Q.A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.
Concept understanding — Molecular Mass Calculation
What is Molecular Mass? The Intuition
Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.
Atoms and molecules are unimaginably tiny. A single water molecule (H2O) weighs about 3×10−23 grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, 6.022×1023 molecules, which is called one mole.
Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have 6.022×1023 molecules of this compound, how much would they weigh on a lab balance?"
The number 6.022×1023 is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.
The Precise Definition
Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.
For example:
- Water (H2O): 2 hydrogen atoms + 1 oxygen atom
- Atomic mass of H = 1.008 g/mol
- Atomic mass of O = 16.00 g/mol
- Molecular mass of H2O = 2(1.008)+16.00=18.016 g/mol
Molecular mass=∑(number of atoms of each element×atomic mass of that element)
How to Calculate It — Step by Step
Let's take glucose, C6H12O6, as a worked example.
Step 1: Identify each element and its count
- Carbon (C): 6 atoms
- Hydrogen (H): 12 atoms
- Oxygen (O): 6 atoms
Step 2: Look up atomic masses (from the periodic table)
- C: 12.01 g/mol
- H: 1.008 g/mol
- O: 16.00 g/mol
Step 3: Multiply and add
Molecular mass=6(12.01)+12(1.008)+6(16.00)
=72.06+12.096+96.00
=180.156 g/mol
Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.
Why This Matters
Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:
- Convert grams to moles: moles=molecular massmass in grams
- Convert moles to grams: mass=moles×molecular mass
- Determine the number of molecules: molecules=moles×6.022×1023
Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like CO2 = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).
Common Exam Pitfalls
- Forgetting to multiply by the subscript. In H2SO4, there are 2 hydrogens, not 1.
- Using atomic number instead of atomic mass. Atomic number (protons) is not mass.
- Rounding too early. Keep 2-3 decimal places until the final answer.
- Confusing molecular mass with molecular weight. They mean the same thing — both are in g/mol.
Quick Reference Table
| Substance | Formula | Calculation | Molecular Mass (g/mol) |
|---|---|---|---|
| Oxygen gas | O2 | 2(16.00) | 32.00 |
| Carbon dioxide | CO2 | 12.01+2(16.00) | 44.01 |
| Methane | CH4 | 12.01+4(1.008) | 16.042 |
| Sodium chloride | NaCl | 22.99+35.45 | 58.44 |
The last one is a formula mass (ionic compound), but the calculation is identical.
The Big Picture
Molecular mass is not a property you measure directly — it's a calculated value from the periodic table. Every molecule of a given compound has the same molecular mass. When you weigh out that many grams, you know exactly how many moles (and therefore how many molecules) you have. That's the foundation of all stoichiometry.
Searches like "molecular mass calculation formula chemistry" and "mole concept class 11 chemistry" are extremely common, since this is one of the very first skills taught in the Some Basic Concepts of Chemistry chapter of the NCERT/CBSE Class 11 curriculum. Molecular mass calculations underpin virtually every stoichiometry question in board exams, JEE Main, and NEET.
Why this formula?
Stoichiometry & Mole Calculation: The "Why" Behind the Formula
Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.
1. The Core Question: What is a Mole?
A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains 6.022×1023 particles (Avogadro's number, NA).
Why this number?
It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.
- Example: 1 atom of carbon-12 has mass 12 amu.
- 1 mole of carbon-12 has mass 12 grams.
This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).
2. The Fundamental Relationship
The key formula is:
n=Mm
Where:
- n = number of moles
- m = mass of substance (in grams)
- M = molar mass (in g/mol)
Why does this work?
Think of it as a conversion factor:
If 1 mole of a substance weighs M grams, then m grams contains Mm moles.
Derivation logic:
- Molar mass M tells you: "1 mol = M g"
- So the conversion factor is M g1 mol
- Multiply mass m by this factor: m×M1=Mm moles
3. Connecting to Number of Particles
n=NAN
Where:
- N = number of particles (atoms, molecules, ions)
- NA=6.022×1023 particles/mol
Why?
- 1 mole = NA particles
- So N particles = NAN moles
Combine both formulas:
Mm=NAN
This single equation ties mass, molar mass, number of particles, and Avogadro's number together.
4. The Gas Volume Connection (for gases at STP)
For gases only:
n=22.4 L/molV
Why 22.4 L?
From the ideal gas law: PV=nRT
At STP (Standard Temperature and Pressure: 0°C, 1 atm):
- P=1 atm
- T=273.15 K
- R=0.0821 L·atm/(mol·K)
For n=1 mole:
V=PnRT=11×0.0821×273.15≈22.4 L
So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.
5. The Stoichiometry Chain: From One Substance to Another
In a balanced chemical equation like:
aA+bB→cC+dD
The coefficients tell you the mole ratio:
moles of Bmoles of A=ba
Why this works:
- Coefficients represent relative numbers of molecules (or moles of molecules)
- If a molecules of A react with b molecules of B, then a moles of A react with b moles of B
- The ratio is fixed by the balanced equation
The complete problem-solving chain:
Mass of A÷MAMoles of A×acMoles of C×MCMass of C
Each step uses one of the relationships above.
Summary: The Logical Flow
| What you know | Formula | Why it works |
|---|---|---|
| Mass of substance | n=m/M | Molar mass is the conversion factor between grams and moles |
| Number of particles | n=N/NA | Avogadro's number is the conversion factor between particles and moles |
| Volume of gas (STP) | n=V/22.4 | Derived from ideal gas law at standard conditions |
| Moles of one reactant | nC=nA×(c/a) | Balanced equation gives fixed mole ratios |
The mole is the universal translator — it converts between mass, particle count, and gas volume, allowing you to move seamlessly through a chemical reaction.
The key idea is that mass per cent is simply the mass of the solute divided by the total mass of the solution, multiplied by 100.
Step 1: Identify the masses.
Mass of solute (A) = 2 g
Mass of solvent (water) = 18 g
Step 2: Find the total mass of the solution.
Total mass = 2 g + 18 g = 20 g
Step 3: Apply the formula for mass per cent.
Mass %=total mass of solutionmass of solute×100=202×100=10%
The mass per cent of the solute is 10%.
Mass per cent is the mass of the solute divided by the total mass of the solution, multiplied by 100. Here, the solute is 2 g, the solvent is 18 g, so the total mass is 20 g, giving a mass per cent of 10%.
Why mass per cent works
Mass per cent (or weight/weight percentage) is one of the simplest ways to express concentration. It tells you: out of every 100 grams of solution, how many grams are the solute? The key insight is that the solution's total mass is just the sum of solute and solvent — there's no volume shrinkage or expansion to worry about here, unlike with volume-based units. So the calculation is straightforward: divide the part by the whole, then scale to 100.
Mass per cent=Mass of solutionMass of solute×100
Where mass of solution = mass of solute + mass of solvent.
Step-by-step
-
Identify the given masses.
The solute (substance A) is 2 g. The solvent (water) is 18 g. No other components are present.
-
Find the total mass of the solution.
Add the two:
Mass of solution=2 g+18 g=20 g
- Apply the mass per cent formula.
Mass per cent=20 g2 g×100
- Simplify the fraction.
202=0.1
Then multiply by 100:
0.1×100=10
So the mass per cent of the solute is 10%.
A common mistake is to divide by the mass of the solvent (18 g) instead of the total mass of the solution (20 g). That would give 182×100≈11.1%, which is wrong. Always use the solution mass in the denominator, not the solvent mass.
If you ever forget the formula, just think: "per cent" means "per hundred". So ask yourself: If I had 100 g of this solution, how many grams would be the solute? Since 2 g out of 20 g is the same ratio as 10 g out of 100 g, the answer is 10%.
The mass per cent of the solute is 10%.
Method: Mass Percentage Formula Method
This is the most direct method for calculating concentration when both solute and solvent masses are given.
Concept First (Why this works)
Mass per cent tells us how many grams of solute are present in 100 grams of solution. It’s a ratio scaled to 100 — that’s why we multiply by 100.
Steps
Step 1: Identify the given quantities
- Mass of solute (substance A) = 2g
- Mass of solvent (water) = 18g
Step 2: Calculate the total mass of the solution
Mass of solution=Mass of solute+Mass of solvent
=2g+18g=20g
Step 3: Apply the mass percentage formula
Mass per cent of solute=Mass of solutionMass of solute×100
Step 4: Substitute and compute
=202×100=0.1×100=10
Step 5: Write the final answer with units
10%
Quick Check
- The solute is one-tenth of the total mass → 10% is correct.
- Always ensure the denominator is solution mass, not solvent mass alone — a common exam mistake.
Here are the common mistakes students make when calculating mass per cent (also called mass percentage or weight/weight percentage), along with clear, exam-focused corrections.
Mistake 1: Using the wrong formula
What students do wrong:
They calculate mass per cent as:
mass of solventmass of solute×100
Why it’s wrong:
Mass per cent is defined as the mass of the solute divided by the total mass of the solution (solute + solvent), not just the solvent.
Correct formula:
Mass %=Mass of solutionMass of solute×100
How to avoid:
Always write the formula before plugging numbers. Remember: solution = solute + solvent.
Mistake 2: Forgetting to add the masses
What students do wrong:
They directly use 18 g (mass of water) as the denominator.
Example of error:
182×100≈11.11%
Why it’s wrong:
The denominator should be 2+18=20 g, not 18 g.
Correct calculation:
2+182×100=202×100=10%
How to avoid:
Always compute total mass of solution first:
Mass of solution=mass of solute+mass of solvent.
Mistake 3: Confusing solute and solvent
What students do wrong:
They treat water as the solute and substance A as the solvent.
Why it’s wrong:
In a solution, the solute is the substance present in smaller amount (here, 2 g of A). Water (18 g) is the solvent.
How to avoid:
Identify:
- Solute = substance being dissolved (usually smaller mass)
- Solvent = substance doing the dissolving (usually larger mass)
Mistake 4: Not simplifying or rounding incorrectly
What students do wrong:
They leave the answer as 20200=10 without the % sign, or round to 10.0% when the question expects 10%.
How to avoid:
- Always include the % symbol in the final answer.
- Follow the significant figures given in the question (here, 2 g and 18 g → 1 or 2 significant figures → 10% is fine).
Mistake 5: Using volume instead of mass
What students do wrong:
If the question gave volume (e.g., 18 mL water), they might treat mL as grams without checking density.
Why it’s wrong:
Mass per cent requires mass, not volume. For water, 18 mL ≈ 18 g only at room temperature, but the concept must be clear.
How to avoid:
If volume is given, convert to mass using density (mass=density×volume) before applying the formula.
Quick Summary – How to Get It Right Every Time
| Step | Action |
|---|---|
| 1 | Identify solute (smaller mass) and solvent (larger mass) |
| 2 | Compute total mass of solution = solute + solvent |
| 3 | Apply formula: total masssolute mass×100 |
| 4 | Write answer with % sign |
Final correct answer for this question:
Mass %=2+182×100=202×100=10%
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL2 marksQ.Pressure of 1g of an ideal gas 'A' at 27°C is found to be 2 bar. When 2g of another ideal gas 'B' is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses.
›Reveal solutionSolution
Setting up PV = nRT for each gas separately and dividing the equations gives M_B = 4 x M_A.
Given: 1 g of ideal gas A at 27 deg C (300 K) gives a pressure of 2 bar in a flask of volume V. Then 2 g of ideal gas B is added at the SAME temperature and volume, and the total pressure becomes 3 bar.
Step 1 -- Partial pressure of B (Dalton's Law of partial pressures):
P(total) = P(A) + P(B)
3 = 2 + P(B) => P(B) = 1 bar
Step 2 -- Apply the ideal gas equation to gas A:
P(A) V = n(A) R T = (1 g / M_A) R T
2V = (1/M_A) R T ... (i)
Step 3 -- Apply the ideal gas equation to gas B:
P(B) V = n(B) R T = (2 g / M_B) R T
1.V = (2/M_B) R T ... (ii)
Step 4 -- Divide equation (i) by equation (ii):
(2V) / (1V) = [(1/M_A) R T] / [(2/M_B) R T]
2 = (1/M_A) x (M_B/2)
2 = M_B / (2 M_A)
M_B = 4 M_A
✓Final answerM_B = 4 x M_A -- the molar mass of gas B is four times that of gas A.
- Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2020Set ANNUAL2 marksQ.At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
›Reveal solutionSolution
Using density = PM/RT at the same temperature for both gases, equal densities mean P1·M1 = P2·M2; solving with the known values gives a molar mass of 70 g/mol for the oxide.
From the ideal gas equation, PV = nRT = (mass/M)RT, so density (d = mass/V) is:
d = PM/RT
Given both gases are at the same temperature (0°C = 273 K):
For the oxide: d_oxide = P_oxide × M_oxide / RT, with P_oxide = 2 bar
For N2: d_N2 = P_N2 × M_N2 / RT, with P_N2 = 5 bar, M_N2 = 28 g/mol
Since d_oxide = d_N2 and RT is the same for both (same temperature):
P_oxide × M_oxide = P_N2 × M_N2
2 × M_oxide = 5 × 28
2 × M_oxide = 140
M_oxide = 70 g/mol
✓Final answerThe molecular mass of the oxide is 70 g/mol.
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