(b) Mass % of N in NH4NO3 = 35.00 % Mass of N in sample = 35.00 100 x 35.8 x 103 g = 1.25 x 104 g N
Elemental analysis of a sample an ionic compound gave the following results; 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula of the compound? EXERCISE – 16 1.2 SK015
Element Na Cl O Mass Mole Mole ratio 2.82 g 4.35 g 7.83 g 2.82 g 22.99 g/mol = 0.123 mol 4.35 g 35.45 g/mol = 0.123 mol 7.83 g 16.00 g/mol = 0.489 mol 0.123 mol 0.123 mol 0.123 mol 0.123 mol 0.489 mol 0.123 mol 1.00 1.00 3.98 Empirical Formula = NaClO4
During physical activity, lactic acid (molar mass = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass % C, 6.7 mass % H and 53.3 mass % O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. EXERCISE – 17 1.2 SK015
Mass % Mass Mole Mole ratio 40.0 g 6.7 g 53.3 g 40.0 g 12.01 g/mol = 3.33 mol 6.7 g 1.008 g/mol = 6.65 mol 53.3 g 16.00 g/mol = 3.33 mol 3.33 mol 3.33 mol 6.65 mol 3.33 mol 3.33 mol 3.33 mol 1.00 2.00 1.00 (a) C H O 40 % 6.7 % 53.3% Empirical Formula = CH2O
(b) Molar mass of lactic acid Molar mass of the empirical formula = n (whole number) 90.08 g/mol 30.03 g/mol = 3 Molecular Formula = (CH2O)3 = C3H6O3
Ascorbic acid (vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.92 % C, 4.58 % H and 54.50 % O by mass. Determine its empirical formula. EXERCISE – 18 1.2 SK015
Element C H O Mass % Mass Mole Mole Ratio 40.92 % 4.58 % 54.50 % 40.92 g 4.58 g 54.50 g 40.92 g 12.01 g/mol = 3.407 mol 4.58 g 1.01 g/mol = 4.535 mol 54.50 g 16.00 g/mol = 3.406 mol 3.407 3.406 = 1.00 4.535 3.406 = 1.33 3.407 3.406 x 3 = 1.00 = 3.00 3 x 3 = 3.99 4 x 3 = 3.00 3 Empirical formula = C3H4O3
Vitamin C (molar mass = 176.12 g/mol) is a compound of C, H and O found in many natural. sources, especially citrus fruits. When a 1.000 g sample of vitamin C is placed in combustion chamber and burned, the following data are obtained: Mass of CO2 absorber after combustion = 85.35 g Mass of CO2 absorber before combustion = 83.85 g Mass of H2O absorber after combustion = 37.96 g Mass of H2O absorber before combustion = 37.55 g What is the molecular formula of vitamin C? EXERCISE – 19 1.2 SK015
Mass of CO2 absorber after combustion = 85.35 g Mass of CO2 absorber before combustion = 83.85 g Mass of CO2 produced = 1.50 g CO2
Mass of C = mass of C in 1 mol CO2 mass of 1 mol CO2 x mass of CO2 (produced) = 12.01 g C 44.01 g CO2 x 1.50 g = 0.409 g C
Mass of H2O absorber after combustion = 37.96 g Mass of H2O absorber before combustion = 37.55 g Mass of H2O produced = 0.41 g H2O
Mass of H = mass of H in 1 mol H2O mass of 1 mol H2O x mass of H2O (produced) = 2 x 1.008 g H 18.02 g H2O x 0.41 g = 0.046 g H
Mass of C = 0.409 g Mass of H = 0.046 g Mass of vitamin C sample = 1.000 g (contains C, H and O) Mass of O = 1.000 g – ( 0.409 g + 0.046 g) = 0.545 g O
Element C H O Mass Mole Mole Ratio 0.409 g 12 g/mol = 0.034 mol 0.409 g 0.046 g 1 g/mol = 0.046 mol 0.046 g 0.545 g 16 g/mol = 0.034 mol 0.545 g 0.034 0.034 = 1.00 0.046 0.034 = 1.35 0.034 0.034 x 3 = 1.00 = 3.00 3 x 3 x 3 = 4.05 4 = 3.00 3 Empirical formula = C3H4O3
Empirical formula = C3H4O3 Molar mass of Vitamin C Molar mass of empirical formula = n (whole number) 176.12 g/mol 88.03 g/mol = 2 Molecular Formula = (C3H4O3 )2 = C6H8O6
A caproic acid, which is responsible for the foul odor of dirty socks. Combustion of a 0.225 g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. What is the empirical formula of caproic acid? Caproic acid has a molar mass of 116 g /mol. What is its molecular formula? EXERCISE – 20 1.2 SK015
Mass of C = mass of C in 1 mol CO2 mass of 1 mol CO2 x mass of CO2 (produced) = 12.01 g C 44.01 g CO2 x 0.512 g = 0.1397 g C
Mass of H = mass of H in 1 mol H2O mass of 1 mol H2O x mass of H2O (produced) = 2 x 1.008 g H 18.02 g H2O x 0.209 g = 0.0234 g H
Mass of C = 0.1397 g Mass of H = 0.0234 g Mass of caproic acid sample = 0.225 g (contains C, H and O) Mass of O = 0.225 g – (0.1397 g + 0.0234 g) = 0.0619 g O
Element C H O Mass Mole Mole Ratio 0.1397 g 12.01 g/mol = 0.0116 mol 0.1397 g 0.0234 g 1.008 g/mol = 0.0232 mol 0.0234 g 0.0619 g 16.00 g/mol = 0.0038 mol 0.0619 g 0.0117 0.0038 = 3.08 0.0232 0.0038 = 6.10 0.0038 0.0038 = 1.00 3 6 1 Empirical formula = C3H6O
Empirical formula = C3H6O Molar mass of caproic acid Molar mass of empirical formula = n (whole number) 116 g/mol 58.03 g/mol = 2 Molecular Formula = (C3H6O)2 = C6H12O2
A student prepared a solution of NaCl by dissolving 1.461 g of NaCl in a 250 mL volumetric flask. What is the molarity of this solution EXERCISE – 21 1.2 SK015
Mole of NaCl = Mass of NaCl (g) Molar mass of NaCl (g/mol) 1.461 g 58.44 g/mol = 0.0250 mol NaCl =
Molarity = Moles of NaCl (mol) Volume of NaCl solution (L) 0.0250 mol 0.250 L = 0.100 M NaCl solution =
What mass of KI is required to make 500 mL of a 2.80 M KI solution? EXERCISE – 22 1.2 SK015
Mole of KI = Molarity x Volume of KI solution (in L) = 2.80 mol/L x 0.50 L = 1.400 mol KI Mass of KI = Mole of KI x Molar mass of KI = 1.400 mol x 166.00 g/mol = 232.4 g KI
A 15.00−mL sample of 0.450 M K2CrO4 is diluted to 100.00 mL. What is the concentration of the new solution? EXERCISE – 23 1.2 SK015
MiVi = MfVf Mi = 0.450 M Mf = ? Vi = 15.00 mL Vf = 100.00 mL Mf = Mi x Vi Vf = 0.450 M x 15.00 mL 100.00 mL = 0.0675 M K2CrO4 solution
How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? dilution add solvent EXERCISE – 24 1.2 SK015
MiVi = MfVf Mi = 4.00 M Mf = 0.200 M Vi = ? Vf = 0.060 L Vi = Mf x Vf Mi = 0.200 M x 0.060 L 4.00 M = 0.003 L = 3 mL
Tranfer 3 mL of HNO3 stock solution into a 60 mL volumetric flask and add sufficient water (57 mL) until the total volume of the solution is 60 mL. 3 mL + 57 mL of water = 60 mL of solution
Calculate the molality of the following: (a) A solution containing 88.4 g of glycine (NH2CH2COOH) dissolved in 1.250 kg H2O (b) A solution containing 8.89 g glycerol (C3H8O3 ) in 75.0 g of ethanol (C2H6O) EXERCISE – 25 1.2 SK015
Mole of NH2CH2COOH = Mass of NH2CH2COOH Molar mass of NH2CH2COOH 88.4 g 75.07 g/mol = 1.17757 mol NH2CH2COOH = (a)
(a) Molality = Moles of NH2CH2COOH Mass of solvent (H2O) (in kg) = 1.17757 mol 1.250 kg = 0.942 m NH2CH2COOH solution
(b) Mole of C3H8O3 = Mass of C3H8O3 Molar mass of C3H8O3 8.89 g 92.09 g/mol = 0.09654 mol C3H8O3 =
(b) Molality = Moles of C3H8O3 Mass of solvent (C2H6O) (in kg) = 0.09654 mol 0.075 kg = 1.29 m C3H8O3 solution
How many grams of glucose (C6H12O6 ) must be dissolved in 563 g of ethanol (C2H5OH) to prepare a 2.40 x 10–2 m solution? EXERCISE – 26 1.2 SK015
Molality = Moles of C6H12O6 Mass of solvent (C2H5OH) (in kg) Moles of C6H12O6 = Molality x Mass of C2H5OH (in kg) = 2.40 x 10–2 mol/kg x 0.563 kg = 0.01351 mol C6H12O6
Mass of C6H12O6 = Mole of C6H12O6 x Molar mass of C6H12O6 = 0.01351 mol x 180.16 g/mol = 2.43 g C6H12O6
Calculate the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL. EXERCISE – 27 1.2 SK015
Molarity (5.86 M) = Moles of C2H5OH (mol) Volume of solution (L) Assume: volume of solution = 1.0 L Moles of C2H5OH = 5.86 mol Mass of C2H5OH = Mole of C2H5OH x Molar mass of C2H5OH = 5.86 mol x 46.07 g/mol = 270.0 g C2H5OH
Density of solution = Mass of solution Volume of solution Mass of solution = 0.927 g/mL x 1000 mL = 927.0 g solution
Mass of solution = mass of C2H5OH + mass of solvent Mass of solvent = 927.0 g – 270.0 g = 657.00 g solvent = 0.657 kg solvent
Molality = Moles of C2H5OH Mass of solvent (in kg) = 5.85 mol 0.657 kg = 8.90 m C2H5OH solution
A sample of 0.892 g of potassium chloride (KCl) was dissolved in 54.6 g of water. What is the % by mass of KCl in this solution? EXERCISE – 28 1.2 SK015
Mass % of KCl Mass of KCl Mass of solution = x 100 = 0.892 g (0.892 g + 54.6 g) x 100% = 1.61 %
A 200 mL of perfume contains 28 mL of alcohol. What is the % concentration of alcohol by volume in this solution? EXERCISE – 29 1.2 SK015
Volume % = Volume of alcohol Volume of perfume x 100 = 28 mL 200 mL x 100 = 14 % Volume of solute Volume of solution = x 100
A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water? EXERCISE – 30 1.2 SK016