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Gravimetric Analysis Gravimetric analysis is an analytical technique based on the measurement of mass of solid substances and/or volume of gaseous species. The gravimetric analysis is broadly classified into three heads.
Gravimetric analysis is an analytical technique based on the measurement of mass of solid substances and/or volume of gaseous species.
The gravimetric analysis is broadly classified into three heads.
Mass – mass relationship
Volume - volume relationship
Mass- volume relationship
This relates the mass of a species (reactant or product) with the mass of another species (reactant or product) involved in a chemical reaction.
Let us consider a chemical reaction,
CaCO_{3}(s) CaO(s) + CO_{2}(g)
Let the mass of CaCO_{3} taken be x g and we want to calculate the mass of CaO obtained by heating x g of CaCO_{3}. Then the moles of CaCO_{3} taken would be (where M_{1} represents the molar mass of CaCO_{3}).
According to the balanced reaction, the molar ratio of CaCO_{3} and CaO is 1 : 1, so same number of moles ii.e. , of CaO would be formed.
Now for converting the moles of CaO into mass of CaO obtained, we need to multiply the moles of CaO with the molar mass of CaO. Let the molar mass of CaO be M_{2}, so the mass of CaO obtained by heating x g of CaCO_{3} would be g.
We want to find out the volume of CO_{2} evolved at STP by heating x g of CaCO_{3}. Then
Moles of CaCO_{3} =
Moles of CO_{2} evolved = (since molar ratio of CaCO_{3} and CO_{2} is 1 : 1)
Hence Volume of CO_{2} evolved at STP = L
But, if the volume of CO_{2} evolved is to be calculated at pressure P atm and temperature T K. Then,
Volume of CO_{2} evolved at pressure P and temperature T = ( Using PV = nRT)
This relationship deals with the volume of a gaseous species (reactant or product) with the volume of another gaseous species (reactant or product) involved in a chemical reaction.
Let us consider the reaction, N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g).
We are given x L of N_{2} at pressure P atm and temperature TK and we want to know the volume of H_{2} required to react with it at the same pressure and temperature.
Moles of N_{2} =
Moles of H_{2} required = (since molar ratio of N_{2} & H_{2} is 1 :3)
Hence Volume of H_{2} required at same pressure & temperature = = 3 x L.
This result could also have been obtained by knowing that for a gaseous relation, at the same pressure and temperature, the moles of gas is directly proportional to volume of the gas (V a n since P & T are constant) or molar ratio and volume ratio are same.
Moles of H_{2} required = Hence Volume of H_{2} required at pressure P’ atm & temperature T’K.
=
Refer to the following video for problems on stoichiometry
Now let us see the usefulness of these concepts in solving problems.
How much zinc should be added to 0.01 mol AgNO_{3} solution to displace all the silver in the solution?
Solution:
The involved balanced reaction would be
Zn + 2AgNO_{3} → Zn(NO_{3})_{2} + 2Ag
Moles of AgNO_{3} in the solution = 0.01
Moles of Zn to be added to solution = 0.005
(since AgNO_{3}) and Zn are reacting the molar ratio of 2 : 1)
Hence Mass of Zn to be added to solution = 0.005 × 65.4 = 0.327 g
_______________________________________________________
NaCI of 95% purity is used to prepare salt cake (Na_{2}SO_{4}) by the reaction,
2NaCI + H_{2}SO_{4} → Na_{2}SO_{4} + 2HCI
If the product (Na_{2}SO_{4}) is only % pure, what weight of NaCI is used up in producing 1 kg of the impure salt cake?
Let the weight of NaCI used up in producing 1 kg of impure product be x g.
Mass of pure NaCI =
Moles of pure NaCI =
Moles of pure Na_{2}SO_{4} =
Mass of pure Na_{2}SO_{4} =
Hence x = 737.2 g
_________________________________________
KCIO_{3} decomposes on heating in two possible ways as
KCIO_{3} KCI + 3/2 O_{2}
2KCIO_{3} KCI + 3KCIO_{4}
when 15 g of KCIO_{3} were heated in an experiment, analysis of the product showed that 5.21 g of KCI and 4.59 g of KCIO_{4} were formed. What was the weight of KCIO_{3} remaining undecomposed? What fraction of KCIO_{3} decomposed yielded KCIO_{4}?
In order to calculate the weight of KCIO_{3} remaining undecomposed, we need to find out eh total weight of KCIO_{3} that decomposes. The weight of KCIO_{3}, that decomposes in second reaction would be known using the KCIO_{4} data. Using KCIO_{4} data, we would also calculate the weight of KCI obtained in second reaction. Since, total weight of KCI obtained is known to us, so the weight of KCI obtained in first reaction would also be known, using this we will find out the weight of KCIO_{3} decomposed in first reaction.
Moles of KCIO_{4} obtained =
Moles of KCI obtained in IInd reaction = (since molar ratio of KCIO_{4} and KCI in IInd reaction is 3 : 1).
Moles of KCIO_{3} that decomposes in IInd reaction = (since molar ratio of KCIO_{4} and KCIO_{3} in IInd reaction is 3 : 4)
Hence Mass of KCIO_{3} that decomposes in IInd reaction = 122.5 = 5.41 g
Moles of KCI obtained in 1^{st} reaction = = 0.07 – 0.011 = 0.059
Moles of KCIO_{3} that decomposes in 1^{st} reaction = 0.059
(since molar ratio of KCI and KCIO_{3} in 1^{st} reaction is 1 : 1)
Mass of KCIO_{3} that decomposes in 1^{st} reaction = 0.059 × 122.5 = 7.23 g
Total mass of KCIO_{3} that decomposes in both the reaction = (5.41 +
7.23) = 12.64 g
Hence Mass of KCIO_{3} that remain undecomposed = 15 – 12.64 = 2.36 g
Moles of O_{2} evolved in 1^{st} reaction = 0.059 ×
Volume of O_{2} evolved at STP in 1^{st} reaction = 0.059 × × 22.4 = 1.98 L
Fraction of KCIO_{3} decomposed yielding KCIO_{4} = = 0.36
This relationship deals with the volume of a gaseous species (reactant or product) with the mass of a solid species (reactant or product) involved in a chemical reaction.
Let’s take an example of combustion of Mg in air
2Mg + O_{2} → MgO
In this reaction, one can find out the volume of O_{2} required to react with given mass of Mg.
Let the mass of Mg taken is x g. So, the number of moles of Mg taken would be , where M is the molar weight of Mg.
From the balanced equation, we know that the moles of O_{2} required is half of the number of moles of Mg taken i.e.
Now we know that one mole of O2 gas at S.T.P =22.4L
So, mole of O_{2} = L
Question 1: What is the molar ratio of reactant and one of the products in the following reaction?
a. 1:1
b. 2:1
c. 3:1
d. 1:2
Question 2: Volume of 1 mole O_{2} gas at S.T.P is
a. 22.4 L
b. 2.24 L
c.2.24 L
d. 24.22 L
Question 3: How many moles of H_{2}O are required to react with 46 g Na ?
Na + H_{2}O → NaOH + H_{2}
a. 1
b. 2
c. 3
d. 4
Question 4: What is the mass of NH3 produced from 22.4 L N_{2} gas at STP
N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g) ?
a. 36 g
b. 26 g
c. 17 g
d. 61 g
Q.1
Q.2
Q.3
Q.4
a
b
Look here for past year papers of IIT JEE
Click here to refer syllabus of chemistry for IIT JEE
Have a look at stoichiometry
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