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Calculate the molecular mass and molar mass of each of these compounds: (a) NaCl (b) MgO (c) K2S (d) Ca(NO3 )2 (e) Mg3 (PO4 )2 EXERCISE – 01 1.2 SK015
(a) NaCl Molecular / formula mass: 1 Na 1 x 22.99 amu = 22.99 amu 1 Cl 1 x 35.45 amu = 35.45 amu = 58.44 amu Molar mass of NaCl = 58.44 g/mol
(b) MgO Molecular / formula mass: 1 Mg 1 x 24.31 amu = 24.31 amu 1 O 1 x 16.00 amu = 16.00 amu = 40.31 amu Molar mass of MgO = 40.31 g/mol
(c) K2S Molecular / formula mass: 2 K 2 x 39.10 amu = 78.20 amu 1 S 1 x 32.07 amu = 32.07 amu = 100.27 amu Molar mass of K2S = 100.27 g/mol
(d) Ca(NO3 )2 Molecular / formula mass: 1 Ca 1 x 40.08 amu = 40.08 amu 2 N 2 x 14.01 amu = 28.02 amu = 164.10 amu 6 O 6 x 16.00 amu = 96.00 amu Molar mass of Ca(NO3 )2 = 164.10 g/mol
(e) Mg3 (PO4 )2 Molecular / formula mass: 3 Mg 3 x 24.31 amu = 72.93 amu 2 P 2 x 30.97 amu = 61.97 amu = 262.87 amu 8 O 8 x 16.00 amu = 128.00 amu Molar mass of Mg(PO4 )2 = 262.87 g/mol
How many atoms are in 0.551 g of potassium (K) ? Molar mass of K = 39.10 g/mol. EXERCISE – 02 1.2 SK015
1 mol K = 6.02 x 1023 atoms K x 6.02 x 1023 atoms K 1 mol K = 8.49 x 1021 atoms K Mole of K = Mass of K (g) Molar mass of K (g/mol) 0.551 g 39.10 g/mol = 0.0141 mol of K = 0.0141 mol K Number of atoms K: =
EXERCISE – 03 1.2 What is the mass of 2.35 x 1024 atoms of Cu. SK015
1 mol Cu = 6.02 x 1023 atoms Cu 2.35 x 1024 atoms Cu x 6.02 x 1023 atoms Cu 1 mol Cu = 3.9037 mol Cu Mass of Cu = Mole of Cu x Molar mass of Cu = 3.9037 mol x 63.55 g/mol = 248.1 g Cu
EXERCISE – 04 1.2 Aspartame is a widely used artificial sweetener (Nutrasweet) that is almost 200 times sweeter than sucrose. One sample of aspartame, C14H18N2O5 , has a mass of 1.80g; another contains 0.220 mol aspartame. To see which sample is larger, answer these questions: (a) What is the molar mass of aspartame? (b) How many moles of aspartame are in the 1.80−g sample? (c) How many grams of aspartame are in the 0.220−mol sample? SK015
(a) Molar mass of C14H18N2O5 = 294.3 g/mol (b) Mole of C14H18N2O5 = Mass of C14H18N2O5 Molar mass of C14H18N2O5 1.8 g 294.3 g/mol = 6.12 x 10–3 mol C14H18N2O5 =
The 0.220−mol sample of aspartame is the larger one Mole of C14H18N2O5 = Mass of C14H18N2O5 Molar mass of C14H18N2O5 Mass of C14H18N2O5 : = 0.220 mol x 294.3 g/mol = 64.6 g C14H18N2O5 = Mole of C14H18N2O5 x Molar mass of C14H18N2O5 (c)
Calculate each the following quantities: (a) Mass in grams of 2.04 x 1021 molecules of dinitrogen pentaoxide (N2O5 ) (b) Number of moles and formula units in 57.9 g of sodium perchlorate (NaClO4 ) EXERCISE – 05 1.2 SK015
(a) 1 mol N2O5 = 6.02 x 1023 molecules N2O5 2.04 x 1021 molecules N2O5 x 6.02 x 1023 molecules N2O5 1 mol N2O5 = 3.39 x 10–3 mol N2O5 Mass of N2O5 = Mole of N2O5 x Molar mass of N2O5 = 3.39 x 10–3 mol x 108.02 g/mol = 0.366 g N2O5
(b) Mole of NaClO4 = Mass of NaClO4 Molar mass of NaClO4 57.9 g 122.44 g/mol = 0.473 mol NaClO4 =
(b) x 6.02 x 1023 formula units NaClO4 1 mol NaClO4 = 2.85 x 1023 formula units NaClO4 = 0.473 mol NaClO4 Number of formula units NaClO4 :
Ciplatin (below), or Platinol, is a powerful drug used in the treatment of certain cancers. Calculate (a) The moles of compound in 285.3 g of cisplatin (b) The number of hydrogen atoms in 0.98 mol of cisplatin. EXERCISE – 06 1.2 SK015
Formula = Pt(NH3 )2Cl2 Molar mass = 300.1 g/mol
Mole of Pt(NH3 )2Cl2 = Mass of Pt(NH3 )2Cl2 Molar mass of Pt(NH3 )2Cl2 285.3 g 300.1 g/mol = 0.9507 mol Pt(NH3 )2Cl2 = (a)
(b) 0.9507 mol Pt(NH3 )2Cl2 x 1 mol Pt(NH3 )2Cl2 6 mol H = 5.7042 mol H x 6.02 x 1023 H atoms 1 mol H = 5.7042 mol H Number of H atoms: = 3.4 x 1024 H atoms
What is the volume (in liters) occupied by 49.8 g of HCl gas at STP? EXERCISE – 07 1.2 SK015
= 49.8 g 36.45 g/mol = 1.366 mol HCl Molar mass Mole = Mass 1.366 mol x 22.4 L 1 mol = 30.6 L HCl At STP, Volume of HCl gas:
What is the volume (in liters) occupied by 49.8 g of HCl gas at room temperature? EXERCISE – 08 1.2 SK015
= 49.8 g 36.45 g/mol = 1.366 mol HCl Molar mass Mole = Mass 1.366 mol x 24.0 L 1 mol = 32.8 L HCl At room temperature, Volume of HCl gas:
EXERCISE – 09 1.2 A 27.6−mL sample of PH3 (g) (used in the manufacture of flame−retardant chemicals) is obtained at STP. (a) What is the mass of this gas, in milligrams? (b) How many molecules of PH3 are present? SK015
ANS: (a) 41.8 mg PH (b) 7.41 x 10 27.6 x 10–3 L x 1 mol 22.4 L = 1.23 x 10–3 mol PH3 At STP, Molar mass Mole = Mass Mass of PH3 = mole PH3 x molar mass PH3 = 1.23 x 10–3 mol x 33.97 g/mol = 41.8 x 10–3 g PH3 (a) = 4.18 mg PH3
ANS: (a) 41.8 mg PH (b) 7.41 x 10 1.23 x 10–3 mol PH3 (b) 1.23 x 10–3 mol x 6.02 x 1023 molecules 1 mol = 7.41 x 1020 molecules Number of PH3 molecules:
A home remedy calls for teaspoons (20 g) Epsom salt (magnesium sulfate heptahydrate ( MgSO4 • 7H2O). Calculate the number of moles of the hydrate represented by this mass. EXERCISE – 10 1.2 SK015
Molar mass of MgSO4 • 7H2O = 246.38 g/mol 20 g 246.38 g/mol = Molar mass of MgSO4 • 7H2O Mole of MgSO4 • 7H2O = Mass of MgSO4 • 7H2O = 0.081 MgSO4 • 7H2O
What mass of chromium contained in 35.8 g of (NH4 )2Cr2O7? EXERCISE – 11 1.2 SK015
35.8 g 252.06 g/mol = = 0.14203 mol (NH4 )2Cr2O7 Molar mass of (NH4 )2Cr2O7 Mole of (NH4 )2Cr2O7 = Mass of (NH4 )2Cr2O7
1 mol (NH4 )2Cr2O7 contains 2 mol Cr So, 0.14203 mol (NH4 )2Cr2O7 contains 0.14203 x 2 1 = 0.28406 mol Cr
Mass of Cr = mole Cr x molar mass Cr = 0.28406 mol x 51.99 g/mol = 14.8 g Cr Molar mass of Cr Mole of Cr = Mass of Cr 0.28406 mol Cr
How many H atoms are in 72.5 g of C3H8O ? EXERCISE – 12 1.2 SK015
Molar mass of C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g/mol C3H8O Mole of C3H8O = Mass of C3H8O (g) Molar mass of C3H8O (g/mol) = 72.5 g 60 g/mol = 1.208 mol of C3H8O
1 mol C3H8O contains 8 mol H So, 1.208 mol C3H8O contains: 1.208 x 8 1 = 9.66 mol H
Number of H atoms: 1 mol H = 6.02 x 1023 atoms H = 9.66 mol H 6.02 x 1023 atoms H 1 mol H x = 5.82 x 1024 atoms H
EXERCISE – 13 1.2 What is the molecular formula of each compound? (a) Empirical formula CH2 (M = 42.08 g/mol) (b) Empirical formula NH2 (M = 32.05 g/mol) (c) Empirical formula NO2 (M = 92.02 g/mol) (d) Empirical formula CHN (M = 135.14 g/mol) SK015
(a) Empirical formula CH2 (M = 42.08 g/mol) Empirical formula mass Molar mass = 14.03 42.0 = 2.99 So, molecular formula = C3H6
(b) Empirical formula NH2 (M = 32.05 g/mol) Empirical formula mass Molar mass = 16.03 32.05 = 1.999 So, molecular formula = N2H4
(c) Empirical formula NO2 (M = 92.02 g/mol) Empirical formula mass Molar mass = 46.01 92.02 = 2.00 So, molecular formula = N2O4
(d) Empirical formula CHN (M = 135.14 g/mol) Empirical formula mass Molar mass = 27.03 135.14 = 5.00 So, molecular formula = C5H5N5
Calculate the mass % (percent composition) of each element in ethanol, C2H6O. EXERCISE – 14 1.2 SK015
Molecular formula = C2H6O % C = mass of C in 1 mol C2H6O mass of 1 mol C2H6O x 100 % = 2 x 12.01 g C 46.07 g C2H6O x 100 % = 52.14 %
Molecular formula = C2H6O % H = mass of H in 1 mol C2H6O mass of 1 mol C2H6O x 100 % = 6 x 1.008 g H 46.07 g C2H6O x 100 % = 13.13 %
Molecular formula = C2H6O % O = mass of O in 1 mol C2H6O mass of 1 mol C2H6O x 100 % = 1 x 16.00 g O 46.07 g C2H6O x 100 % = 34.73 %
EXERCISE – 15 1.2 Ammonium nitrate is used as fertilizer and to manufacture explosives. Agronomists base the effectiveness of fertilizers on their nitrogen content. (a) Calculate the mass percent of N in ammonium nitrate (NH4NO3 ). (b) How many grams of N are in 35.8 kg of ammonium nitrate? SK015
Formula: NH4NO3 % N = mass of N in 1 mol NH4NO3 mass of 1 mol NH4NO3 x 100 % = 2 x 14.01 g N 80.05 g NH4NO3 x 100 % = 35.00 % (a) Ammonium nitrate