Structure of the Atom – Numerical Problems with Stepwise Solutions

Class 9 Chemistry — Chapter 4: Structure of the Atom — 20 Numerical Problems with Stepwise Solutions

Content Bank — Key formulas & constantsAtomic mass unit (u): 1 u = 1/12 × mass of ¹²C atom ≈ 1.6605×10⁻²⁷ kg.
Avogadro's number (NA): 6.022 × 10²³ particles per mole.
Relation (moles ↔ particles): Number of particles = n × NA.
Molar mass & atomic mass: Atomic mass in u ≈ molar mass in g mol⁻¹ (e.g., C ≈ 12 u → 12 g mol⁻¹).
Mass per atom (kg): mass(atom) = (atomic mass in g mol⁻¹ × 1e-3) / NA.
Neutrons from A & Z: neutrons = A − Z (A = mass number, Z = atomic number).
Number of atoms in given mass: n = mass (g) / M (g mol⁻¹), then atoms = n × NA.

Topics covered: Atomic number & mass number calculations, isotopic mass & average mass, mole & Avogadro calculations, mass per atom, number of atoms/molecules, valency-related counts, mass conversions.

Topic: Atomic number & mass number

Q1. For the isotope written as ^{23}_{11}Na, calculate the number of protons, neutrons and electrons in a neutral atom.

Step 1: Atomic number Z = 11 → number of protons = 11.

Step 2: Mass number A = 23 → neutrons = A − Z = 23 − 11 = 12.

Step 3: Neutral atom → electrons = protons = 11.

Answer: Protons = 11, Neutrons = 12, Electrons = 11.

Topic: Number of neutrons

Q2. An atom has 17 protons and mass number 35. How many neutrons does it have? Write the isotope notation.

Step 1: Neutrons = A − Z = 35 − 17 = 18.

Step 2: Z = 17 → element is chlorine symbol Cl. Notation: ^{35}_{17}Cl.

Answer: Neutrons = 18; isotope = ^{35}_{17}Cl.

Topic: Mass of one atom (kg)

Q3. Estimate the mass of one carbon atom in kilograms. (Atomic mass of C ≈ 12.0 g mol⁻¹)

Step 1: Molar mass M = 12.0 g mol⁻¹ → mass per mole = 12.0 × 10⁻³ kg.

Step 2: Mass per atom = (12.0 × 10⁻³ kg) / N_A where N_A = 6.022×10²³.

Step 3: Compute: (12.0×10⁻³) / (6.022×10²³) ≈ 1.993×10⁻²⁶ kg.

Answer: ≈ 1.99 × 10⁻²⁶ kg per carbon atom.

Topic: Number of atoms from moles

Q4. How many oxygen atoms are there in 0.250 mol of O atoms? (Use N_A=6.022×10²³)

Step 1: Number of atoms = n × N_A = 0.250 × 6.022×10²³.

Step 2: Compute: 0.250 × 6.022×10²³ = 1.5055×10²³ atoms.

Answer: ≈ 1.5055 × 10²³ oxygen atoms.

Topic: Mass ↔ moles

Q5. What is the mass (in grams) of 0.0200 mol of helium (He)? (Molar mass He = 4.00 g mol⁻¹)

Step 1: mass = n × M = 0.0200 × 4.00 g = 0.0800 g.

Answer: 0.0800 g of He.

Topic: Number of atoms in a given mass

Q6. How many hydrogen atoms are present in 2.0 g of hydrogen (H)? (M(H)=1.008 g mol⁻¹)

Step 1: Moles of H atoms = mass / M = 2.0 / 1.008 ≈ 1.9841 mol.

Step 2: Number of atoms = n × N_A ≈ 1.9841 × 6.022×10²³ ≈ 1.195×10²4 atoms.

Answer: ≈ 1.20 × 10²⁴ hydrogen atoms (rounded).

Topic: Mass per atom (g)

Q7. Calculate the mass of a single sodium atom in grams. (Atomic mass Na ≈ 23.0 g mol⁻¹)

Step 1: Mass per atom = (23.0 g mol⁻¹) / (6.022×10²³) ≈ 3.819×10⁻²³ g.

Answer: ≈ 3.82 × 10⁻²³ g per Na atom.

Topic: Isotopic average mass (weighted average)

Q8. Natural element X has two isotopes: X-35 (abundance 75.77%) and X-37 (abundance 24.23%). Calculate the average atomic mass of X in u (use masses 34.9689 and 36.9659 respectively — chlorine example).

Step 1: Weighted average = (fraction₁×mass₁) + (fraction₂×mass₂).

Step 2: Convert % to fraction: 0.7577 and 0.2423.

Step 3: Average = 0.7577×34.9689 + 0.2423×36.9659 ≈ 26.5 + 8.95 ≈ 35.45 u.

Answer: ≈ 35.45 u (explains why chlorine's atomic mass ≈35.45).

Topic: Atoms in molecules

Q9. How many oxygen atoms are there in 0.100 mol of CO₂?

Step 1: Each CO₂ molecule contains 2 oxygen atoms.

Step 2: Molecules in 0.100 mol = 0.100 × N_A.

Step 3: Oxygen atoms = 2 × (0.100 × N_A) = 0.200 × N_A ≈ 1.2044×10²³ atoms.

Answer: ≈ 1.204 × 10²³ oxygen atoms.

Topic: Mass of single atom (kg)

Q10. Find the mass (in kg) of one helium atom. (M(He)=4.00 g mol⁻¹)

Step 1: Mass per atom in grams = 4.00 / 6.022×10²³ ≈ 6.644×10⁻²⁴ g.

Step 2: Convert to kg: ×10⁻³ ⇒ 6.644×10⁻²⁷ kg.

Answer: ≈ 6.64 × 10⁻²⁷ kg per He atom.

Topic: Moles from particle count

Q11. How many moles are there in 3.011×10²³ atoms of a substance?

Step 1: n = number of particles / N_A = 3.011×10²³ / 6.022×10²³ = 0.5 mol.

Answer: 0.50 mol.

Topic: Mass of sample given atom count

Q12. What mass of carbon (in grams) contains 1.2044×10²³ atoms? (M(C) = 12.0 g mol⁻¹)

Step 1: Moles = particles / N_A = 1.2044×10²³ / 6.022×10²³ = 0.2 mol.

Step 2: mass = n × M = 0.2 × 12.0 = 2.4 g.

Answer: 2.4 g of carbon.

Topic: Electrons from ions

Q13. How many electrons are there in the ion ^{27}_{13}Al^{3+}?

Step 1: Neutral Al has 13 electrons. The 3+ ion has lost 3 electrons → electrons = 13 − 3 = 10.

Answer: 10 electrons.

Topic: Mass of many atoms

Q14. What mass of hydrogen (in grams) contains 6.022×10²² hydrogen atoms? (M(H)=1.008 g mol⁻¹)

Step 1: Moles = particles / N_A = 6.022×10²² / 6.022×10²³ = 0.10 mol.

Step 2: mass = n × M = 0.10 × 1.008 ≈ 0.1008 g.

Answer: ≈ 0.1008 g (≈ 0.101 g).

Topic: Atoms in compound sample

Q15. How many hydrogen atoms are present in 18 g of water (H₂O)? (M(H₂O) = 18 g mol⁻¹)

Step 1: Moles of H₂O = mass / M = 18 / 18 = 1.00 mol.

Step 2: Each molecule has 2 H atoms → total H atoms = 2 × N_A × 1.00 = 2 × 6.022×10²³ = 1.2044×10²4.

Answer: ≈ 1.204 × 10²⁴ hydrogen atoms.

Topic: Average atomic mass (simple)

Q16. Element Y has two isotopes: Y-10 (mass 10 u, abundance 20%) and Y-11 (mass 11 u, abundance 80%). Calculate the average atomic mass.

Step 1: Fractional abundances: 0.20 and 0.80.

Step 2: Average = 0.20×10 + 0.80×11 = 2.0 + 8.8 = 10.8 u.

Answer: 10.8 u.

Topic: Number of atoms in small mass

Q17. How many carbon atoms are there in a 12 mg diamond sample? (M(C)=12.0 g mol⁻¹)

Step 1: Convert mass: 12 mg = 0.012 g.

Step 2: Moles = 0.012 / 12.0 = 0.001 mol.

Step 3: Atoms = 0.001 × 6.022×10²³ = 6.022×10²0 atoms.

Answer: ≈ 6.02 × 10²⁰ carbon atoms.

Topic: Relating atomic mass to grams

Q18. If atomic mass of element Z is 40 u, what mass in grams contains 3.011×10²³ atoms of Z?

Step 1: 3.011×10²³ atoms = 0.5 mol (since it's half of N_A).

Step 2: Molar mass = 40 g mol⁻¹ → mass = n × M = 0.5 × 40 = 20 g.

Answer: 20 g.

Topic: Electrons and ions (count)

Q19. How many electrons does the ion ^{35}_{17}Cl^{-} have?

Step 1: Neutral Cl (Z=17) has 17 electrons. The 1− ion has gained 1 electron → electrons = 17 + 1 = 18.

Answer: 18 electrons.

Topic: Mass & atoms combined practice

Q20. A sample contains 0.30 g of neon gas (Ne, M = 20.18 g mol⁻¹). How many neon atoms are there in the sample?

Step 1: Moles n = mass / M = 0.30 / 20.18 ≈ 0.01486 mol.

Step 2: Atoms = n × N_A ≈ 0.01486 × 6.022×10²³ ≈ 8.95×10²¹ atoms.

Answer: ≈ 8.95 × 10²¹ neon atoms.

Tip: keep a small clipboard with N_A, 1 u and common molar masses (H=1.008, C=12.0, O=16.0, Na=23.0) when doing quick calculations in exams. If you'd like these problems converted into a printable PDF, worksheet (two-column) or a version with hidden solutions for testing, tell me which format and I will produce it.

Structure of the Atom – Numerical Problems with Stepwise Solutions

Content Bank — Key formulas & constants

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