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Some more Practice Questions
Example 1
What is the atomic symbol for bromine, and what is its atomic number?
Why isn't the symbol
for bromine just the first letter of its name? What other element preempts
the symbol B?
Solution
Bromine's atomic number is 35, and its symbol is Br; B is the symbol for
boron.
Example 2
How many protons, neutrons, and electrons are there in an atom of the most
stable isotope of uranium, uranium-238? Write the symbol for this isotope.
Refer to Figure. 1-1.
Solution
The atomic number of uranium (see the inside back cover) is 92, and the
mass number of the isotope is given as 238. Hence it has 92 protons, 92
electrons,and 238 - 92 = 146 neutrons. Its symbol is
U (or
238
U).
Example 3
Calculate the mass that is lost when an atom of carbon-12 is formed from
protons, electrons, and neutrons.
Solution
Since the atomic number of every carbon atom is 6, carbon-12 has 6 protons
and therefore 6 electrons. To find the number of neutrons, we subtract the
number of protons from the mass number: 12 - 6 = 6 neutrons. We can use
the data in Table 1-1 to calculate the total mass of these particles:
Protons: 6 X 1.00728 amu = 6.04368 amu
Neutrons: 6 X 1.00867 amu = 6.05202 amu
Electrons: 6 X 0.00055 amu = 0.00330 amu
Total particle mass: 12.09900 amu
But by the definition of the scale of atomic mass units, the mass of one
carbon-12 atom is exactly 12 amu. Hence 0.0990 amu of mass has
disappeared in the process of building the atom from its particles.
Example 4
Calculate the expected atomic weight of the isotope of chlorine that has 20
neutrons. Compare this with the actual atomic weight of this isotope as
given in Table 1-2.
Solution
The chlorine isotope has 17 protons and 20 neutrons:
Protons: 17 X 1.00728 amu =
17.1238 amu
Neutrons: 20 X 1.00867 amu =
20.1734 amu
Electrons: 17 X 0.00055 amu =
0.0094 amu
Total particle mass:
37.3066 amu
Actual observed atomic weight: 36.966 amu
Mass Loss:
0.341 amu
Example 5
Magnesium (Mg) has three significant natural isotopes: 78.70% of all
magnesium atoms have an atomic weight of 23.985 amu, 10.13% have an
atomic weight of 24.986 amu, and 11.17% have an atomic weight of 25.983
amu. How many protons and neutrons are present in each of these three
isotopes? How do we write the symbols for each isotope? Finally, what is the
weighted average of the atomic weights?
Solution
There are 12 protons in all magnesium isotopes. The isotope whose atomic
weight is 23.985 amu has a mass number of 24 (protons and neutrons), so
24 - 12 protons gives 12 neutrons. The symbol for this isotope is 24Mg.
Similarly, the isotope whose atomic weight is 24.986 amu has a mass number
of 25, 13 neutrons, and 25Mg as a symbol. The third isotope (25.983 amu)
has a mass number of 26, 14 neutrons, and 26Mg as a symbol. We calculate
the average atomic weight as follows:
(0.7870 X 23.985) + (0.1013 X 24.986) + (0.1117 X 25.983) =
24.31 amu
Example 6
Boron has two naturally occurring isotopes, lOB and 11B. We know that
80.22% of its atoms are 11B, atomic weight 11.009 amu. From the natural
atomic weight given on the inside back cover, calculate the atomic weight of
the lOB isotope.
Solution
If 80.22% of all boron atoms are
11
B, then 100.00 - 80.22, or 19.78%, are
the unknown isotope. We can use W to represent the unknown atomic
weight in our calculation:
(0.8022 X 11.009) + (0.1978 X W) = 10.81 amu (natural atomic
weight)
W=
= 10.01 amu
Example 7
Calculate the molecular weight of methyl alcohol.
Solution
The molecular formula is CH30H or CH4O. Then:
1 carbon:
1 X 12.011 amu = 12.011 amu
4 hydrogens:
4 X 1.008 amu = 4.032 amu
1 oxygen:
1 X 15.999 amu = 15.999 amu
Total particle mass:
32.04 amu
(If you wonder why the last figure has been dropped, see the discussion of
significant figures in Appendix 4.)
In Example 7 notice that the natural atomic weight of carbon is not 12.000
amu but 12.011 amu, since carbon occurs as a mixture of 98.89% carbon12 and 1.11% carbon-1 3, with trace amounts of carbon-14.
Example 8
What is the molecular weight of pure octane?
Solution
Since the molecular formula is C8H18", the molecular weight is:
(8 X 12.011) + (18 X 1.008) = 114.23 amu
Example 9
How many grams of each of the following substances are there in 1 mole of
that substance: H2, H20 , CH3OH, octane (C8H18), and neon gas (Ne)?
Solution
The molecular weights (in atomic mass units) of most of these substances
have been given in previous examples, and the atomic weight of neon is
listed on the inside back cover. One mole of each substance is therefore:
H2
2.0160 g break C8H18 114.23 g
H2O
18.0154 g break Ne
CH3OH 32.04 g
20.179 g
break
Example 10
One molecule of H2 reacts with one molecule of Cl2 to form two molecules of
hydrogen chloride gas, HCl. What weight of chlorine gas should be used in
order to react completely with 1 kilogram (kg) of hydrogen gas?
Solution
The molecular weights of H2 and Cl2 are 2.0160 g mole-1 and 70.906 g mole1
, respectively. * Hence 1000 g of H2 contains:
= 496.0 moles of H2 molecules
Without knowing how many molecules there are in a mole, we can still be
sure that 496.0 moles of Cl2 will have the same number of molecules as
496.0 moles or 1000 g of H2. How many grams of Cl2 are there in 496.0
moles? Since the molecular weight of Cl2 is 70.906 g mole-1,
496.0 moles X 70.906 g mole-1 = 35,170 g of Cl2
One kilogram equals 1000 g, so 35,170 g is 35.17 kg. If 1.00 kg H2 is made
to react with 35.17 kg of C12, the reaction will be complete and none of
either starting material will be left over.
* The expression "g mole-1" should be read as "grams per mole." In this
notation, a speed in miles per hour is written with units of "miles hr-I."
Example 11
How many molecules of H2 and Cl2 would be present in the experiment of
Example 10?
Solution
In 496.0 moles of any substance, there will be 496.0 moles X 6.022 X 1023
molecules mole-1, which equals 2.99 X 1026 molecules.
Example 12
Is chlorine oxidized or reduced in forming the chloride ion? What is the
oxidation state of the ion?
Solution
Chlorine is reduced, since one electron per chlorine atom is added to form
the ion. The chloride ion, Cl- , is in the - 1 oxidation state.
Example 13
When metals are converted into their ions, are they oxidized or reduced?
What is the oxidation state of the aluminum ion?
Solution
Metals are oxidized to their ions, since electrons are removed. The aluminum
ion, AP+, is in the +3 oxidation state.
Example 14
When the ferric ion is converted to the ferrous ion, is this an oxidation or
reduction? Write the equation for the process.
Solution
The equation is Fe3+ + e- → Fe2+ . The process is a reduction since an
electron is added.
Example 15
What is the molecular weight of ammonium sulfate?
Solution
The chemical formula of ammonium sulfate is (NH4)2SO4, so the molecular
weight (actually the formula weight) is
2 nitrogens: tt 2 X 14.007 amu= tt 28.014 amu
8 hydrogens: tt 8 X 1.008 amu= tt 8.064 amu
1 sulfur:
tt 1 X 32.06 amu= tt 32.06 amu
4 oxygen:
tt 4 X 15.999 amu= tt 63.996 amu
Total:
tt
tt 132.13 amu
Example 16
Write equations for the reactions that occur when current is passed through
molten NaCl. How many grams of sodium and chlorine are released when 1
of charge is passed through the cell?
Solution
The cathode reaction is Na+ + e- → Na, and the anode reaction is Cl-→ Cl2
+ e-. When 1 mole of electrons (1
) passes through molten NaCI, each
electron reduces one sodium ion, so 1 mole of sodium atoms is produced.
Hence 22.990 g of Na are deposited at the cathode. At the anode, 1 mole of
electrons is removed from 1 mole of chloride ions, leaving 1 mole of chlorine
atoms, which combine pairwise to make
mole of Cl2 molecules. Hence the
weight of chlorine gas released is 35.453 g (the atomic weight of Cl, half the
molecular weight of Cl2).
Example 17
How many grams of magnesium metal and chlorine gas are released when
1
of electricity is passed through an electrolytic cell containing molten
magnesium chloride, MgC12?
Solution
The cathode reaction is Mg2+ + 2e- → Mg, and the anode reaction is this:
2Cl-→ Cl2 + 2e-. Since two electrons are required to reduce each ion of
Mg2+, 1 mole of electrons will be sufficient to reduce half a mole of
magnesium ions, depositing 12.153 g of magnesium. (The atomic weight
of magnesium is 24.305 g mole-1.) As in Example 16, 1 mole of Cl- ions is
oxidized, liberating half a mole or 35.453 g of Cl2 gas.
Example 18
The main commercial source of aluminum metal is the electrolysis of
molten salts of Al3+. How many faradays of charge, and how many
coulombs, must be passed through the melt to deposit 1 kg of metal?
Solution
One kilogram of aluminum is 1000 g/26.98 g mole-1, or 37.06 moles. Since
each atom of aluminum deposited requires three electrons, 37.06 moles
will require 3 X 37.06, or 111.2, moles of electrons. Hence 111.2
or
10,730,000 coulombs will be needed.
Example 19
Electron flow at the rate of 1 coulomb per second (coulomb sec-1) is a
current of 1 ampere (A). Currents in industrial electrolytic production of
aluminum are ordinarily in the range of 20,000 to 50,000 A. If a cell is
operated at 40,000 A (40,000 coulombs sec-1), how long will it take to
produce the kilogram of aluminum metal mentioned in Example 18?
Solution
The time required will be
= 268 sec or 4.5 min
Example 20
Assume that Thomson's cathode ray particles are in fact the same as
Stoney's and Faraday's electrons, and that 1
Calculate the mass of one electron.
is a mole of electrons.
Solution
The charge on one electron is
e=
=
= 1.602x 1019 coulomb
m=
= 0.910 x 10-27 g
Example 21
Assume that you do not know the value of Avogadro's number, but that
you recognize that the faraday is the charge necessary to reduce 1 mole of
Na+ ions, with one of Millikan's electrons combining with each ion.
Calculate the number of ions in a mole, or Avogadro's number.
Solution
The charge on one electron is
N=
= 6.022 x 1023 ions mole-1