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%
- CBSE 2025Set ANNUAL1 markMCQQ.The molecular weight of glucose (C6H12O6) molecule is(a) 90 U(b) 120 U(c) 180 U(d) 360 U
›Reveal solutionSolution
Glucose (C6H12O6) has a molecular mass of 180 u.
Atomic masses: C = 12 u, H = 1 u, O = 16 u.
Molecular mass of C6H12O6 = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 u.
✓Final answer(C) 180 U.
- CBSE 2025Set sz1 markMCQQ.Select the correct one: Which of the following is the standard for atomic mass?(a) 1/1 H(b) 12/6 C(c) 14/6 C(d) 16/8 O
›Reveal solutionSolution
The modern standard for atomic mass is the carbon-12 isotope; 1 amu = 1/12 the mass of a 12/6 C atom.
Before 1961, both oxygen-16 and hydrogen-1 standards were tried, but chemists and physicists used slightly different oxygen-based scales, causing confusion. In 1961 IUPAC adopted a single unified standard: the carbon-12 isotope (12/6 C) was assigned a mass of exactly 12 atomic mass units (amu), and 1 amu is defined as 1/12th of the mass of one atom of carbon-12. All other atomic masses are expressed relative to this standard.
✓Final answer(B) 12/6 C — one atomic mass unit (amu) is defined as exactly 1/12th the mass of a carbon-12 atom.
- CBSE 2024Set ANNUAL1 markMCQQ.What is the molar mass of H2O in gm/mol?(a) 44(b) 18(c) 17(d) 60
›Reveal solutionSolution
Molar mass of H₂O = 2 × (atomic mass of H) + 1 × (atomic mass of O) = 2(1) + 16 = 18 g/mol.
Atomic mass of H ≈ 1 u, atomic mass of O ≈ 16 u.
M(H2O)=2(1)+16=18 g/mol
✓Final answerThe molar mass of H₂O is 18 g/mol (option b).
- CBSE 2024Set ANNUAL1 markMCQQ.Molecular mass of volatile substance is determined by:(a) Kjeldahl's method(b) Duma's method(c) Victor Mayer's method(d) Leibig's method
›Reveal solutionSolution
Victor Meyer's method determines the molecular mass of a volatile substance by measuring the volume of air displaced when a known mass of the substance is vaporised.
Each method listed determines something different:
-
Kjeldahl's method — estimates the percentage of nitrogen in an organic compound, not molecular mass.
-
Dumas' method — also estimates % nitrogen (by converting it to N₂ gas and measuring its volume), not molecular mass of a volatile substance directly.
-
Victor Meyer's method — a known mass of a volatile liquid is vaporised in a heated tube; the volume of air it displaces gives the vapour density, from which molecular mass = 2 × vapour density.
-
Liebig's method — estimates carbon and hydrogen percentage by combustion analysis.
✓Final answerMolecular mass of a volatile substance is determined by Victor Meyer's method (option c).
-
- CBSE 2023Set ANNUAL1 markMCQQ.Molar mass of CO2 is:(a) 22(b) 38(c) 44(d) 28
›Reveal solutionSolution
Adding one carbon (12 u) and two oxygens (16 u each) gives the molar mass of CO2 as 44 g/mol.
Molar mass = sum of atomic masses of all atoms in the formula.
M(CO2)=1×M(C)+2×M(O)=1(12)+2(16)=12+32=44 g mol−1
✓Final answerMolar mass of CO2 = 44 g/mol (option c).
- CBSE 2022Set TERM11 markMCQQ.The molar mass of CH4 is(a) 16 u(b) 20 u(c) 10 u(d) 24 u
›Reveal solutionSolution
Add up the atomic masses of all atoms in one CH4 molecule: 1 carbon + 4 hydrogens.
Molar mass is the sum of the atomic masses of every atom in the formula.
CH4 has 1 carbon atom and 4 hydrogen atoms.
Molar mass = (1 x 12 u) + (4 x 1 u) = 12 u + 4 u = 16 u.
✓Final answer(a) 16 u.
- CBSE 2022Set ANNUAL1 markQ.Write right or wrong: Molecular mass of water is 18.
›Reveal solutionSolution
The statement is Right: the molecular mass of water (H2O) is 18 u.
Molecular mass is the sum of the atomic masses of all atoms in the molecular formula. Water's formula is H2O: two hydrogen atoms (average atomic mass about 1 u each) plus one oxygen atom (average atomic mass about 16 u): Molecular mass = 2 x 1 + 16 = 18 u. This matches the statement exactly.
✓Final answerRight - the molecular mass of water is 18 u.
- CBSE 2022Set sz1 markQ.What is the relation between vapour density and molecular mass of a gas?
›Reveal solutionSolution
Molecular mass equals twice the vapour density, because vapour density is defined relative to hydrogen (M = 2 g/mol).
Vapour density of a gas is defined as:
VD = density of the gas / density of hydrogen (at the same temperature and pressure)
At the same temperature and pressure, density is directly proportional to molar mass (from the ideal gas equation, PM = dRT, so d is proportional to M for fixed P, T). Therefore:
VD = M(gas) / M(H2)
Since M(H2) = 2 g/mol:
VD = M(gas) / 2
Rearranging:
M(gas) = 2 x VD
This relation lets chemists determine the molar mass of a gas experimentally just by measuring how many times denser it is than hydrogen.
✓Final answerMolecular mass (M) = 2 x Vapour Density (VD).
- CBSE 2018Set ANNUAL1 markQ.Calculate the molecular weight of the following compounds:(i) C6H12O6(ii) H2SO4
›Reveal solutionSolution
The molecular weight of C6H12O6 (glucose) is 180 g/mol and of H2SO4 (sulphuric acid) is 98 g/mol, found by summing the atomic weights of each constituent atom.
Using standard atomic weights C = 12, H = 1, O = 16, S = 32:
- C6H12O6: C: 6 × 12 = 72 H: 12 × 1 = 12 O: 6 × 16 = 96 Total = 72 + 12 + 96 = 180 g/mol
- H2SO4: H: 2 × 1 = 2 S: 1 × 32 = 32 O: 4 × 16 = 64 Total = 2 + 32 + 64 = 98 g/mol
✓Final answer(i) C6H12O6 = 180 g/mol (ii) H2SO4 = 98 g/mol.
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