Gravimetric Analysis: Definition, Formula, Questions and Examples

One of the oldest and most reliable techniques in analytical chemistry is gravimetric analysis. This technique is based on the measurement of mass, relying on precise and accurate weighing to determine the concentration of an analyte. At its core, gravimetric analysis involves the conversion of the analyte into a stable, pure compound that can be weighed. This often entails a series of carefully controlled chemical reactions, followed by filtration, washing, drying, or ignition of the precipitate to a constant mass.

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This Story also Contains

1. Gravimetric Analysis
2. Some Solved Examples
3. Conclusion

Gravimetric Analysis

It is an analytical technique based on the measurements of the mass of solid substances or the volume of the gaseous species. It is divided into three categories:

Mass-Mass (weight-weight) relation

This relationship relates to the mass of a reactant or product with the mass of another reactant or product.

1. Write down the balanced equation to represent the chemical change.

2. Write the number of moles below the formula of reactants and products.

3. Finally, apply the unitary method to calculate the unknown factor.

Mass-volume relation

This relationship relates the mass of a reactant or product with the volume of another gaseous reactant or product involved in a chemical reaction.

1. Write down the relevant balanced chemical equation.

2. Write the weights of various solid reactants and products.

3. Gases are normally expressed in terms of volume. In case the volume of the gas is measured at room temperature and pressure, convert it into N.T.P. by applying gas laws.

4. The volume of a gas at any temperature and pressure can be converted into its weight and vice versa by using the relation,
   PV = (g/M) x RT
   Here g is the weight of the gas, M is the molecular weight of gas and R is the gas constant.

5. Finally, calculate the unknown factor (n or s) by unitary method.

Volume-Volume relation

This relationship relates the volume of a gaseous reactant or product with the volume of another gaseous reactant or product involved in a chemical reaction.

1. First, write down the relevant balanced chemical equation.

2. Now write down the volume of the reactants and the products below the formula to each reactant and product using the fact that one gram molecules of every gaseous substance occupies 22.4 liters at N.T.P. (22.7 Litre at STP)

3. If the volume of the gas is measured at a particular or room temperature, convert it to N.T.P. with the help of the ideal gas equation.

4. Now use Avogadro’s hypothesis that gases under similar conditions of temperature and pressure contain the same number of molecules. Thus under similar conditions of temperature and pressure, the number of moles of the gases in the balanced equation.

Recommended topic video on (Gravimetric Analysis)

Some Solved Examples

Qu 1. What is the mass (in kg) of the solution when the mass of solvent is 3.5 kg and that of solute is 0.5 kg?

1) 1.25

2) 7

3) 3

4) 4

Solution

Mass of solution = mass of solute + mass of solvent

= 3.5 + 0.5

= 4 kg

Hence, the answer is (4 kg).

Qu 2. What volume of O₂ is obtained at STP, when 12.25 g of KClO₃ is decomposed according to the reaction (Molar mass of KClO₃= 122.5 g/mol)

$\mathrm{KClO}_3 \xrightarrow{\Delta} \mathrm{KCl}+\frac{3}{2} \mathrm{O}_2$

1) 22.4 L

2) 11.2 L

3) 3.36 L

4) 44.8 L

Solution

1 mole of KClO₃ produces 1.5 moles of O₂

So, 0.1 mole of KClO₃ will produce 0.15 moles of O₂

Thus, volume = 22.4 X 0.15 = 3.36L

Hence, the answer is the option (3).

Qu 3. 6 g of ethane (C₂H₆) is burnt in excess of O₂. The moles of H₂O formed would be

1) 0.2

2) 0.4

3) 0.6

4) 0.8

Solution

The reaction occurs as follows:

$2 \mathrm{C}_2 \mathrm{H}_6+7 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}$

Molar Mass of Ethane = 30 g/mole.
The molar mass of water = 18 g/mole.
From the Equation,
∵ 2 x 30 grams of Ethane produces 6 x 18 grams of Water.
∴ 1 gram of the Ethane produces (9/5) grams of Water.
∴ 6 grams of Ethane produces 10.8 grams of Water.
Now, Using the Formula,
No. of Moles of Water = Mass/Molar Mass
= 10.8/18
= 0.6 moles.

Hence, the answer is the option (3).

Qu 4. At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O₂ for complete combustion, and 40 mL of CO₂ is formed. The formula of the hydrocarbon is:

1) C₄H₁₀

2) $\mathrm{C}_4 \mathrm{H}_6$

3) $\mathrm{C}_4 \mathrm{H}_7 \mathrm{Cl}$

4) $\mathrm{C}_4 \mathrm{H}_8$

Solution

The equation of hydrocarbon can be written as :

$\mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{y}{2} \mathrm{H}_2 \mathrm{O}$

Since , 10mL of $C_x H_y$ produces 40mL of CO₂ &

1mL of $C_x H_y$ produces mL of CO₂

$\therefore x=\frac{40}{10}=4$

Now,

10mL of $C_x H_y$ requires 55mL of O₂

1mL of $C_x H_y$ requires $\left(\frac{55}{10}\right)$mL of O₂

$\therefore x+\frac{y}{4}=\frac{55}{10}$

=> $\frac{y}{4}=\frac{55}{10}-x$

=> $y=4(5.5-x)$

=>$y=4(5.5-4)$

=>$y=6$

$\therefore$ Hydrocarbon $=C_4 H_6$

Hence, the answer is the option (2).

Qu 5. 25g of an unknown hydrocarbon upon burning produces 88 g of $\mathrm{CO}_2$ and 9 g of H₂O. This unknown hydrocarbon contains:

1) 20 g of carbon and 5 g of hydrogen

2) 22 g of carbon and 3 g of hydrogen

3) 24 g of carbon and 1 g of hydrogen

4) 18 g of carbon and 7 g of hydrogen

Solution

$\mathrm{C}_x \mathrm{H}_y+\mathrm{O}_2 \rightarrow x \mathrm{CO}_2+y \mathrm{H}_2 \mathrm{O}$

$\eta_{\mathrm{co}_2}=\frac{88}{44}=2 \mathrm{~mole}$

$\eta_c=2 \mathrm{~mole}$

weight of carbon $W_c=2 \times 12=24 g$

Mole of water $\eta_{H_2} O=\frac{9}{18}=\frac{1}{2}$

weight of water $W_{\mathrm{H}_2 \mathrm{O}}=\frac{1}{2} \times 18=9 g$

Thus, the weight of Hydrogen $W_g=1 g$

Therefore, the hydrocarbon contains 24g of carbon and 1g of hydrogen

Hence, the answer is the option (3).

Conclusion

The importance of gravimetric analysis spans various fields, from environmental science, where it is used to determine pollutant levels in air and water, to pharmaceuticals, ensuring the purity and dosage of active ingredients. Its application in materials science and metallurgy further underscores its versatility and robustness. By exploring the strengths and limitations of this technique, we aim to highlight its enduring relevance and the critical role it plays in contemporary chemical analysis.