Department of Chemistry

CARNEGIE MELLON UNIVERSITY

Introduction to Modern Chemistry (09-105)

MASTERY EXAM III Name KEY

October 20, 1998

Instructions:

Show all work.

Use significant figures correctly.

Double-check your results!

This EXAM MUST BE RETURNED BY THE END OF CLASS.

1. When aluminum metal reacts with copper oxide, metallic copper is produced along with aluminum oxide according to the following reaction (which is unbalanced).

Al + CuO --> Al₂O₃ + Cu

If 500.0 g of copper oxide are mixed with 80.00 g of aluminum, how many grams of what material(s) do you expect to find when the reaction has concluded?

 (balanced) 2 Al + 3 CuO --> Al₂O₃ + 3 Cu

Mol.wts. (Al=26.98); (CuO=79.55); (Al₂O₃=101.96); (Cu=63.55)

We have 80.00g/26.98 g/mol = 2.968 mols Al which would consume

2.968(1.5)=4.452 mols CuO, producing 4.452 mols Cu

and 2.968/2=1.482 mols Al₂O₃ leaving behind unconsumed CuO. Al is the limiting reagent.

We have 500.0g/ 79.55 g/mol = 6.285 mol CuO of which 4.452 get consumed, leaving

Behind 6.282-4.452=1.833 mol CuO.

Our final composition is 1.833 mol CuO, 1.482 mol Al₂O₃, and 4.452 mol Cu

Corresponding to (1.833 mol)(79.55 g/mol) = 145.8 g CuO

Plus (1.482 mol)(101.96 g/mol) = 151.2 g Al₂O₃

Plus (4.452 mol)(63.55 g/mol) = 282.9 g Cu

CHECK! Initial mass = 80.00 g + 500.0 g = 580.0 g

Final mass = 145.8 g + 151.2 g + 282.9 g = 579.9 g (CHECKED!)

2. You are given 25.96 grams of lithium metal and determine that the atomic weight of lithium in this sample is 6.5000. Lithium consists of two stable isotopes. ⁷Li has an atomic weight of 7.0160 and ⁶Li has an atomic weight of 6.0151. What is the percent composition by mass of these two isotopes in your lithium sample?

 Let x = fraction that is ⁷Li

 6.500 = 7.0160(x) + 6.0151(1-x)

 0.4849 = 1.0009x

 0.4845 = x

 ⁷Li = 48.45%

 ⁶Li = 51.55%

3. An unknown element combines with oxygen to form a compound with the formula MO₃. If 122.6 g of the element combines with 32.0 g of oxygen, what is the atomic mass of M?

 (There is more than one way to approach this....)

 MO₃ means the moles "M" to moles "O" is 1:3

 The number of moles of O is 32.0g/16.0 g/mol = 2.00 moles O

 Then the number of moles of "M" is one-third that = 0.667 moles M

 0.667 moles M = 122.6 g M/ x g M/mol M

 x = 183.9 g/mol

4. Determine the empirical formula of a compound having the following percent composition by mass: K, 27.53%; Mn, 38.68%; O, 33.79% (Show all work.)

 Assume you have 100.0 g

K: 27.53 g/ 39.10 g/mol = 0.7041 mol K

Mn: 38.68 g/ 54/94 g/mol = 0.7040 mol Mn

O: 33.79 g/16.00 g/mol = 2.112 mol O

K:Mn:O = 0.7041:0.7040:2.112 --> 1.00:1:3.00

Empirical formula is KMnO₃

5. What is the ethanol molarity of a solution whose volume is 85.0 mL and which contains 1.77 g of ethanol (C₂H₅OH)?

 Molecular weight of ethanol is 2(12.01) + 6(1.008) + 16.00 = 46.07

 1.77 g --> 1/77 g/46.07 g/mol = 0.0384 mols

 molarity = 0.0384 mols/ 0.0850 L = 0.452 mols/L or 0.452 M
(not 0.000452 mols/mL!)

How should you prepare 200.0 mL of a 0.866 M ethanol solution starting with a 5.00 M ethanol solution.

 200.0 mL (0.2000 L) of 0.866 M ethanol contains

 (0.2000 L)(0.866 mol/L) = 0.173 mol ethanol

 If we take "V" liters of the 5.00 M solution, it contains 5.00V moles of ethanol which should amount to 0.173 moles.

0.173 moles = (5.00 mols/L)(V)

V = 0.173/5.00 = .035 L = 35 mL.

We then need to take that 35 mL of 5.00 M ethanol and add ~165 mL of water to bring the final volume to 200 mL.