CSEC Mole Concept pptx 2 Useful Notes.pptx

CSEC Chemistry

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THE MOLE CONCEPT
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Bonous Points Question
How many moles of molecules of water are present in 5g of water?
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Bonous Points Question Answer
How many moles of molecules of water are present in 5g of water?
RMM of H
2O = (1x2) + 16 = 18
Mol of H
2O = _5_= 0.28 mol of H
2O
18
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Objectives
●Perform calculations for molar volume.
●Perform calculations to determine percentage composition.
●State the law of conservation of matter.
●Perform calculations involving chemical equations.
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Mole and Gases
In addition to mass and number of particles, the quantity of a gas
can be expressed in terms of volume.
At the same temperature and pressure, 1 mole of any gas has a fixed
volume. This volume is known as Molar Volume.
Two standard temperatures and pressures are used when quoting
molar volume:
●Room temperature and pressure (r.t.p) –20 ∘C, 1 atmosphere
pressure.
●Standard temperature and pressure (s.t.p) -0 ∘C, 1 atmosphere
pressure.
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Calculating Molar Volume
NEED TO KNOW
Molar volume at:
●R.t.p. = 24 dm
3
●S.t.p= 22.4 dm
3
In other words, 1 mole of any gas has:
●A volume of at 24 dm
3
r.t.p
●A volume of at 22.4 dm
3
s.t.p
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Calculating Molar Volume
To calculate moles from volume of gas or volume of gas from moles,
use the following formulae:
Moles = Given volume of gas
Molar volume at rtpor stp
Volume of gas = mole xmolar volume at rtpor stp
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Calculating Molar Volume
Sample calculation:
●How many moles are there in 1.12 dm
3
of NH
3 gas at s.t.p?
Moles = Given volume of gas
Molar volume at rtpor stp
Moles of NH
3 = 1.12= 0.05 mol
22.4
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Calculating Molar Volume
Sample calculation:
●Calculate the number of molecules of oxygen in 0.03 dm
3
of
oxygen gas at rtp.
Moles = Given volume of gas
Molar volume at rtpor stp
Mol of O
2= 0.03= 0.0013 mol
24
Moles = Given number of particles
6 x 10
23
# of molecules of O
2= mol x 6 x 10
23
= 0.0013 x 6 x 10
23
= 7.8 x 10
20
molecules of O
2
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Percentage Composition
If the molar mass of a compound is known, the percentage
composition by mass of each element in the compound can be
calculated.
Percentage composition = Mass of element in compoundx 100
Molar mass of compound
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Percentage Composition
Sample calculation:
●Calculate the percentage composition in each element in CuSO
4.
Molar mass of CuSO
4 = 64 + 32 + (16x4) = 160 g
Mass of Cu = 64g Mass of O = 16 x 4 = 64g
% of Cu = 64x 100 = 40% % of O = 64x 100 = 40%
160 160
Mass of S = 32g
% of S = 32x 100 = 20%
160
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The Mole and Chemical Equations
The Law of Conservation of Matter states that matter can neither be
created nor destroyed. In a chemical equation the total mass of
reactants must be equal to the amount of products.
This means that if the amount of any one substance is known in a
chemical reaction, the mass of the other substances can be
calculated using simple proportion and the mole ratio from the
balanced equation.
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The Mole and Chemical Equations
When performing a calculation involving an equation, follow these
steps:
●Write the balanced chemical equation for the reaction.
●Decide which reactants/products the question is concerned with.
●Determine the simplest mole ratio between these from the
balanced equation.
●Write a standard statement connecting the 2 substances.
●Use simple proportion to perform the calculation.
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The Mole and Chemical Equations
Examples of Mole ratios
Reactants Products
Pb(NO
3)
2(aq)+2NaCl
(aq)→PbCl
2(s)+2NaNO
3(aq)
1 2
Mole ratio Pb(NO
3)
2(aq): 2NaCl
(aq)
1 : 2
Mole ratio 2NaCl
(aq): PbCl
2
2 : 1
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The Mole and Chemical Equations
Sample calculation:
●1) o.o5 mol of copper (II) carbonate was heated. Calculate:
a)The mass of copper (II) oxide formed.
b)The volume of carbon dioxide produced at stp.
a)CuCO
3(s) CuO
(s) + CO
2(g)
Mole ratio 1 : 1
1 mol of CuCO
3 produces 1 mol of CuO
Molar mass of CuO= 64 + 16 = 80g
If 1 mol of CuCO
3 produces 80g of CuO
Therefore 0.05 mol of CuCO
3 produces x g of CuO
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The Mole and Chemical Equations
Sample calculation:
●1) o.o5 mol of copper (II) carbonate was heated. Calculate:
a)The mass of copper (II) oxide formed.
b)The volume of carbon dioxide produced at stp.
a) 1mol = 80g
0.05 mol = x g
x = 0.05 x 80 = 4g
1
Mass of CuOproduced is 4g.
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The Mole and Chemical Equations
Sample calculation:
●1) o.o5 mol of copper (II) carbonate was heated. Calculate:
a)The mass of copper (II) oxide formed.
b)The volume of carbon dioxide produced at stp.
b) Mole ratio 1 : 1
1 mol of CuCO
3 produces 1 mol of CO
2
Molar volume of CO
2at stp= 22.4 dm
3
If 1 mol of CuCO
3 produces 22.4 dm
3
of CO
2
Therefore 0.05 mol of CuCO
3 produces x dm
3
of CO
2
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The Mole and Chemical Equations
Sample calculation:
●1) o.o5 mol of copper (II) carbonate was heated. Calculate:
a)The mass of copper (II) oxide formed.
b)The volume of carbon dioxide produced at stp.
b)1 mol = 22.4 dm
3
0.05 mol = x dm
3
X = 0.05 x 22.4= 1.12 dm
3
1
Volume of CO
2produced = 1.12 dm
3
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See you next class!
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