Fundamentals%20FtoK%20Solns

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

This preview shows pages 1–4. Sign up to view the full content.

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:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online