SQSM-01 - Chapter 1 Stoichiometry 1. Determine molar masses...

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Chapter 1 Stoichiometry 1-1 1. Determine molar masses for the following: a) C 22 H 10 O 2 (22 mol C)(12.011 g . mol -1 ) + (10 mol H)(1.008 g . mol -1 ) + (2 mol O)(15.999 g . mol -1 ) = 306.320 g . mol -1 b) Ca(NO 3 ) 2 (1 mol Ca)(40.078 g . mol -1 ) + (2 mol N)(14.007 g . mol -1 ) + (6 mol O)(15.999 g . mol -1 ) = 164.086 g . mol -1 c) P 2 O 5 (2 mol P)(30.974 g . mol -1 ) + (5 mol O)(15.999 g . mol -1 ) = 141.943 g . mol -1 d) Al 2 (CO 3 ) 3 (2 mol Al)(26.982 g . mol -1 ) + (3 mol C)(12.011 g . mol -1 ) + (9 mol O)(15.999 g . mol -1 ) = 233.988 g . mol -1 3. Determine mass of each of the following: a) 0.694 mol C 22 H 10 O 2 (0.694 mol)(306.320 g . mol -1 ) = 213 g b) 2.84 mol Ca(NO 3 ) 2 (2.84 mol)(164.086 g . mol -1 ) = 466 g c) 0.00652 mol P 2 O 5 (0.00652 mol)(141.943 g . mol -1 ) = 0.925 g d) 8.44 mol Al 2 (CO 3 ) 3 (8.44 mol)(233.988 g . mol -1 ) = 1.97x10 3 g = 1.97 kg 5. Determine mass of each of the following: a) 2.24x10 20 molecules of CO 2 20 22 23 2 1 mol CO 44.01 g CO 2.24 10 molecules = 0.0164 g 1 mol CO 6.02 10 molecules ×× × × b) 2.24x10 24 molecules of H 2 24 23 2 1 mol H 2.016 g H 2.24 10 molecules = 7.53 g 1 mol H 6.02 10 molecules × × c) 12 C atoms -22 23 1 mol C 12.011 g C 12 C atoms = 2.39 10 g 1 mol C 6.02 10 C atoms × × d) 8.66x10 18 Pt atoms 18 -3 23 1 mol Pt 195 g Pt 8.66 10 Pt atoms = 2.81 10 g = 2.81 mg 1 mol Pt 6.02 10 Pt atoms × × × 7. How many moles of people are on earth if world population is 6.3 billion (6.3x10 9 ) people? 9- 1 4 23 1 mol people 6.3 10 people = 1.0 10 mol people 6.02 10 people × × 9. A bottle contains 12.6 g of (NH 4 ) 3 PO 4 . a) How many moles of (NH 4 ) 3 PO 4 does it contain? 43 4 4 4 1 mol (NH ) PO 12.6 g (NH ) PO = 0.0845 mol = 84.5 mmol 149.1 g (NH ) PO × b) How many oxygen atoms does it contain? 23 23 4 4 4 mol O atoms 6.02 10 O atoms 0.0845 mol (NH ) PO = 2.03 10 O atoms 1 mol (NH ) PO 1 mol O atoms × × c) What mass of nitrogen atoms does it contain? 4 4 3 mol N 14.007 g N 0.0845 mol (NH ) PO = 3.55 g N 1 mol (NH ) PO 1 mol N d) How many moles of H does it contain? 4 4 12 mol H atoms 0.0845 mol (NH ) PO = 1.01 mol H atoms 1 mol (NH ) PO × 11. What is the simplest formula of each of the following compounds? a) C 22 H 10 O 2 b) C 3 H 6 O c) C 6 H 6 d) C 3 H 6 O 3 C 11 H 5 O C 3 H 6 O CH CH 2 O
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Stoichiometry 1-2 13. What is the elemental composition of each of the molecules in Exercise 11? Express your answer as percents? a) C 22 H 10 O 2 First calculate the molar mass (M m ) of the molecule. M m (C 22 H 10 O 2 ) = 22 x (12.011g / mol C) = 264.24 g C 10 x (1.008 g / mol H) = 10.08 g H + 2 x (15.999 g / mol O) = 31.998 g O 306.32 g / mol C 22 H 10 O 2 The fraction of each element’s contribution will be that element’s mass in the compound over the molar mass of the compound. Multiplying this fraction by 100 gives the percent. 22 10 2 264.24 g C 100% 86.263% C 306.32 g C H O ×= 22 10 2 10.08 g H 100% 3.291% H 306.32 g C H O 22 10 2 31.998 g O 100%=10.446% O 306.32 g C H O × Parts b, c, & d are solved in the same way. b) C 3 H 6 O M m (C 3 H 6 O) = 58.080 g/mol C: 36.033 g C / 58.080 g C 3 H 6 O x 100% = 62.040% C H: 6.048 g H / 58.080 g C 3 H 6 O x 100% = 10.41% H O: 15.999 g O / 58.080 g C 3 H 6 O x 100% = 27.546% O c) C 6 H 6 M m 6 H 6 ) = 78.114 g/mol C: 72.066 g C / 78.114 g C 6 H 6 x 100% = 92.257% H: 6.048 g H / 78.114 g C 6 H 6 x 100% = 7.743% H d) C 3 H 6 O 3 M m (C 3 H 6 O 3 ) = 90.078 g/mol C: 36.033 g C / 90.078 g C 3 H 6 O 3 x 100% = 40.002% C H: 6.048 g H / 90.078 g C 3 H 6 O 3 x 100% = 6.714% H O: 47.997 g O / 90.078 g C 3 H 6 O 3 x 100% = 53.284% O 15. How many moles of magnesium are present in a sample of each of the following that contains 3.0 moles of oxygen atoms?
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This note was uploaded on 01/27/2011 for the course CHEM 201 taught by Professor Wilson during the Spring '10 term at N.C. Central.

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SQSM-01 - Chapter 1 Stoichiometry 1. Determine molar masses...

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