Atomic Mass and Mass Defect

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This supplementary reading for Chem 12H describes atomic mass and mass defect. ... We can calculate the nuclear binding energy for 1 mol of 12C atoms as ... Problems: 1.) a.) In principle, is it possible to know Avogadro's number exactly? b.

Atomic Mass and Mass Defect This supplementary reading for Chem 12H describes atomic mass and mass defect. You should begin by re-reading Section 2.4 of the text on atomic weights, which describes the atomic mass scale and average atomic mass. Pay special attention to the ‘A Closer Look’ section that describes the mass spectrometer. Be familiar with how a mass spectrometer works and how to interpret a mass spectrum With regards to the atomic mass scale, the most important concept is the relationship between grams and amu. The atomic mass unit, or amu, is defined with regards to the pure isotope, 12C. We say that 1 atom of 12C has a mass of exactly 12 amu. You can think of this as 12 with an infinite number of zeroes after it. In addition, 1 mol of 12C, which has exactly Avogadro’s number of atoms in it, has a mass of exactly 12 g (infinite number of zeroes). In other words NA atoms of 12C 1 atom of 12C

12 g (exactly) 12 amu (exactly)

(where NA = Avogadro’s number)

Dividing these two lines, we get NA = 12 g/ 12 amu or 1 amu = 1 g / NA = 1 g / 6.022 x1023 = 1.661 x 10-24 g Now lets consider the concept of average atomic mass (also known as atomic weight). The average atomic mass is the value that you find for an element on the periodic table on the inside cover of your text. The average atomic mass is the weighted average of all isotopes for an element. Weighted average is defined as the abundance of a given isotope of an element times the mass of that isotope, summed over all the isotopes. (Note that the number of isotopes an element has is highly variable and unpredictable; e.g. Co has only 1 isotope, Cl has 2, and Hg has 7. You can find this information for your favorite element by using the Periodic Table link on the course website, clicking on the element of interest, and then looking under ‘nuclear properties’ in the left-hand column.) If there is more than one isotope for a given element, then the average atomic mass is not equal to the mass of any given isotope. If this sounds a bit foreign to you, you can think about this way. Imagine an exam is given to 10 students and the breakdown is as follows: # Students 3 5 2

Grade 90 80 70

A typical weigh to calculate the average is

(3*90+5*80+2*70)/(3+5+2) = 81

But this is equivalent to = 0.3*90 + 0.5*80 + 0.2*70, or = ΣfiVi, where fi is the fraction of the population with the ith value, and Vi is the ith value. Thus, we can see that weighted average (= ΣfiVi) is another way of taking an average.


We’ve been discussing the weighted average of the mass of all the isotopes of an element. Now let’s give some consideration to the mass of a given isotope. We said that 1 atom of 12C has a mass of exactly 12 amu. But where does that come from? If we look at the sum of the mass of the constituent subatomic particles that make up the 12C isotope, we see that it is more than 12: Particle proton neutron electron 12

Mass (amu) 1.0073 1.0087 5.486 x 10-4

C = 6p + 6n + 6e- = 6(1.0073) +6(1.0087) + 6(5.486 x 10-4) = 12.0993 (not 12.000000000) !?!

Is this discrepancy due to sig figs? No! There are 6 sig figs in the answer, so clearly it is something else. Some of the mass is converted to energy in binding the nucleons (the protons and neutrons) into the nucleus (the strong nuclear force that keeps the same-charged protons next to each other). Einstein in his Special Theory of Relativity (1905) formulated the equation to describe this phenomenon, E=mc2, which says that mass and energy are exchangeable. If we think of this as reaction, which it is, then ΔE is Efinal -Einit = mfinalc2 – minitc2 = (mfinal-minit)c2 = Δmc2, or

ΔE = Δmc2

We can calculate the nuclear binding energy for 1 mol of 12C atoms as = (12.0000 – 12.992 amu/atom) (1g/NA amu) (0.001 kg/g) (2.9979 x 108 m/s)2 = –1.48 x 10–11 kg m2/s2 = –14.8 pJ /atom This is exothermic, and on a per-atom basis, not very large). But what about on a per-mol basis? Multiplying by NA, we have 8.92 x1012 J/mol, or 8.92 x109 kJ/mol, a very large energetic value!!! Problems: 1.) a.) In principle, is it possible to know Avogadro’s number exactly? b.) If so, how many digits would it have in it? c.) In practice, how many significant figures do we know Avogadro’s number to? (You can use the Internet to help here, but give a reference.) 2.) a.) Given that 35Cl = 34.969 amu and 37Cl = 36.966 amu, calculate the natural abundance of 35 Cl. (Hint, you will need to use your periodic table.) b.) Does your answer make sense? Briefly explain.


3.) a.) Based on your answer to question 2, calculate a mass spectrum of dichlorine gas, Cl2 (g). b.) You should sketch the mass spectrum of dichlorine and pay attention to the mass values and their amplitudes. 4.) a.) Use the Periodic Table website to calculate the atomic weight of Hg from its isotopes. Express your answer in amu. Use the weighted average technique and show all your work. b.) Is there an isotope with the same mass as the atomic weight of Hg? 5.) This exercise has you compare the energy of nuclear reactions to chemical reactions. Go back to your notes and look up the energy associated with the reaction in which we exploded a hydrogen-oxygen balloon. Assume that the balloon held about 2.4 L of gas. How many times more energetic would the nuclear binding energy of 1 mol of 12C atoms be than the (very loud) balloon explosion we did in class? 6.) a.) How much energy must be supplied to break a single 20Ne nucleus into its constituent nucleons? b.) What is the nuclear binding energy for 1 mol of 20Ne? (You should use the Periodic Table website as needed on this problem.)