Mole Concept: Calculating the Molar Mass of an Unknown Carbonate
One of the most classic question formats in the Mole Concept & Stoichiometry chapter involves an unknown metal reacting with an acid. These questions test two fundamental skills: writing a balanced chemical equation and mapping mole-to-mole ratios.
The Question: 1 g of a metal carbonate (M2CO3) on treatment with excess HCl produces 0.01 mol of CO2. What is the molar mass of M2CO3?
Let's break down the logic exactly as an examiner would expect you to solve it.
Video Tutorial: Solving the Unknown Carbonate
Watch Abhishek Sengar sir from CHEMCA expertly decode the chemical formula, balance the equation, and build the stoichiometric bridge to find the final answer.
Step 1: Writing the Balanced Equation
Before we can do any math, we must know exactly how the molecules interact. The question gives us the formula M2CO3.
We know the Carbonate ion is CO32-. Since there are two metal atoms (M2) balancing that -2 charge, each metal atom 'M' must have a charge of +1.
Therefore, when 'M' reacts with Chlorine (Cl-), the resulting salt will simply be MCl.
Any metal carbonate reacting with an acid produces Salt + Water + Carbon Dioxide. Let's write and balance it:
Step 2: The Stoichiometric Bridge
Look strictly at the coefficients of the balanced equation. They tell us the universal "recipe" for this reaction.
- Coefficient of M2CO3 = 1
- Coefficient of CO2 = 1
This means: 1 mole of M2CO3 produces exactly 1 mole of CO2.
The question tells us that the experiment produced 0.01 moles of CO2. Because of our 1:1 ratio, this means exactly 0.01 moles of M2CO3 must have reacted!
Step 3: Calculating Molar Mass
We now have two critical pieces of information about our pile of reactant:
- We know its physical mass: 1 gram (given in the question).
- We know how many moles are in that pile: 0.01 moles (calculated in Step 2).
The definition of Molar Mass is the mass of exactly 1 mole of the substance. If 0.01 moles weighs 1 gram, how much does 1 mole weigh?
Molar Mass = 1 g / 0.01 mol
Molar Mass = 100 g/mol
Fig: By finding the moles of the product, we instantly know the moles of the reactant due to the balanced 1:1 ratio.
Practice Questions for JEE & NEET
Let's take this problem one step further to test your atomic mass calculation skills!
Question 1: Now that we know the entire M2CO3 molecule has a molar mass of 100 g/mol, calculate the exact Atomic Mass of the unknown metal 'M'.
Answer: The atomic mass of M is 20 g/mol.
Reasoning:
The total molar mass is the sum of the atomic masses of all the atoms in the formula.
Formula: M2CO3
Total Mass = (2 × Mass of M) + (Mass of C) + (3 × Mass of O)
100 = 2M + 12 + (3 × 16)
100 = 2M + 12 + 48
100 = 2M + 60
2M = 40
M = 20.
(Note: While 20 is the atomic mass of Neon, Neon is a noble gas and does not form carbonates. In exam problems like this, 'M' is often just a hypothetical mathematical placeholder!)
Question 2: What if the question stated the metal carbonate was MCO3 (implying an M2+ metal like Calcium) instead of M2CO3? Would the number of moles of CO2 produced change if 0.01 moles of MCO3 reacted?
Answer: No. The mole ratio remains exactly 1:1.
Reasoning:
Let's write the balanced equation for a +2 metal carbonate reacting with HCl:
MCO3 + 2HCl → MCl2 + H2O + CO2
Look at the coefficients: 1 mole of MCO3 still produces exactly 1 mole of CO2. The stoichiometry regarding carbon dioxide generation from a single carbonate unit is always a 1:1 ratio, regardless of the metal's valency!
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