1
SOLUTIONS MANUAL
,
, 1
1 Basic concepts: atoms
1.1 The notation:
50
24
Cr
shows that the atomic number, Z, is 24 and the mass number for the isotope
is 50.
Number of protons = Number of electrons = Z = 24
Number of neutrons = Mass number – Z = 50 – 24 =
26
For each isotope, Z = 24 and so there are 24 electrons and 24 protons.
For mass numbers 52, 53 and 54, there are 28, 29 and 30 neutrons,
respectively.
1.2 ‘Monotopic’ means that the element possesses only one isotope. Examples
See Appendix 5 in H&S ▶ other than As include P, Na and Be.
1.3 (a) Al is monotopic, i.e. there is only one naturally occurring isotope.
Z = 13 Mass number = 27
Number of electrons = Number of protons = 13
Notation: ▶ Number of neutrons = 27 – 13 = 14
27
13 Al (b) Br (Z = 35) has 2 naturally occurring
isotopes.
Each isotope has 35 electrons and 35 protons.
79 81 For the isotope with mass number 79: number of neutrons = 79 –
80Br
Br
35 35 35 = 44 For the isotope with mass number 81: number of neutrons
57
54 56 Fe Fe 5 Fe = 81 – 35 = 46
26 Fe 8 (c) Fe (Z = 26) has 4 naturally occurring isotopes.
2 26
6 2 Each isotope has 26 electrons and 26 protons.
6 For the isotope with mass number 54: number of neutrons = 54 –
26 = 28 For the isotope with mass number 56: number of neutrons
= 56 – 26 = 30 For the isotope with mass number 57: number of
neutrons = 57 – 26 = 31 For the isotope with mass number 58:
1.4 number of neutrons = 58 – 26 = 32
Assume that 3H can be ignored since abundance is so low; error introduced
by this assumption is negligible. The mass numbers of 1H and 2H are 1 and
2 respectively. Let % 1H = x, and % 2H = 100 – x
Then:
x 1 100 x
A r = 1.008 =
100 2 100
+
100.8 = x + – 2x
200
x = 99.2
This result gives 99.2 % 1H and 0.8 % 2H. The values do not agree with
those in Appendix 5 (99.985 % 1H and 0.015 % 2H) because we have used
, 2 Basic concepts: atoms
integral atomic masses for the isotopes. The accurate masses (5 sig. fig.)
are 1.0078 and 2.0141, and if you work through the above calculation
again, this gives 99.98 % 1H and
0.02 % 2H.
SOLUTIONS MANUAL
,
, 1
1 Basic concepts: atoms
1.1 The notation:
50
24
Cr
shows that the atomic number, Z, is 24 and the mass number for the isotope
is 50.
Number of protons = Number of electrons = Z = 24
Number of neutrons = Mass number – Z = 50 – 24 =
26
For each isotope, Z = 24 and so there are 24 electrons and 24 protons.
For mass numbers 52, 53 and 54, there are 28, 29 and 30 neutrons,
respectively.
1.2 ‘Monotopic’ means that the element possesses only one isotope. Examples
See Appendix 5 in H&S ▶ other than As include P, Na and Be.
1.3 (a) Al is monotopic, i.e. there is only one naturally occurring isotope.
Z = 13 Mass number = 27
Number of electrons = Number of protons = 13
Notation: ▶ Number of neutrons = 27 – 13 = 14
27
13 Al (b) Br (Z = 35) has 2 naturally occurring
isotopes.
Each isotope has 35 electrons and 35 protons.
79 81 For the isotope with mass number 79: number of neutrons = 79 –
80Br
Br
35 35 35 = 44 For the isotope with mass number 81: number of neutrons
57
54 56 Fe Fe 5 Fe = 81 – 35 = 46
26 Fe 8 (c) Fe (Z = 26) has 4 naturally occurring isotopes.
2 26
6 2 Each isotope has 26 electrons and 26 protons.
6 For the isotope with mass number 54: number of neutrons = 54 –
26 = 28 For the isotope with mass number 56: number of neutrons
= 56 – 26 = 30 For the isotope with mass number 57: number of
neutrons = 57 – 26 = 31 For the isotope with mass number 58:
1.4 number of neutrons = 58 – 26 = 32
Assume that 3H can be ignored since abundance is so low; error introduced
by this assumption is negligible. The mass numbers of 1H and 2H are 1 and
2 respectively. Let % 1H = x, and % 2H = 100 – x
Then:
x 1 100 x
A r = 1.008 =
100 2 100
+
100.8 = x + – 2x
200
x = 99.2
This result gives 99.2 % 1H and 0.8 % 2H. The values do not agree with
those in Appendix 5 (99.985 % 1H and 0.015 % 2H) because we have used
, 2 Basic concepts: atoms
integral atomic masses for the isotopes. The accurate masses (5 sig. fig.)
are 1.0078 and 2.0141, and if you work through the above calculation
again, this gives 99.98 % 1H and
0.02 % 2H.
