Chapter_3_ISM - Chapter Three 31 Practice Exercises 3.1 mol...

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Unformatted text preview: Chapter Three 31 Practice Exercises 3.1 mol Al = 3.47 g Al 1 mol Al 26.98 g Al ! " # $ % & = 0.129 mol Al 3.2 Uncertainty in moles = 0.002 g 1 mol Si 28.0855 g Si ! " # $ % & = 7.12 10 5 mol Si 3.3 Find the mass of 5.64 10 18 molecules of Ca(NO 3 ) 2 (MW = 164.09 g/mo) g = 5.64 10 18 ( ) ( ) ( ) ( ) 3 3 2 2 23 3 3 2 2 1 mol Ca NO 164.09 g Ca NO 1 mol Ca NO 6.022 10 molecules Ca NO ! " ! " # $ # $ # $ # $ % & & = 1.54 10 3 g g = 1.54 10 3 g = 0.00154 g Many laboratory balances can measure 1 mg (0.001 g); therefore, it is possible to weigh 5.64 10 18 molecules of Ca(NO 3 ) 2 . 3.4 Formula mass of sucrose = (12 C)(12.011 g/mol) + (22 H)(1.0079 g/mol) + (11 O)(15.9994 g/mol) = 342.299 g/mol 0.002 g of uncertainty = ? mol of sucrose mol of sucrose = 0.002 g 1 mol sucrose 342.299 g ! " # $ % & = 5.8 10 6 mol sucrose molecules of sucrose = 5.8 10 6 mol sucrose 23 6.022 10 molecules sucrose 1 mol sucrose ! " # $ % $ % & = 3.5 10 18 molecules of sucrose 3.5 Aluminum sulfate: Al 2 (SO 4 ) 3 , the aluminum is Al 3+ mole Al 3+ = 0.0774 mol SO 4 2 3+ 2 4 2 mol Al 3 mol SO ! " # $ % $ % & = 0.0516 mol Al 3+ 3.6 ( ) 2 mol N mol N = 8.60 mol O 5 mol O ! " # $ % & = 3.44 mol N atoms 3.7 ( ) 1 mol O 2 mol Fe 55.85 g Fe g Fe = 25.6 g O = 59.6 g Fe 16.00 g O 3 mol O 1 mol Fe ! " ! "! " # $# $# $ % &% & % & 3.8 g Fe = ( ) 2 3 2 3 2 3 2 3 1 mol Fe O 2 mol Fe 55.85 g Fe 15.0 g Fe O 10.5 g Fe 159.7 g Fe O 1 mol Fe O 1 mol Fe ! "! " ! " = # $# $ # $ % & % &% & 3.9 g Fe = (12.0 g O) 2 3 2 3 1 mol Fe O 1 mol O 2 mol Fe 55.85 g Fe 16.00 g O 3 mol O 1 mol Fe O 1 mol Fe ! " ! "! " ! " # $ # $# $ # $ % & % & % & % & = 27.9 g Fe Chapter Three 32 3.10 % H = mass H 0.0870 g H 100% 100% total mass 0.6672 g total ! " ! " # = # $ % $ % & & = 13.04% % C = mass C 0.3481 g C 100% 100% total mass 0.6672 g total ! " ! " # = # $ % $ % & & = 52.17% It is likely that the compound contains another element since the percentages do not add up to 100%. 3.11 % N = 0.2012/0.5462 100% = 36.84% N % O = 0.3450/0.5462 100% = 63.16% O Since these two values constitute 100%, there are no other elements present. 3.12 We first determine the number of grams of each element that are present in one mol of sample: 2 mol N 14.01 g/mol = 28.02 g N 4 mol O 16.00 g/mol = 64.00 g O The percentages by mass are then obtained using the formula mass of the compound (92.02 g): % N = (28.02/92.02) 100% = 30.45% N % O = (64.00/92.02) 100% = 69.55% O 3.13 N 2 O: Formula mass = 44.02 g/mol 2 mol N 14.01 g/mol = 28.02 g N % N = (28.02/44.02) 100% = 63.65% N 1 mol O 16.00 g/mol = 16.00 g O % O = (16.00/44.02) 100% = 36.34% O NO: Formula mass = 30.01 g/mol 1 mol N 14.01 g/mol = 14.01 g N % N = (14.01/30.01) 100% = 46.68% N 1 mol O 16.00 g/mol = 16.00 g O % O = (16.00/30.01) 100% = 53.32% O NO 2 : Formula mass = 46.01 g/mol 1 mol N 14.01 g/mol = 14.01 g N % N = (14.01/46.01) 100% = 30.45% N 100% = 30....
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This note was uploaded on 02/13/2011 for the course CHEMISTRY 155 taught by Professor Bullock during the Spring '11 term at UCSB.

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Chapter_3_ISM - Chapter Three 31 Practice Exercises 3.1 mol...

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