Chem 50 – Exam #3 (Chapter 9,10)
Name: __________________
1. If one (1) molecule of NH3 has a mass of 17.04 amu, what is the mass of one (1) mole of NH3 molecules? (5)
17.04 g
2. What is the molar volume for helium gas? (5)
22.4 L / mol
3. What is the value of Avogadro’s number? (5)
6.02 x 1023
4. What does STP stand for? (3)
Standard Temperature and Pressure
5. What are the values associated with STP? (5)
0oC and 1 atm
6. What is a limiting reagent? (5)
A reactant that limits the yield (product produced) because it runs out first in a reaction.
7. Given the following chemical formulas, which one is an empirical formula?
(c) CH2
(5)
(a) C6H12
(b) C3H12
8. Find the molar mass of boron oxide (B2O3, B=10.81 amu, O=16.00 amu). (6)
MM = 2(10.81 g/mol) + 3(16.00 g/mol) = 69.62 g / mol
9. Find the molar mass of chlorine gas (Cl2, Cl=35.45 amu). (6)
MM = 2(35.45 g/mol) = 70.90 g / mol
10. Find the mass of 0.222 mole of boron oxide. (6)
0.222 mol (69.62 g/mol) = 15.5 g
11. Find the number of molecules in 0.222 mole of boron oxide. (6)
0.222 mol (6.02 x 1023 molecules / mol) = 1.34 x 1023 molecules
12. Find the volume of 0.222 mole of chlorine gas at STP. (8)
0.222 (22.4 L / mol) = 4.97 L
13. Find the number of molecules of in 0.222 grams of boron oxide. (8)
[0.222 g / (69.62 g/mol)] (6.02x1023 molecules/mol) = 1.92 x 1021 molecules
14. Find the mass of 0.222 liters of chlorine gas at STP. (8)
[0.222 L / (22.4 L/mol)] (70.90g/mol) = 0.703 g
15. Find the percent composition for all elements in boron oxide. (8)
B = mass of B / mass of boron oxide
= 2(10.81) / (69.62) = 31.05%
O = 3(16.00) / (69.62) = 68.95%
16. A compound was found to contain 43.64% phosphorous and 56.36% oxygen. Find its empirical formula.
(10)
assume 100 g sample, then P = 43.64 g, O = 56.36 g
P = 43.64 g / 30.97 g/mol = 1.409 mol
O = 56.36 g / 16.00 g/mol = 3.523 mol
O/P ratio = 3.523/1.409 = 2.5
E.F = P1O2.5 = P10O25 = P2O5
17. A compound with an empirical formula of CH2 was found to have molar mass of 84 g. What is its
molecular formula? (5)
MF / EF = MMMF / MMEF
84 / (12.01 + 2*1.01) = 6
6 = MF / EF
MF = 6 (CH2) = C6H12
Given the balanced equation:
2 P (s) + 5 Cl2 (g) → 2 PCl5 (s)
18. For every 2 mole of phosphorous, it will react with 5 mole chlorine gas and produce 2 mole phosphorous
pentachloride. (4)
19. Given 0.222 mol of phosphrous, how many moles of chlorine gas is needed react with it? (5)
0.222 mol P ( 5 Cl2 / 2 P) = 0.555 mol Cl2
20. Given 0.222 g of phosphorous (P = 30.97 amu), how many moles of chlorine gas is needed to react fully? (8)
0.222 g / 30.97 g/mol = 7.17 x 10-3 mol P
7.17 x 10-3 mol P (5 Cl2 / 2 P) = 1.79 x 10-2 mol Cl2
21. Given 0.222 L of chlorine gas, how many grams of phosphorous pentachloride will be produced at STP if
phosphorous is provided in excess? (8)
0.222L / (22.4 L /mol) = 9.91 x 10-3 mol
9.91 x 10-3 mol Cl2 (2 PCl5 / 5 Cl2) = 3.96 x 10-3 mol PCl5
3.96 x 10-3 mol PCl5 (208.22 g/mol) = 0.825 g
22. Given 8.88 mol of phosphorous and 17.77 mol of chlorine gas, which reactant is the limiting reagent? Must
show work for full credit. (6. 2/4)
8.88 mol P / 2P = 4.44 mol eq
17.77 mol Cl2 / 5Cl2 = 3.554 mol eq
3.554 < 4.44, therefore Cl2 is the limiting reagent.
23. A student plans to conduct an experiment based on this reaction using 0.222 g of phosphorous and 0.222 L
of chlorine gas, what is the theorectical yield (in grams)? (10)
theoretical yield = calculated product yield.
0.222 g P = 7.17 x 10-3 mol (Q#20)
0.222 L Cl2 = 9.91 x 10-3 mol Cl2 (Q#21)
7.17 x 10-3 mol P / 2 = 3.59 x 10-3 P
9.91 x 10-3 mol Cl2 / 5 = 1.98 x 10-3 Cl2
Cl2 is limiting reagent.
9.91 x 10-3 mol Cl2 (2 PCl5 / 5 Cl2) = 3.96 x 10-4 mol PCl5
3.96 x 10-4 mol PCl5 (208.22 g / mol) = 0.825 g
24. If the student performs the experiment and found 0.333 gram of product was produced. What is the percent
yield? (5)
Actual yield = 0.333 g / 0.825 g * 100% = 40.3%