CHEMISTRY 2090 PRELIM I
October 8, 2015
Answers
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1. (8 points) An unknown atom Z forms multiple chlorides with the generic formula Z x Cly . One such compound is analyzed and found to contain 77.5% Cl by mass; a second, 85.1% Cl by mass. Show that these compounds conform to Dalton’s law of multiple proportions.
Other than the substitution of Z where it used to say P, this was a clicker question; the answer given in class is shown below.
For those of you with answers for the ratio of 10:6, 6:10, 3:2, 2:3 or anything similar, if you did not receive full credit you might wish to submit your exam for reconsideration.
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2. (a) (5 points) If 7.45 × 1021 atoms of an unknown element X has mass 1.10 g, what is the element X ?
One mole of atoms contains 6.02 × 1023 atoms; so our sample of 7.45 × 1021 atoms represents 7.45 × 1021 = 1.238 × 10 2 moles of atoms, with mass 1.10 g. One mole therefore has mass 23 6.02 × 10 1.10 g = 88.9 g/mole. Suggesting the atom is Y. 1.238 × 10 2 mole −
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(b) (5 points) Radio signals sent from the first NASA flight to Jupiter, in the late 1970’s, were transmitted at a frequency of 8 .34 × 109 s 1 . The receiving antenna on earth was capable of detecting 4.11 × 10 21 watts (1 W = 1 J/s = 1 kg-m 2 /s3 ). What was the minimum number of photons/second which were required for a signal to register at the antenna? −
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Energy/photon is hν = (6.626 × 10 J/photon.
34
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J − s)(8.34 × 109 /s) = 55.26 × 10
25
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= 5.526 × 10
4.11 × 10 21 J/s So the antenna requires an input of = 744 photons/s. 5.526 × 10 24 J/photon −
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3
24
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3. (a) (4 points) An electron is found to have a velocity such that its wavelength, λ, is equal to the distance it travels in one second. What is its velocity?
We are told that λ = v × 1 s. We also know that the DeBroglie wavelength is given by h h h = ; we can combine what we know to give v × 1 s = . Rearranging, we λ= p me v me v 6.626 × 10 34 J − s h 2 find v × 1 s = = ; cancelling common units (remember that the J 9.109 × 10 31 kg me is a kg-m2 /s2 !) leads to v2 = 7.274 × 10 4 m2 /s2 . After taking the square root, we find v = .02697 m. Call it 2.7 cm. −
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(b) (6 points) An intense beam of green light shines on a copper metal surface whose threshold energy for the photoelectron effect is 0.753 aJ. i. What is observed? hc Nothing. The threshold energy of 0.753 aJ/photon (E = ) corresponds to a photon λ wavelength of hc (6.626 × 10 34 J − s)(2.998 × 108 m/s) = = 26.38 × 10 λ = E 0.753 × 10 18 J −
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m = 264 nm
264 nm is found in the ultraviolet–where green light is at longer wavelengths (500-600 nm), at lower energy. ii. The intensity of the beam is doubled. What will be observed? Still nothing. Output of photoelectrons requires higher energy photons, not more low energy photons
iii. The green light of part (i) is replaced with an equally intense beam (same number of photons/second) of red light. What will be observed? Still nothing. Output of photoelectrons requires higher energy photons; red photons carry less energy/photon than is found in green photons.
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4. (a) (8 points) Nitric acid is produced commercially by the Ostwald process, whose first step involves the reaction of ammonia (NH3 ) with oxygen (O 2 ) via 4 NH 3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2 O(g) Assume the reaction described above is carried out in the apparatus shown below; all measurements of pressure are made at T = 137 C. ◦
Flask A contains 2.00 of NH3 at 0.500 atm; Flask B contains 1.00 of O2 at 1.50 atm. The valve is opened and the reaction proceeds as the gases mix. What is the partial pressure of NO produced, if the final container volume is 3.00 and we assume 100% yield in the reaction as written?
The ideal gas law PV = nRT tells us that at fixed T, the product PV is proportional to the number of moles. We could calculate n directly but it is less work to just find PV. Flask A contains 1.00 -atm of NH3 ; B contains 1.50 -atm of O2 . 4 moles of NH3 react with 5 moles of O 2 ; so the NH3 is limiting, and we use up all the NH 3 5 (1.00 -atm) and as much of the O 2 , or 1.25 -atm of O2 . Producing 1.00 -atm of NO 4 and 1.50 -atm of H2 O. 1 1 1 Final volume is 3 liters; so we have atmosphere of NO; atm of water, and atm of O2 . 3 2 12
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(b) (8 points) Even in small quantities nicotine as found in tobacco is addictive. In large quantities, it is a deadly poison. Find the empirical formula of nicotine (that is, x, y and z for Cx Hy Nz ) if upon combustion of 4.40 mg of nicotine one collects 11.9 mg of CO 2 and 3.41 mg of H2 O.
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5. (10 points) 10.0 grams of a mixture of Ca(ClO) 2 and Ca(ClO3 )2 are placed in 10.0 vessel and heated to 700 C where the two compounds decompose via ◦
Ca(ClO)2 (s) → CaCl2 (s) + O2 (g) and Ca(ClO3 )2 (s) → CaCl2 (s) + 3O2 (g) The final pressure inside the vessel at T = 700 C is 0.91 atm. What is the mass of each compound in the orginal mixture? ◦
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6. (a) (5 points) No object can travel faster than the speed of light, c; so the uncertainty in the velocity, ∆v can never be greater than 3 × 108 m/s. What is the minimum uncertainty, ∆x, in the position of an electron, if we know only that its velocity is less than c?
For those of you who left out m from the expression, and simply wrote h something that looks like c∆x ≥ ; (or if you plugged in the value of 4π c = ∆v) you should have received 3 points in partial credit. If you didn’t please submit for a regrade. With our apologies. e
(b) (5 points) An atom is observed to emit light at 100. nm, 125. nm, and 500. nm. Theory predicts that only the ground state and two excited states are involved. In aJ, how far above the ground state are the excited states?
A small amount of partial credit was given for calculating the energies of transition. More was provided for the energy level diagram showing the relationship between the states, and why there were three transitions with only two excited states. 8
7. (a) (7 points) The element arsenic (As) forms an oxide with formula As x Oy which can be studied as a gas. Under the same conditions of temperature (T ) and pressure (P ) it is found that (i) the density of the gaseous As x Oy is 12.4 times the density of gaseous O2 , and that (ii) six volumes of gaseous hydrogen H 2 react with one volume of gaseous Asx Oy to form elemental arsenic and water. i. The formula of the gaseous As x Oy (i.e. identify as best you can x and y): If the density of our unknown gas is 12.4 times that of O 2 (with molecular weight 32.0) then its molecular weight must be (12.4 ± 0.1) × 32.0 = 397 ± 3 g/mole. We also have the result Asx Oy + 6 H2 → x As + y H2 O. The ratio of volumes is the same as the ratio of moles of reactants, therefore y = 6 so as to keep balance of the hydrogen atoms. The atomic mass of As is 74.9; that of O, 16.0. The only combination of x and y = 6 that comes close is As4 O6 with mass 395.6–which is within the uncertainty bounds 397 ± 3.
ii. Is this an empirical or molecular formula? (Explain your answer). This is molecular. The density measures the mass of the units as they exist in the gas phase, as we know that the number of gas units is identical to the number of O 2 units as the pressure and volume are identical to that of the oxygen (b) (4 points) You have two closed containers as shown below; one contains gas X , the second, gas Y . The two gases behave as ideal gases and are at the same temperature, T . The molecular mass of Y is double that of X . The density of gas Y , in g/, is half that of X . Find the ratio of the pressure of X to the pressure of Y .
Gas Y has double the molecular mass of gas X. Container Y has half the density in g/liter of container X. As each unit of gas Y has double the mass, there must be one-quarter as many molecules of gas Y in container Y as there are or molecules of X in container X. As the pressure is proportional to the number of molecules (for fixed temperature and identical volume), the pressure in Y is one-quarter that in X. (The same result holds even if you don’t assume identical volumes, as the density is measured per liter; so the molecule density in any specified volume in Y is also one-quarter the density in the same volume in X.)
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8. (a) (4 points) How many electrons can have the sets of quantum numbers indicated below? i. n = 3, m = 1 We are asked how many electrons can have the four quantum numbers {n = 3, , m = 1, ms }. There are three possible values for n = 3; that is, = 0, 1, or 2. For m = 1 we are limited to only = 1 or = 2. Each of the two orbitals 1 {3, 2, 1, ms } or {3, 1, 1, ms } can house two electrons, with spin m s = ± Four electrons 2 can have these quantum numbers. 1 ii. n = 2, ms = − Following the logic above, we seek allowed orbitals with quantum 2 1 numbers {n = 2, , m , ms = − }. can take on values = 0, with m = 0; or = 1, 2 with m = −1, 0 or m = 1. There are four possible orbitals, and four possible electrons with the specified quantum numbers. iii. n = 2, = 2 There are no possible orbitals; = 2 requires n ≥ 3. These quantum numbers describe one specific orbital; it can 1 house two electrons, with ms = ± . 2
iv. n = 4, = 2, m = 2
(b) (4 points) Below are shown pictures of Ψ2 for two H-atom orbitals; white areas correspond to low probabilities, and black to high probabilites Which values of n and do these orbitals represent? (Put your answers on the lines below the illustrations.)
4d
4p
(c) (2 points) Consider the orbitals listed below. Enter the total number of nodes, the number of radial nodes and the number of angular nodes found, in each in the appropriate columns. orbital all nodes radial nodes angular nodes 3s
2 total nodes
2 radial nodes
0 angular nodes
6f
5 total nodes
2 radial nodes
3 angular nodes
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(d) (3 points) Below are drawn a number of possible electronic configurations for the 8 electrons found in an oxygen atom (O). Electrons are drawn as arrows, whose direction indicates the value of the spin quantum number m s . For each describe on the blank line at right whether the configuration is ”excited” if it represents a possible excited state of the oxygen atom; “ground” if it represents the ground state, or “forbidden” if the configuration cannot exist.
9. (a) (5 points) A series of emission lines are found in the spectrum of the one-electron atom He1+ corresponding to transitions starting from the n = 6 electronic state. What are the longest and shortest values of λ which can be observed in this series of lines?
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(b) (3 points) In the spectrum of the H atom, there is a blue line observed at 434.0 nm corresponding to the emission of light as an excited H atom relaxes between the ni = 5 and nf = 2. The same line is observed in the He+ spectrum. For which pair of energy levels n i and n f does the 434.0 nm line appear in He+ ? Z2 The energies of the various levels vary as 2 in the one-electron atom with nuclear charge Z. n So to get transitions at the same energy, we need to simply keep the ratio Z /n fixed; that is, if we double Z in going from H to He + , we also need to double n. So the same line at 434.0 nm is found in He + for ni = 10 and nf = 4.
(c) (4 points) You are provided with He in a doubly excited state corresponding to an electronic configuration 5s2 . The doubly-excited He atom emits a photon at λ0 when it undergoes the transition 5s2 → 1s1 5s1 . Would you expect λ0 to be larger, smaller, or the same as λ1 observed for the transition from the singly excited state 1 s1 5s1 → 1s2 ? (You should not try to be quantitative; but, instead, describe how Z eff varies for the various electrons in the different states.) For the transition 1s1 5s1 → 1s2 we need to consider the energies of the electron undergoing the transition, in the excited and ground state. In the excited state, it should be approximately hydrogen-like with Zeff ≈ 1 because the much closer 1s electron is very effective at shielding the nuclear charge of one proton; the energy of the electron in the 5s orbital should 12 be approximately −2.18 aJ( 2 ) ≈ −0.09 aJ. In the ground state, as we said in class, the 5 ionization energy of the He atom is about 3.9 aJ; or, if you prefer; Z eff ≈ 1.35 because each of the 1s electrons partially shields the other, and therefore the transition energy should be the difference between the energies of the two states: about 4 aJ (I’m being loose as all we really need are rough estimates!). For the transition 5s2 → 1s1 5s1 one electron is making the same transition; but in a very different background. In the doubly-excited state, it should be approximately hydrogen-like with Zeff > 1 because the two 5s electrons partially shield one another; for purposed of argument I’ll choose Zeff = 2, the worst case, where there is no shielding; it won’t really matter, and then the energy of either electron in the 5s orbital should be approximately 22 −2.18 aJ( 2 ) ≈ −0.35 aJ. In the 1s orbital, the electron sees effectively the full nuclear 5 charge Zeff ≈ 2 because the 5s electron is ineffective at shielding the 1s electron, therefore 22 the energy of the electron when it is in the 1s orbital is about − 2.18 aJ( 2 ) ≈ −8.7 aJ, and 1 energy released on emission is > 8 aJ. Then λ1 represents a photon with energy less than 4 aJ; λ0 represents a photon with more than 8 aJ. λ0 is therefore shorter than λ 1 –by about a predicted factor of two.
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Some possibly useful facts: For the one-electron atom with nuclear charge Z , the Bohr energy of the nth electronic state is Z 2 E n = −2.18 aJ n2 and the Rydberg formula for the wavelength of emitted/absorbed light in a one-electron atom is 1 1 1 = Z 2 (1.097 × 107 m 1 ) ( 2 − 2 ) λ n1 n2 −
for n1 and n2 the two energy levels of the one-electron connected by the photon at wavelength λ, where n 1 < n2 . 2 (n, ) Z eff For all atoms, E (n, ) = −2.18 aJ n2 (∆ p)(∆x) ≥
h ; 4π
λ =
h mv
or
λ =
h p
The mass of the neutron is m n = 1.675 × 10 27 kg; the mass of the proton is m p = 1.673 × 10 The mass of the electron is m e = 9.109 × 10 31 kg; N A = 6.022 × 1023 −
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P V = nRT where
R = 0.0821
− atm − torr = 63.36 mol − K mol − K
Kelvin degrees = 273.15 + Celsius Degrees E photon = hν = h = 6.626 × 10
hc ; 1 watt = 1 Joule/sec = 1 kg − m2 /sec3 λ
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Joule − sec
c = 2.998 × 108 meter/sec
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27
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kg
