1. Define "Mole" and explain why it is useful and convenient in Chemistry. [6%]

Several acceptable answers which should include idea that the mass of a mole, in grams, equals molecular weight, and that mole contains 6.02 x 10²³ molecules.

What is the difference between the "molar mass" and the "molecular weight" of a substance (use N₂O₃ as an example). [4%]
molar mass of N₂O₃ = 2(14) + 3(16) = 76 grams
molecular weight = 76 (either no units, or amu, or 76 times the mass of 1/12th of a ¹²C atom)

2. A powdered sample (A) weighs 45 grams.E , P
    The sample is green. I, P
    "A" is found to be soluble in water. I or E, P or C (both answers acceptable)
    The density of the green solid is 2.87 g/mL. I, P
    The solution of "A" in water conducts electricity. I, P
    Addition of a solution of hydrochloric acid to the solid causes the solid to dissolve and a gas to be evolved. I, C

For each of the six preceding statements indicate whether the property so described is intensive (I) or extensive (E), and whether it is chemical (C) or physical (P). [6%]

Describe any two measurements or experiments that you could carry out in order to try to discover whether "A" a pure substance or a mixture ? [4%]

1. Add water to see if one component will dissolve, and the other remain insoluble
2. Heat solid to see if one component will sublime, or melt before the other.
3. Test with magnet to see if one component is magnetic
4. Examine with magnifying glass or microscope to see if all crystals have same shape and color.

3. The listed atomic weight of chlorine is 35.453. The two stable isotopes of chlorine are ³⁵Cl with an atomic mass of 34.969, and ³⁷Cl (mass 36.966) . Determine the relative percent abundances of these isotopes in normal chlorine. [Hint: let the fraction of ³⁵Cl atoms in normal chlorine be "x".] [4%]
fraction of ³⁷Cl atoms is therefore (1-x)
35.453 = 34.969(x) + 36.966(1-x)
36.966x - 34.969x = 36.966 - 35.453
1.997x = 1.513
x = 0.7576
Answer: 75.76% ³⁵Cl and 24.24% ³⁷Cl

4. Name the following compounds: [6%]
As₄S₅, tetraarsenic pentasulfide

CsBrO₂, cesium bromite

HgCr₂O₇, mercuric (or mercury(II)) dichromate

Ba₃P₂ barium phosphide (tribarium diphosphide also OK)

5.  A 500.0-mg tablet of vitamin C (ascorbic acid, C_xH_yO_z) was combusted in oxygen. The water produced was condensed and weighed (204.5 mg) and the carbon dioxide was bubbled into a solution of barium hydroxide to give a precipitate of BaCO₃. This was filtered off, washed, dried, and weighed (3.357 g).
(A) Calculate the mass of carbon contained in the barium carbonate [4%]

Molwt BaCO₃ = 137.3+12+48 = 197.3
12 g C in 197.3 g BaCO₃
x g C in 3.357 g BaCO₃
x = 12(3.357)/197.3 = 0.2042 g (or 204.2 mg)

(B) Determine the empirical formula of ascorbic acid [6%]

0.5000 g ascorbic acid contain 0.2042 g C (part A) and 0.2045(2)/18 = 0.0227 g H
Therefore the mass of O in the ascorbic acid is 0.5000 - 0.2042 - 0.0227 = 0.2731 g
moles of C = 0.2042/12 = 0.0170
moles of H = 0.0227/1 = 0.0227
moles of O = 0.2731/16 = 0.0171
ratio of moles
C:H:O = 0.017/0.017 : 0.0227/0.017 : 0.017/0.0171 = 1 : 1.33 : 1 = 3 : 4 : 3

Empirical formula = C₃H₄O₃

6.  The fizz produced when an Alka-Seltzer^® tablet is dissolved in water is due to the reaction between sodium bicarbonate, NaHCO₃, and citric acid, H₃C₆H₅O₇:
3NaHCO₃ (aq) + H₃C₆H₅O₇ (aq) --> 3CO₂ (g) + 3H₂O (l) + Na₃C₆H₅O₇ (aq)
How many grams of citric acid should be used for each 1.00 g of sodium bicarbonate? [10%]

3(23+1+12+48)=252 g NaHCO₃ react with 3+6(12)+5+7(16) = 192 g citric acid
252 g/1.00 g = 192 g/x g
x = 192/252 = 0.76 g citric acid

7. (A) Contrast the models of atomic structure proposed by J.J. Thomson and Ernest Rutherford. What experiment demonstrated that one of these models was incorrect? How? [4%]

Thomson: electrons embedded in sphere of positive charge. ("Plum Pudding" model)
Rutherford: electrons surrounding very small positively charged nucleus.
Experiment: scattering of a stream of alpha particles by a thin sheet of gold; most alpha particles were undeflected, but a small number were deflected by interaction with the atomic nuclei

(B) Uranium-238 (²³⁸U) is a nuclide that is radioactive. From its position in the Periodic Table determine how many protons and how many neutrons are contained in the nucleus of ²³⁸U? [2%]

Atomic number of U is 92. Nucleus contains 92 protons and 238-92=146 neutrons

(C) What is the mass of 1000 atoms of ²³⁸U. Give your answer in picograms (pg). [2%]

1000(238)/(6.02 x 10²³) g = 3.95 x 10^-19 g = 3.95 x 10^-7 pg

(D) When ²³⁸U undergoes radioactive decay it expels an alpha particle from the nucleus.
Write the atomic number, mass number, and symbol of the resulting ("daughter") nuclide. [2%]

An alpha particle contains 2 protons and 2 neutrons
Atomic number of daughter nuclide = 92-2 = 90
Mass number = 238 - 4 = 234
Symbol = Th

8. Write and balance chemical equations for the following processes: [2% each]

(A) Combustion of trimethylphosphine oxide, (CH₃)₃PO, in oxygen

2(CH₃)₃PO + 12O₂    ==>  6CO₂ + 9H₂O + P₂O₅

(B) The decomposition of ammonium permanganate (s) to nitrogen (g), manganese(IV) oxide (s), and water (g)

2NH₄MnO₄   ==>  N₂ + 2MnO₂ + 4H₂O

(C) The direct combination of barium metal with fluorine gas

Ba (s) + F₂ (g) ==> BaF₂ (s)

(D) The reaction of aqueous solutions of calcium nitrate and sodium phosphate to give a precipitate of calcium phosphate and a solution of sodium nitrate.

3Ca(NO₃)₂ (aq) + 2Na₃PO₄ (aq) ==>  6NaNO₃ (aq) + Ca₃(PO₄)₂ (s)

(E) The decomposition of solid mercury(I) nitrate to metallic mercury, dinitrogen
tetroxide gas, and oxygen gas.

Hg₂(NO₃)₂ (s)  ==> 2Hg (l) + N₂O₄ (g) + O₂(g)

9.  Reaction (D) in question 8 was carried out by mixing 200 mL of 2.50 M calcium nitrate with 250 mL of 1.60 M sodium phosphate.
Which is the limiting reagent? Show calculations justifying your answer [5%]

200 x 2.50 = 500 mmol calcium nitrate are present and would need 500(2/3) = 333 mmol of sodium phosphate for complete reaction.
250 x 1.60 = 400 mmol sodium phosphate are present and would need 400(3/2) = 600 mmol of calcium nitrate for complete reaction.

Therefore calcium nitrate is the limiting reagent.

How many millimoles of calcium phosphate are formed? [5%]

from equation, 3 mol calcium nitrate form 1 mol calcium phosphate.
500 mmol calcium nitrate will form 500/3 = 167 mmol calcium phosphate

10.  What is the molarity of "concentrated" (65.0%) nitric acid, density = 1.40 g/mL? [4%]
1000 mL of nitric acid weigh 1000(1.40) = 1400 g and contain 1400(0.65) = 910 g HNO₃.
910 g HNO₃ = 910/(1+14+48) = 14.4 mol
Molarity: 14.4 M

Assuming that you have no balance available, describe how you would prepare 500 mL 3.00M HNO₃ starting with the concentrated acid. [6%]

500 mL 3.00 M HNO₃ contain 3(500) = 1500 mmol = 1.5 mol
14.4 mol/1000mL = 1.5 mol/V mL
V = 1.5(1000)/14.4 = 104 mL
Take 104 mL concentrated acid and add water to make total volume 500 mL

{Alternative approach: M₁V₁ = M₂V₂
14.4(V₁) = 1.5(500)
V₁ = 104 mL, etc.}

11.  Oxides of main group metals (e.g. CaO) obey the law of constant composition, but those of some transition elements do not. "Ferrous oxide" has the formula range Fe_0.946O to Fe_0.875O.
What is a name given to this class of compounds? [2%]

"Non-stoichiometric compound" or "berthollide"

Why can iron, but not calcium, form such compounds? [2%]

Fe has two valence states (2+ and 3+) in "ferrous oxides"  to make them neutral. Calcium can only be 2+.

12.  Write empirical formulas for
(A) the heteropolytungstate Na₁₅P₅W₃₀O₁₁₀ [2%]         Na₃PW₆O₂₂

(B) alpha-D-glucose [2%]        CH₂O

(C) L-lysine [2%]       C₃H₇NO