Chemistry 31A Autumn 2004 Professors Chidsey & Zare Exam 2 Name:

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1 Chemistry 31A Autumn 2004 Professors Chidsey & Zare Exam 2 Name: Honor Code Observed: (Signature) Circle your section 9:00am 10:00am 2:15pm 3:15pm 7:00pm 8:00pm S02 OC103 S04 OC103 S06 OC110 S S12 Ricker S15 Ricker S03 OC110 S05 OC110 S P S13 Lagunita S16 Lagunita S14 Stern S18 Stern S17 Wilbur S19 Wilbur Instructions: This is a closed book and closed notes exam. You may use a calculator. For more space, use the back of a page and so indicate below the problem and in the boxes below. Box your answers! Show your work for full credit. Crossed out or erased material will not be graded. Be careful of units and significant figures. This exam contains 5 pages, none of them blank. COUNT YOUR PAGES. 1A 8A 1 H A 3A 4A 2 5A 6A 7A He Li Na K Rb Cs Be Mg Ca Sr Ba B 4B 5B 6B 7B 8B 8B 8B 1B 2B 21 Sc Y La Ti Zr Hf V Nb Ta Cr Mo W Mn Tc Re Fe Ru Os Co Rh Ir Ni Pd Cu Ag Zn Cd B Al Ga In C Si Ge Sn N P As Sb O S Se Te F Cl Br STUDENT FILL THIS OUT Question Grade Initials I have used the back of the following pages Total Pt Au Hg Tl Pb Bi Po I At Ne Ar Kr Xe Rn 222 page 1 of 5 e2.doc 10/21/2004

2 1. (15 pts). Tetrasilane (Si 4 H 10 ) is a liquid with a density of g cm -3. It reacts with oxygen to give solid silica (SiO 2 ) and water. Calculate the maximum mass of silica that could be obtained if 75.0cm 3 of tetrasilane reacted completely with excess gaseous oxygen. 2. (18 pts). Two Lewis structures of the phosphate ion (PO 4 3- ) are shown below: (a) (10 pts). Calculate the formal charge on all the atoms in both Lewis structures. Put your answer on the above diagrams. (b) (4 pts). Which structure (left or right) satisfies the Lewis octet rule for each atom? (c) (4 pts). Which structure (left or right) has the preferred Lewis structure based on minimizing the formal charge on the atoms? page 2 of 5 e2.doc 10/21/2004

3 3. Environmental concerns have prompted the development of methods for the production of copper from CuFeS 2 ore that reduce energy consumption and that avoid the release of toxic gaseous or solid by-products. Through a series of clever steps involving treating powdered ore and various solid intermediates with aqueous solutions, the following overall reaction is accomplished at near ambient temperature: CuFeS2 ( s ) + 2 FeCl3 ( aq) Cu( s) + 2 S( s) + 3 FeCl2 ( aq) Some of the FeCl 2 (aq) is converted back into FeCl 3 (aq) in another multistep process which overall consumes O 2 (g) from the air, S(s) from the preceding reaction and CaCl 2 (s). The byproduct is CaSO 4 (s), also known as gypsum, the white powder used to make wallboard. (a) (10 pts) Provide a balanced chemical equation for the overall reaction that converts FeCl 2 (aq) back into FeCl 3 (aq). (b) (22 pts). Imagine starting up a new copper mill that conducts these two overall reactions in batches, one batch a day. First thing in the morning, the FeCl 2 (aq) from the previous day is converted to FeCl 3 (aq) which is then reacted with ore to produce copper metal. One particular morning, there is 4.0x10 4 L of 2.5 M FeCl 2 (aq) in the holding tank. The processing capacity of the machinery is 11 Mg of ore per day. Assuming the ore is pure CuFeS 2 (s) (formula mass = g/mol), what will be the mass of Cu(s) produced that day? (c) (3 pts) Will the amount of FeCl 2 at the end of the day be greater or less than at the beginning of the day? Explain why in 2 sentences or less. page 3 of 5 e2.doc 10/21/2004

4 4. (6 pts). Using the periodic table and your understanding of valency, write down the values of x and y for the most stable forms of the following compounds: Li x (CH 3 ) y Ga x P y H x Se y 5. Sodium chloride forms a crystal that has a cubic structure of Na + ions and Cl - ions, as shown below: This figure represents the unit cell of NaCl. One unit cell is stacked after another to form the entire crystal. Each unit cell contains a net of 4 sodium cations and 4 chloride anions. These values are calculated by keeping track of the sharing between neighboring unit cells: Eight sodium cations at the corners contribute 1/8 each to the cell for a total of ONE. Six sodium cations at face-centers contribute 1/2 each to the cell for a total of THREE. So, the net number of sodium cations per unit cell is FOUR. Twelve chloride anions at the center of the cell edges contribute 1/4 each to the cell for a total of THREE. An additional chloride anion is present in the center of the cell; it contributes all of itself to the cell, i.e., ONE. So, the net number of chloride anions per unit cell is also FOUR. Thus, equal numbers of cations and anions are present in the unit cell as required by the chemical formula. Careful x-ray measurements tell us that the edge of the unit cell of the NaCl crystal has a length of x 10-8 cm. It is also determined that the density of crystalline NaCl is 2.18 g/cm 3. page 4 of 5 e2.doc 10/21/2004

5 (a) (6 pts). Determine the volume of the NaCl unit cell in units of cm 3. (b) (8 pts). Determine the volume of one mole of NaCl (formula mass = 58.45g/mol). (c) (8 pts). Historically, one of the earliest determinations of Avogadro s number was based on the volumes you just calculated. Determine Avogadro s number with the appropriate number of significant figures using your answers to (a) and (b)?. (d) (4 pts). Assuming the largest contribution to the systematic error of this determination is that some small fraction of the sodium and chloride ions are missing in the crystal lattice, explain in no more than two sentences why your determination of Avogadro s number may be too high or too low. page 5 of 5 e2.doc 10/21/2004