Step 3
Calculate!!!!
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 (𝑔.𝑚𝑜𝑙−1)
𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 (𝑔.𝑚𝑜𝑙−1)× 100 =
𝑁𝑎 = 27.38% 𝐻 = 1.19% 𝐶 = 14.29%
𝑂 = 57.14%
Homework EXERCISE 22 PG. 205-206
Step 1
Find the formula mass of 𝑁𝑎𝐻𝐶𝑂3.
Step 2Find the atomic mass of each element in the compound.
Step 3
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 (𝑔.𝑚𝑜𝑙−1)
𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 (𝑔.𝑚𝑜𝑙−1)× 100 =
Compound Composition
Used to determine the amount of mol water from
crystallization.
Example 1
Weigh out 2.5g of 𝐶𝑢𝑆𝑂4. 𝑛𝐻2𝑂 crystals.
Dry the crystals by heating them gently in a crucible. This is done to remove the water.
Weigh the crystals again.
Example 1
At start: 2.5g of 𝐶𝑢𝑆𝑂4. 𝑛𝐻2𝑂
After drying: 1.6g of 𝐶𝑢𝑆𝑂4
So, in 2.5g of 𝐶𝑢𝑆𝑂4. 𝑛𝐻2𝑂
1.6g was 𝐶𝑢𝑆𝑂4 0.9g was 𝑯𝟐𝑶and
Example 1
Just like Empirical Formula
𝑛𝐶𝑢𝑆𝑂4 =𝑚
𝑀=
1.6
159.5= 0.01𝑚𝑜𝑙
𝑛𝐻2𝑂 =𝑚
𝑀=0.9
18= 0.05𝑚𝑜𝑙
Example 1
Ratio
𝐶𝑢𝑆𝑂4: 𝐻2𝑂
0.01: 0.05
1: 5
Example 1
SO!!!! 1 𝐶𝑢𝑆𝑂4. 5𝐻2𝑂 crystals.
Example 2
Weigh out 6.25g of 𝑀𝑔𝑆𝑂4. 𝑛𝐻2𝑂 crystals.
Dry the crystals by heating them gently in a crucible. This is done to remove the water.
Weigh the crystals again.
Example 2
At start: 6.25g of 𝑀𝑔𝑆𝑂4. 𝑛𝐻2𝑂
After drying: 4g of 𝑀𝑔𝑆𝑂4
So, in 6.25g of 𝑀𝑔𝑆𝑂4. 𝑛𝐻2𝑂
4g was 𝑀𝑔𝑆𝑂4 2.25g was 𝑯𝟐𝑶and
Example 1
Just like Empirical Formula
𝑛𝑀𝑔𝑆𝑂4 =𝑚
𝑀=
4
120= 0.03𝑚𝑜𝑙
𝑛𝐻2𝑂 =𝑚
𝑀=2.25
18= 0.13𝑚𝑜𝑙
Example 2
Ratio
𝑀𝑔𝑆𝑂4 ∶ 𝐻2𝑂
0.03: 0.13
1: 4
Example 2
SO!!!! 1 𝑀𝑔𝑆𝑂4. 4𝐻2𝑂 crystals.
Determine the number of moles of water of
crystallization in 5.4g of 𝐴𝑙𝐶𝑙3. 𝑛𝐻2𝑂.
Workbook Exercise
Write down and complete the following
example in your workbook.
Molar gas volume
1 mole of a gas occupies a volume of 22.4 𝑑𝑚3 at STP.
𝑶𝟐
𝑶𝟐𝑶𝟐
𝑶𝟐
𝑶𝟐
𝑶𝟐
𝟐𝟐. 𝟒𝒅𝒎𝟑
•What is STP???
Standard temperature
and pressure𝑜℃ 𝑜𝑟 273𝐾
101.3𝑘𝑃𝑎
𝑛 =𝑉
𝑉𝑚
mol
volume
Molar gas volume
( )
(𝒅𝒎𝟑)
(𝟐𝟐. 𝟒𝒅𝒎𝟑)
TB. PG 229
The equation # 1
Molar gas volume
𝑛 =𝑚
𝑀
mol
mass
Molar mass
( )
(𝒈)
(𝒈.𝒎𝒐𝒍−𝟏)
TB. PG 229
The equation #2
Molar gas volume
𝑛 =𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠
𝑁𝐴
mol
Avogardo’s number:
𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑
( )
TB. PG 229
The equation #3
Molar gas volume
•Calculate the mass of 5.6𝑑𝑚3 of 𝐶𝑂2 at STP.
Example 1
What are you given?
What do you want to find?
Homework EXERCISE 23 pg. 207-209
1.1
2.1
3.15.1
4.1
ConcentrationThis is the quantity of a dissolved solute in mole per 𝑑𝑚3 of
solution.
𝑐 =𝑛
𝑉
mol
Volume
(𝒎𝒐𝒍. 𝒅𝒎−𝟑)
( )
(𝒅𝒎𝟑)
USE IF YOU ARE GIVEN
MOLE
The equation # 1
Concentration
concentration
𝑐 =𝑚
𝑀𝑉
mass
Volume
(𝒎𝒐𝒍. 𝒅𝒎−𝟑)
( 𝒈 )
(𝒅𝒎𝟑)
USE IF YOU ARE GIVEN
MASS
The equation # 1
Concentration
concentration
Molar mass
(𝒈.𝒎𝒐𝒍−𝟏)
Oh! But converting is so VERY vital
𝑚𝑚3 𝑐𝑚3 𝑑𝑚3 𝑚3
× 1000× 1000× 1000
÷ 1000 ÷ 1000 ÷ 1000
AND…
1c𝑚3 = 1𝑚𝑙1d𝑚3 = 1 𝑙𝑖𝑡𝑟𝑒
•Calculate the concentration of a solution that contains 0.2 mol copper
(II) sulphate in 250𝑐𝑚3 water.
Example 1
What are you given?
What do you want to find?
Homework EXERCISE 24 pg. 211-212
2.1
Stoichiometric calculations
•A chemical reaction occurs when substances
(reactants) react chemically and produce
new substances (products).
•We represent such a reaction with a
chemical formula.
2𝐻2 + 𝑂2 → 2𝐻2𝑂
• 2 mole hydrogen reacts with 1 mole of oxygen to form 2 moles of water.
• 4 hydrogen atoms react with 2 oxygen atoms to form 2 water molecules.
• 2 hydrogen molecules react with 1 oxygen molecule to form 2 water molecules.
• 4g of 𝑯𝟐 reacts with 32g of 𝑶𝟐 to form 36g of 𝑯𝟐𝑶
+ →
𝐶 + 𝑂2 → 𝐶𝑂2
• 1 mole carbon reacts with 1 mole of oxygen to form 1 mole of carbon dioxide.
• 1 carbon atom react with 2 oxygen atoms to form 1 carbon dioxide molecule..
• 12g of 𝑪 reacts with 32g of 𝑶𝟐 to form 44g of 𝑪𝑶𝟐
+ →
Calculations
Example 1 – mass-mass calculation
•Calculate the mass of oxygen obtained when 14.7g
of potassium chlorate decomposes completely to
form potassium chloride.
Example 2 – mass-volume calculation
•What mass of potassium chlorate must be heated to
release 90𝒅𝒎−𝟑 of oxygen at STP?
HOMEWORK
EXERCISE 25 PG. 215-217
