2. Give short answer.
(i) What is mole?
Answer: A mole is the SI unit for the amount of a substance. It is defined as the amount of substance that contains the same number of particles (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. This number is Avogadro's number, approximately 6.022 × 10²³. One mole of any substance has a mass equal to its molar mass in grams.
(ii) Differentiate between empirical formula and molecular formula.
Answer:
| Feature | Empirical Formula | Molecular Formula |
| Definition | The simplest whole-number ratio of atoms in a
compound. | The actual number of atoms of each element in a
molecule. |
| Relationship | It is a reduced form. | It is a multiple of the empirical formula. |
| Example | CH₂O (for glucose) | C₆H₁₂O₆ (for glucose) |
(iii) What is the number of molecules in 9.0 g of steam?
Answer: Molar mass of steam (H₂O) = 18 g/mol. Moles = 9.018 = 0.5 moles. Number of molecules = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules.
(iv) What are the molar masses of uranium-238 and uranium-235?
Answer: The molar mass of uranium-238 is 238 g/mol. The molar mass of uranium-235 is 235 g/mol. Molar mass in grams is numerically equal to the atomic mass in amu.
(v) Why one mole of hydrogen molecules and one mole of H-atoms have different masses?
Answer: One mole of hydrogen molecules (H₂) has a mass of 2 g (since each molecule has 2 atoms). One mole of hydrogen atoms (H) has a mass of 1 g (since each atom has 1 amu). The difference is because a hydrogen molecule consists of two hydrogen atoms, so its mass is twice that of a single hydrogen atom.
3. Define ion, molecular ion, formula unit, free radical, atomic number, mass number, atomic mass unit.
Answer:
Ion: An atom or group of atoms that has gained or lost electrons, resulting in a positive (cation) or negative (anion) charge. Molecular Ion: An ion formed when a molecule loses or gains electrons (e.g., CH₄⁺). Formula Unit: The simplest ratio of ions in an ionic compound (e.g., NaCl). Free Radical: An atom or molecule with an unpaired electron, making it highly reactive (e.g., Cl•). Atomic Number: The number of protons in the nucleus of an atom. Mass Number: The sum of protons and neutrons in the nucleus. Atomic Mass Unit (amu): One-twelfth the mass of a carbon-12 atom.
4. Describe how Avogadro's number is related to a mole of any substance.
Answer: Avogadro's number (6.022 × 10²³) is the number of particles (atoms, molecules, ions, etc.) present in one mole of any substance. It is a constant that bridges the macroscopic world (grams) and the microscopic world (atoms/molecules). One mole of any substance contains exactly 6.022 × 10²³ particles. For example, one mole of carbon contains 6.022 × 10²³ carbon atoms, and one mole of water contains 6.022 × 10²³ water molecules.
5. Calculate the number of moles of each substance in samples with the following masses:
(a) 2.4g of He
Answer: Moles = 2.44 = 0.6 moles
(b) 250mg of carbon
Answer: 250mg = 0.25g. Moles = 0.2512 = 0.0208 moles
(c) 15g of sodium chloride
Answer: Moles = 1558.5 = 0.256 moles
(d) 40g of sulphur
Answer: Moles = 4032 = 1.25 moles (e) 1.5kg of MgO
Answer: 1.5kg = 1500g. Moles = 150040 = 37.5 moles
6. Calculate the mass in grams of each of the following samples:
(a) 1.2 moles of K
Answer: Mass = 1.2 × 39 = 46.8 g
(b) 75moles of H₂
Answer: Mass = 75 × 2 = 150 g
(c) 0.25 moles of steam
Answer: Mass = 0.25 × 18 = 4.5 g
(d) 1.05 moles of CuSO₄·5H₂O
Answer: Mass = 1.05 × 249.5 = 261.975 g (e) 0.15moles of H₂SO₄
Answer: Mass = 0.15 × 98 = 14.7 g
7. Calculate the number of molecules present in each of the following samples:
(a) 2.5 moles of carbon dioxide
Answer: 2.5 × 6.022 × 10²³ = 1.5055 × 10²⁴ molecules
(b) 3.4 moles of ammonia, NH₃
Answer: 3.4 × 6.022 × 10²³ = 2.047 × 10²⁴ molecules
(c) 1.09 moles of benzene, C₆H₆
Answer: 1.09 × 6.022 × 10²³ = 6.56 × 10²³ molecules
(d) 0.01 moles of acetic acid, CH₃COOH
Answer: 0.01 × 6.022 × 10²³ = 6.022 × 10²¹ molecules
8. Decide whether or not each of the following is an example of empirical formula:
(a) Al₂Cl₆
Answer: No, it is a molecular formula. Empirical formula is AlCl₃.
(b) Hg₂Cl₂
Answer: No, it is a molecular formula. Empirical formula is HgCl.
(c) NaCl
Answer: Yes, it is an empirical formula (simplest ratio 1:1).
(d) C₂H₄O
Answer: No, it is a molecular formula. Empirical formula is CH₂O.
9. TNT or trinitrotoluene is an explosive compound used in bombs. It contains 7 C-atoms, 5 H-atoms, 3 N-atoms and 6 O-atoms. Write its empirical formula.
Answer: The molecular formula is C₇H₅N₃O₆. The empirical formula is the simplest whole-number ratio, which is C₇H₅N₃O₆ (since the numbers 7, 5, 3, 6 have no common divisor other than 1).
10. A molecule contains four phosphorus atoms and ten oxygen atoms. Write the empirical formula of this compound. Also determine the molar mass of this molecule.
Answer: The molecular formula is P₄O₁₀. The empirical formula is P₂O₅ (simplest ratio 2:5). Molar mass of P₄O₁₀ = 4(31) + 10(16) = 124
+ 160 = 284 g/mol.
11. Indigo (C₁₆H₁₀N₂O₂), the dye used to colour blue jeans is derived from a compound known as indoxyl (C₈H₇ON). Calculate the molar masses of these compounds. Also write their empirical formulas.
Answer:
Indigo (C₁₆H₁₀N₂O₂): Molar mass = 16(12) + 10(1) + 2(14) + 2(16) = 192 + 10 + 28 + 32 = 262 g/mol. Empirical formula
= C₈H₅NO.
Indoxyl (C₈H₇ON): Molar mass = 8(12) + 7(1) + 16 + 14 = 96 + 7 + 16 + 14 = 133 g/mol. Empirical formula = C₈H₇ON.
12. Identify the substance that has formula mass of 133.5amu.
(a) MgCl₂
Answer: 24 + 2(35.5) = 95 amu
(b) S₂Cl₂
Answer: 2(32) + 2(35.5) = 64 + 71 = 135 amu
(c) BCl₃
Answer: 11 + 3(35.5) = 11 + 106.5 = 117.5 amu
(d) AlCl₃
Answer: 27 + 3(35.5) = 27 + 106.5 = 133.5 amu
Answer: (d) AlCl₃
13. Calculate the number of atoms in each of the following samples:
(a) 3.4 moles of nitrogen atoms
Answer: 3.4 × 6.022 × 10²³ = 2.047 × 10²⁴ atoms
(b) 23g of Na
Answer: Moles = 2323 = 1. Atoms = 6.022 × 10²³ atoms
(c) 5g of H atoms
Answer: Moles = 51 = 5. Atoms = 5 × 6.022 × 10²³ = 3.011 × 10²⁴ atoms
14. Calculate the mass of the following:
(a) 3.24 × 10²⁰ atoms of iron
Answer: Moles = 3.24 × 10²⁰ / 6.022 × 10²³ = 5.38 × 10⁻⁴. Mass = 5.38 × 10⁻⁴ × 56 = 0.0301 g
(b) 2 x 10²⁵ molecules of nitrogen gas
Answer: Moles = 2 × 10²⁵ / 6.022 × 10²³ = 33.2. Mass = 33.2 × 28 = 929.6 g
(c) 1 x 10¹⁰ molecules of water
Answer: Moles = 1 × 10¹⁰ / 6.022 × 10²³ = 1.66 × 10⁻¹⁴. Mass = 1.66 × 10⁻¹⁴ × 18 = 2.99 × 10⁻¹³ g
(d) 3 x 10⁶ atoms of Al
Answer: Moles = 3 × 10⁶ / 6.022 × 10²³ = 4.98 × 10⁻¹⁸. Mass = 4.98 × 10⁻¹⁸ × 27 = 1.34 × 10⁻¹⁶ g
15. Balance the following chemical equations:
(a) Na + H₂O [source symbol unavailable] NaOH + H₂
Answer: 2Na + 2H₂O [source symbol unavailable] 2NaOH + H₂
(b) NH₃ [source symbol unavailable] N₂ + H₂
Answer: 2NH₃ [source symbol unavailable] N₂ + 3H₂
16. Potassium is Group 1 element. It is silvery white metal. It burns in air and forms both potassium oxide and potassium nitride. The nitride ion is N³⁻.
(a) Predict the formula of potassium oxide and potassium nitride.
Answer: Potassium oxide: K₂O. Potassium nitride: K₃N.
(b) A 0.5g sample of K was added in 100cm³ of water. Show that 1.28x10⁻² mole of K were added to the water.
Answer: Moles of K = 0.5g / 39 g/mol = 0.0128 moles = 1.28 × 10⁻² moles.
(c) Balance above chemical equation.
Answer: 2K + 2H₂O [source symbol unavailable] 2KOH + H₂
(d) Transform above chemical equation into ionic equation.
Answer: 2K(s) + 2H₂O(l) [source symbol unavailable] 2K⁺(aq) + 2OH⁻(aq) + H₂(g) (e) Calculate the number of atoms present in the sample of K.
Answer: 0.0128 × 6.022 × 10²³ = 7.71 × 10²¹ atoms (f) Predict period number of potassium in the periodic table.
Answer: Period 4.
