CH 101/Keys 5
From WolfWikis
| CH 101 |
|---|
| Exercises 1 |
| Exercises 2 |
| Exercises 3 |
| Exercises 4 |
| Exercises 5 |
| Exercises 6 |
| Exercises 7 |
| Exercises 8 |
| Exercises 9 |
| Exercises 10 |
| Exercises 11 |
| Exercises 12 |
| Exercises 13 |
| Exercises 14 |
| Exercises 15 |
|
Back to CH 101 |
Contents |
Exercises
Exercise 1
What is the number of protons, electrons and neutrons in the following particles.
- We need the periodic table, because we need the atomic numbers
- 234Th37+ See[1]
- At. number 90 (equals the # of protons); # of electrons= 90-37= 53; # of neutrons= 234-90=144
- 13C
- At. number 6 (# of protons); # of electrons= 6-0=6; # of neutrons= 13-6=7
- 210Pb4+
- At. number 82 (# of protons); # of electrons= 82-4=78; # of neutrons= 210-82=128
- 85Br-
- At. number 35 (# of protons); # of electrons= 35-(-1)=36; # of neutrons= 85-35=50
- 40K+
- At. number 19 (# of protons); # of electrons= 19-1=18; # of neutrons= 40-19=21
- 60Co3+
- At. number 27 (# of protons); # of electrons= 27-3=24; # of neutrons= 60-27=33
- 55Fe2+
- At. number 26 (# of protons); # of electrons= 26-2=24; # of neutrons= 55-26=29
- 56Fe21+[1]
- At. number 26 (# of protons); # of electrons= 26-21=5; # of neutrons= 56-26=30
- 31P3-
- At. number 15 (# of protons); # of electrons= 15-(-3)=18; # of neutrons= 31-15=16
Exercise 2
Carbon and oxygen occur mostly as 12C and 16O. Other stable isotopes are 13C and 17O, they occur only in trace amounts. However there are (costly) ways of enriching the elements in the heavier isotopes. Departing from 50:50 mixtures of 12C/13C and 16O/17O carbon dioxide is synthesized.
- What is the molar mass of the reaction product?
- When in doubt balance an equation...
- C + O2 ==> CO2
- The molar mass of the carbon will be the average of 12 and 13 = 12.5 [g/mol] because we have equal amounts of each isotope. Likewise the molar mass of oxygen will be 2*16.5= 33 [g/mol]. Thus the molar mass of the carbon dioxide will be 12.5+33= 45.5 [g/mol].
- This is a bit heavier than the normal 12+32=44 [g/mol]
- Assuming the atoms of the isotopes react randomly with each other, what are the molar weights of a various combined species that are formed?
- We will get all combinations:
- 12C16O2: M=12+16+16= 44 [g/mol]
- 12C16O17O: M=45 [g/mol]
- 12C17O2: M=46 [g/mol]
- 13C16O2: M=45 [g/mol]
- 13C16O17O: M=46 [g/mol]
- 13C17O2: M=47 [g/mol]
- (If you like statistics you can even figure out how much of each you get. I did not ask that, did I? But it is 12.5%,25%,12.5%,12.5%,25%,12.5%)
Exercise 3
How many moles of protons, neutrons and electrons are there in a mole of 39Ca19F2?
- This is an interesting calculation, but also a pretty useless one, because the protons, the neutrons and most of the electrons are so tightly bound that they are not available for any chemistry...
- 1 39Ca atom has 20 protons, 39-20=19 neutrons and 20 electrons in total.
- 1 19F atom has 9 protons, 19-9=10 neutrons and 9 electrons in total.
- Thus in one mole 39Ca19F2 we have 20+2*9=38 moles of protons and electrons and 19+2*10= 39 moles of neutrons.
- ...However...
- The chemistry is actually just done by the two valence electrons of the Ca atom: two moles of valence electrons do all the work....
- The rest just adds to the mass (protons and neutrons) or to the size or volume (the other 'core' electrons).
Exercise 4
In diffraction experiments the scattering power is a measure for how strongly the particle will contribute to the scattering of the radiation.
In X-ray diffraction the scattering power of an atom (or ion) is equal to the number of electrons of the particle.
In neutron diffraction the scattering power depends not on the number of electrons but differs from isotope to isotope, because it is the nucleus that scatters, not the electron.
Some neutron scattering powers:
| nucleus | absolute scattering power |
|---|---|
| 2H | 6.671 |
| 7Li | 2.22[2] |
| 11B | 6.65 |
| 238U | 8.402 |
- What percentage of the total scattering power of the compound 238U2H3 is the result of the hydrogen for X-ray scattering? For neutron scattering?
- Whoa... X-ray diffraction? Neutron scattering? Never heard of? Don't be scared! You actually know enough to solve this pretty easily and yes it is a real life problem.
- First let's do the X-ray one. We need the number of electrons. If we think in atoms U has 92 and H only one electron. The isotopes don't enter into that at all. This means that UH3 has 92+3=95 electrons. The hydrogens only contribute 3/95* 100% = 3.15%. That means that some 97% of the scatter comes from the uranium. This makes X-ray diffraction pretty much blind for the hydrogens..
- This is actually why people use neutron diffraction instead. Here we do need to look at the isotopes. The total scattering power for the compound becomes 8.402+3*6.671 =28.415, of which no less than 3*6.671=20.013 comes from the deuterium. So 20.013/28.415*100%= 70% of our information pertains to the deuteriums.
- I have assumed here that the U and H are atoms and not ions. For neutron diffraction that does not matter, they are scattered by the nucleus, not the electrons. For X-rays it matters a little bit. If I assume U3+ and 3 H- hydride ions, the percentage becomes 6/95*100% because each hydide ion has two electrons. X-ay would still be as good as blind....
- What percentage of the total scattering power of the compound 7Li11B2H4 is the result of the hydrogen for X-ray scattering? For neutron scattering?
- Here X-ray is much more competitive. Li has 3, B has 5 and four hydrogens have 4 electrons. Total: 12. Four out of 12 is 33.3%. (If I do it in ions it even becomes 66%)
- Neutrons are doing OK as well: total is 2.22+6.65+4*6.67 =35.55. The deuteriums contribute 4*6.67=26.68 or about 75% to the total scattering.
- ..However..
- In order to make neutrons work we need to use only these three isotopes, because other isotopes like 1H or 10B are very unsuitable. They absorb neutrons very strongly... We need to take them out of our material! This makes the neutron experiment rather costly. For X-rays the isotopes do not matter and I can take off-the-shelf cheep lithium borohydride.
- Neutron and X-ray diffraction are very important ways to study the three dimensional structure of materials. As you see you do need to have the kind of skills you learn in CH 101 before choosing which experiment you need to do!