Calculate the molar mass of each of the following: (a) SnO 2. Calculate the molar mass of each of the following: (a) N 2 O 4
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- Abner Neal
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1 Calculate the molar mass of each of the following: (a SnO (b BaF (c Al (SO 3 (d MnCl Calculate the molar mass of each of the following: (a N O (b C 8 H 10 (c MgSO. 7 H O (d Ca(C H 3 O (a Mass in grams of 0.57 mol of KMnO (b Moles of O atoms in 8.18 g of Mg(NO 3 (c Number of O atoms in 8.1 x 10-3 g of CuSO 5H O (a Mass in kilograms of 3.8x10 0 molecules of NO (b Moles of Cl atoms in 0.05 g of C H Cl
2 (c Number of H - ions in.9 g of SrH (a Mass in grams of 0.6 mol of MnSO (b Moles of compound in 15.8 g of Fe(ClO 3 (c Number of N atoms in 9.6 g of NH NO (a Total number of ions in 38.1 g of CaF (b Mass in milligrams of 3.58 mol of CuCl?H O (c Mass in kilograms of.88 x 10 formula units of Bi(NO 3 3 5H O (a Mass in grams of 8.1 mol of copper(i carbonate (b Mass in grams of.0x10 1 molecules of dinitrogen pentaoxide (c Number of moles and formula units in 57.9 g of sodium perchlorate (d Number of sodium ions, perchlorate ions, Cl atoms, and O atoms in the mass of compound in part (c
3 a M = (1 mol Sn (118.7 g Sn/mol Sn + ( mol O (16.00 g O/mol O = g/mol of SnO b M = (1 mol Ba (137.3 g Ba/mol Ba + ( mol F (19.00 g F/mol F = g/mol of BaF c M = ( mol Al (6.98 g Al/mol Al + (3 mol S (3.07 g S/mol S + (1 mol O (16.00 g O/mol O = 3.17g/mol of Al (SO 3 d M = (1 mol Mn (5.9 g Mn/mol Mn + ( mol Cl (35.5 g Cl/mol Cl = 15.8 g/mol of MnCl 3.11 Plan: The mass of a substance and its number of moles are related through the conversion factor of M, the molar mass expressed in g/mol. The moles of a substance and the number of entities per mole are related by the conversion factor, Avogadro s number. Solution: a M of KMnO = ( x = g/mol of KMnO Mass of KMnO = 0.57 mol KMnO g KMnO 1 mol KMnO = = 9.0 x 10 1 g KMnO b M of Mg(NO 3 =.31 + ( x (6 x = g/mol Mg(NO 3 3 Moles of O atoms = 8.18 g Mg(NO 3 1 mol Mg(NO 6 mol O atoms g Mg(NO 1 mol Mg(NO 3 3 = = mol O atoms c M of CuSO 5 H O = (9 x (10 x = 9.70 g/mol (Note that the waters of hydration are included in the molar mass mol Cu Cmpd 9 mol O atoms 6.0 x 10 O atoms O atoms = 8.1 x 10 g Cu Cmpd 9.70 g Cu Cmpd 1 mol Cu Cmpd 1 mol O atoms = x 10 0 = 1.8 x 10 0 O atoms 3.1 a Mass NO = 3.8 x10 molecules NO 1 mol NO 6.01 g NO 1 kg 6.0 x 10 molecules NO 10 g mol NO =.9033 x 10 5 =.9 x 10 5 kg NO
4 b Moles Cl atoms = 0.05 g C H Cl = x 10 = 8.59 x 10 mol Cl atoms 1 mol C H Cl mol Cl atoms g C H Cl 1 mol C H Cl c Number of H =.9 g SrH x 10 = 6.61 x 10 H ions 3 1 mol SrH mol H 6.0 x 10 H ions 89.6 g SrH 1 mol SrH 1 mol H 3.13 Plan: Determine the molar mass of each substance, then perform the appropriate molar conversions. Solution: a M of MnSO = (5.9 g Mn/mol Mn + (3.07 g S/mol S + [( mol O (16.00 g O/mol O] = g/mol of MnSO Mass of MnSO = g MnSO ( 0.6 mol MnSO = = 97 g MnSO 1mol MnSO b M of Fe(ClO 3 = (55.85 g Fe/mol Fe + [(3 mol Cl (35.5 g Cl/mol Cl] + [(1 mol O (16.00 g O/mol O] = 35.0 g/mol of Fe(ClO 3 Moles Fe(ClO 3 = 1mol Fe(ClO 3 ( 15.8 g Fe(ClO g Fe(ClO 3 = =.6 x 10 mol Fe(ClO 3 c M of NH NO =[( mol N (1.01 g N/mol N] + [( mol H (1.008 g H/mol H] + [( mol O (16.00 g O/mol O] = 6.05 g/mol of NH NO N atoms = 3 1 mol mol N 6.0 x 10 N atoms 9.6g NH NO 6.05 g 1 mol NH NO 1 mol N = x 10 = 1.7 x 10 N atoms 3.1 FU = formula unit
![Solution: a M of MnSO = (5.9 g Mn/mol Mn + (3.07 g S/mol S + [( mol O (16.00 g O/mol O] = 151.01 g/mol of MnSO Mass of MnSO = 151.01g MnSO ( 0.6 mol MnSO = 96.](/docs-images/44/17850912/images/page_4.jpg "65 = 97 g MnSO 1mol MnSO b M of Fe(ClO 3 = (55.85 g Fe/mol Fe + [(3 mol Cl (35.5 g Cl/mol Cl] + [(1 mol O (16.00 g O/mol O] = 35.0 g/mol of Fe(ClO 3 Moles Fe(ClO 3 = 1mol Fe(ClO 3 ( 15.")
5 a Ions = 38.1 g CaF 3 1 mol CaF 6.0 x 10 FU CaF 3 ions g CaF 1 mol CaF 1 FU CaF = x 10 3 = 8.8 x 10 3 ions g CuCl? H O 1 mg 1 mol CuCl? H O 10 g b Mass CuCl H O = 3.58 mol CuCl? H O = = 6.10 x 10 5 mg CuCl H O c Mass Bi(NO 3 3 5H O =.88 x 10 FU = = 0.03 kg Bi(NO 3 3 5H O 3 1 mol 85.1 g Bi(NO? H O 1 kg 6.0 x 10 FU 10 g FU Bi(NO 3 3? HO 3.15 Plan: The formula of each compound must be determined from its name. The molar mass for each formula comes from the formula and atomic masses from the periodic table. Avogadro s number is also necessary to find the. number of particles. Solution: a Carbonate is a polyatomic anion with the formula, CO 3. Copper (I indicates Cu +. The correct formula for this ionic compound is Cu CO 3. M of Cu CO 3 = ( x (3 x = g/mol 3 Mass Cu CO 3 = 8.1 mol Cu CO g Cu CO 1 mol CuCO3 = = 1.57 x 10 3 g Cu CO 3
6 b Dinitrogen pentaoxide has the formula N O 5. Di indicates N atoms and penta indicates 5 O atoms. M of N O 5 = ( x (5 x = g/mol Mass N O 5 =.0 x 10 N O molecules 1 mol N O g N O x 10 N 1 mol N O5molecules O5 = = g N O 5 c The correct formula for this ionic compound is NaClO. There are Avogadro s number of entities (in this case, formula units in a mole of this compound. M of NaClO = ( x = 1. g/mol Moles NaClO = 57.9 g NaClO 1 mol NaClO 1. g NaClO = = 0.73 mol NaClO FU = formula units FU NaClO = 57.9 g NaClO 3 1 mol NaClO 6.0 x 10 FU NaClO 1. g NaClO 1 mol NaClO = x 10 3 =.85 x 10 3 FU NaClO d The number of ions or atoms is calculated from the formula units given in part c. Note the unrounded initially calculated value is used to avoid intermediate rounding x 10 3 mol NaClO 1 Na ion 1 FU NaClO =.85 x 10 3 Na + ions x ClO ion mol NaClO 1 FU NaClO =.85 x 10 3 ClO ions x 10 3 mol NaClO 1 Cl atom 1 FU NaClO =.85 x 10 3 Cl atoms x 10 3 mol NaClO O atoms 1 FU NaClO = 1.1 x 10 O atoms
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