Atomic & Molecular Masses (Relative atomic mass, Molecular mass, Formula mass, Molar mass, Percentage composition, Empirical formula, Molecular formula: NEET MCQs with Answers

15 free questions on Atomic & Molecular Masses (Relative atomic mass, Molecular mass, Formula mass, Molar mass, Percentage composition, Empirical formula, Molecular formula for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
If 2.6 mol of a compound X weighs 312 g, what is its molar mass? [Re-NEET 2024]
  1. 120 g mol⁻¹
  2. 60 g mol⁻¹
  3. 811.2 g mol⁻¹
  4. 312 g mol⁻¹
Show answer
Answer: (A) 120 g mol⁻¹
Explanation: Molar mass = mass/moles = 312 g / 2.6 mol = 120 g mol⁻¹.

Question 2

Previous-year question
A metal oxide has the formula Z₂O₃. It can be reduced by hydrogen to give free metal and water. If 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction, the atomic weight of the metal is [AIPMT 1989]
  1. 27.9
  2. 159.6
  3. 79.8
  4. 55.8
Show answer
Answer: (D) 55.8
Explanation: Z₂O₃ + 3H₂ ⟶ 2Z + 3H₂O n(H₂) = 0.006/2 = 0.003 mol. From the stoichiometry, n(Z₂O₃) = 0.003/3 = 0.001 mol. Molar mass of Z₂O₃ = 0.1596/0.001 = 159.6 g/mol. So 2A_Z + (3 × 16) = 159.6 ⇒ 2A_Z = 159.6 − 48 = 111.6 ⇒ A_Z = 55.8 u. The metal is iron and the oxide is Fe₂O₃ (haematite).

Question 3

Previous-year question
Boron has two stable isotopes, ¹⁰B (19%) and ¹¹B (81%). Calculate the average atomic weight of boron as listed in the periodic table. [AIPMT 1990]
  1. 10.8
  2. 10.2
  3. 11.2
  4. 10.0
Show answer
Answer: (A) 10.8
Explanation: Average atomic mass = [(% of ¹⁰B × mass of ¹⁰B) + (% of ¹¹B × mass of ¹¹B)] / 100 = [(19 × 10) + (81 × 11)] / 100 = (190 + 891)/100 = 1081/100 = 10.81 ≈ 10.8 u. This is exactly the value of 10.81 u listed for boron in the periodic table.

Question 4

Previous-year question
Which of the following fertilizers has the highest nitrogen percentage by mass? [AIPMT 1993]
  1. Ammonium sulphate
  2. Calcium cyanamide
  3. Urea
  4. Ammonium nitrate
Show answer
Answer: (C) Urea
Explanation: Urea, NH₂CONH₂ (M = 60): %N = (28/60) × 100 = 46.67%. Ammonium sulphate, (NH₄)₂SO₄ (M = 132): %N = (28/132) × 100 = 21.2%. Calcium cyanamide, CaCN₂ (M = 80): %N = (28/80) × 100 = 35.0%. Ammonium nitrate, NH₄NO₃ (M = 80): %N = (28/80) × 100 = 35.0%. Urea is highest, which is precisely why it is the most widely used nitrogenous fertilizer in agriculture.

Question 5

Previous-year question
The weight of one molecule of the compound C₆₀H₁₂₂ is [AIPMT 1995]
  1. 1.2 × 10⁻²⁰ g
  2. 1.4 × 10⁻²¹ g
  3. 5.025 × 10²³ g
  4. 6.023 × 10²³ g
Show answer
Answer: (B) 1.4 × 10⁻²¹ g
Explanation: Molar mass of C₆₀H₁₂₂ = 720 + 122 = 842 g mol⁻¹. Mass of one molecule = 842/(6.022 × 10²³) = 1.4 × 10⁻²¹ g; values of the order of 10²³ g are impossible for a single molecule.

Question 6

Previous-year question
An organic compound contains C = 40%, O = 53.34% and H = 6.60%. The empirical formula of the compound is [AIPMT 1994]
  1. CH₂O
  2. CHO
  3. CH₄O₂
  4. C₂H₂O
Show answer
Answer: (A) CH₂O
Explanation: Relative moles: C = 40/12 = 3.33; H = 6.60/1 = 6.60; O = 53.34/16 = 3.33. Dividing by 3.33: C : H : O = 1 : 2 : 1. Empirical formula = CH₂O. This is the empirical formula of a large family of carbohydrates (e.g. glucose, C₆H₁₂O₆) as well as of formaldehyde, HCHO.

Question 7

Previous-year question
An organic compound containing C, H and N gave the following results on analysis: C = 40%, H = 13.33%, N = 46.67%. Its empirical formula would be [AIPMT 2002, 1999, 1998]
  1. C₂H₇N₂
  2. CH₅N
  3. CH₄N
  4. C₂H₇N
Show answer
Answer: (C) CH₄N
Explanation: Relative moles: C = 40/12 = 3.33; H = 13.33/1 = 13.33; N = 46.67/14 = 3.33. Dividing by the smallest (3.33): C : H : N = 1 : 4 : 1. Empirical formula = CH₄N. (The corresponding molecular formula C₂H₈N₂ is ethylenediamine, H₂N–CH₂–CH₂–NH₂.)
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Question 8

Previous-year question
In an experiment it was found that 10 mL of a 0.05 M solution of a chloride required 10 mL of 0.1 M AgNO₃ solution for complete precipitation. Which of the following will be the formula of the chloride? (X stands for the symbol of the element other than chlorine.) [AIPMT 2000]
  1. X₂Cl
  2. X₂Cl₃
  3. XCl₂
  4. XCl₄
Show answer
Answer: (C) XCl₂
Explanation: Millimoles of the chloride compound = 10 × 0.05 = 0.5 mmol. Millimoles of AgNO₃ = 10 × 0.1 = 1.0 mmol. Each Ag⁺ ion precipitates exactly one Cl⁻ ion as AgCl. Since twice as many moles of AgNO₃ are consumed, each formula unit of the compound must supply 2 chloride ions. XCl₂ + 2AgNO₃ ⟶ 2AgCl↓ + X(NO₃)₂ Hence the formula of the chloride is XCl₂.

Question 9

Previous-year question
The specific volume of a cylindrical virus particle is 6.02 × 10⁻² cc/g, and its radius and length are 7 Å and 10 Å respectively. If Nₐ = 6.02 × 10²³, find the molecular weight of the virus. [AIPMT 2001]
  1. 3.08 × 10³ kg/mol
  2. 3.08 × 10⁴ kg/mol
  3. 1.54 × 10⁴ kg/mol
  4. 15.4 kg/mol
Show answer
Answer: (D) 15.4 kg/mol
Explanation: Convert dimensions: r = 7 Å = 7 × 10⁻⁸ cm; l = 10 Å = 1 × 10⁻⁷ cm. Volume of one cylindrical particle = πr²l = 3.14 × (7 × 10⁻⁸)² × (10⁻⁷) = 154 × 10⁻²³ cc. Specific volume is the volume occupied per gram, so mass of one particle = volume ÷ specific volume = (154 × 10⁻²³)/(6.02 × 10⁻²) g. Molecular weight = mass of one particle × Nₐ = [(154 × 10⁻²³)/(6.02 × 10⁻²)] × 6.02 × 10²³ = 154 × 10² = 15400 g/mol. Converting: 15400 g/mol = 15.4 kg/mol.

Question 10

Previous-year question
An element X has the following isotopic composition: ²⁰⁰X : 90%, ¹⁹⁹X : 8.0%, ²⁰²X : 2.0%. The weighted average atomic mass of the naturally occurring element X is closest to [AIPMT 2007]
  1. 201 u
  2. 202 u
  3. 199 u
  4. 200 u
Show answer
Answer: (D) 200 u
Explanation: Average atomic mass = Σ (isotopic mass × fractional abundance). = (200 × 0.90) + (199 × 0.08) + (202 × 0.02) = 180.00 + 15.92 + 4.04 = 199.96 u ≈ 200 u. Note that the answer is dominated by the most abundant isotope (²⁰⁰X, 90%), which is why the average lies so close to 200.

Question 11

Previous-year question
An organic compound contains carbon, hydrogen and oxygen only. Its elemental analysis gave C = 38.71% and H = 9.67%. The empirical formula of the compound would be [AIPMT 2008]
  1. CH₃O
  2. CH₂O
  3. CHO
  4. CH₄O
Show answer
Answer: (A) CH₃O
Explanation: Percentage of O = 100 − (38.71 + 9.67) = 51.62%. Relative moles: C = 38.71/12 = 3.23; H = 9.67/1 = 9.67; O = 51.62/16 = 3.23. Dividing by 3.23: C : H : O = 1 : 3 : 1. Empirical formula = CH₃O. (The corresponding molecular compound is ethylene glycol, C₂H₆O₂, which has exactly this composition.)

Question 12

Previous-year question
Suppose the elements X and Y combine to form two compounds, XY₂ and X₃Y₂. If 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂ weighs 9 g, the atomic weights of X and Y respectively are [NEET Phase II 2016]
  1. 40, 30
  2. 60, 40
  3. 20, 30
  4. 30, 20
Show answer
Answer: (A) 40, 30
Explanation: Molar mass of XY₂ = 10/0.1 = 100 g/mol ⇒ Aₓ + 2Aᵧ = 100 …(i) Molar mass of X₃Y₂ = 9/0.05 = 180 g/mol ⇒ 3Aₓ + 2Aᵧ = 180 …(ii) Subtracting (i) from (ii): 2Aₓ = 80 ⇒ Aₓ = 40 g/mol. Substituting in (i): 40 + 2Aᵧ = 100 ⇒ Aᵧ = 30 g/mol.

Question 13

Previous-year question
An organic compound is made up of 78% carbon by mass, the remaining percentage being hydrogen. The correct option for the empirical formula of this compound is (At. wt. of C is 12, H is 1) [NEET 2021]
  1. CH
  2. CH₂
  3. CH₃
  4. CH₄
Show answer
Answer: (C) CH₃
Explanation: Percentage of H = 100 − 78 = 22%. Relative number of moles: C = 78/12 = 6.5; H = 22/1 = 22. Dividing by the smallest: C = 6.5/6.5 = 1; H = 22/6.5 = 3.38 ≈ 3. Empirical formula = CH₃ (this is the empirical formula of ethane, C₂H₆). Cross-check on composition: CH₃ corresponds to 12/15 = 80.0% C, whereas CH₄ would require 75.0% C. The given 78% is much closer to CH₃.

Question 14

Previous-year question
A compound X consists of 32% of A, 20% of B and the remaining percentage of C. Given the atomic masses A = 64, B = 40 and C = 32 u, the empirical formula of X is [NEET 2024]
  1. A₂BC₂
  2. ABC₃
  3. AB₂C₂
  4. ABC₄
Show answer
Answer: (B) ABC₃
Explanation: Percentage of C = 100 − (32 + 20) = 48%. Relative number of moles: A = 32/64 = 0.5; B = 20/40 = 0.5; C = 48/32 = 1.5. Dividing each by the smallest value (0.5): A : B : C = 1 : 1 : 3. Hence the empirical formula is ABC₃.

Question 15

Previous-year question
Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂. When 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂ weighs 9 g, the atomic weights of X and Y are: (AIPMT 2016)
  1. 30, 20
  2. 40, 30
  3. 60, 40
  4. 20, 30
Show answer
Answer: (B) 40, 30
Explanation: x + 2y = 100 and 3x + 2y = 180. Subtracting: 2x = 80 → x = 40; then y = 30.

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