Solutions to General Chemistry (Linus Pauling)/Elements and Compounds Atomic and Molecular Masses

4-5
Define isotope, isobar, nuclide, mass number, N, A, Z, the dalton.


 * Nuclide: either a nucleus with a certain value of Z and A (see below) or an atom containing such a nucleus.
 * Isotope: a nuclide which shares in common with a different nuclide the same value of Z, but differs in N (see below)
 * Isotone: a nuclide which shares in common with a different nuclide the same value of N, but differs in Z (see below)
 * Isobar: a nuclide which shares in common with a different nuclide the same combined number of protons and neutrons, for example, boron-12 (an exotic radioactive isotope of boron) and carbon-12 (the most common stable form of carbon).
 * Mass number: same as A?
 * N: the number of neutrons in a nuclide
 * Z: the number of protons in a nuclide, also known as atomic number
 * A: the sum of neutrons and protons in a nuclide, $$N + Z$$
 * Dalton: exactly one-twelfth the mass of a neutral carbon atom, or approximately, 1.66033 × 10-27 kg. Used as the standard for atomic mass measurement.

4-6
'''What are the atomic number and approximate atomic weight of the element each of whose nuclei contains 81 protons and 122 neutrons? Give the complete symbol for this nuclide, including chemical symbol, atomic number, mass number.'''

The atomic number would be 81; the atomic weight, approximately equal to the sum of protons and neutrons in the nuclide, so what again will work out 203. The complete symbol for this nuclide is then $$\displaystyle ^{203}_{81}\mathrm{Tl}$$.

4-8
'''An atom of 90Sr emits a beta ray. What are the atomic number and mass number of the resulting nucleus? What element is it? This nucleus also emits a beta ray. What nucleus does it produce?'''

Because strontium-90 undergoes beta decay, specifically, beta minus decay, one of its neutrons becomes a proton and a free electron. As such it becomes yttrium-90, with an atomic number of 39, and a mass number of 90. In turn, yttrium-90 undergoes a further beta minus decay, resulting in zirconium-90, with an atomic number of 40, and a mass number of 90.

4-9
'''Argon, potassium, and calcium all have nuclides with mass number 40. How many protons and how many neutrons constitute each of the three nuclei?'''

The number of protons in each nuclide is given by the atomic number. Then the neutrons are whatever number is left over. Argon has 18 protons and 22 neutrons; potassium has 19 and 21; and calcium, 20 and 20.

4-11
'''What would be the implications of defining Avogadro's number as 1.00000 × 1024? What units would have to be changed?'''

The redefinition of Avogrado's number, arbitrary as it is, would upset a great many chemical formulae. Because atomic mass would no longer be convenient for use with the dalton, which would have to be altered to restore the relationship between moles and atomic mass. is it possible to elaborate here?

4-13
'''The molecule of the anesthetic agent nitrous oxide, N2O, contains two atoms of nitrogen and one atom of oxygen. Using Avogrado's number and the atomic weights of nitrogen and oxygen, calculate the weight in kilograms of one atom of oxygen and that of two atoms of nitrogen. Also calculate the weight of the nitrous oxide molecule. What is the percentage composition by weight of nitrous oxide in terms of nitrogen and oxygen?'''

The atomic weight (or, better said, atomic mass) of oxygen is 15.9994. In other words, a mole of oxygen is 15.9994 g. Dividing this number by Avogadro's constant, the number of molecules of a substance in a mole, or simply N, the mass of an oxygen atom is 2.65676255 × 10-23 grams. (Remember units.) So, the mass of an oxygen atom is 2.65676255 × 10-26 in kilograms. Through virtually identical calculations, the mass in kilograms of a nitrogen atom is 2.32586697 × 10-26, and of two, it is 4.65173394 × 10-26. The mass of a nitrous oxide molecule is then 7.30849649 × 10-26 kg. The percentage (by mass) of oxygen in a quantity of nitrous oxide is then the quotient of the mass of an oxygen atom and the mass of nitrous oxide molecule, approximately 36.35%. Likewise, the percentage by mass of nitrogen in nitrous oxide is approximately 63.65%.

4-14
'''Balance the following equations of chemical reactions. The molecular formulas are correct.'''


 * Fe2O3 + C → Fe + CO2
 * Ag + S8 → Ag2S
 * C12H22O11 + O2 → CO2 + H2O
 * H3PO4 + NaOH → Na3PO4 + H2O
 * Li + H2O → LiOH + H2
 * HCl + Ba(OH)2 → BaCl2 + H2O
 * CuO + NH3 → N2 + Cu + H2O
 * N2 + H2 → NH3
 * H3BO3 → B2O3 + H2O
 * Fe + F2 → FeF3

The author informs the reader that the only things that need to be changed are the amounts of each substance. The most general way to do this is through the solution of a system of linear equations. Like most algorithms, the method is best shown by example.

First, assign an unknown variable to each substance in the reaction. For instance:

wFe2O3 + xC → yFe + zCO2

Then, for each element mentioned in the reaction, write a linear equation equating the relative amount of each element on each side and relating the variables given above:

$$ \begin{align} 2w & = & y \\ 3w & = & 2z\\ x & = & z \end{align} $$

The equations are written for the balances of iron, oxygen, and carbon, respectively. Taking iron, for example, the appropriate equation is $$2w = y$$, because ferric oxide (Fe2O3) on the left contains two iron atoms, whereas its product on the right is pure iron. So, for a given amount w of reactant ferric oxide molecules, there are 2w iron atoms on the left side, which must be equal to the number of iron atoms on the right side, y.

Finally, these equations must be combined into a system of linear equations which is then solved to give the relative amounts of each substance in the reaction

$$ \begin{align} 2w &  &    & - & y &   &    & = & 0 \\ 3w &  &    &   &   & - & 2z & = & 0 \\ &  &  x &   &   & - &  z & = & 0 \end{align} $$

The solution leaves z free, parameterizing the other variables like so:

$$ \begin{align} x & = & z \\ w & = & \frac{2}{3}z \\ y & = & \frac{4}{3}z \end{align} $$

Because the reaction coefficients are necessarily integers, let z equal 3. x then is three; w, 2; and y, 4. The reaction is 2Fe2O3 + 3C → 4Fe + 3CO2.

The remaining reactions, which can be determined in the same manner, are:


 * 16Ag + S8 → 8Ag2S
 * C12H22O11 + 12O2 → 12CO2 + 11H2O
 * H3PO4 + 3NaOH → Na3PO4 + 3H2O
 * 2Li + 2H2O → 2LiOH + H2
 * 2HCl + Ba(OH)2 → BaCl2 + 2H2O
 * 3CuO + 2NH3 → N2 + 3Cu + 3H2O
 * N2 + 3H2 → 2NH3
 * 2H3BO3 → B2O3 + 3H2O
 * 2Fe + 3F2 → 2FeF3

4-15
'''How much coal (assumed to be pure carbon) is needed to reduce one ton of Fe2O3 to iron? How much iron is produced?'''

First, notice that the reaction is already given: 2Fe2O3 + 3C → 4Fe + 3CO2. Because the amount of ferric oxide given in the equation is in tons, it is necessary to convert it to grams. Then there are about 907,184.74 grams of ferric oxide in the reaction. In itself, however, the mass is not useful for the purposes of this problem. What is needed is the number of molecules of ferric oxide in the reaction, a number manageable in the form of moles. The atomic mass of ferric oxide is 2 × 55.845 + 3 × 15.9994 = 159.6882. There are then about 5,680.98 moles of ferric oxide in the reaction. The reaction specifies that for every two moles of ferric oxide, there must also be three moles of carbon. So there 8,521.47 moles of carbon are needed, or 8,521.47 × 12.0107 = 102,348.82 g. Note that, although the molar amount of carbon needed for the reaction is greater by half than the amount of ferric oxide, a chemist ordering the coal in Imperial units would ask for a little more than 225 pounds, due to the lower molar mass of coal.

4-16
Exactly what is meant by the statement that the atomic weight of samarium is 150.33?

In the broadest sense, this means that that the average mass of naturally occurring samarium atoms is 150.33 daltons (see above), as the mass of an individual samarium atom varies according to its isotope. Specifically, this number is the sum of the masses of the 62 protons, 88 electrons, and (to a much lesser extent) the 62 electrons in the average samarium atom.