F10HW4-CH3-4

# F10HW4-CH3-4 - HOMEWORK FOR WEEK 4: CHAPTERS 3 AND 4 3.6...

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HOMEWORK FOR WEEK 4: CHAPTERS 3 AND 4 3.6 The empirical formula is the formula of a substance written with the smallest integer (whole number) subscripts. Each of the subscripts in the formula C 6 H 12 O 2 can be divided by two, so the empirical formula of the compound is C 3 H 6 O. 010. The answer is a, 3.27 g of NH 3 . (i) One approach is to calculate the number of molecules in 15.0g of C 6 H 6 (1.2 x 10 23 ) and then find the mass of this number of molecules of NH 3 . # of molecules of C 6 H 6 = 15.0g x 23 1 mol 6.02 x 10 molec x 78.1g 1 mol = 1.16 x 10 23 molec Mass of NH 3 = 1.16 x 10 23 molec x 23 1 mol 17.03g x 6.02 x 10 molec 1 mol = 3.27g (ii) A slightly more cunning approach would be to find the number of moles of C 6 H 6 , then find the mass of this number of moles of NH 3 . # moles C 6 H 6 = 15.0g x 6 6 1 mol C H 78.1g = 0.192 mol Mass of 0.192mol NH 3 = 0.192mol x 3 17.03g 1 mol NH = 3.27g (iii) Finally(?), if you understand that there is a direct proportionality between mass and molar mass for constant number of molecules, then you can multiply the 15.0g of C 6 H 6 by the ratio of molar masses, 17.0g NH 3 / 78.0g C 6 H 6 = 3.27g. 3.25 a. {OCL} Focus The problem is that Avogadro’s number was inadvertently used for the molar mass of calcium, which should be 40.08 g/mol. The correct calculation is 27.0 g Ca x 1 mol Ca 40.08 g Ca = 0.673 6 = 0.674 mol Ca b. In attempting to calculate the number of potassium ions in 2.5 mol of K2SO4, Betty used the incorrect mole ratio. She should have used a 2:1 ratio, not a 1:1 ratio Again, her thought process was correct: she started with what she was given, tried to convert to moles of K+ ions, and then multiply by Avogadro’s number to obtain the number of K+ ions within the number of moles of K2SO4 she started with. 2.5 mol K 2 SO 4 x + 2 4 2 mol K ions 1 mol K SO x 23 + + 6.022 x 10 K ions 1 mol K ions = 3.0 1 x 10 24 = 3.0 x 10 24 K + ions **The most important concept to remember in all of these examples is that for a conversion factor to be appropriate: 1. the units of the denominator must be the same as the units of the quantity being converted. 2. The conversion factor must be equal to 1. That is, the

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numerator, though it has different units, should be equivalent to the denominator in some way. e.g. 1mol = 6.02x10^23 molecules. 3.38 From the table of atomic weights, we obtain the following molar masses for parts a through d: C = 12.01 g/mol; O = 16.00 g/mol; K = 39.10 g/mol; Cr = 52.00 g/mol; Fe = 55.85 g/mol; and F = 19.00 g/mol. a. 0.205 mol Fe x Fe mol 1 Fe g 55.85 = 11.4 4 = 11.4 g Fe d. Using molar mass 194.20 g/mol for K 2 CrO 4 , we obtain 1.02 4 93 = CrO K mol 1 CrO K g 194.20 x CrO K mol 48.1 4 2 4 2 4 2 = 9.34 x 10 3 g K 2 CrO 4 3.46 All the masses are typical of lab amounts. We are asked for the number of atoms or molecules in these samples and these have to be very large numbers. a) Number of atoms in 25.7g Al. (O) First we need to realize that we
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## This note was uploaded on 02/20/2011 for the course CHEM 110 taught by Professor Dr.huston during the Spring '06 term at Pittsburgh.

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F10HW4-CH3-4 - HOMEWORK FOR WEEK 4: CHAPTERS 3 AND 4 3.6...

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