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Gravimetric Analysis

Table of Content

Mass- Mass Relationship

Volume-Volume Relationship

Mass – Volume Relationship

Related Resources

Gravimetric analysis is an analytical technique based on the measurement of mass of solid substances and/or volume of gaseous species.

Mass – mass relationship

Volume - volume relationship

Mass- volume relationship

Mass- Mass 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,

CaCO3(s)  CaO(s) + CO2(g)

Let the mass of CaCO3 taken be x g and we want to calculate the mass of CaO obtained by heating x g of CaCO3. Then the moles of CaCO3 taken would be  (where M1 represents the molar mass of CaCO3).

According to the balanced reaction, the molar ratio of  CaCO3 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 M2, so the mass of CaO obtained by heating x g of CaCO3 would be g.

CaCO3(s)  CaO(s) + CO2(g)

We want to find out the volume of CO2 evolved at STP by heating x g of CaCO3. Then

Moles of CaCO3 =

Moles of CO2 evolved =  (since molar ratio of CaCO3 and CO2 is 1 : 1)

Hence Volume of CO2 evolved at STP = L

But, if the volume of CO2 evolved is to be calculated at pressure P atm and temperature T K. Then,

Volume of CO2 evolved at pressure P and temperature T =  ( Using PV = nRT)

Volume-Volume Relationship

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, N2(g) + 3H2(g) →  2NH3(g).

We are given x L of N2 at pressure P atm and temperature TK and we want to know the volume of H2 required to react with it at the same pressure and temperature.

Moles of N2 =

Moles of H2 required =  (since molar ratio of N2 & H2 is 1 :3)

Hence Volume of H2 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 H2 required =
Hence Volume of H2 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.

Solved Examples

Question 1

How much zinc should be added to 0.01 mol AgNO3 solution to displace all the silver in the solution?

Solution:

The involved balanced reaction would be

Zn + 2AgNO3 → Zn(NO3)2 + 2Ag

Moles of AgNO3 in the solution = 0.01

Moles of Zn to be added to solution = 0.005

(since AgNO3) 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

_______________________________________________________

Qestion 2

NaCI of 95% purity is used to prepare salt cake (Na2SO4) by the reaction,

2NaCI + H2SO4 →  Na2SO4 + 2HCI

If the product (Na2SO4) is only % pure, what weight of NaCI is used up in producing 1 kg of the impure salt cake?

Solution:

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 Na2SO4 =

Mass of pure Na2SO4 =

Hence x = 737.2 g

_________________________________________

Question 3:

KCIO3 decomposes on heating in two possible ways as

KCIO3  KCI + 3/2 O2

2KCIO3  KCI + 3KCIO4

when 15 g of KCIO3 were heated in an experiment, analysis of the product showed that 5.21 g of KCI and 4.59 g of KCIO4 were formed. What was the weight of KCIO3 remaining undecomposed? What fraction of KCIO3 decomposed yielded KCIO4?

Solution:

In order to calculate the weight of KCIO3 remaining undecomposed, we need to find out eh total weight of KCIO3 that decomposes. The weight of KCIO3, that decomposes in second reaction would be known using the KCIO4 data. Using KCIO4 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 KCIO3 decomposed in first reaction.

Moles of KCIO4 obtained =

Moles of KCI obtained in IInd reaction =  (since molar ratio of KCIO4 and KCI in IInd reaction is 3 : 1).

Moles of KCIO3 that decomposes in IInd reaction =  (since molar ratio of KCIO4 and KCIO3 in IInd reaction is 3 : 4)

Hence Mass of KCIO3 that decomposes in IInd reaction =  122.5 = 5.41 g

Moles of KCI obtained in 1st reaction = = 0.07 – 0.011 = 0.059

Moles of KCIO3 that decomposes in 1st reaction = 0.059

(since molar ratio of KCI and KCIO3 in 1st reaction is 1 : 1)

Mass of KCIO3 that decomposes in 1st reaction = 0.059 × 122.5 = 7.23 g

Total mass of KCIO3 that decomposes in both the reaction = (5.41 +

7.23) = 12.64 g

Hence Mass of KCIO3 that remain undecomposed = 15 – 12.64 = 2.36 g

Moles of O2 evolved in 1st reaction = 0.059 ×

Volume of O2 evolved at STP in 1st reaction = 0.059 ×  × 22.4 = 1.98 L

Fraction of KCIO3 decomposed yielding KCIO4 =  = 0.36

Mass – Volume Relationship

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 + O2 →  MgO

In this reaction, one can find out the volume of O2 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 $\frac{x}{M}$, where M is the molar weight of Mg.

From the balanced equation, we know that the moles of O2 required is half of the number of moles of Mg taken i.e. $\frac{x}{2M}$

Now we know that one mole of O2 gas at S.T.P =22.4L

So, $\frac{x}{2M}$ mole of O2 = $\frac{22.4x}{2M} = \frac{11.2x}{M}$ L

Question 1: What is the molar ratio of reactant and one of the products in the following reaction?

CaCO3(s)  CaO(s) + CO2(g)

a. 1:1

b. 2:1

c. 3:1

d. 1:2

Question 2: Volume of 1 mole O2 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 H2O are required to react with 46 g Na ?

Na + H2O → NaOH + H2

a. 1

b. 2

c. 3

d. 4

Question 4:  What is the mass of NH3 produced from 22.4 L N2 gas at STP

N2(g) + 3H2(g) →  2NH3(g) ?

a. 36 g

b. 26 g

c. 17 g

d. 61 g

Q.1

Q.2

Q.3

Q.4

a

a

b

a

Related Resources

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