Fundamentals%20FtoK%20Solns

Fundamentals%20FtoK%20Solns - F.1 C10H16O 10 12.01 g =...

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F.1 C 10 H 16 O 10 × 12.01 g = 120.1 g × ( 100 152.2264 ) = 78.90% 16 × 1.0079 g = 16.1264 g × ( 100 152.2264 ) = 10.59% 1 × 16.00 g = 16.00 g × ( 100 152.2264 ) = 10.51% F.5 (a) 2 M O 88.8% M For 100 g of compound, 88.8 g is M, 11.2 g is O. 2 2 2 2 1 2 ? g M O 100 g M O 16.00 g O 1 mol O = mole M O 11.2 g O 1 mol O 1 mol M O = 143 g mol M O - Therefore, 143−16 = 127 g/mol are due to M. Since there are 2 moles of M per mole of 2 M O , the molar mass of M = 63.4 g/mol. That molar mass matches Cu. (b) copper(I) oxide F.7 (a) For 100 g of compound, moles of 1 32.79 g Na 1.426 mol 22.99 g mol - = = moles of 1 13.02 g Al 0.4826 mol 26.98 g mol - = = moles of 1 54.19 g F 2.852 mol 19.00 g mol - = = Dividing each number by 0.4826 gives a ratio of 1 Al : 2.95 Na : 5.91 F. The formula is 3 6 Na AlF . (b) For 100 g of compound, moles of 1 31.91 g K 0.8161 mol 39.10 g mol - = = moles of 1 28.93 g Cl 0.8161 mol 35.45 g mol - = = mass of O is obtained by difference:
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moles of 1 100 g 31.91 g 28.93 g O 2.448 mol 16.00 g mol - - - = = Dividing each number by 0.8161 gives a ratio of 1.00 K : 1 Cl : 3.00 O. The formula is 3 KClO . (c) For 100 g of compound, moles of 1 12.2 g N 0.871 mol 14.01g mol - = = moles of 1 5.26 g H 5.22 mol 1.0079 g mol - = = moles of 1 26.9 g P 0.869 mol 30.97 g mol - = = moles of 1 55.6 g O 3.475 mol 16.00 g mol - = = Dividing each number by 0.869 gives a ratio of 1.00 N : 6.01 H : 1.00 P : 4.00 O. The formula is 6 4 4 2 4 NH PO or [NH ][H PO ], ammonium dihydrogen phosphate. F.15 For 100 g of caffeine, moles of 1 49.48 g C 4.12 mol 12.01g mol - = = moles of 1 5.19 g H 5.15 mol 1.0079 g mol - = = moles of 1 28.85 g N 2.059 mol 14.01g mol - = = moles of 1 16.48 g O 1.03 mol 16.00 g mol - = = Dividing each number by 1.03 mol gives a ratio of 4.00 C : 5.00 H : 2.00 N : 1.00 O. The formula is 4 5 2 C H N O with a molar formula mass of 1 97.10 g mol . - Because the molecular molar mass is twice this value, the actual formula will be 8 10 4 2 C H N O .
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F.23 This problem requires that we relate unknowns to each other appropriately by writing a balanced chemical equation and using other information in the problem. + 2 3 2 4 3 4 + + 1 + + 1 1 NaNO Na SO ( 2 ) Na NO + SO 1.61 g Na 0.07003 mol Na 2 22.99 g mol Na 5.37 g total 1.61 g Na =3.76 g =(62.01 g mol ) + (96.07 g mol ) x y x y x y x y x y - - - - - + + + = = + - Rearrange and substitute: 3.76 = 62.01(0.07003 2 ) 96.07 0.06065 0.07003 2 1.548 0.009385 0.4516 0.02078 moles sulfate, = 0.02847 moles nitrate y y y y y y x - + = - + = = Therefore, the mass of sodium nitrate in the mixture was 3 3 3 3 85.00 g NaNO 0.02847 mol 2.42 g NaNO 1 mol 2.42 g NaNO 100 45.1% NaNO . 5.37 g total = = G.1 (a) density; (b) the abilities of the components to adsorb; (c) boiling points G.5 mass of 1 1 3 AgNO (0.179 mol L )(0.5000 L)(169.88 g mol ) 15.2 g - - = = G.9 (a) Weigh 1.6 g (0.010 mol, molar mass of 1 4 KMnO 158.04 g mol ) - = into a 1.0-L volumetric flask and add water to give a total volume of 1.0 L. Smaller (or larger) volumes could also be prepared by using a
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Fundamentals%20FtoK%20Solns - F.1 C10H16O 10 12.01 g =...

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