Department of Chemistry, Geosciences, and Environmental Science
Chemistry 1054-010 College Chemistry Fall 2006
Exam 1-A-Key
Name:_______________________________________________________
(Please write your name legibly)
Read all directions and questions carefully!! This exam consists of two parts. The first part consists of 10 multiple choice questions worth four points each for a total of 40 points. The second part consists of five numerical problems worth either 10 or 15 points per question for a total of 60 points. Show all your work necessary for the numerical problems as partial credit will be given for those problems.
Possibly Useful Constants
Avogadro’s Number: NA = 6.022 × 1023 particles/mol
Score
Part 1 (40 points):_____________________
Part 2 (60 points):_____________________
Total (100 points):_____________________
Don’t forget to put your name on this test!
Good Luck!!
Part 1
Multiple Choice
Please indicate the answer to each question by putting your choice in the space provided. There is only one correct answer for each question. There will be 10 multiple choice questions worth 4 points each.
1. If matter has uniform composition throughout and can be separated into its components by physical methods, then this form of matter is:
(a) a compound (c) a homogeneous mixture
(b) an element (d) a heterogeneous mixture
(e) either an element or a compound
Answer: C
2. Which of the following is an intensive property?
(a) volume (c) mass
(b) number of moles of a substance (d) density
Answer: D
3. Precision refers to:
(a) how close a measured value is to other measured values.
(b) how close a measured value is to the true value of the measurement.
(c) how close a measured value is to its calculated value.
(d) how close a measured value is to zero.
Answer: A
4. The charge on the electron was determined by:
(a) Thomson’s cathode ray tube experiment. (c)
(b) Millikan’s oil drop experiment. (d)
Answer: B
5. How many protons, neutrons, and electrons are there in one atom of 63Cu?
(a) 29 protons, 29 neutrons, 63 electrons (c) 29 protons, 63 neutrons, 29 electrons
(b) 29 protons, 34 neutrons, 29 electrons (d) 29 protons, 29 neutrons, 34 electrons
(e) 34 protons, 29 neutrons, 34 electrons
Answer: B
The atomic number of Cu is 29. Therefore, an atom of Cu has 29 protons and 29 electrons. The number of neutrons is 63 – 29 = 34.
6. What is the formula of the ionic compound formed by the monatomic ions of magnesium and nitrogen assuming each ion has the group charge?
(a) MgN (c) MgN2
(b) Mg2N3 (d) Mg3N2
(e) Mg2N
Answer: D
Magnesium, being in group 2A, forms +2 cations. Nitrogen, being in group 5A, forms –3 (=5–8) anions. The ionic compound formed from Mg2+ and N3- is Mg3N2.
7. The definition of an atomic mass unit is based on which isotope?
(a) oxygen-16 (c) hydrogen-1
(b) helium-4 (d) carbon-12
Answer: D
8. Of the reactions below, which one is not a combination reaction?
(a) C(s) + O2(g) ¾¾® CO2(g) (c) CaO(s) + H2O(g) ¾¾® Ca(OH)2(s)
(b) 2 Mg(s) + O2(g) ¾¾® 2 MgO(s) (d) N2(g) + 3 H2(g) ¾¾® 2 NH3(g)
(e) CH4(g) + 2 O2(g) ¾¾® CO2(g) + 2 H2O(l)
Answer:E
(e) is a combustion reaction.
9. The mass percent of Al in the compound Al2(SO4)3 is equal to:
(a) 7.866% (c) 21.93%
(b) 15.77% (d) 45.70%
(e) 35.94%
Answer: B
10. The correct name for SO is:
(a) sulfur oxide (c) sulfur monoxide
(b) sulfoxide (d) sulfate
(e) sulfide
Answer: C
Part 2
Numerical Problems
Solve the following problems, keeping track of significant figures where applicable. Please show all the work necessary to obtain your answer in order to receive partial credit for possibly wrong answers. Generally, full credit will not be given for the correct answer without any of the work performed to obtain the answer being shown on the paper. Each question is worth either 10 or 15 points.
11. (10 points) Balance the following chemical reactions with the lowest whole number ratio between all of the coefficients:
(a) CO(g) + O2(g) ¾¾® CO2(g)
2 CO + O2 ¾¾® 2 CO2
(b) Al4C3(s) + H2O(l) ¾¾® Al(OH)3(s) + CH4(g)
Al4C3 + 12 H2O ¾¾® 4 Al(OH)3 + 3 CH4
(c) N2O5(g) + H2O(l) ¾¾® HNO3(aq)
N2O5 + H2O ¾¾® 2 HNO3
(d) Fe(OH)3(s) + H2SO4(aq) ¾¾® Fe2(SO4)3(aq) + H2O(l)
2 Fe(OH)3 + 3 H2SO4 ¾¾® Fe2(SO4)3 + 6 H2O
12. (15 points) A certain organic compound consists of 50.70%C, 4.26%H and 45.03%O by mass. It has a molar mass of 213.2 g/mol. What are the empirical and molecular formulas for this compound?
Assuming exactly 100 grams of compound:
The empirical formula is C3H3O2. It has an empirical formula weight of 71.05 g/mol. Dividing this into the molar mass:
The molecular formula is 3 times the empirical formula or C9H9O6.
13. (10 points) What is the volume occupied by 1.15 kg of copper? The density of copper is equal to 8.96 g/mL.
Convert kg of copper to g of copper: 
Use the density to calculate the volume occupied by 1150 g of copper:
14. (10 points) How many moles are there in 48.3 g of NH4Cl?
Calculate the molar mass of NH4Cl:
Converting from grams to moles:
15. (15 points) (a) (10 points) What is the maximum amount, in grams, of MgCl2 that can be prepared by the reaction of 16.14 g of Mg(OH)2 with 10.97 g of HCl. The balanced chemical reaction is:
Mg(OH)2 + 2 HCl ¾¾® MgCl2 + 2 H2O
knowns: 16.14 g Mg(OH)2 and 10.97 g HCl
unknown: maximum amount in grams of MgCl2
LIMITING REAGENT PROBLEM!!!
Calculate how much MgCl2 that each reactant could possible produce if it reacted fully:
The reactant that produces the lesser amount of product is the limiting reactant and the amount of MgCl2 it produces is the maximum amount. Therefore, 0.1504 mol MgCl2 is produced. Converting from moles to grams:
___ (b) (5 points) Suppose that when the reaction is performed in lab, 13.24 g of MgCl2 are collected. What is the %yield for this reaction?
For 10 points extra credit, no partial credit given for this part: Calculate how many grams of the excess reactant is left over after the limiting reactant in part (a) has been totally consumed.
Calculate how many grams of Mg(OH)2 is used up by reaction with 10.97 g of HCl
Convert to grams:
Grams of Mg(OH)2 consumed = (0.1504 mol)(58.32 g/mol) =8.771 g
Subtract from the initial amount of Mg(OH)2 to calculate grams of excess reactant:
Grams of excess Mg(OH)2 = 16.14 g – 8.771 g = 7.37 g
In order to get full credit on this last extra credit, everything must be correct including the number of significant figures in your answer!!