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Atoms and Molecules
1. In a reaction, 5.3g of sodium carbonate reacted with 6g of acetic acid. The products were 2.2g of carbon dioxide, 0.9g of water and 8.2g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass. Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water Solution: Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water 5.3g 6g 8.2g 2.2g 0.9g As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products. As per the above reaction, L.H.S. = R.H.S. i.e., 5.3g + 6g = 2.2g + 0.9g + 8.2g = 11.3g Hence, the observations are in agreement with the law of conservation of mass. 2. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass? Solution: The relative number and types of atoms are constant in a given composition, says Dalton’s atomic theory, which is based on the rule of conservation of mass. “Atoms cannot be created nor be destroyed in a chemical reaction.” 3. Which postulate of Dalton’s atomic theory can explain the law of definite proportions? Solution:
The postulate of Dalton’s atomic theory that can explain the law of definite proportions is that the relative number and kinds of atoms are equal in given compounds. 4. Define the atomic mass unit. Solution: An atomic mass unit is a unit of mass used to express the weights of atoms and molecules where one atomic mass is equal to 1/12th the mass of one carbon-12 atom. 5. Write down the formulae of (i) sodium oxide (ii) aluminium chloride (iii) sodium sulphide (iv) magnesium hydroxide Solution: The following are the formulae: (i) sodium oxide – Na2O (ii) aluminium chloride – AlCl (iii) sodium sulphide – Na2S (iv) magnesium hydroxide – Mg (OH) 6. Write down the names of compounds represented by the following formulae: (i) Al2(SO4) (ii) CaCl (iii) K2SO (iv) KNO (v) CaCO Solution:
Na = 23u, K=39u, C = 12u, and O=16u. Solution: Given: The atomic mass of Zn = 65u The atomic mass of Na = 23u The atomic mass of K = 39u The atomic mass of C = 12u The atomic mass of O = 16u The formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = ( x 23)u + 16u = 46u +16u = 62u The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u +12u + (3 x 16)u = 78u + 12u + 48u = 138u 10. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon? Solution: Given: 1 mole of carbon weighs 12g 1 mole of carbon atoms = 6.022 x 1023 The molecular mass of carbon atoms = 12g = an atom of carbon mass Hence, mass of 1 carbon atom = 12 / 6.022 x 1023 = 1.99 x 10-23g 11. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given the atomic mass of Na = 23u, Fe = 56 u)? Solution:
(a) In 100 grams of Na: m = 100g, Molar mass of Na atom = 23g, N0 = 6.022 x 1023, N =? N = (Given mass x N0)/Molar mass N = (100 x 6.022 x 1023)/ 23 N = 26.18 x 1023 atoms (b) In 100 grams of Fe: m = 100 g, Molar mass of Fe atom = 56 g, N0 = 6.022 x 1023, N =? N = (Given mass x N0)/ Molar mass N = (100 x 6.022 x 1023)/ 56 N = 10.75 x 1023 atoms Therefore, the number of atoms is more in 100g of Na than in 100g of Fe. 12. A 0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight. Solution: Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g To calculate the percentage composition of the compound, Percentage of boron = mass of boron / mass of the compound x 100 = 0.096g / 0.24g x 100 = 40% Percentage of oxygen = 100 – percentage of boron = 100 – 40 = 60%
(a) Quick lime – Calcium and oxygen (CaO) (b) Hydrogen bromide – Hydrogen and bromine (HBr) (c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3) (d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4) 16. Calculate the molar mass of the following substances. (a) Ethyne, C2H (b) Sulphur molecule, S (c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31) (d) Hydrochloric acid, HCl (e) Nitric acid, HNO Solution: 17. What is the mass of? (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (Atomic mass of aluminium =27)? (c) 10 moles of sodium sulphite (Na2SO3)? Solution: The mass of the above-mentioned list is as follows: (a) Atomic mass of nitrogen atoms = 14u Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen atoms
Therefore, the mass of 1 mole of nitrogen atom is 14g. (b) Atomic mass of aluminium =27u Mass of 1 mole of aluminium atoms = 27g 1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g (c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3x Mass of O = (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g Therefore, mass of 10 moles of Na2SO3 = 10 x 126 = 1260g 18. Convert into a mole. (a) 12g of oxygen gas (b) 20g of water (c) 22g of carbon dioxide Solution: Conversion of the above-mentioned molecules into moles is as follows: (a) Given: Mass of oxygen gas=12g Molar mass of oxygen gas = 2 Mass of Oxygen = 2 x 16 = 32g Number of moles = Mass given / molar mass of oxygen gas = 12/32 = 0.375 moles (b) Given: Mass of water = 20g Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g Number of moles = Mass given / molar mass of water = 20/18 = 1.11 moles (c) Given: Mass of carbon dioxide = 22g
= 3.763 x 1022 molecules
