S99.MIw

Department of Chemistry

CARNEGIE MELLON UNIVERSITY

09-105 MASTERY EXAM I
January 19, 1999 Section

Show all work.

This EXAM must be returned.

N_Avog. = 6.022 X 10²³ mol^-1

1 L = 1000 cm³ = 10^-3 m³

	Element	Atomic Weight

	H	1.008
	Li	6.941
	B	10.811
	C	12.01
	N	14.01
	O	16.00
	Na	22.99
	Cl	35.45
	K	39.10
	As	74.92
	I	126.90
	Hg	200.59

	1.	__________/25
	2.	__________/25
	3.	__________/25
	4.	__________/25
	Total	__________/100

1. Lithium borate is a solid compound that is used in the preparation of certain enamels. Its chemical analysis shows it to be 25.57% boron (B), 8.21% lithium (Li) and 66.22% oxygen (O) by mass. Consistent with the precision of these values, what is the empirical formula for lithium borate? (Show all work.)

Assuming you have, for example, 100 g of compound, you then have 25.57 g of B --> 25.57 g of B/10.811 g of B per mol of B = 2.365 moles of B 8.21 g of Li --> 8.21 g/6.941 g per mol = 1.183 moles of Li 66.22 g of O --> 66.22 g/16.00 g per mol = 4.139 moles of O	10 pts
The mole (number) ratio of boron:lithium:oxygen is 2.365:1.183:4.139	5 pts
To convert these to integers for an empirical formula, first divide by the lowest value, 1.183 giving B:Li:O = 2.00:1:3.50	5 pts
Next, double these to get integers of a precision (significant figures!) consistent with the original information. B:Li:O = 4:2:7 -->	5 pts

Li₂B₄O₇

2. The atomic masses of ¹⁰B and ¹¹B are 10.013 amu (atomic mass units) and 11.009 amu, respectively. A sample of boron from a nuclear reactor laboratory is analyzed and found to have an average atomic weight of 10.887 amu. Calculate the percent abundances by mass of each boron isotope in this sample. (Show all work.)

Letting x = the fractional abundance of ¹⁰B, we have 10.013x + 11.009(1 - x) = 10.887	10 pts
10.013x + 11.009 - 11.009x = 10.887 0.122 = 0.996x x = 0.122 = fractional abundance of boron 10	10 pts
¹⁰B percent abundance = 12.2%	3 pts
¹¹B percent abundance = 100% - 12.2% = 87.8%	2 pts

3. Potassium nitrate (KNO₃), also known as saltpeter, is employed in fertilizer, fireworks, pickling meats and treating tobacco among other uses. Ingestion of large quantities may cause violent gastroenteritis. It is produced through the reaction, unbalanced, shown below.

KCl + HNO₃ + O₂ KNO₃ + Cl₂ + H₂O

Calculate the minimum mass of chlorine gas (Cl₂) that will be produced during the production of each 1000 g of KNO₃.

First, the equation must be balanced or everything that follows is incorrect!

4 KCl +4 HNO₃ + O₂ 4 KNO₃ +2 Cl₂ + 2 H₂O	10 pts
This means that 2 moles of chlorine are produced for every four moles of potassium nitrate produced, or the number of moles of chlorine will be one-half that of potassium nitrate. The number of moles of KNO₃is 1000 g KNO₃/101.11 g KNO₃ per mole KNO₃ = 9.89 moles of KNO₃.	5 pts
Hence, (1/2)(9.89 moles) = 4.94 moles of Cl₂ must have been produced. This corresponds to	5 pts
(4.94 moles)(70.9 g Cl₂ per mole Cl₂) = 351 g Cl₂	5 pts

4. Veterinarians use Donovan's solution to treat skin diseases in animals. The solution is prepared by mixing 1.00 g of AsI₃, 1.00 g of HgI₂, and 0.900 g of NaHCO₃ in enough water to make a total volume of 100.0 mL. Compute the total mass of iodine per liter of Donovan's solution, in grams per liter.

1.00 g AsI₃corresponds to 1.00 g/455.62 g per mole = 0.00219 moles AsI₃	3 pts
1.00 g HgI₂corresponds to 1.00 g/454.39 g per mole = 0.00220 moles HgI₂	3 pts
0.00219 moles AsI₃corresponds to (3 moles I/mole AsI₃)(0.00219 moles AsI₃) = 0.00658 moles I	3 pts
0.00220 moles HgI₂corresponds (2 moles I/mole HgI₂)(0.00220 moles HgI₂) =0.00440 moles I	3 pts
Total number of moles of I is 0.00648 + 0.00440 = 0.0110 moles iodine	3 pts
corresponding to (0.0110 moles)(126.9 g per mol) = 1.39 g iodine (in the 100 mL)	3 pts
For 1 L (as requested), you have 13.9 g of iodine.	2 pts

How many atoms of iodine are in one liter of Donovan's solution?

0.011 moles iodine per 100 mL --> 0.110 moles per L	1 pts
0.110 mol/L corresponds to (0.110 moles)(6.02 X 10²³ atoms/mole) = 6.62 X 10²² iodine atoms	4 pts