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03_Petrucci10e_SSM

# 03_Petrucci10e_SSM - CHAPTER 3 CHEMICAL COMPOUNDS PRACTICE...

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35 CHAPTER 3 CHEMICAL COMPOUNDS PRACTICE EXAMPLES 1A First we convert the number of chloride ions to the mass of MgCl 2 . 2 23 1 22 2 MgCl 2 23 2 2 1 MgCl 1mol MgCl 95.211g MgCl mass 5.0 10 Cl =4.0 10 g MgCl 1mol MgCl 2Cl 6.022 10 MgCl  1B First we convert mass Mg(NO 3 ) 2 to moles Mg(NO 3 ) 2 and formula units Mg(NO 3 ) 2 then finally to NO 3 - ions. 22 32 - 15 3 3 1 g Mg(NO ) 1 mol Mg(NO ) 6.022×10 formula units Mg(NO ) 1.00 μ g Mg(NO ) × × × 1,000,000 μ g Mg(NO ) 148.3148 g Mg(NO ) 1 mol Mg(NO ) 2 NO ions = 4.06×10 formula units Mg(NO ) 1 formula unit Mg(NO ) 15 - 3 2 = 8.12×10 NO ions Next, determine the number of oxygen atoms by multiplying by the appropriate ratio. 15 16 6 atoms O # atoms O = 4.06×10 formula units Mg(NO ) = 2.44×10 O 1 formula unit Mg(NO ) 2A The volume of gold is converted to its mass and then to the amount in moles.  23 2 3 21 1 cm 19.32 g 1 mol Au 6.022 10 atoms #Au atoms= 2.50 cm 0.100 mm 10 mm 196.97 g Au 1 mol Au 1 cm = 3.69 10 Au atoms     2B We need the molar mass of ethyl mercaptan for one conversion factor.   26 = 2 12.011 g C + 6 1.008 g H + 1 32.066 g S = 62.136g/mol C H S M Volume of room: 62 ft × 35 ft × 14 ft = 3.04 ×10 4 ft 3 . We also need to convert ft 3 to m 3 . 33 3 43 23 12 in 2.54 cm 1 m 3.04 10 ft 8.6 10 m 1 ft 1 in 100 cm 6 6 34 3 1.0 L C H S 1mol C H S 1L 1000mL 0.84g 10 mol [C H S] = 8.6 10 m 1 10 L 1mL 62.136g 1mol = 0.016 mol/m 9.0 10 mol/m = the detectable limit   Thus, the vapor will be detectable.

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Chapter 3: Chemical Compounds 36 3A The molar mass of halothane is given in Example 3-3 in the textbook as 197.4 g/mol. The rest of the solution uses conversion factors to change units. 23 2 22 2 3 1mol C HBrClF 1.871g C HBrClF 1 mol Br mass Br = 25.0mL C HBrClF × × × 1mL C HBrClF 197.4g C HBrClF 1mol C HBrClF 79.904 g Br × = 18.9 g Br 1 mol Br 3B Again, the molar mass of halothane is given in Example 3-3 in the textbook as 197.4 g/mol. 24 halothane 23 1mol C HBrClF 197.4g C HBrClF 1mol Br 1mL V = 1.00 10 Br 1mol Br 1mol C HBrClF 1.871g 6.022 10 Br = 175 mL C HBrClF  4A We use the same technique as before: determine the mass of each element in a mole of the compound. Their sum is the molar mass of the compound. The percent composition is determined by comparing the mass of each element with the molar mass of the compound.       10 12.011gC 16 1.008g H 5 14.01g N 3 30.97g P 13 15.999gO =120.11g C +16.13g H + 70.05g N +92.91g P + 207.99g O = 507.19g ATP/mol M   120.11gC 16.13g H %C= 100% 23.681%C %H = 100% 3.180% H 507.19g ATP 507.19g ATP  70.05g N 92.91g P %N = 100% 13.81% N %P= 100% 18.32% P 507.19g ATP 507.19g ATP 207.99gO %O= 100% 41.008%O 507.19g ATP (NOTE: the mass percents sum to 99.999%) 4B Both (b) and (e) have the same empirical formula, that is, CH 2 O. These two molecules have the same percent oxygen by mass. 5A Once again, we begin with a 100.00 g sample of the compound. In this way, each elemental mass in grams is numerically equal to its percent. We convert each mass to an amount in moles, and then determine the simplest integer set of molar amounts. This determination begins by dividing all three molar amounts by the smallest.
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03_Petrucci10e_SSM - CHAPTER 3 CHEMICAL COMPOUNDS PRACTICE...

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