CHEM101, B2
HMWK 01, Answers
Early Developments - Rutherford
2008 01 16
HT
1.) Provide a critique of Dalton’s “atomic statements”.
Give at least 3 instances where he wasn’t quite correct.
a.) In high energy physics, atoms can be smashed into much smaller pieces.
b.) The atoms in an element are actually not all the same, but are a mixture of isotopes.
c.) Electrons can be knocked off atoms, e.g. in the Thomson and Millikan experiments.
d.) In large molecules, e.g. drugs, protein, DNA ... the ratio of atoms is in fact not that
simple.
e.) etc. etc.
2.) Do the following unit conversions:
a.) 453 nm to µm (typical wavelength of visible light).
106 µm
1m
453 nm × 109 nm × 1 m = 453×10–3 µm = 0.453 µm
b.) 134 pm to nm (typical atomic radius).
1m
109 nm
134 pm × 1012 pm × 1 m = 134×10–3 nm = 0.134 nm
3.) A 15.4 kg piece of wood is burnt in excess oxygen (O2).
Ash with a mass of 2.7 kg is left over.
The mass of oxygen is reduced from 21.2 kg to 9.6 kg.
a.) What is the mass of gas produced during the burning process?
initial mass: 15.4 kg(wood) + 21.2 kg (oxygen) = 36.6 kg
final mass: 2.7kg (ashes) + 9.6kg (oxygen) + x kg (gases) = 36.6 kg ( Conservation of Mass)
x = 36.6kg - 2.7 kg - 9.6 kg = 24.3 kg (produced gases)
b.) What do you think are the main components of the produced gas?
You might know that the main component of wood is cellulose, a complex carbohydrate,
composed of carbon, hydrogen and oxygen. So we expect carbon dioxide and water (in vapor
form) as main combustion components.
4.) Methane and propane are 2 examples of hydrocarbons.
(consisting of hydrogen and carbon only).
A methane sample was analyzed and found to consist of 5.70 g carbon and 1.90 g hydrogen.
Similarly, a propane sample was found to consist of 4.47 g carbon and 0.993 g hydrogen.
Is this in agreement with the Law of Multiple Proportions?
Explain by doing appropriate calculations.
The C/H mass ratio in methane is = 5.70/1.90 = 3.00; in propane it is 4.47/0.993 = 4.50
The ratio of mass ratios in the two C/H compounds is 3.00/4.50 which is equivalent to 2 : 3.,
a simple whole number ratio. Thus the Law of Multiple Proportions is satisfied.
5.) In an instrument called a mass spectrometer the mass to charge (m/q) ratio of a
positively charged particle was determined to be 1.983 10–3g/C.
a.) Assuming that this particle has a single positive charge what is its atomic mass?
A single positive charge has the same numerical value, but opposite in sign,
as the charge of an electron.
We look up the value for the charge of an electron (Millikan): q = – 1.6022 x 10–19 C
Our particle has m/q = 1.983 x 10–3 g/C,
therefore
m = (1.983 x 10–3 g/C) x q
= (1.983 x 10–3 g/C) x (1.6022 x 10–19 C) = 3.178 x 10–22 g
This is the mass of our particle in absolute terms.
b.) Assuming that the particle is a single atom, what is its identity?
With the approximation that protons and neutrons have an atomic mass of
about 1.674 x10–24 g and with the further approximation the mass change is small when
protons and neutrons combine to form nuclei of heavier atoms we can estimate that the mass
number of our particle is 3.178 x 10–22 g / 1.674 x10–24 g = 189.8 ≈ 190 (no units).
The numerical values of atomic masses and mass numbers are generally close
(of course, the exact composition of the isotopic mixture can be quite variable).
The element that has an atomic mass close to 190 (see Periodic Table) is Os (osmium), but
we could not exclude a heavy isotope of Re (rhenium) or a light isotope of Ir (iridium) etc. ...
So our answer has to remain somewhat vague:
Our particle is possibly an isotope of Os, Re or Ir or something similar.
(The point of this exercise to illustrate the terms: atomic charge, atomic mass, mass number,
isotope, isotopic mixtures, ...)
6.) Natural boron (B) is a mixture of 2 isotopes, B-10 and B-11.
The atomic mass of natural boron is 10.81 u.
Pure B-10 has an atomic mass of 10.0129 u and B-11 has an atomic mass of 11.0093 u.
What is the percentage of B-10 and B-11 in natural boron?
First of all, we “ballpark” that there is more B-11 than B-10 because the atomic mass of the
natural mixture is closer to that of B-11.
More detailed calculations:
The general formula for weighted average calculations:
(x1) (I1) + (x2) (I2) = E (symbols as defined in class)
(x1) 10.0129 + (x2) 11.0093 = 10.81
x1 + x2 =1.0000...
x2 = 1.0000 - x1
(x1) 10.0129 + (1.0000 - x1) 11.0093 = 10.81
10.0129 x1 - 11.0093 x1 + 11.0093 = 10.81
- 0.9964 x1 = - 0.1993
x1 = 0.1993 / 0.9964 = 0.2000 ; x2 = 0.8000
In words, 20% of the boron atoms are of the B-10 type and 80% are of the B-11 type.
7.) Using phosphorus as an example describe the difference between the terms “atomic mass”
and “mass number”. Natural phosphorus is isotopically pure (contains only isotope)
Phosphorus has the atomic number 15, therefore has 15 protons. Since there is only 1
isotope in phosphorus, an atomic mass of 30.97 u indicates there should be 16 neutrons.
Therefore the mass number is 31 ( count of protons and neutrons, no units),
while the atomic mass (actually “relative atomic mass”) is 30.97 u (atomic mass units).
It really means that a phosphorus atom has a mass 30.97/12 = 2.581 times as large as that of
the C-12 atom.
8.) What are the chances that in a molecule of propane all 3 C atoms are C-13?
A rough estimate is sufficient. The formula for propane is H3C-CH2-CH3.
I emphasized in class that in natural carbon, 1% of the atoms are C-13 while the rest is
C-12.
The probability that the first C is C-13 is 0.01 , the probability that this is connected to
another C-13 is 0.01, and, finally, that these are connected to yet another C-13 is again
0.01.
We combine the probabilities and get a factor of 0.01 x 0.01 x 0.01 = 0.000 001.
This means in a collection of a million molecules of propane one (1) will have all C-13
isotopes.