Q: Question 1From the equation for burning of hydrogen and oxygen2H2 + O2 ⟶ 2H2O (Steam)Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.

2H2+O2⟶2H2O2 mole1 mole2 mole\begin{matrix} 2\text{H}_2 & + &\text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} \\ 2\text{ mole} & & 1\text{ mole}& & 2\text{ mole} \end{matrix}2H22 mole+O21 mole⟶2H2O2 mole1 mole of oxygen gives 2 moles of steam∴ 0.5 mole of oxygen will give 21\dfrac{2}{1}12 × 0.5= 1 mole of steam

Q: Question 2From the equation3Cu + 8HNO3 ⟶ 3Cu(NO3)2 + 4H2O + 2NO(At. mass Cu=64, H=1, N=14, O=16)Calculate:(a) Mass of copper needed to react with 63 g of HNO3(b) Volume of nitric oxide at S.T.P. that can be collected.

3Cu+8HNO3⟶3Cu(NO3)2+4H2O+2NO3(64)8(1+14+3(16))=192g=8(63)=504g\begin{matrix} 3\text{Cu} & + &8\text{HNO}_3 & \longrightarrow & 3\text{Cu(NO}_3)_2 + 4\text{H}_2\text{O} + 2\text{NO} \\ 3(64) && 8(1 + 14 + 3(16)) \\ = 192 \text{g} & & = 8(63) \\ &&= 504 \text{g} \\ \end{matrix}3Cu3(64)=192g+8HNO38(1+14+3(16))=8(63)=504g⟶3Cu(NO3)2+4H2O+2NO(a) 504 g nitric acid reacts with 192 g of copper∴ 63 g of nitric acid reacts with 192504\dfrac{192}{504}504192 x 63 = 24 g of copperHence, 24 g of copper is required.(b) 504 g of nitric acid gives 2 × 22.4 litre volume of NO∴ 63 g of nitric acid gives 2×22.4504\dfrac{2 × 22.4}{504}5042×22.4 x 63= 5.6 litre of NO5.6 L of NO is collected.

Q: Question 4Some of the fertilizers are sodium nitrate NaNO3, ammonium sulphate (NH4)2SO4 and urea CO(NH2)2. Which of these contains the highest percentage of nitrogen?

(i) Molar mass of NaNO3 = 23 + 14 + 3(16) = 23 + 14 + 48 = 85 gNitrogen in 85 g NaNO3 = 14 g∴ Percentage of Nitrogen in NaNO3 = 1485\dfrac{14}{85}8514 x 100 = 16.47%(ii) Molar mass of (NH4)2SO4 = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132 gNitrogen in 132 g of (NH4)2SO4 = 28 g∴ Percentage of Nitrogen in (NH4)2SO4 = 28132\dfrac{28}{132}13228 x 100 = 21.21%(iii) Molar mass of CO(NH2)2 = 12 + 16 + 2[14 + (2 x 1)]= 28 + 2(16)= 28 + 32= 60 gNitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g∴ Percentage of Nitrogen in CO(NH2)2 = 2860\dfrac{28}{60}6028 x 100 = 46.66%So, the highest percentage of Nitrogen is in Urea.

Q: Question 5200 cm3 of CO2 is collected at STP when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in the original mixture.2C2H2 (g) + 5O2 (g) ⟶ 4CO2 (g)+ 2H2O (g)

According to Gay lussac's law,2C2H2+5O2⟶4CO2+2H2O2 vol.:5 vol.⟶4 vol.\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}2C2H22 vol.+:5O25 vol.⟶⟶4CO24 vol.+2H2O4 Vol of CO2 is collected with 2 Vol. of C2H2So, 200 cm3 CO2 will be collected with=24×200=100 cc=\dfrac{2}{4} \times 200 \\[0.5em] = 100 \text{ cc}=42×200=100 ccSimilarly, 4 Vol of CO2 is produced by 5 Vol of O2So, 200 cm3 CO2 will be produced by = 54\dfrac{5}{4}45 x 200 = 250 cm3Hence, acetylene = 100 cm3 and oxygen = 250 cm3

Q: Question 6Urea (CO(NH2)2) is an important nitrogenous fertilizer, and is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?

Molar mass of urea [CO(NH2)2] = 12 + 16 + 2[14 + (2 x 1)]= 28 + 2(16)= 28 + 32= 60 g50 kg of urea = 50,000 g of ureaNitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g∴ Nitrogen in 50,000 g urea = 28×50,00060\dfrac{28 \times 50,000 }{60}6028×50,000= 23,333 g = 23.3 kg

Q: Question 7Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon80128012\dfrac{80}{12}1280 = 6.666.666.66\dfrac{6.66}{6.66}6.666.66 = 1Hydrogen201201\dfrac{20}{1}120 = 20206.66\dfrac{20}{6.66}6.6620 = 3Simplest ratio of whole numbers = C : H = 1 : 3Hence, empirical formula is CH3Empirical formula weight = 12 + 3(1) = 15V.D. = 15Molecular weight = 2 x V.D. = 2 x 15n=Molecular weightEmpirical formula weight=2×1515=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 15}{15} = 2n=Empirical formula weightMolecular weight=152×15=2∴ Molecular formula = n[E.F.] = 2[CH3] = C2H6

Q: Question 8The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.0.145 g of X was heated with dry copper (II) oxide and 224 cm3 of carbon dioxide was collected at S.T.P.(a) Which elements does X contain?(b) What was the purpose of copper (II) oxide?(c ) Calculate the empirical formula of X by the following steps:(i) Calculate the number of moles of carbon dioxide gas.(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.(iii) Calculate the mass of hydrogen in sample X.(iv) Deduce the ratio of atoms of each element in X (empirical formula).

(a) X contains carbon and hydrogen as it is a hydrocarbon.(b) The purpose of copper (II) oxide is to act as an oxidizing agent.(c) (i) 22400 cm3 CO2 has mass = 44 g∴ 224 cm3 CO2 will have mass = 4422400\dfrac{44}{22400}2240044 x 224 = 0.44 gMolar mass of CO2 = 12 + 2(16) = 12 + 32 = 44 gMoles of CO2 = MassMolecular mass\dfrac{\text{Mass}}{\text{{Molecular mass}}}Molecular massMass = 0.4444\dfrac{0.44}{44}440.44 = 0.01 moles(ii) Mass of carbon in 44 g CO2 = 12 g∴ Mass of carbon in 0.44 g CO2 = 1244\dfrac{12}{44}4412 x 0.44 = 0.12 gAs X contains carbon and hydrogen, so sample X has 0.12 g of carbon(iii) Mass of Hydrogen in X = 0.145 - 0.12 = 0.025 g(iv) Ratio of moles of C : H= 0.010.01\dfrac{0.01}{0.01}0.010.01 : 0.0250.01\dfrac{0.025}{0.01}0.010.025= 1 : 2510\dfrac{25}{10}1025= 1 : 52\dfrac{5}{2}25= 2 : 5Hence, ratio of C : H = 2 : 5so, the empirical formula of hydrocarbon is C2H5

Q: Question 9Compound is formed by 24 g of X and 64 g of oxygen. If atomic mass of X = 12 and O = 16, calculate the simplest formula of compound.

ElementMass (g)Atomic massMolesSimplest ratioElement X24122412\dfrac{24}{12}1224 = 222\dfrac{2}{2}22 = 1Oxygen64166416\dfrac{64}{16}1664 = 442\dfrac{4}{2}24 = 2Simplest ratio of whole numbers = X : O = 1 : 2Hence, simplest formula of compound is XO2

Q: Question 10A gas cylinder filled with hydrogen holds 5 g of the gas. The same cylinder holds 85 g of gas X under the same temperature and pressure. Calculate:(a) Vapour density of gas X.(b) Molecular weight of gas X.

(a ) V.D. = mass of certain volume of gas at S.T.P.mass of equal volume of H2 at S.T.P.\dfrac{\text{mass of certain volume of gas at S.T.P.}}{\text{mass of equal volume of H}_2\text{ at S.T.P.}}mass of equal volume of H2 at S.T.P.mass of certain volume of gas at S.T.P. = 855\dfrac{85}{5}585 = 17(b) Molecular weight = 2 x V.D. = 2 x 17 = 34 a.m.u.

Question 1

Question 3The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at S.T.P., which gas will contain the least number of molecules and which gas the most?

Accepted Answer

According to Avogadro's law:Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal numbers of molecules.So, 1 mole of each gas contains = 6.02 × 1023 moleculesMol. Mass of :H2 = 2 g,O2 = 32 g,CO2 = 12 + 2(16) = 44 g,SO2 = 32 + 2(16) = 64 g,Cl2 = 2(35.5) = 71 g(i) 2 g of hydrogen contains molecules = 6.02 × 1023So, 8 g of hydrogen contains molecules= 6.02×10232\dfrac{6.02 × 10^{23}}{2}26.02×1023​ x 8= 4 × 6.02 × 1023molecules= 24.08 × 1023molecules(ii) 32 g of oxygen contains molecules = 6.02 × 1023So, 8 g of oxygen contains molecules= 6.02×102332\dfrac{6.02 × 10^{23}}{32}326.02×1023​ x 8= 1.505 × 1023molecules(iii) 44 g of carbon dioxide contains molecules = 6.02 × 1023So, 8 g of carbon dioxide contains= 6.02×102344\dfrac{6.02 × 10^{23}}{44}446.02×1023​ x 8= 1.09 × 1023molecules(iv) 64 g of sulphur dioxide contains molecules = 6.02 × 1023So, 8g of sulphur dioxide contains= 6.02×102364\dfrac{6.02 × 10^{23}}{64}646.02×1023​ x 8= 0.75 × 1023molecules(v) 71 g of chlorine contains molecules = 6.02 × 1023So, 8g of chlorine contains= 6.02×102371\dfrac{6.02 × 10^{23}}{71}716.02×1023​ x 8= 0.67 × 1023moleculesThus Cl2 < SO2 < CO2 < O2 < H2(i) Least number of molecules in Cl2(ii) Most number of molecules in H2

Question 2

Question 1From the equation for burning of hydrogen and oxygen2H2 + O2 ⟶ 2H2O (Steam)Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.

Accepted Answer

2H2+O2⟶2H2O2 mole1 mole2 mole\begin{matrix} 2	ext{H}_2 & + &	ext{O}_2 & \longrightarrow & 2	ext{H}_2	ext{O} \ 2	ext{ mole} & & 1	ext{ mole}& & 2	ext{ mole} \end{matrix}2H2​2 mole​+​O2​1 mole​⟶​2H2​O2 mole​1 mole of oxygen gives 2 moles of steam∴ 0.5 mole of oxygen will give 21\dfrac{2}{1}12​ × 0.5= 1 mole of steam

Question 3

Question 2From the equation3Cu + 8HNO3 ⟶ 3Cu(NO3)2 + 4H2O + 2NO(At. mass Cu=64, H=1, N=14, O=16)Calculate:(a) Mass of copper needed to react with 63 g of HNO3(b) Volume of nitric oxide at S.T.P. that can be collected.

Accepted Answer

3Cu+8HNO3⟶3Cu(NO3)2+4H2O+2NO3(64)8(1+14+3(16))=192g=8(63)=504g\begin{matrix} 3	ext{Cu} & + &8	ext{HNO}_3 & \longrightarrow & 3	ext{Cu(NO}_3)_2 + 4	ext{H}_2	ext{O} + 2	ext{NO} \ 3(64) && 8(1 + 14 + 3(16)) \ = 192 	ext{g} & & = 8(63) \ &&= 504 	ext{g} \ \end{matrix}3Cu3(64)=192g​+​8HNO3​8(1+14+3(16))=8(63)=504g​⟶3Cu(NO3​)2​+4H2​O+2NO(a) 504 g nitric acid reacts with 192 g of copper∴ 63 g of nitric acid reacts with 192504\dfrac{192}{504}504192​ x 63 = 24 g of copperHence, 24 g of copper is required.(b) 504 g of nitric acid gives 2 × 22.4 litre volume of NO∴ 63 g of nitric acid gives 2×22.4504\dfrac{2 × 22.4}{504}5042×22.4​ x 63= 5.6 litre of NO5.6 L of NO is collected.

Question 4

Question 4Some of the fertilizers are sodium nitrate NaNO3, ammonium sulphate (NH4)2SO4 and urea CO(NH2)2. Which of these contains the highest percentage of nitrogen?

Accepted Answer

(i) Molar mass of NaNO3 = 23 + 14 + 3(16) = 23 + 14 + 48 = 85 gNitrogen in 85 g NaNO3 = 14 g∴ Percentage of Nitrogen in NaNO3 = 1485\dfrac{14}{85}8514​ x 100 = 16.47%(ii) Molar mass of (NH4)2SO4 = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132 gNitrogen in 132 g of (NH4)2SO4 = 28 g∴ Percentage of Nitrogen in (NH4)2SO4 = 28132\dfrac{28}{132}13228​ x 100 = 21.21%(iii) Molar mass of CO(NH2)2 = 12 + 16 + 2[14 + (2 x 1)]= 28 + 2(16)= 28 + 32= 60 gNitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g∴ Percentage of Nitrogen in CO(NH2)2 = 2860\dfrac{28}{60}6028​ x 100 = 46.66%So, the highest percentage of Nitrogen is in Urea.

Question 5

Question 5200 cm3 of CO2 is collected at STP when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in the original mixture.2C2H2 (g) + 5O2 (g) ⟶ 4CO2 (g)+ 2H2O (g)

Accepted Answer

According to Gay lussac's law,2C2H2+5O2⟶4CO2+2H2O2 vol.:5 vol.⟶4 vol.\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \ \end{matrix}2C2​H2​2 vol.​+:​5O2​5 vol.​⟶⟶​4CO2​4 vol.​+2H2​O4 Vol of CO2 is collected with 2 Vol. of C2H2So, 200 cm3 CO2 will be collected with=24×200=100 cc=\dfrac{2}{4} \times 200 \[0.5em] = 100 \text{ cc}=42​×200=100 ccSimilarly, 4 Vol of CO2 is produced by 5 Vol of O2So, 200 cm3 CO2 will be produced by = 54\dfrac{5}{4}45​ x 200 = 250 cm3Hence, acetylene = 100 cm3 and oxygen = 250 cm3

Question 6

Question 6Urea (CO(NH2)2) is an important nitrogenous fertilizer, and is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?

Accepted Answer

Molar mass of urea [CO(NH2)2] = 12 + 16 + 2[14 + (2 x 1)]= 28 + 2(16)= 28 + 32= 60 g50 kg of urea = 50,000 g of ureaNitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g∴ Nitrogen in 50,000 g urea = 28×50,00060\dfrac{28 	imes 50,000 }{60}6028×50,000​= 23,333 g = 23.3 kg

Question 7

Question 7Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon80128012\dfrac{80}{12}1280​ = 6.666.666.66\dfrac{6.66}{6.66}6.666.66​ = 1Hydrogen201201\dfrac{20}{1}120​ = 20206.66\dfrac{20}{6.66}6.6620​ = 3Simplest ratio of whole numbers = C : H = 1 : 3Hence, empirical formula is CH3Empirical formula weight = 12 + 3(1) = 15V.D. = 15Molecular weight = 2 x V.D. = 2 x 15n=Molecular weightEmpirical formula weight=2×1515=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{2 \times 15}{15} = 2n=Empirical formula weightMolecular weight​=152×15​=2∴ Molecular formula = n[E.F.] = 2[CH3] = C2H6

Question 8

Question 8The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.0.145 g of X was heated with dry copper (II) oxide and 224 cm3 of carbon dioxide was collected at S.T.P.(a) Which elements does X contain?(b) What was the purpose of copper (II) oxide?(c ) Calculate the empirical formula of X by the following steps:(i) Calculate the number of moles of carbon dioxide gas.(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.(iii) Calculate the mass of hydrogen in sample X.(iv) Deduce the ratio of atoms of each element in X (empirical formula).

Accepted Answer

(a) X contains carbon and hydrogen as it is a hydrocarbon.(b) The purpose of copper (II) oxide is to act as an oxidizing agent.(c) (i) 22400 cm3 CO2 has mass = 44 g∴ 224 cm3 CO2 will have mass = 4422400\dfrac{44}{22400}2240044​ x 224 = 0.44 gMolar mass of CO2 = 12 + 2(16) = 12 + 32 = 44 gMoles of CO2 = MassMolecular mass\dfrac{	ext{Mass}}{	ext{{Molecular mass}}}Molecular massMass​ = 0.4444\dfrac{0.44}{44}440.44​ = 0.01 moles(ii) Mass of carbon in 44 g CO2 = 12 g∴ Mass of carbon in 0.44 g CO2 = 1244\dfrac{12}{44}4412​ x 0.44 = 0.12 gAs X contains carbon and hydrogen, so sample X has 0.12 g of carbon(iii) Mass of Hydrogen in X = 0.145 - 0.12 = 0.025 g(iv) Ratio of moles of C : H= 0.010.01\dfrac{0.01}{0.01}0.010.01​ : 0.0250.01\dfrac{0.025}{0.01}0.010.025​= 1 : 2510\dfrac{25}{10}1025​= 1 : 52\dfrac{5}{2}25​= 2 : 5Hence, ratio of C : H = 2 : 5so, the empirical formula of hydrocarbon is C2H5

Question 9

Question 9Compound is formed by 24 g of X and 64 g of oxygen. If atomic mass of X = 12 and O = 16, calculate the simplest formula of compound.

Accepted Answer

ElementMass (g)Atomic massMolesSimplest ratioElement X24122412\dfrac{24}{12}1224​ = 222\dfrac{2}{2}22​ = 1Oxygen64166416\dfrac{64}{16}1664​ = 442\dfrac{4}{2}24​ = 2Simplest ratio of whole numbers = X : O = 1 : 2Hence, simplest formula of compound is XO2

Question 10

Question 10A gas cylinder filled with hydrogen holds 5 g of the gas. The same cylinder holds 85 g of gas X under the same temperature and pressure. Calculate:(a) Vapour density of gas X.(b) Molecular weight of gas X.

Accepted Answer

(a ) V.D. = mass of certain volume of gas at S.T.P.mass of equal volume of H2 at S.T.P.\dfrac{	ext{mass of certain volume of gas at S.T.P.}}{	ext{mass of equal volume of H}_2	ext{ at S.T.P.}}mass of equal volume of H2​ at S.T.P.mass of certain volume of gas at S.T.P.​ = 855\dfrac{85}{5}585​ = 17(b) Molecular weight = 2 x V.D. = 2 x 17 = 34 a.m.u.

Question 11

Question 11(a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :CO2 + C ⟶ 2COWhat volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?(b) 60 cm3 of oxygen was added to 24 cm3 of carbon monoxide and mixture ignited. Calculate:(i) volume of oxygen used up and(ii) Volume of carbon dioxide formed.

Accepted Answer

CO2+C⟶2CO1 vol.:1 vol.⟶2 vol.\begin{matrix} 	ext{CO}_2 & + &	ext{C} & \longrightarrow & 2	ext{CO} \ 1	ext{ vol.} & : & 1 	ext{ vol.} & \longrightarrow & 2 	ext{ vol.} \ \end{matrix}CO2​1 vol.​+:​C1 vol.​⟶⟶​2CO2 vol.​[By Gay Lussac's law]1 Vol. of C produces 2V of CO12 g of C produces 2 x 22.4 L of CO∴ 3 g of C will produce 2×22.412\dfrac{2 	imes 22.4}{12}122×22.4​ x 3= 11.2 L of CO(b)O2+2CO⟶2CO21 vol.:2 vol.⟶2 vol.\begin{matrix} 	ext{O}_2 & + &	ext{2CO} & \longrightarrow & 2	ext{CO}_2 \ 1	ext{ vol.} & : & 2 	ext{ vol.} & \longrightarrow & 2 	ext{ vol.} \ \end{matrix}O2​1 vol.​+:​2CO2 vol.​⟶⟶​2CO2​2 vol.​(i) 2 Vol. of CO requires 1 Vol. of oxygen∴ 24 cm3 CO will require 12\dfrac{1}{2}21​ x 24= 12 cm3 of oxygen(ii) 2 Vol. of CO gives 2 vol. of CO2∴ 24 cm3 of CO will give 24 cm3 of CO2

Question 12

Question 12How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO2 evolved:2Ca(NO3)2 ⟶ 2CaO + 4NO2 + O2

Accepted Answer

2Ca(NO3)2⟶2CaO+ 4NO2 +O22[40+2[14+3(16)]]2[40+16]=2[40+28+96]=2(56)=328 g=112 g\begin{matrix} 2	ext{Ca(NO}_3)_2 & \longrightarrow & 2	ext{CaO} & + &\space 4	ext{NO}_2 \space + & 	ext{O}_2 \ 2[40 + 2[14 + 3(16)]] & & 2[40 + 16] \ = 2[40 + 28 + 96] & & = 2(56) \ = 328 	ext{ g} & & = 112 	ext{ g} \end{matrix}2Ca(NO3​)2​2[40+2[14+3(16)]]=2[40+28+96]=328 g​⟶​2CaO2[40+16]=2(56)=112 g​+ 4NO2​ +O2​328 g of calcium nitrate produces 112 g of CaO∴ 82 g of calcium nitrate will produce 112328\dfrac{112}{328}328112​ x 82= 28 g of CaO328 g of calcium nitrate produces 4 x 22.4 L of NO2∴ 82 g of calcium nitrate will produce 4×22.4328\dfrac{4 	imes 22.4}{328}3284×22.4​ x 82= 22.4 L of NO2

Question 13

Question 13The equation for the burning of octane is:2C8H18 + 25O2 ⟶ 16CO2 + 18H2O(i) How many moles of carbon dioxide are produced when one mole of octane burns?(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane?(iv) What is the empirical formula of octane?

Accepted Answer

2C8H18+25O2⟶16CO2+18H2O2 Vol.25 Vol.16 Vol.18 Vol.16[12+2(16)]16[12+32]16[44]=704g\begin{matrix} 2	ext{C}_8	ext{H}_{18} & + &25	ext{O}_2 & \longrightarrow & 16	ext{CO}_2 & + & 18	ext{H}_2	ext{O} \ 2	ext{ Vol.}&& 25	ext{ Vol.} && 16	ext{ Vol.}&& 18	ext{ Vol.} \ &&&&16[12 + 2(16)] \ &&&&16[12 + 32] \ &&&&16[44] = 704 	ext{g} \ \end{matrix}2C8​H18​2 Vol.​+​25O2​25 Vol.​⟶​16CO2​16 Vol.16[12+2(16)]16[12+32]16[44]=704g​+18H2​O18 Vol.(i) 2 moles of octane produces 16 moles of CO2∴ 1 mole octane produces 162\dfrac{16}{2}216​ x 1= 8 moles of CO2(ii) 1 mole CO2 occupies volume = 22.4 LAs 2 moles of octane produces 8 moles of carbon dioxide which will occupy volume 22.41\dfrac{22.4}{1}122.4​ x 8= 179.2 dm3 of CO2(iii) 1 mole CO2 has mass = 44 g∴ 16 moles will have mass 441\dfrac{44}{1}144​ x 16= 704 g of CO2(iv) Molecular formula is C8H18∴ Ratio of C and H is 8 : 18Simple ratio is 4 : 9Hence, empirical formula = C4H9

Question 14

Question 14Ordinary chlorine gas has two isotopes 3517Cl and 3717Cl in the ratio of 3:1. Calculate the relative atomic mass of chlorine.

Accepted Answer

The relative atomic mass of Cl = (35×3)+(1×37)4\dfrac{(35 × 3) + (1 × 37)}{4}4(35×3)+(1×37)​ = 35.5

Question 15

Question 15Silicon (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 the compound.

Accepted Answer

ElementMassAt. wt.gram atomsSimplest ratioSilicon5.6285.628\dfrac{5.6 }{28}285.6​ = 0.20.20.2\dfrac{0.2 }{0.2}0.20.2​ = 1Chlorine21.335.521.335.5\dfrac{ 21.3}{35.5}35.521.3​ = 0.60.60.2\dfrac{ 0.6 }{0.2}0.20.6​ = 3Simplest ratio of whole numbers = Si : Cl = 1 : 3Hence, empirical formula is SiCl3

Question 16

Question 16An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of the acid and it's molecular formula, given that it's relative molecular mass is 162.

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioPhosphorus38.273138.2731\dfrac{38.27 }{31}3138.27​ = 1.2341.2341.234\dfrac{ 1.234}{1.234 }1.2341.234​ = 1Hydrogen2.4712.471\dfrac{2.47 }{1}12.47​ = 2.472.471.234\dfrac{2.47 }{1.234 }1.2342.47​ = 2Oxygen59.261659.2616\dfrac{59.26}{16}1659.26​ = 3.703.701.234\dfrac{3.70}{ 1.234 }1.2343.70​ = 3Simplest ratio of whole numbers = P : H : O = 1 : 2 : 3Hence, empirical formula is H2PO3Empirical formula weight = 31 + 2(1) + 3(16) = 31 + 2 + 48 = 81Molecular weight = 162n=Molecular weightEmpirical formula weight=16281=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{162}{81} = 2n=Empirical formula weightMolecular weight​=81162​=2So, molecular formula = 2(H2PO3) = H4P2O6

Question 17

Question 17(a) Calculate the mass of substance 'A' which in gaseous form occupies 10 litres at 27°C and 700 mm pressure. The molecular mass of 'A' is 60.(b) A gas occupied 360 cm3 at 87°C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find it's relative molecular mass.

Accepted Answer

(a) Given, molecular mass of 'A' is 60V1 = 10 LT1 = 27 + 273 K = 300 KP1 = 700 mmT2 = 273 KP2 = 760 mmV2 = ?Using formula:P1V1T1\dfrac{	ext{P}_1	ext{V}_1}{	ext{T}_1}T1​P1​V1​​ = P2V2T2\dfrac{	ext{P}_2	ext{V}_2}{	ext{T}_2}T2​P2​V2​​Substituting in the formula,700×10300\dfrac{700 	imes 10}{300}300700×10​ = 760×V2273\dfrac{760 	imes	ext{V}_2}{273}273760×V2​​V2 = 700×10×273300×760\dfrac{700 	imes10 	imes 273}{300 	imes760}300×760700×10×273​ = 1911228\dfrac{1911}{228}2281911​ = 8.38 LAs, 22.4 L of A weighs 60 g∴ 8.38 L of A will weigh 6022.4\dfrac{60}{22.4}22.460​ x 8.38= 22.446 = 22.45 g(b) V1 = 360 cm3T1 = 87 + 273 K = 360 KP1 = 380 mm Hg pressureT2 = 273 KP2 = 760 mm Hg pressureV2 = ?Using formula:P1V1T1\dfrac{	ext{P}_1	ext{V}_1}{	ext{T}_1}T1​P1​V1​​ = P2V2T2\dfrac{	ext{P}_2	ext{V}_2}{	ext{T}_2}T2​P2​V2​​Substituting in the formula,380×360360\dfrac{380	imes 360}{360}360380×360​ = 760×V2273\dfrac{760 	imes	ext{V}_2}{273}273760×V2​​V2 = 380×273760\dfrac{380 	imes 273}{760}760380×273​ = 10,374760\dfrac{10,374}{760}76010,374​ = 136.5 cm3136.5 cm3 of gas weigh = 0.546 g22400 cm3 of gas weigh 0.546136.5\dfrac{0.546}{136.5}136.50.546​ x 22400= 89.6 amu

Question 18

Question 18A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?(b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result.

Accepted Answer

(a) Given, cylinder can hold = 1 kg of hydrogenMolar mass of hydrogen = 2[1] = 2gNumber of moles of hydrogen present in cylinder = 10002\dfrac{1000}{2}21000​ = 500According to avogadro's law : cylinder will hold 500 moles of carbon dioxide gasMolar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g1 mole of carbon dioxide = 44 g∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.Hence, Weight of carbon dioxide in cylinder = 22 kg(b) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.∴ molecules in the cylinder of carbon dioxide = XAvogadro's law helped to arrive at this result.

Question 19

Question 19Following questions refer to one mole of chlorine gas.(a) What is the volume occupied by this gas at S.T.P.?(b) What will happen to the volume of gas, if pressure is doubled?(c) What volume will it occupy at 273°C?(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?

Accepted Answer

(a) According to Avogadro's law : The volume occupied by 1 mole of chlorine is 22.4 dm3(b) According to Boyle's Law: PV = constantHence, if pressure is doubled then volume will become half i.e. 22.42\dfrac{22.4}{2}222.4​ = 11.2 dm3(c) V1 = 22.4 dm3T1 = 273 KT2 = 273 + 273 K = 546 KV2 = ?According to charles law:V1T1\dfrac{	ext{V}_1}{	ext{T}_1}T1​V1​​ = V2T2\dfrac{	ext{V}_2}{	ext{T}_2}T2​V2​​Substituting to we get,22.4273\dfrac{ 22.4}{273}27322.4​ = V2546\dfrac{	ext{V}_2}{546}546V2​​Hence, V2 = 22.4273\dfrac{ 22.4}{273}27322.4​ x 546 = 44.8 dm3(c) Mass of 1 mole Cl2 gas = 35.5 × 2 = 71 g

Question 20

Question 20(a) A hydrate of calcium sulphate CaSO4.xH2O contains 21% water of crystallisation. Find the value of x.(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.(c) How much volume will be occupied by 2g of dry oxygen at 27°C and 740 mm pressure?(d) What would be the mass of CO2 occupying a volume of 44 litres at 25°C and 750 mm pressure?(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.AgNO3 (aq) + NaCl (aq) ⟶ AgCl (s) + NaNO3Calculate the percentage of NaCl in the mixture.

Accepted Answer

Relative molecular mass of CuSO4.xH2O= 40 + 32 + (4×16) + [x(2+16)]= 40 + 32 + 64 + 18x= 136 + 18x∴ 21% water of crystallization = 18 x⇒18x136+18x=211001800x=21(136+18x)1800x=2856+378x1800x−378x=28561422x=2856x=28561422x=2\Rightarrow \dfrac{18x}{136 + 18x } = \dfrac{21}{100} \[1em] 1800x = 21(136 + 18x) \[1em] 1800x = 2856 + 378x \[1em] 1800x - 378x = 2856 \[1em] 1422x = 2856 \[1em] x = \dfrac{2856}{1422} \[1em] x = 2⇒136+18x18x​=10021​1800x=21(136+18x)1800x=2856+378x1800x−378x=28561422x=2856x=14222856​x=2Hence, water of crystallization = 2(b) Molar mass of H2O = 2(1) + 16 = 18 gFor 18 g water, vol. of hydrogen needed = 22.4 litre∴ For 1.8 g, vol. of H2 needed = 22.418\dfrac{22.4}{18}1822.4​ x 1.8 = 2.24 LAccording to Gay Lussac's Law, 2 vol of hydrogen requires 1 vol of oxygenWhen 2 vol of hydrogen in 1.8 g H2O is 2.24 L, then one vol. of oxygen will be:2.242\dfrac{2.24}{2}22.24​ = 1.12 L(c) Gram molecular mass of O2 = 2 x 16 = 32 g1 mole of O2 weighs 32 g and occupies 22.4 lit. vol.∴ 2 g of O2 occupies = 22.432×2=1.4 lit.\dfrac{22.4}{32} \times 2 = 1.4 \text{ lit.}3222.4​×2=1.4 lit.Volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure:S.T.P.Given ValuesP1 = 760 mm of HgP2 = 740 mm of HgV1 = 1.4 litV2 = x litT1 = 273 KT2 = 27 + 273 KUsing the gas equation,P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}T1​P1​V1​​=T2​P2​V2​​Substituting the values we get,760×1.4273=740×x300x=760×1.4×300740×273x=3,19,2002,02,020x=1.58 lit\dfrac{760 \times 1.4}{273} = \dfrac{740 \times x}{300} \[0.5em] x = \dfrac{760 \times 1.4 \times 300}{740 \times 273 } \[0.5em] x = \dfrac{3,19,200}{2,02,020} \ \[0.5em] x = 1.58 \text{ lit}273760×1.4​=300740×x​x=740×273760×1.4×300​x=2,02,0203,19,200​x=1.58 litHence, the volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure is 1.58 lit.(d) Gram molecular mass of CO2 = 12 + 2(16) = 12 + 32 = 44 gGiven ValuesS.T.P.P1 = 750 mm of HgP2 = 760 mm of HgV1 = 44 litV2 = x litT1 = 25 + 273 K = 298 KT2 = 273 KUsing the gas equation,P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}T1​P1​V1​​=T2​P2​V2​​Substituting the values we get,750×44298=760×x273x=750×44×273760×298x=9,009,000226,480x=39.78 lit\dfrac{750 \times 44}{298} = \dfrac{760 \times x}{273} \[0.5em] x = \dfrac{750 \times 44 \times 273}{760 \times 298 } \[0.5em] x = \dfrac{9,009,000}{226,480} \ \[0.5em] x = 39.78 \text{ lit}298750×44​=273760×x​x=760×298750×44×273​x=226,4809,009,000​x=39.78 lit22.4 litre of CO2 at S.T.P. has mass = 44 g39.78 litre of CO2 at S.T.P. has mass 4422.4\dfrac{44}{22.4}22.444​ x 39.78= 78.14 g(e)AgNO3+NaCl⟶AgCl+NaNO323+35.5108+35.5=58.5g=143.5g\begin{matrix} \text{AgNO}_3 & + &\text{NaCl} & \longrightarrow & \text{AgCl} & + & \text{NaNO}_3 \ && 23 + 35.5 && 108 + 35.5 \ && = 58.5 \text{g}&& = 143.5\text{g} \ \end{matrix}AgNO3​​+​NaCl23+35.5=58.5g​⟶​AgCl108+35.5=143.5g​+NaNO3​(i) 143.5 g AgCl is formed by 58.5 g NaCl∴ 1.435 g of AgCl will be formed by 58.5143.5\dfrac{58.5}{143.5}143.558.5​ x 1.435 = 0.582 gPercentage of NaCl = 0.5821\dfrac{0.582}{1}10.582​ x 100 = 58.5%Hence, percentage of NaCl is 58.5%

Mole Concept & Stoichiometry Miscellaneous Exercises

Miscellaneous Exercises Very Short Answer Type

Numerical Problems