Gravimetric Analysis

Gravimetric Analysis

Calculation involving in mass – mass relationship

In general, the following steps are adopted in making necessary calculations:

1. Write down balanced molecular equation for the chemical change

2. Write down the no of moles below the formula of each of the reactant and product

3. Write down the relative masses of the reactants and the products with the help of formula below the respective formula. These shall be the theoretical amounts of reactants and product.

4. By the applications of unitary method, mole concept or proportionality method, the unknown factor or factors are determined.

Calculation involving percentage yield :

In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield.

Knowing the actual yield and theoretical yield the percentage yield can be calculated as

$\large  Percentage \;  yield = \frac{Actual \; yield}{Theoretical \; yield} \times 100 $

Weight  – Volume Relationship

2Mg(s) + O2(g) → 2MgO(s)

In the above reaction, one can find out the volume of O2 at STP required to react with 10 gm of Mg. The moles of Mg is 10/24 .

The moles of O2 required would be 1/2 the moles of Mg.

Therefore moles of O2 = (1/2)× (10/24) .

Since 1 mole of a gas (ideal) occupies 22.4L at STP,

Therefore , $\large \frac{1}{2} \times \frac{10}{24} $ moles of O2 would occupy

$\large \frac{1}{2} \times \frac{10}{24} \times 22.4 L$

= 4.67 L.

Volume  – Volume Relationship

Let us consider the reaction

H2(g) + ½O2(g) → H2O(l).

We are given 10L of H2 at a given temperature and pressure. How many liters of O2 would react with hydrogen at the same temperature and pressure?

From the ideal gas equation [PV = nRT] it is clear that the volume of an ideal gas is directly proportional to its no. of moles. Therefore under the same conditions of P and T,

$\large \frac{V_{H_2}}{V_{O_2}} = \frac{n_{H_2}}{n_{O_2}} $

Since the molar ratio is 2:1 (H2:O2),

∴ the volume ratio would also be 2:1.

Therefore the volume of O2 required would be 5L.

On the other hand if we need to calculate the volume of O2 at a different T and P, then

$\large P_{H_2} V_{H_2} = n_{H_2} R T_{H_2}$

$\large P_{O_2} V_{O_2} = n_{O_2} R T_{O_2}$

on dividing we get ;

$\large \frac{P_{H_2} V_{H_2}}{P_{O_2} V_{O_2}} = \frac{n_{H_2} T_{H_2}}{n_{O_2} T_{O_2}}$

Illustration  : A sample of CaCO3 and MgCO3 weighed 2.21 gm is ignited to constant weight of 1.152 gm. What is the composition of the mixture. Also calculate the volume of CO2 evolved at 0°C and 76 cm of pressure.

Solution:

CaCO3 → CaO + CO2
x gm

MgCO3 → MgO + CO2
y gm

∴ (x + y) = 2.21 gm  …(i)

100 gm of CaCO3 gives 56 gm of CaO

∴ x gm of CaCO3 ≡$\large \frac{56}{100}\times (x ) gm CaO $

Similarly 84 gm of MgCO3 gives 40 gm of MgO

∴ y gm of MgCO3 = $\large \frac{40 \times y}{84} gm $

∴ Wt. of residue =$\large \frac{56 x}{100} + \frac{40 y}{84} = 1.152 $   …(ii)

Solving equations (i) and (ii)

x = 1.19 gm

y = 1.02 gm

Mole of CO2 formed = Mole of CaCO3 + Mole of MgCO3

$\large = \frac{1.19}{100} + \frac{1.02}{84} $

= 0.0241

∴ Volume of CO2 at STP = 0.0421 × 22.4 litre

= 539.8 ml

Illustration : A mixture of FeO and FeO3 when heated in air to constant weight gains 5% in weight. Find out composition of mixture.

Solution:

2FeO + (1/2) O2 → Fe2O3

2Fe3O4 + (1/2)O2 → 3Fe2O3

Let, weight of FeO = x

Weight of Fe3O4 = y

∴ x + y = 100   ……….(i)

2 × 72 gm of FeO give Fe2O3= 160 gm

∴ x gm FeO gives Fe2O3 $\large = \frac{160 x}{144} gm $

2 × 232 gm of Fe3O4 gives Fe2O3 = 3 × 160 gm

∴ y gm Fe3O4 gives Fe2O3 $\large = \frac{3 \times 160 \times y}{2 \times 232} gm $

$\large \frac{160 x}{144} + \frac{3 \times 160 \times y}{464} = 105 $ …(ii)

Solving equation (i) & (ii)

x = 20.25 gm

y = 79.95 gm

Illustration : Calculate the weight of FeO from 2 gm VO and 5.75 gm of Fe2O3 . Also report the limiting reagent.
VO + Fe2O3 → FeO + V2O5

Solution: Balanced equation

2VO + 3Fe2O3 → 6 FeO + V2O5

Moles before reaction = 0.0298   0.0359   0    0

Moles after reaction = (0.0298 – 0.0359)   0  [(6/3)×0.0359 ]   [(1/3)× 0.0359]

As 2 mole of VO react with 3 mole of Fe2O3

0.0298 gm mole of VO = (3/2)× 0.0298 = 0.0447 mole of Fe2O3

Moles of Fe2O3 available = 0.0359 only

Hence, Fe2O3 is the limiting reagent

Mole of FeO formed = (6/3)× 0.0349

Weight of FeO formed = 0.0359 × 2 × 72 = 5.17 gm

$\large \frac{n_{FeO}}{n_{Fe_2 O_3}} = \frac{6}{3} $

$\large n_{FeO} = \frac{6}{3} \times n_{Fe_2 O_3}$

$\large W_{FeO} = \frac{6}{3} \times n_{Fe_2 O_3} \times M_{Fe_2 O_3}$

Also Read :

→ Oxidation and Reduction
→ Balancing Redox Equations
→ Volumetric Analysis
→ Law of Equivalence
→ ‘ n ‘ Factor
→Applications of the Law of Equivalence : Simple titration , Back titration
→ Double titration
→ Volume of Strength of H2O2

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