Hints for Assignment 3

 
 

1, 2. Hints:

There is no substitute for knowing the names and formulas (including charges) for common cations and anions.

Make yourself a set of flashcards with the name of a cation or anion on one side, and its formula on the other. Not high-tech but it works!

Also remember that you do not name the number of cations or anions! CaI2 is calcium iodide, not calcium diiodide. Fe2(SO4)3 is iron(III) sulfate, not diiron trisulfate.

For lots of practice, go to the ChemTeam page and work through the sections on naming compounds.

http://dbhs.wvusd.k12.ca.us/ChemTeamIndex.html

3. Hints: Review the rules for naming binary compounds between nonmetals on page 118 of Kotz and Treichel. Note that the compounds given at the bottom of the page are called by their common names.
In contrast to ionic compounds, the numbers of each kind of atom are given for binary compounds of nonmetals. Thus P4S10 is tetraphosphorus decasulfide, not phosphorus(V) sulfide.
4. Hints: This question is a straightforward calculation involving mass to moles, moles to mass, and atoms to moles. See Examples 3.6, 3.7, and 3.8 on p. 123, Kotz and Treichel, 4th ed.
 
 

5. Sample Problem. This question refers to the compound potassium aluminum sulfate dodecahydrate (commonly known as alum, used for making pickles). Its formula is

KAl(SO4)2.12H2O.
(a) Calculate its formula weight. Answer: The overall formula is KAlS2O20H24. The formula weight is thus
39.0983 + 26.9815 + 2 x 32.066 + 20 x 15.9994 + 24 x 1.0079 = 474.38 (units are amu/formula unit or, for one mole, g/mol)
(b) What is the % water in the compound? Answer:
% water = 100 x (weight of water in 1 mole alum) / (weight of 1 mole of alum)

= [100 x 12 mol H2O/mol alum x 18.015 g H2O/mol H2O ] / [474.38 g alum/mol alum]

= 45.57%

(c) What is the % oxygen in the compound? Answer:
% oxygen = 100 x (weight of oxygen in 1 mole alum) / (weight of 1 mole of alum)

= [100 x 20 mol O/mol alum x 15.9994 g O/mol O] / [474.38 g alum/mol alum]

= 67.45 %

(d) How many atoms of hydrogen are present in 1.00 g of alum?
Hint. Each formula unit of alum contains 24 hydrogen atoms. Therefore one mole of alum contains 24 moles of hydrogen atoms. The problem solving sequence is:

weight alum Æ moles alum Æ moles H atoms Æ number of H atoms

6. Hint 1: What weight of the element do you need? How is this related to the volume needed, and thus to the length?

Hint 2: Wire has a circular cross section, so a piece of wire is a cylinder. The volume of a cylinder is:

V = pi  r2 l, where r is the radius of the wire and l is its length.

Remember that the radius of a circle is half the diameter.

7. Hints: See Example 3.10, p. 130, Kotz and Treichel, 4th ed.
It's useful to know the whole-number ratios for some decimal fractions. A few examples of the more common ratios seen in formulas:
0.2500 = 1:4  0.3333 = 1:3  0.2000 = 1:5 
0.7500 = 3:4  0.6666 = 2:3  0.4000 = 2:5 
1.500 = 3:2  1.3333 = 4:3  0.6000 = 3:5 
2.500 = 5:2  1.6667 = 5:3  0.8000 = 4:5 
8. Hint: this question links percentage of a mineral in an ore with percent composition of the mineral. Watch units carefully (especially: how do you represent % in terms of dimensional analysis) and you should be ok.
Sample problem. An ore contains 15.35 % of the mineral powellite, CaMoO4, which is a source of the element molybdenum. How much ore must be processed in order to obtain 50.0 kg of Mo?
The ore contains 15.35% CaMoO4; therefore 15.35 g CaMoO4 in 100 g ore:

CaMoO4 has 1 mole of Mo in each mole of CaMoO4, or 95.94 g Mo in 200.01 g of CaMoO4.

So now that we have defined the information we've been given, we can use dimensional analysis to solve the problem:

= 6.79 x 105 g or 679 kg of ore is needed to yield 50.0 kg of molybdenum metal
 
 

9. Hint: See Example 3.13 on p. 137 of Kotz and Treichel, 4th ed.
 
 

Back to Chem 1P80 Home Page

Back to Assignment Schedule

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem180/CHEM_1P80_Assign_3.html
Last revised: August 20 2000 by M. F. Richardson
© Brock University, 1999