CH107 SPECIMEN EXAM-TYPE QUESTIONS (5)
Chapters 7/8
The questions are worth equal points
1. A 10.0 g sample of potassium chlorate is decomposed by heating to
potassium chloride and oxygen gas. The mass of O2 produced is 3.0 g.
(a) Write a balanced equation of this reaction.
(b) Determine the purity of the sample, assuming all the potassium chlorate
present decomposes. Show working.
2. A research chemist synthesizes a new solid organic compound. She uses
a newly discovered reaction to give an (impure) product as part of a
suspension in water. The compound is extracted from the aqueous
suspension by extracting with diethyl ether, followed by drying over MgSO 4,
filtering and then distilling off the ether. The residue is purified by
recrystallization from ethyl acetate to give the pure product, but only in 62%
yield.
Suggest reasons why the new compound is produced in much less than
100% yield.
3. Magnetic nuclei (such as 1H and 13C) when subjected to a strong
homogeneous magnetic field, can absorb electromagnetic radiation and
hence undergo excitation or transition (which involves change of spin). This is
the basis of the very important analytical technique known as nuclear
magnetic resonance (NMR).
In a field of 7.04 Tesla, a typical value of the frequency of electromagnetic
radiation absorbed by 1H is 300 MHz.
(a) In which part of the electromagnetic spectrum does this frequency lie?
(b) Calculate the energy in J mol-1 associated with 300 MHz radiation.
(c) How does the energy in (b) compare with those associated with transitions
that occur in the visible region (i.e. in visible spectroscopy)?
[Planck’s constant h = 6.63 x 10-34 Js; Avogadro number = 6.02 x 1023 mol-1]
4. Atoms or ions that contain unpaired electrons spins are said to be
paramagnetic, those with all paired electron spins are said to be diamagnetic.
Write ground state electronic configurations for (a) Mg 2+ (b) C4-, (c) Mg+, (d)
P3+, (e) Se, and state which are paramagnetic and which are diamagnetic.
5. Predict the group in the periodic table in which an element with the
following ionization energies would most likely be found. IE 1 = 1060 kJ/mol;
IE2 = 1900 kJ/mol; IE3 = 2920 kJ/mol; IE4 = 4960 kJ/mol; IE5 = 6280 kJ/mol
IE6 = 21200 kJ/mol; IE7 = 25900 kJ/mol.
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Solutions
1.
(a) 2KClO3 (s)
2KCl(s) +
3O2(g)
(b) We need to determine the mass of KClO 3 needed to give 3.0 g of O2
This is
3.0 g O 2 x
1mol O 2
x
32.00 g O2
2 mol KClO 3
3 mol O2
x
122.6 g KClO 3
mol KClO3
= 7.66 g
% purity
=
7.66 g
x 100 = 76.6%
10.0 g
2.
There is not enough information to tell us exactly why the yield is low, but we
can say that it is due to one or more of the following reasons:
Incomplete reaction (e.g. reversible reaction or insufficient reaction time)
Competing reaction(s) (which give unwanted products)
Manipulation losses (e.g. in the solvent extraction, drying, filtration and
recrystallization steps)
3.
(a) Radiofrequency region (radio and TV waves)
(b) The energy of the photon absorbed is necessarily equivalent to the
difference in energy levels of the transition:
E = h (in J) or h x Avogadro number (in J/mol)
Hence energy = 6.62 x 10-34 Js x 3 x 108 /s x 6.02 x 1023 /mol = 0.120 J/mol
(or 120 mJ/mol)
(b) This energy is much smaller (~ 1 millionth) than that associated with
visible spectroscopy.
4.
(a) Mg2+
(b) C4(c) Be+
(d) P3+
(e) Se
1s22s22p6 (or [Ne])(diamagnetic)
1s22s22p6 (or [Ne])(diamagnetic)
1s22s1 (paramagnetic)
[Ne] 3s23p1 (paramagnetic)
[A] 3d10 4s24p4 (paramagnetic)
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5.
Since there is a huge jump in ionization energy between the 5 th and 6th
ionization, the element most likely can be found in Group 5A.
(ns2np3 : five ionizations lead to electronic configuration of preceding noble
gas. The sixth ionization energy will be comparatively high because an
electron will be lost from a specially stable noble gas electronic configuration.)
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