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EXAMPLE 12.1

EXAMPLE 12.1. Lewis Structures. Solution.

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EXAMPLE 12.1

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  1. EXAMPLE 12.1 Lewis Structures Solution Although these molecules both look complicated, it isn’t hard to draw skeletal structures and Lewis structures. Each molecule has two carbon atoms, and all the remaining atoms tend to have a single bond each. That means that the two carbon atoms are connected, and each carbon atom has three more atoms connected to it. Now we count the valence electrons: H = 1, C = 4, Cl = 7, F = 7. Subtract a pair of electrons for each bond in the skeletal structure to find the number of electrons remaining. In structure (a) the hydrogen atoms and the carbon atom have filled shells. The four fluorine atoms need six more electrons each, so we place 24 electrons on the fluorine atoms (six each). In structure (b) the three fluorine and two chlorine atoms each need six more electrons, so we place six electrons on each. Cl F F F (a) H C C F (b) H C C F H F Cl F 38 – 14 = 24 44 – 14 = 30 Cl F F F H C C F H C C F H F Cl F Draw Lewis structures for (a) the hydrofluorocarbon CH2FCF3; (b) the hydrochlorofluorocarbon CHCl2CF3.

  2. EXAMPLE 12.1 Lewis Structures Exercise 12.1A Exercise 12.1B Draw a Lewis structure for Freon-12, which has the formula CCl2F2. Freon-12 is unreactive in part because it is almost nonpolar. Explain the low polarity of the Freon-12 molecule. (Hint: What are the electronegativities of fluorine and chlorine?)

  3. EXAMPLE 12.2 Stoichiometry Solution The problem gives us a mass of a reactant, and we are to find the mass of a product, so this is a stoichiometry problem, as covered in Chapter 6 (see Example 6.13 for a similar example). We begin by converting the mass of the given substance, C8H18, to moles. Now we use the relationship of 2 mol C8H18 16 mol CO2 to find moles of CO2. The final step is to use the molar mass of CO2 to determine the mass of CO2. The mass of CO2 is actually greater than the mass of gasoline consumed, because the carbon of the gasoline is combined with oxygen from the air. 1 mol C8H18 2670 g C8H18x = 23.4 mol C8H18 114.0 g C8H18 16 mol CO2 23.4 mol C8H18x = 187 mol CO2 2 mol C8H18 44.0 g CO2 187 mol CO2x = 8240 g CO2 1 mol CO2 What mass of carbon dioxide (molar mass = 44.0 g/mol) is produced by the burning of 2670 g (1 gallon) of gasoline? Gasoline may be represented by the formula C8H18 (molar mass = 114.0 g/mol). 2 C8H18(I) + 25 O2(g)  18 H2O(g) + 16 CO2(g)

  4. EXAMPLE 12.2 Stoichiometry continued Exercise 12.2A Exercise 12.2B Calculate the mass in grams of water vapor produced from burning 2670 g of gasoline. Calculate the mass in kilograms of carbon dioxide produced by a class of 60 students in one week, if each student uses 10 gallons of gasoline in a week.

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