Chapter 3

# Chapter 3 - CHAPTER 3 STOICHIOMETRY 3.5 3.6(34.968...

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CHAPTER 3 STOICHIOMETRY 3.5 ( 34.968 amu)(0.7553) + (36.956 amu)(0.2447) = 35.45 amu 3.6 Strategy: Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope. It would seem that there are two unknowns in this problem, the fractional abundance of 6 Li and the fractional abundance of 7 Li. However, these two quantities are not independent of each other; they are related by the fact that they must sum to 1. Start by letting x be the fractional abundance of 6 Li. Since the sum of the two abundance’s must be 1, we can write Abundance 7 Li = (1 x ) Solution: Average atomic mass of Li = 6.941 amu = x (6.0151 amu) + (1 x )(7.0160 amu) 6.941 = 1.0009 x + 7.0160 1.0009 x = 0.075 x = 0.075 x = 0.075 corresponds to a natural abundance of 6 Li of 7.5 percent . The natural abundance of 7 Li is (1 x ) = 0.925 or 92.5 percent . 3.7 23 6.022 10 amu The conversion factor required is 1g ⎛⎞ × ⎜⎟ ⎝⎠ 23 13.2 amu 6.022 10 amu = × 23 ? g 2.19 10 g × 3.8 23 6.022 10 amu The unit factor required is × 23 6.022 10 amu 8.4 g = × 24 ? amu 5.1 10 amu × 3.11 In one year: 91 7 365 days 24 h 3600 s 2 particles (6.5 10 people) 4.1 10 particles/yr 1 yr 1 day 1 h 1 person ×× × × ×= × 23 17 6.022 10 particles 4.1 10 particles/yr × == × 6 Total time 1.5 10 yr × 3.12 The thickness of the book in miles would be: 23 16 0.0036 in 1 ft 1 mi (6.022 10 pages) = 3.4 10 mi 1 page 12 in 5280 ft × × ×

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CHAPTER 3: STOICHIOMETRY 31 The distance, in miles, traveled by light in one year is: 8 12 365 day 24 h 3600 s 3.00 10 m 1 mi 1.00 yr 5.88 10 mi 1 yr 1 day 1 h 1 s 1609 m × ×× × × ×= × The thickness of the book in light-years is: 16 12 1light-yr (3.4 10 mi) 5.88 10 mi = × 3 5.8 10 light - yr × It will take light 5.8 × 10 3 years to travel from the first page to the last one! 3.13 23 6.022 10 S atoms 5.10 mol S 1molS × 24 3.07 10 S atoms × 3.14 9 23 1molCo (6.00 10 Co atoms) = 6.022 10 Co atoms × 15 9.96 10 mol Co × 3.15 1mol Ca 77.4 g of Ca 40.08 g Ca 1.93 mol Ca 3.16 Strategy: We are given moles of gold and asked to solve for grams of gold. What conversion factor do we need to convert between moles and grams? Arrange the appropriate conversion factor so moles cancel, and the unit grams is obtained for the answer. Solution: The conversion factor needed to covert between moles and grams is the molar mass. In the periodic table (see inside front cover of the text), we see that the molar mass of Au is 197.0 g. This can be expressed as 1 mol Au = 197.0 g Au From this equality, we can write two conversion factors. 1 mol Au 197.0 g Au and 197.0 g Au 1 mol Au The conversion factor on the right is the correct one. Moles will cancel, leaving the unit grams for the answer. We write 197.0 g Au = 15.3 mol Au = 1molAu × 3 ? g Au 3.01 10 g Au × Check: Does a mass of 3010 g for 15.3 moles of Au seem reasonable? What is the mass of 1 mole of Au? 3.17 (a) 23 200.6 g Hg 1 mol Hg 1 mol Hg 6.022 10 Hg atoms × 22 3.331 10 g/Hg atom × (b) 23 20.18 g Ne 1 mol Ne 1molNe 6.022 10 Ne atoms × 23 3.351 10 g/Ne atom ×
CHAPTER 3: STOICHIOMETRY 32 3.18 (a) Strategy: We can look up the molar mass of arsenic (As) on the periodic table (74.92 g/mol). We want to find the mass of a single atom of arsenic (unit of g/atom).

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## This note was uploaded on 10/11/2009 for the course CHE 102 taught by Professor Maggie during the Fall '09 term at Wisc Whitewater.

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Chapter 3 - CHAPTER 3 STOICHIOMETRY 3.5 3.6(34.968...

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