__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) + O_{2}(g) → 2MgO(s)

In the above reaction, one can find out the volume of O_{2} at STP required to react with 10 gm of Mg. The moles of Mg is 10/24 .

The moles of O_{2} required would be 1/2 the moles of Mg.

Therefore moles of O_{2} = (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

H_{2}(g) + ½O_{2}(g) → H_{2}O(l).

We are given 10L of H_{2} at a given temperature and pressure. How many liters of O_{2} 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 (H_{2}:O_{2}),

∴ the volume ratio would also be 2:1.

Therefore the volume of O_{2} required would be 5L.

On the other hand if we need to calculate the volume of O_{2} 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 CaCO_{3} and MgCO_{3} 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 CO_{2} evolved at 0°C and 76 cm of pressure.

**Solution:**

CaCO_{3} → CaO + CO_{2}

x gm

MgCO_{3} → MgO + CO_{2}

y gm

∴ (x + y) = 2.21 gm …(i)

100 gm of CaCO_{3} gives 56 gm of CaO

∴ x gm of CaCO_{3} ≡$\large \frac{56}{100}\times (x ) gm CaO $

Similarly 84 gm of MgCO_{3} gives 40 gm of MgO

∴ y gm of MgCO_{3} = $\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 CO_{2} formed = Mole of CaCO_{3} + Mole of MgCO_{3}

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

= 0.0241

∴ Volume of CO_{2} at STP = 0.0421 × 22.4 litre

= 539.8 ml

**Illustration :** A mixture of FeO and FeO_{3} when heated in air to constant weight gains 5% in weight. Find out composition of mixture.

**Solution:**

2FeO + (1/2) O_{2} → Fe_{2}O_{3}

2Fe_{3}O_{4} + (1/2)O_{2} → 3Fe_{2}O_{3}

Let, weight of FeO = x

Weight of Fe_{3}O_{4} = y

∴ x + y = 100 ……….(i)

2 × 72 gm of FeO give Fe_{2}O_{3}= 160 gm

∴ x gm FeO gives Fe_{2}O_{3} $\large = \frac{160 x}{144} gm $

2 × 232 gm of Fe_{3}O_{4} gives Fe_{2}O_{3} = 3 × 160 gm

∴ y gm Fe_{3}O_{4} gives Fe_{2}O_{3} $\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 Fe_{2}O_{3} . Also report the limiting reagent.

VO + Fe_{2}O_{3} → FeO + V_{2}O_{5}

Solution: Balanced equation

2VO + 3Fe_{2}O_{3} → 6 FeO + V_{2}O_{5}

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 Fe_{2}O_{3}

0.0298 gm mole of VO = (3/2)× 0.0298 = 0.0447 mole of Fe_{2}O_{3}

Moles of Fe_{2}O_{3} available = 0.0359 only

Hence, Fe_{2}O_{3} 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} $

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

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