Chem 14A Midterm Answers - University of California, Los Angeles CHEM 14A

FALL 2018 MIDTERM (Total number of pages = 10) (Total points = 110) (Total time = 110 minutes) **Carefully remove the last two pages: Constants and Formulas, and Periodic Table.** YOUR DISCUSSION SECTION ...................................... YOUR TA’s NAME ……………………………….. WRITE IN PEN Do not use white-out. Show all your work. Check units and significant figures. Box your final This is a closed book exam: Only a pen and simple scientific calculator are allowed. No other material allowed. Good Luck answer. 2 Do not write on this page. QUESTION SCORE 1 2 3 4 5 6 7 8 TOTAL (max 110) 3 Q1. Compounds are being developed that can store hydrogen in vehicles. One reaction being studied for hydrogen storage is: Li3N(s) + H2(g) → LiNH2(s) + LiH(s) (Concept question) A. Why do we balance chemical equations? (2pt) Conservation of mass in chemical reactions. Therefore the number (or moles) of atoms must be the same for the reactants and products. (Concept question) B. Write the balanced chemical equation for the above reaction. (4pt) Li3N(s) + 2 H2(g) → LiNH2(s) + 2 LiH(s) C. How many moles of H2 are needed to react with 1.5 mg of Li3N? (4pt) Moles of H2 needed to react with 1.5 mg Li3N: = 1.5 × 10-3 g Li3N                 mol Li N mol H g Li N mol Li N 3 2 3 3 1 2 34.83 1 = 8.6 × 10-5 mol H2 1pt 1pt 1pt 1pt D. Calculate the mass of Li3N that will produce 0.650 mol LiH. (4pt) Mass of Li3N to produce 0.650 mol LiH: = 0.650 mol LiH                 mol Li N g Li N mol LiH mol Li N 3 3 3 1 34.83 2 1 = 11.3 g Li3N 1pt 1pt 2pt -1sf -1units 4 Q2. Gallic acid can be used to prevent dysentery and is known to contain carbon, hydrogen, and oxygen. A chemist wanting to determine the molecular formula of gallic acid burned 1.000 g of the compound in an elemental analyzer. The products of the combustion were 1.811 g of CO2 and 0.3172 g of H2O. A. Determine the empirical formula of the compound. (12pt) 2 1 2 2 1.811 g CO 12.01 g C 0.041 15 mol CO 0.4942 g C 44.01 g mol 1 mol CO −   =   = ⋅   1pt 1pt 1pt 2 1 2 2 0.3172 g H O 2.0158 g H 0.017 62 mol H O 0.03552 g H 18.02 g mol 1 mol H O −   =   = ⋅   1pt 1pt 1pt 1.000 g – (0.4942 + 0.03552)g = 0.4703 g O 1pt 1 0.4703 g O 0.029 39 mol O 16.00 g mol− = ⋅ 1pt Dividing each amount of moles by 0.02939 gives a ratio of C:H:O of 1.4:1.2:1 1pt Multiplying through by 5 gives whole numbers as required for an empirical formula: CHO 765 3pt B. The molar mass was found to be 170.12 –1. What is the molecular formula of gallic acid? (2pt) Since the molar mass of this empirical formula matches the given value of about 170 g/mol, the molecular formula is the same as the empirical formula. CHO 765 5 Q3. When ultraviolet radiation is directed at an atom or a molecule, electrons can be ejected from the valence shell, and their kinetic energies are measured. Because the energy of the incoming ultraviolet photon is known and the kinetic energy of the outgoing electron is measured, the ionization energy, I, can be deduced from the fact that the total energy is conserved. (Concept question) A. Explain, comment, and show that the speed, V, of the ejected electron and the frequency, ν, of the incoming radiat