MOLE. Formula: 1. n = M / Mr or, n = M / Ar

MOLE Mole is the SI unit of amount of substance. 1 mole is the amount of substance which has 6.02 * 10 23 particles. ( Avogadro's constant L ) Note: 1m 3 = 1000 dm 3 = 10 6 cm 3 Formula: 1. n = M / Mr or, n = M / Ar 2. For gases; n = (Volume in dm 3 ) / molar volume, n = V / V m At RTP, V m = 24.0 dm 3 At STP, V m = 22.4 dm 3 3. For solution; n = c * v 4. n = (no. of particles ) / (avogadro s constant ) No. of particles = n * L 5. For dilution; n 1 = n 2 C 1 * V 1 = C 2 * V 2 Q.1) 15.6 gram of MgCO 3 is heated. Calculate: I. Mass of MgO and CO 2 produced. II. Volume occupied by CO 2 (g) at RTP. III. No. of particles of MgO. I. MgCO 3 MgO + CO 2 n(mg) = M / Mr = 15.6 / (24.3 + 12.0 + (16.0 * 3) ) = 15.6 / 84.3 = 0.185 mol n(mg) = n(mgo) = 0.185 mol n(mg) = n(co 2 ) = 0.185 mol

M( MgO) = n(mgo) * (24.3 + 16) = 0.185 * 40.3 = 7.46 g M(CO 2 ) = n(co 2 ) * (12.0 + 32) = 0.185 * 44.0 = 8.14 g II. volume (CO 2 ) = n * V m = 0.185 * 24.0 = 4.44 dm 3 III. no. of particles (MgO) = n * 6.02 * 10 23 = 0.185 * 6.02 * 10 23 = 1.11* 10 23 particles Q.2) 12.4 gram of ZnCO 3 is heated. Calculate I. Mass of ZnO and CO 2. II. Volume occupied by CO 2 at STP. III. No. of particles of ZnO. I. ZnCO 3 ZnO + CO 2 n(znco 3 ) = M / Mr = 12.4 / 125.4 = 0.0989 mol n( ZnO) = n(znco 3 ) = 0.0989 mol n(co 2 ) = n(znco 3 ) = 0.0989 mol M( ZnO) = n( ZnO) * Mr(ZnO) = 0.0989 * 81.4 = 8.05 g M(CO 2 ) = n(co 2 ) * Mr(CO 2 ) = 0.0989 * 44 = 4.35 g II. volume = n * V m = 0.0989 * 22.4 = 2.21 dm 3 III. no. of particles = n(zno) * Avogadro s constant = 0.0989 * 6.02 * 10 23 = 5.95 * 10 22 particles Q.3) 5.85 g of KClO 3 is heated. Calculate I. Mass of KCl and O 2 produced. II. Volume occupied by O 2 (g). III. no. of particles of KCl and O 2.

I. KClO 3 KCl + 3/2 O 2 n(kclo 3 ) = 5.85 / (39.1 + 35.5 + (3 * 16.0) ) = 5.85 / 122.6 = 0.0477 mol n(kcl) = n(kclo 3 ) = 0.0477 M = n * Mr = 0.0477 * (39.1 + 35.5 ) = 3.56 g n(co 2 ) = 3 / 2 * n(kclo 3 ) = 3 / 2 * 0.0477 = 0.0716 mol M = n * Mr = 0.0716 * 32.0 = 2.29 g II. volume = n * V m = 0.0716 * 24.0 = 1.72 dm 3 III. no. of particles of KCl = n * 6.02 * 10 23 = 0.0477 * 6.02 * 10 23 = 2.87 * 10 22 particles no. of particles of O 2 = n * 6.02 * 10 23 = 0.0716 * 6.02 * 10 23 = 4.31 * 10 22 particles Q.4) 12.3 gram of Ca(NO 3 ) 2 is heated. Calculate I. Mass of CO 2, NO 2, O 2 produced. II. Volume occupied by NO 2 (g), O 2 (g) at STP. III. No. of particles of CaO, NO 2, and O 2. I. Ca(NO 3 ) 2 CaO + 2NO 2 + ½ O 2 n(ca(no 3 ) 2 ) = M / Mr = 12.3 / (40.1 + 2 ( 14.0 + ( 3 * 16.0 ) ) ) = 12.3 / 164.1 = 0.0750 mol n(cao) = n(ca(no 3 ) 2 ) = 0.0750 M(CaO) = n(cao) * Mr = 0.0750 * 56.1 = 4.21 g n(no 2 ) = 2 * n(ca(no 3 ) 2 ) = 0.150 M(NO 2 ) = n(no 2 ) * Mr = 0.150 * 46.0 = 6.90 g n( O 2 ) = ½ * n(ca(no 3 ) 2 ) = 0.0375 M(O 2 ) = n( O 2 ) * Mr = 0.0375 * 32.0 = 1.20 g

II. V(NO 2 ) = n * V m = 0.150 * 22.4 = 3.36 dm 3 V(CaO) = n * V m = 0.0750 * 22.4 = 1.68 dm 3 V(O 2 ) = n * V m = 0.0375 * 22.4 = 0.840 dm 3 III. no. of particles of (CaO) = 0.0750 * 6.02 * 10 23 = 4.52 * 10 22 particles no. of particles of (NO 2 ) = 0.150 * 6.02 * 10 23 = 9.03 * 10 22 particles no. of particles of (O 2 ) = 0.0375 * 6.02 * 10 23 = 2.26 * 10 22 particles Q.5) 33.60 cm 3 of 6.00 mol dm -3 HCl (aq) is completely neutralized by 38.70 cm 3 of NaOH (aq). Calculate; I. Concentration of NaOH(aq), II. Mass of NaCl and H 2 O produced. III. No. of particles of NaCl. I. HCl(aq) + NaOH(aq) NaCl + H 2 O 33.60 cm 3 = 33.60 * 10-3 dm 3 n(hcl) = c * v = 0.600 * 33.60 * 10-3 = 0.0202 mol n(naoh) = n(hcl) = 0.0202 mol C = n / v = 0.0202 / (38.70 * 10-3 ) = 0.521 mol dm -3 II. n(nacl) = n(hcl) = 0.0202 mol M(NaCl) = n * Mr = 0.0202 * 58.5 = 1.18 g n(h 2 O) = n(hcl) = 0.0202 mol M(H 2 O) = n * Mr = 0.0202 * 18.0 = 0.364 g III. No. of particles of NaCl = n * 6.02 * 10 23 = 0.0202 * 6.02 * 10 23 = 1.22 * 10 22 particles Q.6) 24.30 cm 3 of 0.400 mol dm -3 H 2 SO 4 (aq) is completely neutralized by 35.50 cm 3 of NaOH (aq) in gram dm -3. I. Concentration of NaOH (aq) in gram dm -3. II. Mass of Na 2 SO 4 and H 2 O.

III. No. of particles of Na 2 SO 4 and H 2 O. I. 2NaOH(aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2H 2 O n(h 2 SO 4 ) = C * V = 0.400 * 24.30 * 10-3 = 9.72 * 10-3 mol n(naoh) = 2 * n(h 2 SO 4 ) = 2 * 9.72 * 10-3 = 0.0194 mol C = n(naoh) / V = 0.0194 / (35.50 * 10-3 ) = 0.548 mol dm -3 = 0.548 / Mr = 0.548 * 40 = 21.9 g dm -3 II. n(na 2 SO 4 ) = n(h 2 SO 4 ) = 9.72 * 10-3 mol M = n * Mr = 9.72 * 10-3 * 142.1 = 1.38 g n(h 2 O) = 2 * n(h 2 SO 4 ) = 2 * 9.72 * 10-3 mol M = n * Mr = 0.0194 * 18.0 = 0.350 g III. No. of particles (Na 2 SO 4 ) = n * 6.02 * 10 23 = 9.72 * 10-3 * 6.02 * 10 23 = 5.85 * 10 21 No. of particles (H 2 O) = 0.0194 * 6.02 * 10 23 = 1.17 * 10 22 Titration; Titration is a method of quantitative analysis in which a known reagent is reacted with an unknown reagent to obtain different information about the unknown reagents like concentration, purity etc. Q.1) 7.25 g of Na 2 CO 3 solute is completely dissolved in 250 cm 3 of water to form FA3. Calculate the concentration of FA3. 25.0 cm 3 of FA3 is pipetted out in a conical flask and titrated with HCl(aq) is used. Calculate the concentration of HCl(aq). n(na 2 CO 3 ) = 7.25 / 106 = 0.0684 mol C( FA3 or Na 2 CO 3 ) = n / v = 0.0684 / (250 * 10-3 ) = 0.274 mol dm -3 n(na 2 CO 3 ) in 25.0 cm 3 of FA3 = 0.274 * 10-3 * 25 = 0.00685 mol Na 2 CO 3 (aq) + 2HCl (aq) NaCl + H 2 O + CO 2

n(hcl) = 2 * n(na 2 CO 3 ) = 2 * 0.00685 = 0.0137 mol C(HCl) = n / v = 0.0137 / (27.60 * 10-3 ) = 0.496 mol dm -3 Q.2) 6.26 g of NaHCO 3 is completely dissolved in 250 cm 3 of water to form FA3. Calculate the concentration of FA3. 25.0 cm 3 of FA3 is pipetted out in a conical flask and titrated with HCl(aq). 28.70 cm 3 of HCl (aq) is used. Calculate the concentration of HCl(aq). n(nahco 3 ) = 6.26 / 84.0 = 0.0745 mol c(nahco 3 ) = n / v = 0.0745 / (250 * 10-3 ) = 0.298 mol dm -3 n(fa3 or NaHCO 3 (aq) ) in 25.0 cm 3 = 0.298 * 25 * 10-3 = 7.45 * 10-3 NaHCO 3 (s) + HCl (aq) NaCl + CO 2 + H 2 O n(nahco 3 ) = n(hcl) = 7.45 * 10-3 C = n / v = (7.45 * 10-3 ) / (28.70 * 10-3 ) = 0.260 mol dm -3 Redox Titration; Q.3) 6.20 gram of H 2 C 2 O 4 solid is dissolved completely in 250 cm 3 of water to form FA5. 25.0 cm 3 of FA5 is pipetted out in a conical flask and titrated with KMnO 4 is used. Calculate the concentration of KMnO 4 (aq) in gram dm -3. n(h 2 C 2 O 4 ) = 6.20 / 90 = 0.0689 mol C = n / v = 0.0689 / (250 * 10-3 ) = 0.276 mol dm -3 n(h 2 C 2 O 4 ) in 25.0 cm 3 = 0.276 * 25.0 * 10-3 = 6.90 * 10-3 moles 2MnO 4 - + 6H + + 5H 2 C 2 O 4 2Mn 2+ + 10CO 2 + 8H 2 O n(kmno 4 ) = ⅖ * 6.9 * 10-3 = 2.76 * 10-3 mol C(KMnO 4 ) in 26.30 cm 3 = (2.76 * 10-3 ) / (26.30 * 10-3 ) = 0.105 mol dm -3 = 0.105 * 158 = 16.6 gram dm -3

Percentage Purity; Q.1) 12.6 gram of impure sample containing CaCO 3 is reacted with 100 cm 3 of 2.00 mol dm -3 HCl (aq) in excess. After removing the impurities and the solution is transferred to a conical flask and titrated with 0.600 mol dm -3 NaOH (aq). 32.80 cm 3 of NaOH(aq) is used. Calculate I. No. of mole of NaOH used. II. No. of mole of HCl reacted with NaOH. III. No. of mole of HCl total used. IV. No. of mole of CaCO 3. V. Mass of CaCO 3 present in the sample. VI. Percentage of purity of the sample. CaCO 3 + 2HCl CaCl 2 + H 2 O + CO 2 HCl + NaOH NaCl + H 2 O I. n(naoh) = c * v = 0.600 * 32.80 * 10-3 = 0.0197 II. n(hcl) reacted with NaOH = n(naoh) = 0.0197 mol III. n(hcl) total used = C * V = 2.00 * 100 * 10-3 = 0.200 mol IV. n(hcl) reacted with CaCO 3 = 0.200-0.0197 = 0.180 mol V. n(caco 3 ) = ½ * 0.180 = 0.090 mol Vi. Mass of CaCO 3 = n * Mr = 0.090 * (40.1 + 12 + (3 * 16) ) = 9.01 g VII. % purity = ( 9.01 / 12.6 ) * 100% = 71.5%

Q.2) 9.28 gram of impure sample containing MgCO 3 is reacted with 90 cm 3 of 2.20 mol dm -3 HCl (aq). After removing the impurities the solution is transferred to a conical flask and titrated with 0.750 mol dm -3 NaOH(aq). 31.70 cm 3 of NaOH is used. Calculate I. No. of mole of NaOH used. II.No. of mole of HCl total used. III. No. of mole of HCl reacted with MgCO 3. IV. Mass of MgCO 3 present in the sample. V. Percentage purity of sample. I. MgCO 3 + 2HCl MgCl 2 + CO 2 + H 2 O HCl + NaOH NaCl + H 2 O n(naoh) = C * V = 0.750 * 31.70 * 10-3 = 0.0238 mol n(hcl) reacted with NaOH = n(naoh) = 0.0238 mol n(hcl) total used = C * V = 2.20 * 80 * 10-3 = 0.176 mol n(hcl) reacted with MgCO 3 = 0.176-0.0238 = 0.152 mol n(mgco 3 ) = ½ * 0.152 = 0.0760 mol Mass of MgCO 3 = n * Mr = 0.0760 * (24.3 + 12.0 + (3 * 16) ) = 6.41 g % purity = ( 6.41 / 9.28 ) * 100% = 69.1 % Q.3) MnO 4 - + 8H + + 5Fe 2+ Mn 2+ + 4H 2 O + 5Fe 3+ 3.5 gram of impure sample containing iron is reacted with acid to convert all the iron into Fe 2+ ion. After removing the impurities the solution is transferred to a conical flask and titrated with 0.250 mol dm -3 KMnO 4 (aq) 27.20 cm -3 of KMnO 4 is used. Calculate I. No. of mole of KMnO 4. II. No. of mole of Fe 2+. III. No. of mole of Fe. IV. Mass of Fe present in the sample. V. Percentage purity of the sample. I. No. of mole of KMnO 4 = C * V = 0.250 * 27.20 * 10-3 = 0.00680 mol II. No. of mole of Fe 2+ = 0.00680 * 5 = 0.0340 mols

III. No. of mole of Fe = n(fe 2+ ) = 0.0340 mol IV. Mass = n * Ar = 0.0340 * 55.8 = 1.90 gram V. % purity = (1.90 / 3.5 ) * 100% = 54.3 % Dilution; Q.1) FA1 is 2.00 mol dm -3 HCl(aq) 42.30 cm 3 of FA1 is transferred to 250 cm 3 of volumetric flask and diluted up to the mark to form FA2. Calculate I. No. of mole of HCl in 42.30 cm 3 of FA1. II. No. of Mole of HCl in 250 cm 3 of FA2. III. Concentration of FA2. 25.0 cm 3 of FA1 is pipetted out in a conical flask and titrated with NaOH (aq). 27.80 cm 3 of NaOH (aq) is used. Calculate IV. No. of mole of HCl in 25.0 cm 3 of FA2. V. No. of mole of NaOH. VI. Concentration of NaOH. I. n(hcl) in 42.30 cm 3 of FA1 = C * V = 2.00 * 42.30 * 10-3 = 0.0846 mol II. n(hcl) in 250 cm 3 of FA2 = n(hcl) in 42.30 cm 3 of FA1 = 0.0846 mol III. C(FA2) = n/v = 0.0846 / (250 * 10-3 ) = 0.338 mol dm -3

ALTERNATIVE METHOD; n 1 = n 2 C 1 * V 1 = C 2 * V 2 2.00 * 42.30 = C 2 * 250 * 10-3 C 2 = 0.338 mol dm -3 IV. n(hcl) in 25 cm 3 of FA2 = C * V = 0.338 * 25.0 * 10-3 = 0.00845 mol V. HCl + NaOH NaCl + H 2 O n(naoh) = n(hcl) = 0.00845 mol VI. C(NaOH) = n / v = 0.00845 / (27.80 * 10-3 ) = 0.304 mol dm -3 Q.2) FA1 is 2.10 mol dm -3 H 2 SO 4 (aq). 37.80 mol cm 3 of FA1 is transferred to 250 cm 3 volumetric flask and diluted upto the mark to form FA2. Calculate the concentration of FA2. 25.0 cm 3 of FA2 is pipetted out in a conical flask and titrated with NaOH(aq). 31.60 cm 3 of NaOH is used. Calculate the concentration of NaOH in gram dm -3. n 1 = n 2 C 1 * V 1 = C 2 * V 2 2.10 * 37.80 * 10-3 = C 2 * 250 * 10-3 C 2 = 0.318 mol dm -3 H 2 SO 4 + 2NaOH Na 2 SO 4 + 2H 2 O n(h 2 SO 4 ) in 25.0 cm 3 of FA1 = C * V = 0.318 * 25.0 * 10-3 = 0.00795 mol n(naoh) = 2 * 0.00795 = 0.0159 mol C( NaOH) = n / v = 0.0159 / (31.60 * 10-3 ) = 0.503 mol dm -3 = 40 * (0.503) gram dm -3 = 20.1 gram dm -3 Q.3) FA1 is H 2 SO 4 (aq) of unknown concentration. 38.80 cm 3 of FA1 is transferred to 250 cm 3 volumetric flask and diluted upto the mark to form FA2. 25.0 cm 3 of FA2 is pipetted out in a conical flask and titrated with 0.400 mol dm -3 NaOH(aq). 33.50 cm 3 of NaOH(aq) is used. Calculate I. No. of mole of NaOH (aq) used. II. No. of mole of H 2 SO 4 (aq) is used. III. No. of mole of H 2 SO 4 250 cm 3 of FA2.

IV. No. of mole of H 2 SO 4 in 38.80 cm 3 of FA1. V. Concentration of FA1. n(naoh) in 25.0 cm 3 = C * V = 0.400 * 25.0 * 10-3 = 0.0134 mol H 2 SO 4 + 2NaOH Na 2 SO 4 + 2H 2 O n(h 2 SO 4 ) in 25.0 cm 3 = 0.0134 / 2 = 0.0067 mol n(h 2 SO 4 ) in 250.0 cm 3 = (0.0067 / 25.0 ) * 250 = 0.067 mol n(h 2 SO 4 ) in 38.80 cm 3 = n(h 2 SO 4 ) in 250.0 cm 3 C = n / V = 0.067 / (38.80 * 10-3 ) = 1.73 mol dm -3 PERCENTAGE YIELD; * Percentage yield = ( obtained yield / theoretical yield ) * 100% Q.1) 6.75 gram of CaCO 3 is heated and 2.56 gram of CaO is produced. Calculate the percentage yield. CaCO 3 CaO + CO 2 n(caco 3 ) = M / Mr = 6.75 / 100.1 = 0.0674 n(cao) = n(caco 3 ) = 0.0674 M(CaO) = n * Mr = 0.0674 * 56.1 = 3.78 gram % yield = ( 2.56 / 3.78 ) * 100% = 67.7 % Q.2) 7.88 gram of MgCO 3 is heated and 3.25 gram of MgO is produced. Calculate the % yield. MgCO 3 MgO + CO 2 n(mgco 3 ) = ( 7.88 / 84.3 ) = 0.0935 n(mgo) = n(mgco 3 ) = 0.0935 * (24.3 + 16 ) = 3.77 % yield = ( 3.25 / 3.77 ) * 10-3 = 86.2% Q.3) Calculate the mass of zinc carbonate that must be heated to produce 15.0 gram of ZnO. The % yield is 60%. ZnCO 3 ZnO + CO 2

% yield = (obtained yield / theoretical yield ) * 100% 60 % = ( 15.0 / theoretical yield ) * 100% Theoretical yield = ( 15.0 / 0.6 ) = 25.0 gram n(zno) = 25.0 / (65.4 + 16 ) = 0.307 n(znco 3 ) = n(zno) = 0.307 M(ZnCO 3 ) = n * Mr = 0.307 * (65.4 + 12.0 + (3 * 16.0) ) = 38.5 EMPIRICAL AND MOLECULAR FORMULA; Empirical Formula: The simplest ratio of atoms present in the molecules. Molecular Formula: The actual number of atoms present in the molecules. E.g: molecular formula empirical formula C 2 H 6 CH 3 C 6 H 12 O CH 2 O Q.1) A hydrocarbon has 80% of carbon determine its empirical formula and molecular formula if its Mr is 30. Let s take 100 g of compound. C H Mass: 80 g 20 g n = M / Ar 80 / 12.0 20 / 1.0 6.67 20 Simplest whole no. ratio: 6.67 / 6.67 20 / 6.67 1 3 Empirical formula: CH 3 Empirical formula mass: 12 + ( 3 *1 ) = 15 n= Mr / (empirical formula mass ) = 30 / 15 = 2 Molecular formula = n * empirical formula = 2 * CH 3 = C 2 H 6

Q.2) A hydrocarbon has 85.7% of carbon determine its empirical formula and molecular formula if its Mr is 70. Let s take 100 g of compound C H Mass; 85.7 g 14.3 g n = M / Mr 85.7 / 12 14.3 / 1 7.14 14.3 Simplest whole no. ratio: 7.14 / 7.14 14.3 / 7.14 1 2 Empirical formula = CH 2 Empirical formula mass = 12.0 + 2.0 = 14.0 n = Mr / empirical mass = 70 / 14.0 = 5 Molecular formula= empirical formula * n = CH 2 * 5 = C 5 H 10 Q.3) A hydrocarbon has 82.75% of carbon. Determine its empirical and molecular formula if its Mr is 58. C H Mass; 82.75 17.25 n : 82.75 / 12 17.25 / 1 6.90 17.25 Simplest whole no. ratio: 6.90 / 6.90 17.25 / 6.90 1 * 2 2.50 * 2 2 5 Empirical formula = C 2 H 5 Empirical formula mass = (2 * 12) + 5.0 = 29.0 n = 58 / 29 = 2 Molecular formula = 2 * C 2 H 5 = C 4 H 10

Q.4) An organic compound has 40% of carbon and 6.67% of hydrogen. Determine its empirical formula and molecular formula if its Mr is 120. C H O Mass: 40 g 6.67 g 53.33 g n : 40 / 12.0 6.67 / 1.0 53.33 / 16.0 3.33 6.67 3.33 Simplest whole no. ratio: 3.33 / 3.33 6.67 / 3.33 3.33 / 3.33 1 2 1 Empirical formula: CH 2 O Empirical formula mass: 12.0 + 2 +16.0 = 30 n = Mr / empirical formula mass = 120 / 30 = 4 Molecular formula = n * empirical formula = 4 * CH 2 O = C 4 H 8 O 4 Q.5) An organic compound has 26.67% of carbon. 6.67% of hydrogen and 3.11% of nitrogen. Determine its empirical formula and molecular formula if its Mr is 135.0. C H N O Mass: 26.67 g 6.67 g 31.11 g 35.55 g n : 26.67 / 12.0 6.67 / 1.0 31.11 / 14.0 35.55 / 16.0 1.81 6.67 1.22 2.22 Simplest whole: 1 3 1 1 no. ratio Empirical formula: CH 3 NO Empirical mass: 12.0 + 3.0 + 14.0 + 16.0 = 45.0 n = 135.0 / 45.0 = 3 Empirical formula: 3 * CH 3 NO = C 3 H 9 N 3 O 3 Q.6) An organic compound has 20.11 % of carbon 3.91% of hydrogen and 31.28% of nitrogen determine its empirical and molecular formula if its Mr is 358.

C H N O Mass: 20.11 3.91 31.28 44.7 n : 20.11 / 12.0 3.91 / 1.0 31.28 / 14.0 44.7 / 16.0 Simplest whole: 1.68 / 1.68 3.91 / 1.68 2.23 / 1.68 2.78 / 1.68 no. ratio 1 2.33 1.33 1.65 1 * 3 2.33 * 3 1.33 * 3 1.65 * 3 3.0 7.0 4.0 5.0 Empirical formula: C 3 H 7 N 4 O 5 Empirical mass: (3 * 12 ) + 7(1.0) + 4 * (14.0 ) + ( 5 * 16.0 ) = 179 n = 358 / 179 = 2 Molecular formula = 2 * C 3 H 7 N 4 O 5 = C 6 H 14 N 8 O 10 COMBUSTION ANALYSIS: Combustion analysis by mass: Q.1) When a hydrocarbon is burnt completely 11.0 gram of carbon dioxide and 4.50 gram of water vapour are produced. Determine its empirical formula and molecular formula if its Mr is 84.0. C X H Y + (x + y/4 ) O 2 x CO 2 + y / 2 H 2 O Mass of carbon = (12/44) * 11.0 = 3.00 g 44.0 g of CO 2 = 12 g of C 1 g of CO 2 = 12/44 g of C X g of CO 2 = 12/44 * X g of C Mass of hydrogen = (2/18) * 4.50 = 0.500 g 18 g of H 2 O = 2 g of H 1 g of H 2 O = (2/18) g of H Y g of H 2 O = (2/18) * Y g of H

C H Mass: 3.00 g 0.500 g n : 3.00 / 12.0 0.500 / 1.00 0.250 0.500 Simplest whole no. ratio: 0.250 / 0.500 0.500 / 0.500 0.5 1.0 2 * 0.5 2 * 1.0 1.0 2.0 Empirical formula: CH 2 Empirical mass: 12.0 + 2.0 = 14.0 n = 84.0 / 14.0 = 6.00 Molecular formula = 6 * CH 2 = C 6 H 12 Q.2) When a hydrocarbon is burnt completely 17.6 gram of carbon dioxide and 9.00 gram of water vapour are produced. Determine its empirical formula and molecular formula if its Mr is 58.0. C X H Y + (x + y/4 ) O 2 x CO 2 + y / 2 H 2 O Mass of C: (12/44) * 17.6 = 4.80 g Mass of H: (2/18) * 9.00 = 1.00 C H Mass: 4.80 g 1.00 g n : 4.80 / 12.0 1.00 / 1.00 0.400 1.00 Simplest whole no. ratio: 0.400 / 1.00 1.00 / 1.00 0.400 1.00 0.4 * 5 1.00 * 5 2.0 5.00 Empirical formula: C 2 H 5 Empirical mass: ( 2*12 ) + 5 = 29.0

n = 58 / 29.0 = 2.0 Molecular formula: 2 * C 2 H 5 = C 4 H 10 Q.3) When 4.65 gram of organic compound is burned completely 6.60 gram of carbon dioxide and 4.05 gram of water vapour are produced. Determine its empirical formula and molecular formula if its Mr is 62. Mass of carbon = (12/44) * 6.60 = 1.80 g Mass of hydrogen = (2/18) * 4.05 = 0.450 g Mass of oxygen = 4.65-1.80-0.450 = 2.40 g C H O Mass: 1.80 g 0.450 g 2.40 g n : 1.80 / 12.0 0.450 / 1.0 2.40 / 16.0 0.150 0.450 0.15 Simplest whole no: 0.150 / 0.150 0.450 / 0.150 0.150 / 0.150 Ratio 1.00 3.00 1.00 Empirical formula: CH 3 O Empirical mass: 12 + 3 + 16 = 31 n = 62 / 31 = 2 Molecular formula = 2 * CH 3 O = C 2 H 6 O 2 Q.4) When 4.50 gram of organic compound is burn completely. 8.80 gram of carbon dioxide and 4.50 gram of water vapour are produced? Determine its molecular formula and and empirical formula if its Mr is 90. Mass of C = (12 / 44 ) * 8.80 = 2.40 g Mass of H = (2 / 18.0 ) * 4.50 = 0.50 g Mass of O = 4.50-2.40-0.50 = 1.60 g

C H O Mass: 2.40 g 0.50 g 1.60 g n : 2.40 / 12.0 0.50 / 1.0 1.60 / 16.0 0.200 0.50 0.100 Simplest Whole no : 2 5 1 Ratio Empirical formula = C 2 H 5 O Empirical mass = (2 * 12) + 5 +16 = 45 n = 90 / 33 = 2.7 = 3.0 Molecular formula = C 6 H 15 O 3 COMBUSTION ANALYSIS BY GAS VOLUME: C X H Y + (x + y/4 ) O 2 x CO 2 + y / 2 H 2 O CH 4 + 2O 2 CO 2 + 2 H 2 O C 2 H 6 + (7/2) O 2 2CO 2 + 3 H 2 O C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O C 4 H 10 + (13/2) O 2 4 CO 2 + 5 H 2 O C 2 H 4 + 3O 2 2 CO 2 + 2 H 2 O C 3 H 6 + 9/2 O 2 3 CO 2 + 3 H 2 O C 4 H 8 + 6 O 2 4 CO 2 + 4 H 2 O C 2 H 2 + (5/2) O 2 2 CO 2 + H 2 O C 3 H 4 + 4 O 2 3 CO 2 + 2 H 2 O Q.1) 20 cm 3 of a hydrocarbon requires 70 cm 3 of oxygen for complete combustion and 40 cm 3 of carbon dioxide is produced. Determine its molecular formula. C X H Y + (x + y/4 ) O 2 x CO 2 + y / 2 H 2 O 20 cm 3 70 cm 3 40 cm 3 20/20 70/20 40/20 1 1 / 2 2

X = 2 X + (y/4) = 7/2 So, y = 6 Molecular formula = C 2 H 6 Q.2) 30 cm 3 of a hydrocarbon requires 180 cm 3 of oxygen for complete combustion and 120 cm 3 of carbon dioxide. Is produced. Determine its molecular formula. C X H Y + (x + y/4 ) O 2 x CO 2 + y / 2 H 2 O 30 cm 3 180 cm 3 120 cm 3 30/30 180/30 120/30 1 6 4 X = 4 X + (y/4) = 6 So, y = 8 Molecular formula = C 4 H 8 Q.3) 40 cm 3 of a hydrocarbon requires 160 cm 3 of oxygen for complete combustion and 80 cm 3 of water vapour are produced. Determine its molecular formula. C X H Y + (x + y/4 ) O 2 x CO 2 + y / 2 H 2 O 40 cm 3 160 cm 3 80 cm 3 40/40 160/40 80/40 1 4 2 X + ( (y/2)/2 ) = 4 Y/2 = 2 X + ( 2/2 ) = 4 Y = 4 X = 3

* C X H Y + (x + y/4 ) O 2 x CO 2 + y / 2 H 2 O Removed by passing Removed by cooling at Over KOH RTP Q.4) 20 cm 3 of propane C 3 H 8 is reacted with 150 cm 3 of excess oxygen. After the reaction the gas is passed over KOH. Calculate the volume of gas before and after passing over KOH? Volume of gas before passing over KOH = 60 + 80 + 50 = 190 cm 3 Volume of gas after passing over KOH = 80 + 50 = 130 cm 3 Q.5) 30 cm 3 of butene C 4 H 8 is reacted with 160 cm 3 of excess oxygen. After the reaction the gas is passed over KOH. Calculate the volume of gas before and after passing over KOH. Volume of gas before passing over KOH = 120 + 120 + 20 = 260 cm 3 Volume of gas after passing over KOH = 120 + 20 = 140 cm 3

Q.6) 40 cm 3 of ethene C 2 H 4 is reacted with 200 cm 3 of oxygen after the reaction the gas is cooled at RTP. Calculate volume of gas before and after cooling to RTP. Volume of gas before cooling at RTP = 80 + 80 +80 = 240 cm 3 Volume of gas after cooling at RTP = 80 + 80 = 160 cm 3 Q.7) 20 cm 3 of butane C 4 H 10 is reacted with 150 cm 3 of oxygen after the reaction the gas is passed over KOH and cooled to RTP. Calculate the volume of gas before and after conducting the process. Volume of gas before cooling at RTP and before passing over KOH = 20 + 80 +100 = 200 cm 3 Volume of gas after cooling at RTP and after passing over KOH = 204 Please contact us if any further changes are to be made: alevelgeeks@gmail.com