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KEY for Unit 1 – Your Chemical Toolbox:
Scientific Concepts, Fundamentals of Typical Calculations, the
Atom and Much More
The Modern Periodic Table
The Periodic Law - when elements are arranged according to their atomic number, their
chemical and physical properties show repeatable trends. The vertical columns are called groups
and the horizontal rows are called periods. There are four categories of groups (although some
people do not separate the Noble gases from the Main Group elements):
1.
2.
3.
4.
Representative (also Main Group Elements)
Transition elements
Inner transition (also Lanthanide and Actinide)
Noble gasses
The letters A and B in the group distinguish families: A = representative and B = transition.
One of the most useful features of the periodic table is that it summarizes properties in a succinct
manner. Take a look at a periodic table and think about the following.
Where are metals found? See periodic table below – elements in various shades of blue.
Where are nonmetals found? See periodic table below – elements in yellow.
Where are the metalloids found? See periodic table below – elements in green.
Are all main group elements metals, nonmetals, metalloids? All Group 1A (except H) and
2A elements are metals. In groups 3A – 8A degree of metallic character increasing down
the groups (with groups 7A and 8A all being listed as nonmetals). Metalloids are in the
p-block of the periodic table running in a staircase pattern from boron (B) down and to
the right..
Locate the following groups in the periodic table.
Alkali metals (Group 1A or simply Group 1),
alkaline earth metals (Group 2A or simply Group 2),
noble gases [in the past called the inert gases] (Group 8A or simply Group 18),
halogens (Group 7A or simply Group 17),
chalcogens (Group 6A or simply Group 16),
coinage metals (Group 1B or simply Group 11),
and list the semimetals [the metalloids] boron (B), silicon (Si), germanium (Ge), arsenic
(As), antimony (Sb), and tellurium (Te). Polonium (Po) is often considered a metalloid,
too).
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Check three different periodic tables and see if they all list the same metalloids.
Think about why there might not be universal agreement about which elements to call metalloids
(think about the definition of metalloids, or semi-metals as they are sometimes called).
Metalloids are intermediate in behavior between metals and non-metals, and since metallic
behavior varies across a period and up (or down) a group, we should not be surprised if there is
not universal agreement on EXACTLY where the dividing line should be drawn. Some elements
that are listed as metalloids look quite metallic in appearance (like antimony, Sb) whereas others
display properties more similar to nonmetals (example the brittle nature of silicon, Si).
Figure: a reasonably current modern periodic table – new elements are added above atomic
number 109, but that area of the periodic table has no naturally occurring elements.
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Some Practice Problems with Atomic Numbers, etc.
Neutral Atoms
Ions
Mass #
#n
#p
#e
Full
Isotopic
Symbol
C-12
12
6
6
6
55 2 +
26 Fe
55
29
26
24
C-13
13
7
6
6
12
6
C 2−
12
6
6
8
Pb-206
206
124
82
82
16
8
O2−
16
8
8
10
F-19
19
10
9
9
209
83
Bi 3 +
209
126
83
80
Fe-56
56
30
26
26
81
35
Br1−
81
46
35
36
Na-23
23
12
11
11
79
35
Br1−
79
44
35
36
Cu-61
61
32
29
29
235
92
U 6+
235
143
92
86
Cl-35
35
18
17
17
108
47
Ag 1+
108
61
47
46
Cl-37
37
19
17
17
14
7
N 3−
14
7
7
10
O-18
18
10
8
8
32
16
S 2−
32
16
16
18
Isotope
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Mass #
#n
#p
#e
EXAMPLE: Average Atomic Masses
Atomic and Formula Masses
Isotopes and Atomic Masses
Cl-35 75.76 % 34.969 amu
Cl-37 24.24 % 36.966 amu
Ans. Ave. Cl = (34.969)(0.7576)amu + (36.966)(24.24)amu = 35.453 amu
Guided Discovery for Unit Five: Average Atomic Masses
1.
A particular element is found to exist as two isotopes.
(a) One has a mass that is 6.7405 times that of C-12, the other is 6.5787 times the mass of
C-12. What are the masses of these isotopes in atomic mass units (amu)?
ans.: (6.7405)(12 amu exactly) = 80.8860 amu
(6.5787)(12 amu exactly) = 78.9444 amu
(b) The isotope that has the lower mass has a relative abundance of 50.57 percent. What
is the relative abundance of the isotope with the larger mass?
ans.: 100 % – 50.57 % = 49.43 %
(c) What is the average atomic mass for this element?
ans.: (0.5057)(78.9444)amu + (0.4943)(80.8860)amu = 79.904 amu
(d) Write the full isotopic symbols for these isotopes (you know, the elemental symbol
with the mass numbers AND atomic numbers).
ans.:
2.
79
35
Br
81
35
Br
The element discussed in question number one is a halogen. As a pure element, it exists
as a diatomic molecule (at room temperature and 1 atm pressure it is most stable as a
liquid).
(a) What is the average mass of each of these diatomic molecules (in amu)?
ans.: if two average atoms combine they will create an average molecule with an
average mass = 2*79.904 amu = 79.904 amu + 79.904 amu = 159.808 amu
(b) Do any of the individual diatomic molecules have this mass? Ans.: NO !!!
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(c) Calculate the three masses that these diatomic molecules can actually have based on
the masses of the two isotopes used to form the molecules.
ans.:
lightest = 2 Br-79, (2)(78.9444)amu = 157.8888 amu
medium = Br-79+Br-81 or Br-81+Br-79, (78.9444)amu + (80.8860)amu = 159.8304 amu
heaviest = 2 Br-81, (2)(80.8860)amu = 161.7720 amu
(d) 39.95 grams of this element would contain how many atoms of this element?
ans.:
6.022 x10 23 Br atoms 1 mole Br atoms
x
x39.95g Br atoms =
1 mole Br atoms
79.90 4 g Br atoms
3.011x10 23 Br atoms
(e) 39.95 grams of this element would contain how many diatomic molecules of the
element?
Since each diatomic molecule is made up of two atoms, the number of molecules is half
23
the number of atoms from which they can be built = 1.505x10 Br2 molecules
Mass Relations in Chemical Formulas
You can try this one: find the mass percent of each of the elements in the molecule C2HBrClF3
Ans.:
% C = 100*(2*12.011 amu C)/(197.39 amu cmpd.) = 12.17 %
% F = 100*(3*19.00 amu F)/(197.39 amu cmpd.) = 28.88 %
% Br = 100*(1*79.904 amu Br)/(197.39 amu cmpd.) = 40.48 %
% Cl = 100*(1*35.453 amu Cl)/(197.39 amu cmpd.) = 17.96 %
% H = 100*(1*1.008 amu H)/(197.39 amu cmpd.) = 0.510%
check: the sum of these mass % = 100
examples
(a) 7.985 g of an ionic compound that is composed of sodium, chlorine, and oxygen is found to
contain 1.725 g of sodium and 2.659 g of chlorine. What is the simplest (empirical formula)?
strategy: (1) find the grams of oxygen present, (2) convert the mass of each element to moles,
and (3) divide by the smallest number of moles to get a whole number ratio.
Ans.:
(1) 7.985 g cmpd. – 1.725 g Na – 2.659 g Cl = 3.601 g O
(2) mole Na = 1.725 g Na/23.00 gmole-1 = 0.07500 mole Na
mole Cl = 2.659 g Cl/35.453 gmole-1 = 0.07500 mole Cl
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mole O = 3.601 g O/16.00 gmole-1 = 0.2251 mole O
(3) divide by moles of Na (or Cl in this case too) to find the simplest formula is
NaClO3
(b) A molecular compound is found to contain 40.0 % C, 6.73 % H, and 53.3 % O. What is the
simplest formula of the compound AND if it is known to have a molar mass of 181 g/mole, what
is the molecular formula.
strategy: (1) assume 100 g sample and convert % to grams, (2) convert the mass of each element
to moles, and (3) divide by the smallest number of moles to get a whole number ratio. To find
the molecular formula, find the mass of the empirical formula and determine how many of the
empirical units are needed to have a mass equal to the molecular mass.
Ans.:
(1) 40.0 g C, 6.73 g H, and 53.3 g O
(2) mole C = 40.0 g C/12.011 gmole-1 = 3.33 mole C
mole H = 6.73 g H/1.008 gmole-1 = 6.66 mole H
mole O = 53.3 g O/16.00 gmole-1 = 3.33 mole O
(3) divide by the mole of C (or also O in this case)
empirical formula = CH2O [which has an emp. mass of ~30.01 amu]
since 181/30.01 = about 6, the molecule is made up of 6 empirical units giving us a
molecular formula = 6(CH2O) = C6H12O6
(c) Cadaverine is a molecular compound formed in decomposing fish. It has a molar mass of
102.18 and is composed solely of C, H, and N. When 0.03560 g of cadaverine is combusted in an
excess of oxygen, 0.07665 g of CO2 and 0.04392 g of H2O are formed. What are the simplest
formula and molecular formula of cadaverine.
strategy: (1) find moles of carbon dioxide and water, (2) convert to moles of carbon and
hydrogen, (3) find the mass of carbon and hydrogen, (4) find the mass of nitrogen by difference,
(5) find moles of nitrogen, and (6) divide by the smallest number of moles (C, H, or N) to get a
whole number ratio. To find the molecular formula, find the mass of the empirical formula and
determine how many of the empirical units are needed to have a mass equal to the molecular
mass.
Ans.:
(1) mole of CO2 = 0.07665 g CO2/44.01 gmole-1 = 0.001742 mole CO2
mole of H2O = 0.04392 g H2O/18.02 gmole-1 = 0.002437 mole H2O
(2) mole C = mole CO2 = 0.001742 mole C
mole H = 2*mole H2O = (2moleH/mole H2O)*0.002437 mole H2O = 0.004874 mole H
(3) mass C = mole C*12.011g/mole = 0.02092 g C
mass H = mole H*1.008 g/mole = 0.004913 g H
mass N = mass cmpd – mass C – mass H = 0.03560 g – 0.02092 g C – 0.004913 g H
mass N = 0.00977 g N
(4) mole N = 0.00977 g N/14.01 gmole-1 = 0.000697 mole N
(5) divide by the mole of N to find
empirical formula = C2.5H7N1
which must be doubled to find C5H14N2
the empirical mass is 102.19 amu
thus the molecule consists of a single empirical unit giving
molecular formula = C5H14N2
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Some Practice Problems
Molecular Weight and Formula Weight (to find MW or FW add together atomic masses in
formula)
1.
2.
3.
4.
5.
NaCl has a formula weight
K2CO3 has a formula weight
CuSO4·5H2O has a formula weight
CH4 has a molecular weight
C6H13OH has a molecular weight
6. HCl has a molecular weight
=
=
=
=
=
58.44 amu
138.206 amu
249.69 amu
16.04 amu
102.18 amu
= 36.461 amu
and a molar mass
and a molar mass
and a molar mass
and a molar mass
and a molar mass
=
=
=
=
=
58.44 g/mole
138.206 g/mole
249.69 g/mole
16.04 g/mole
102.18 g/mole
and a molar mass = 36.461 g/mole
Mole and Molar Mass
1. 79.63 g NaCl = (1mole NaCl/58.44 g)*(79.63 g) = 1.363 mole NaCl
2. 2.694 mole CH4 = (16.04 g CH4/1 mole)*(2.694 mole) = 43.21 g CH4
3. 476.92 g CuSO4·5H2O = (476.92 g)/(249.69 gmole-1) = 1.910 mole CuSO4·5H2O
5 moles of water per mole of hydrate yields = 5*1.910 mole H2O = 9.550 mole H2O
Mole and Avogadro’s Number (info needed 1 mole = 6.022 x1023 items)
use molar mass for g to mole or mole to gram conversions
use Avogadro’s number for number of things to mole or mole to number of things conversions
1. 475.9 g Na (sodium)
= 20.70 mole Na
2. 2.74 mole Al (aluminum)
= 73.93 g Al
3. 7.46 x1021 Bi (bismuth) atoms = 0.0124 mole Bi
= 1.247x1025 Na atoms
= 1.65x1024 Al atoms
= 2.59 g Bi
End coverage at page 53 . . . . DO NOT FORGET – know ALL chemical symbols and
names through atomic number 30 [that’s from #1 hydrogen (H) through #30 zinc (Zn)]
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