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Mole Concept and Stoichiometry
1. Fill in the blank spaces by choosing appropriate words from the brackets. (i) The relative molecular mass is a number that represents how many times one _________ [atom/molecule] of a substance is heavier than 121 th mass of carbon ⎡⎣^126 C⎤⎦. (ii) Whenever the gases react chemically, they do so in ___________ [weights/volumes] which bear a simple ratio to each other and to the products, if gaseous, provided the temperature and pressure of reacting gases and products remains the same. (iii) An ____________ [atom/molecule] is the smallest unit of matter, which may or may not have an independent existence, but always takes part in a chemical reaction. (iv) Equal volumes of all ____________ (liquids/gases), under similar conditions of temperature and pressure, contain equal number of molecules. (v) The mass of substance containing particles equal to Avogadro’s number is called ____________. [molecule/mole] [5] Ans. (i) molecule (ii) volumes (iii) atom (iv) gases (v) mole.
2. Give one word/words for the following statements : (i) It is a number that represents how many times an atom of an element is heavier than 121 th mass of carbon atom. (ii) The molecular weight of an element expressed in grams. (iii) The number of atoms present in 12 g of carbon ⎡⎣ (^126) C⎤⎦. (iv) Equal volumes of gases under similar conditions of temperature and pressure contain equal number of molecules. (v) A formula of a chemical substance which tells the actual number of atoms in one molecule of a substance. [5] Ans. (i) atomic weight (ii) gram-molecular weight (iii) Avogadro’s number (iv) Avogadro’s law (iv) molecular formula. Numerical Problems on Gay-Lussac’s Law 3. (a) Define atomic weight (relative atomic mass) of an element. [1] (b) From the equation : N 2 + 3H 2 2NH 3. Calculate the volume of ammonia gas formed when 6 litres of hydrogen reacts with excess of nitrogen, all volumes being measured at STP. [2] Ans. (a) It is a number that represents how many times an atom of an element is heavier than 121 th mass of carbon atom. (b) N 2 + 3H 2 —→ 2NH (^3) By Gay-Lussac’s Law : 1 vol 3 vols 2 vols 3 Vols of hydrogen produce ammonia gas = 2 vols ∴ 6 lt of hydrogen produce ammonia gas = 2 6 ×^6 = 4 lt
6. Nitrogen monoxide (NO) reacts with oxygen to form nitrogen dioxide according to the equation : 2NO + O 2 —→ 2NO 2 What volume of nitrogen dioxide would be formed when 100 cm^3 of NO reacts with 50 cm^3 of oxygen, under same conditions of temperature and pressure? [2]
Ans. 2NO + O 2 —→ 2NO (^2) By Gay-Lussac’s Law 2 vols 1 vol 2 vols ∴ 100 cm^3 50 cm^3 100 cm^3 ∴ The volume of nitrogen dioxide formed = 100 cm^3.
7. Hydrogen and oxygen combine to form water according to the following equation : 2H 2 + O 2 —→ 2H 2 O A mixture of 22.4 litres of hydrogen and 22.4 litres of oxygen is ignited at 100 °C. Calculate the (i) volume of steam produced (ii) what volume of oxygen, will be present on cooling to room temperature? [3]
Ans. 2H 2 + O 2 —→ 2H 2 O at 100 °C 2 vols 1 vol 2 vols By Gay-Lussac’s law : 1 vol 1 2
vol 1 vol
∴ 22.4 dm^3 11.2 dm^3 22.4 dm^3 (i) ∴ The volume of steam formed at 100 °C = 22.4 dm^3 (ii) The volume of oxygen left at 100 °C = 22.4 – 11.2 = 11.2 dm^3.
∴ The volume of oxygen left at room temperature will be less than 11.2 dm^3 as volume decreases with the fall in temperature.
8. What volume of propane is burnt for every 100 cm^3 of oxygen used in the reaction? C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O The volumes of gases are measured at room temperature and pressure. [3]
Ans. C 3 H 8 + 5O 2 —→ 3CO 2 + 4H 2 O By Gay-Lussac’s law : 1 vol 5 vols 3 vols 1 5
vol 1 vol 1 5
× 100 cm^3 100 cm 3 20 cm^3 100 cm 3 ∴ The volume of propane burnt = 20 cm^3. Numerical Problems on Mole Concept and Avogadro’s Law
9. (a) Explain what is meant by “molar volume of a gas.” [1] (b) Calculate the number of moles of nitrogen in 7 grams of nitrogen. [2] Ans. (a) The volume occupied by 1 mole (1 gram-molecule) of a gas at STP is called molar volume of the gas. Its experimental value is 22. dm^3 at STP. (b) 1 g-molecule of nitrogen = 2 × 14 = 28 g. Now, 28 g of nitrogen = 1 mole ∴ 7 g of nitrogen = 7 18
moles = 0.25 moles.
13. If 100 cm 3 of oxygen contains Y molecules, how many molecules of nitrogen will be present in 50 cm^3 of nitrogen under the same conditions of temperature and pressure. [2] Ans. 100 cm 3 of oxygen contains number of molecules = Y
∴ By Avogadro’s Law : 100 cm 3 of nitrogen contains number of molecules = Y ∴ 50 cm^3 of nitrogen contains number of molecules = 50Y^ =Y 5100 2
14. (a) “The number of atoms in one mole of hydrogen is twice the number of atoms in 1 mole of helium at the same temperature and pressure.” Why? [2] (b) Calculate the volume occupied by 8 g of sulphur dioxide at STP [S = 32; O = 16] [2] Ans. (a) It is because, hydrogen is a diatomic gas, whereas helium is monoatomic gas. As the number of atoms in one molecule of hydrogen are double, as compared to one molecule of helium, therefore, one mole of hydrogen has double the atoms, as compared to helium. (b) Molecular mass of SO 2 = 32 + 32 = 64 g. 64 g of SO 2 occupies = 22.4 dm^3 at STP ∴ 8 g of SO 2 occupies = 22.4 × 8 64 = 2.8 dm^3 at STP 15. Calculate the mass of 0.2 moles of water. [H = 1; O = 16] [2] Ans. 1 mole of water weighs = 2 + 16 = 18 g
∴ 0.2 moles of water weighs = 18 g × 0.2 = 3.6 g.
16. (a) What do you understand by the statement “vapour density of carbon dioxide gas is 22”? [2] (b) Atomic weight of chlorine is 35.5. What is its vapour density? [2] (c) A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure. (i) What weight of carbon dioxide it can hold under similar conditions of temperature and pressure? (ii) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. Give a reason for your answer. [3] Ans. (a) It means, the ratio between the weight of certain volume of carbon dioxide gas and same volume of hydrogen is 22, provided the carbon dioxide gas and hydrogen are at the same temperature and pressure. (b) Atomic weight of chlorine = 35.5. ∴ Molecular weight of chlorine = 2 × 35.5 = 71 Vapour density of chlorine = Molecular wt 2
=^71
(c) (i) Molecular weight of carbon dioxide = 44 ∴ V.D. of carbon dioxide =^44 2
Now, V.D. = Wt. of carbon dioxide at certain temp. and pressure Wt. of same volume of hydrogen at same temp. and pressure 22 = Wt. of carbon dioxide 1 kg ∴ Wt. of carbon dioxide = 22 kg.
19. Urea is very important fertiliser. The formula is (CON 2 H 4 ). Calculate the percentage of nitrogen in urea. [C = 12; N = 14; O = 16; H = 1] [2] Ans. Molecular weight of urea [CON 2 H 4 ] = [12 + 16 + 28 + 4) = 60 amu Also, weight of nitrogen in urea = 14 + 14 = 28 amu ∴ %age of nitrogen in urea = 28 60
× 100 = 46.67%.
20. Calculate the percentage of water of crystallisation in washing soda [Na 2 CO 3 · 10H 2 O] [Na = 23; C = 12; O = 16; H = 1] [2] Ans. Molecular weight of Na 2 CO 3 .10H 2 O= 46 + 12 + 48 + 180 = 286 amu. Molecular weight of 10 molecules of water= 10 × 18 = 180 amu ∴ %age of water of crystallisation =^182 286
× 100 = 62.93%.
21. Calculate the percentage of ion in K 3 Fe(CN) 6 [K=39, Fe = 56, C = 12, N = 14] Ans. Molecular weight of K 3 Fe(CN) 6 = 3[K] + 1[Fe] +6(C) + 6(N) = 3(39) + 1(56) + 6(12) + 6(14) = 117 + 56 + 72 + 84 = 329 ∴ %age of iron =^56 329
= × 100 = 17.02%
22. Calculate which of the following has higher percentage of nitrogen. Calcium nitrate or ammonium sulphate [Ca = 40, N = 14, O = 16, H = 1, S = 32] Ans. For calcium nitrate Molecular weight of Ca(NO 3 ) 2 = 1(Ca) + 2(N) + 6(O) = 1(40) + 2(14) + 6(16) = 40 + 28 + 96 = 164 ∴ %age of nitrogen in Ca(NO 3 ) 2 = 28 164
× 100 = 17.07%
For ammonium sulphate Molecular weight of (NH 4 ) 2 SO (^4) = 2(N) + 8(H) + 1(S) + 4(O) = 2(14) + 8(1) + 1(32 + 4(16) = 28 + 8 + 32 + 64 = 126 ∴ %age of nitrogen in (NH 4 ) 2 SO 4 = 28 126
× 100 = 21.22%
Thus, ammonium sulphate (21.22%) has higher percentage composition of nitrogen as compared to calcium nitrate (17.07%).
Numerical Problems on V.D., Molecular Formula and Empirical Formula
25. A hydrocarbon contains 82.8 % of carbon and has a relative molecular mass of 58. Write (i) its empirical formula (ii) its molecular formula (iii) its two possible structural formulae. [C = 12; H = 1] [5] Ans. Element %age wt. At. wt. Relative no. of moles Simple ratio of atoms C 82.8 (^12) 82.8 ÷ 12 = 6.9 6.9 ÷ 6.9 = 1 or 2 H 17.2 (^1) 17.2 ÷ 1 = 17.2 17.2 ÷ 6.9 = 2.5 or 5 Empirical formula of hydrocarbon = C 2 H 5 (ii) Empirical formula weight of hydrocarbon = 2 × 12 + 1 × 5 = 29 Molecular wt. of hydrocarbon = 58 Molecular weight = n × Empirical formula weight 58 = n × 29 ∴ n = 2 (iii) Molecular formula of hydrocarbon = 2 × Empirical formula = 2 × C 2 H 5 = C 4 H (^10) Possible structural formulae :
26. The following experiment was performed in order to determine the formula of hydrocarbon. The hydrocarbon X is purified by fractional distillation. 0.145 g of X was heated with dry copper (II) oxide and 224 cm 3 of carbon dioxide was collected at STP. (i) Which elements does X contain? [1] (ii) What was the purpose of copper (II) oxide? [1] (iii) Calculate the empirical formula of X by the following steps : 1. Calculate the number of moles of carbon dioxide gas. [1] 2. Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X. [1] 3. Calculate the mass of hydrogen in sample X. [1] 4. Deduce the ratio of atoms of each element in X (empirical formula). [2] Ans. (i) Sample X contains carbon and hydrogen only. (ii) Copper (II) oxide acts as oxidising agent. (iii) 1. 22400 cm 3 of CO 2 at STP = 1 mole ∴ 224 cm^3 of CO 2 at STP = 224 22400
= 0.01 mole.
- 22400 cm 3 of carbon dioxide contains carbon = 12 g ∴ 224 cm^3 of carbon dioxide contains carbon = 12 × 224 22400 = 0.12g ∴ Wt. of carbon in sample X = 0.12 g.
- Mass of hydrogen in sample X = Mass of hydrocarbon – Mass of carbon = (0.145 – 0.12) g = 0.025 g.
(iii) The compound is ethane. The structural formula of ethane is : H H | | H — C — C — H | | H H
28. A compound is formed by 24 g of X and 64 g of oxygen. If X = 12 and O = 16, calculate the simplest formula of compound. [3] Ans. Element %age wt.
At. wt. Relative no. of moles Simple ratio of atoms X 24 g (^12 24) ÷ 12 = 2 2 ÷ 2 = 1 O 64 g (^16 64) ÷ 16 = 4 4 ÷ 2 = 2 ∴ Simplest formula (Empirical formula) = XO 2.
29. (i) What do you understand by the empirical formula of compound? [2] (ii) If the molecular formula of a compound is C 2 H 6 , write down its empirical formula. [2] (iii) A compound of empirical formula CH 2 O has a vapour density of 30. Write down its molecular formula. [1] Ans. (i) Empirical formula : The formula of a chemical substance which tells the simplest whole number ratio of atoms of different elements present in its one molecule is called empirical formula.
(ii) Molecular formula = C 2 H 6 ∴ Empirical formula = 2 × CH 3 = C 2 H 6. (iii) Empirical formula of compound = CH 2 O ∴ Empirical formula weight of compound = 12 + 2 + 16 = 30 amu. Molecular weight of compound = 2 × V.D. = 2 × 30 = 60 amu. Molecular weight = n × Empirical formula weight 60 = n × 30 ∴ n = 2 ∴ Molecular formula of compound = n × Empirical formula = 2 × CH 2 O = C 2 H 4 O 2.
30. Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of compound. [2] Ans. Element %age wt. At. wt. Relative no. of moles Simple ratio of atoms Si 5.6 (^28) 5·6 ÷ 28 = 0.2 0.2 ÷ 0.2 = 1 Cl 21.3 35.5 (^) 21.3 ÷ 35.5 = 0.6 0.6 ÷ 0.2 = 3 ∴ Empirical formula : SiCl 3. 31. A compound of carbon, hydrogen and oxygen is found to contain 40% of carbon, 6.67% of hydrogen and 53.3% of oxygen. Calculate its empirical formula. Its vapour density is 30, find its molecular formula. [3]
∴ Empirical formula = CHO (^2) ∴ Empirical formula weight = 12 + 1 + 32 = 45. Molecular weight = n × Empirical formula weight 90 = n × 45 ∴ n = 2 ∴ Molecular formula = n × Empirical formula = 2 × CHO 2 = C 2 H 2 O 4.
33. A compound of sodium, sulphur and oxygen has the following percentage composition : Na = 29.11%; S = 40.51%; O = 30.38%. Find its empirical formula. [O = 16; Na = 23; S = 32] Ans. Element %age wt. At. wt. Relative no. of moles Simple ratio of atoms Na 29.11 (^23) 29.11 ÷ 23 = 1.26 1.26 ÷ 1.26 = 1 or 2 S 40.51 (^32) 40.51 ÷ 32 = 1.26 2.26 ÷ 1.26 = 1 or 2 O 30.38 (^16) 30.38 ÷ 16 = 1.89 1.89 ÷ 1.26 = 1.5 or 3 ∴ Empirical formula is Na 2 S 2 O 3. 34. An acid of phosphorus has the following percentage composition; phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of acid and its molecular formula, given that its relative molecular weight is 162. [P = 31; H = 1; O = 16] Ans. Element %age wt. At. wt. Relative no. of moles Simple ratio of atoms P 38.27 (^31) 38.27 ÷ 31 = 1.23 1.23 ÷ 1.23 = 1 H 2.47 (^1) 2.47 ÷ 1 = 2.47 2.47 ÷ 1.23 = 2 O 59.26 (^16) 59.26 ÷ 16 = 3.70 3.70 ÷ 1.23 = 3
∴ Empirical formula of compound = H 2 PO 3 ∴ Empirical formula weight = 2 + 31 + 48 = 81 Molecular weight = 162 Molecular weight = n × Empirical formula weight 162 = n × 81 ∴ n = 2 Molecular formula = n × Empirical formula = 2 × H 2 PO (^3) = H 4 P 2 O 6.
35. An inorganic compound contains 42% of nitrogen, 9% of hydrogen and 48% of oxygen. Calculate its empirical formula. [N = 14; H = 1; O = 16] [2] Ans. Element %age wt. At. wt. Relative no. of moles Simple ratio of atoms Nitrogen (N) 42% (^14 42) ÷ 14 = 3 (^) 3 = 1 3 Hydrogen (H) 9% (^1 9) ÷ 1 = 9 (^) 9 = 3 3 Oxygen (O) 48% (^16 48) ÷ 16 = 3 (^) 3 = 1 3 ∴ Empirical formula is NH 3 O. Numerical Problems on Reacting Weights and Volumes
