Download CHEM 1103 SUMMER, 2003 QUIZ 2 KEY

CHEM 1103 SUMMER, 2003 QUIZ 2 KEY
1. Balance the number of atoms on both sides of the equation. See text, pp134-139.
Answer: B.
2. Add up the atomic masses of each atom. Note that there are 3 atoms of Ca, 3 of P
and 12 of O. You can short cut here by rounding off the atomic masses to whole
numbers. Why? See text, pp 96-97. Answer: B.
3. Find the molar mass, then multiply by the number of moles present. See text pp.
101-104. Answer: C.
4. Find the mass in g of 1 mole of oxygen gas, then multiply by 0.30. See text, pp
101-104. Answer: A.
5. Write a balanced chemical equation. See text, pp 134-139. Then use the mole
ratio (2:1) to find the correct number of moles of product. See text, section 4.2.
Answer: D.
6. Find the molar mass of nitrogen triiodide (392 g/mol). Divide the number of
grams present (2.5x10-3) by this. Shortcut: You are dividing 10-3 by 102 to equal
about 10-5. The only answer close to this is D, which is the correct answer.
7. The empirical formula (CH) mass is 13 and the molar mass of the compound is 52.
There must therefore be 52/13 = 4 units of the empirical formula in the compound
to give C4H4. See text, section 3.6. Answer: D.
8. Molarity is moles solute / liter of solution. See text, section 3.7. Convert 22.0g
MaOH to moles and divide this by the number of liters present. Answer: D.
9. The correct equation has the same number of all atoms on both sides. See text,
section 4.1. Answer: D.
10. Balance the equation: See text, section 4.1. Answer: E.
11. Divide the number of grams of each compound by the molar mass to get the
number of moles. The smallest number of moles gives the smallest number of
atoms. See section 3.4 and 3.5. Shortcut: since the masses given are all the same,
the largest molar mass therefore gives the smallest number of moles, and
therefore the smallest number of atoms. Answer: D.
12. First, find the number of grams of water using the density (text, section 1.6).
Then use the molar mass of water (18) to find the number of moles present and
Avogadro’s number to get the number of molecules. (text, sections 3.4-3.5).
13. See text, section 3.6. First, note that A and B are NOT empirical formulas (why?).
find the number of moles of carbon present from the mass of carbon dioxide (get
moles of carbon dioxide which are equal to moles of carbon (why?)). Next get
grams of carbon from moles of carbon. According to the problem, the sample
contains only C and H, so the difference in mass between the sample and the
carbon is the mass of H. Get moles of both from these masses and find the ratio,
which comes out 1:1. Answer: E.
14. The empirical formula mass is 32. 96/32 = 3. There are three units of the
empirical formula in the compound. See text, section 3.6.
15. See text, section 3.6. First, use the given general formula to eliminate choices C,
D and E. Next, find the combination of C and H that adds up to a mass of 100.
16. See section 3.6. 29.67g of S is 0.95 mol. 70.33g of F is 3.8 mol. The ratio is 4:1,
so the empirical formula is SF4.
Long Answer Problem 1:
1. The mass of Fe present is 56x4=224. The total mass is 6.4x104.
2. The mass percent is the mass of iron divided by the total mass times 100%.
3. Answer: 0.35%.
Long Answer Problem 2:
1. Find the molar mass of sodium chloride (=58).
2. Molarity is concentration in mol of solute per liter of solution.
3. Mol of solute = mass of sodium chloride divided by the molar mass.
4. Divide mol of solute by the number of liters of solution to obtain the molarity in
mol per liter.
5. Answer: 3.1 mol/L.
Long Answer Problem 3:
1. Balance the chemical equation: Fe2O3 + 3CO  2Fe + 3CO2.
2. Obtain the molar mass of each of the two reactants and use the masses provided to
obtain the number of moles of each present. (0.50 mol of Fe2O3, 1.61 mol of CO)
3. The mole ratio is obtained from the coefficients in the balanced chemical equation.
The ratio of Fe2O3 to CO is 1:3.
4. With the mole ratio of 1:3 and the moles present, note that 0.11 mol of CO will
remain when the reaction is complete. This is 2.8 g.