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MOLE CONCEPT
Definitions of Mole:
1. In terms of mass: The mass of one mole of any substance is equal to its relative
atomic or molecular mass taken in grams, e.g.,
1 mole of Fe = 56 g (Atomic mass, Ar)
1 mole of H2O = 18 g (Molecular mass, Mr)
2. In terms of number: One mole of any substance contains particles equal to
6.02 x 1023. For example,
1 mole of Fe = 56 g = 6.02 x 1023 particles
1 mole of H2O = 18 g = 6.02 x 1023 particles
3. In terms of volume (applicable only to gases): One mole of any gas occupies
volume equal to 24 dm3 at r.t.p. For example.
1 mole of H2 = 2g = 24dm3 = 6.02 x 1023 particles
1 mole of NH3 = 17g = 24 dm3 = 6.02 x 1023 particles
Note: The volumes of gas are considered also while defining one mole (def. 3) because
the mass of a gas cannot be that easily determined practically in laboratory.
Worked Examples:
1. Find out the amount of Na2O formed when 50g of Na is made to react completely with
oxygen to form Na2O?
Solution:
4Na(s) + O2(g) →
2Na2O(s)
Na
92g
50g
Na2O
124g
𝑥
𝑥= (124 x 50) / 92 = 67.39 g of Na2O
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2. Find out the amount of CO2 formed when 40g of Carbon reacts completely with O2 to
form CO2.
C(s) + O2(g)
→
C
12g
40g
CO2(g)
CO2
24dm3
𝑥 dm3
𝑥= (24 x 40) / 12 = 80dm3 of CO2
• If amount of a gas is to be calculated in a question without the specification of mass or
volume, it is understood that the amount of a gas is to be determined in volume (dm 3), in
general.
3. Find out the amount of O2 required when it is completely reacted with Carbon
Monoxide to form 65dm3 of CO2.
2CO(g) + O2(g) →
2CO2(g)
O2
24 dm3
𝑥
CO2
48dm3
65
𝑥= (24 x 65) / 48 = 32.5dm 3 of O2
4. Same as in previous one but with the difference that now CO is to be calculated in
place of O2.
2CO(g) + O2(g)
→
CO
2 x 24 dm3
𝑥
2CO2(g)
CO2
2 x 24 dm3
65
𝑥= (24 x 65) / 24 = 65dm3 of O2
Second Method (To be preferred):
2CO(g) + O2(g)
→
2CO2(g)
Since, in case of gases, mole ratio = volume ratio,
Hence, O2 = 65/2 = 32.5 dm3 (because mole ratio – CO2 : O2 = 2 : 1)
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Yet another important rule:
“In gaseous case only, the mole ratio is equal to the volume ratio since one mole of
every gas is equal to the same quantity and same volume i.e. 24dm 3.”
“Moles, if taken in terms of grams – the mole ratio is different from the gram ratio
because Ar or Mr of one species is more or less different from others.”
For example,
Grams:
Volume:
1 mole H2 : 1 mole CO2
1 x 2 : 1 x 44
2 : 44
1 x 24 : 1 x 24
1:1
Excess & Limiting Reagents:
The amount of product formed in any chemical reaction depends upon the limiting
reagent.
1. Find out the limiting and excess reagents when 45g of Fe is made to react with 45g of
Sulphur to form FeS.
Fe(s) + S(s) → FeS(s)
Fe
56g
45g
S
32g
𝑥
𝑥= (45 x 32) / 56 = 25.7g of Sulphur
Since Sulphur required (25.7g) is less than that present (45g), so, Sulphur is the excess
reactant and 19.3g is extra. Hence, Fe is the limiting reagent.
Also,
Fe
56g
𝑥
S
32g
45g
𝑥= (45 x 56) / 32 = 78.75g of Fe
Since Fe required (78.75g) is greater than that present (45g), so, Fe is the limiting
reactant and Sulphur is the excess reagent.
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Note: If any substance present in an experiment is greater than that required, then it is
said to be the excess reagent.
On the other hand, if any substance present in an experiment is lesser than that
required, then it is said to be limiting reagent.
2. Find the amount of FeS formed when 65g of Fe is made to react with 35g of S to form
FeS.
Fe
56g
65g
S
32g
𝑥
𝑥= (65 x 32) / 56 = 37.1g of Sulphur (Limiting reactant)
2nd Step: To find FeS formed.
S
32g
35g
FeS
88g
𝑥
𝑥 = 96.25g of FeS
• Note: Amount of product formed in any reaction always depends upon the limiting
reactant and never on the amount of the excess reactant i.e. amount of product only be
calculated from the amount of limiting reagent.
3. Find the amount of FeS formed when 45g of Fe is made to react with 45g of S.
Fe(s) + S(s) → FeS(s)
45g
45g → 𝑥
Note: If we start solving this question with any of the reactants, then the answer cannot be
guaranteed to be correct as our reactant is not surely to be limiting. So, first calculate the limiting and
the excess reactants.
1st step:
Fe
56g
45g
S
32g
𝑥
𝑥 = 25.71g of S (Excess, since we are given 45g)
Hence, Fe is the limiting reagent.
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2nd step:
Fe
56g
45g
FeS
88g
𝑥
𝑥 = 70.71g of FeS
• Note: Even in Gas-Solid reaction, the product formed could be calculated by the same
“excess and limiting reagent method” and the answer come to be correct, as shown
below.
4. Find the amount of NaCl formed when 65g of Na is made to react with 40dm 3 of Cl2.
2Na(s) + Cl2(g) → 2NaCl
Na
46g
65g
Cl2
24dm3
𝑥 dm3
𝑥 = 33.9dm3 of Cl2
Since, amount of Chlorine given (40dm 3) is greater than that required (33.9dm 3), so,
Chlorine is the limiting reagent.
Na
46g
65g
NaCl
117g
𝑥
𝑥 = 165.3g of NaCl formed.
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