Calculate the radius of a copper atom in the FCC structure if the unit cell dimension is 0.3615 nm
Things to think about: Think of the unit cell as a square (if you are thinking in 2 dimensions) or as a box
(if you want to think in 3 dimensions).
If you want to calculate the radius, you have to make sure that, when given a measurement, it pertains to
the atoms (or ions), not empty space
Let’s draw the FCC structure and see what this 0.3615 nm measurement really means
Haha!! The atoms on the edge of the cell do not
touch. So while the edge length/unit cell
dimension is important, it must be used to
determine information concerning where the
atoms DO touch one another.
The atoms touch each other on the diagonal. So let’s use
that information to determine what the measurement of the diagonal is!
That is where the atoms are touching one another so that is the measurement
that we need in order to determine the radius of any atom in the FCC
structure. If we tried to determine the radius from the edge length, there is a
good portion of the edge length that is empty space, not part of the atoms in
any way. If we said that the edge length was directly correlated to the size
of the atoms, we would be wrong because the edge length also includes
empty space
Let’s examine the FCC structure as a box.
The problem tells us that the edge of the box (unit cell) is
0.3615 nm. Since unit cells are “squares” all edges are
the same length!
Thinking about the layer diagrams on the pre-lab, we have a radius of the of the
top circle contained in the FCC cell, the full diameter (2 radii) contained in the
middle, and again only the radius of the bottom circle contained in the FCC cell
So, we have the edge length, which is helpful, but we need the diagonal! And the diagonal is made of 4
total radii.
Let’s go back to geometry class!
If you have a triangle, that each side is 0.3615 nm, how do you find the length of the hypotenuse??
THINK THINK!!!
4(radii)
0.3615 nm
0.3615 nm
Using our geometry skills: the Pythagorean theorem: a2 + b2 = c2
(4r)2 = (0.3615nm)2 + (03615 nm)2
I’ll let you all do the fun math – but hopefully you come up with r = 0.1278 nm
Calculate Avogadro’s Number for the above Cu structure. The density of Cu is 8.92 g/cm3
Things to think about: FIRST and FOREMOST – you are determining a numerical value for Avogadro’s
number. That means, at no time and under no circumstances do you USE Avogadro’s number!! You are
to calculate it. And therefore, it only appears as the ANSWER, not as a conversion factor in your
problem!
The units for Avo’s number are entities/mole. Since the chemical species in the problem is copper, you
are determining the number of copper atoms per mole of copper.
Let’s examine what we know, because what is comes down to is just organizing a variety of conversion
factors such that when we use them, our units cancel out leaving us atoms Cu/mol Cu!!
We know the edge length of a copper unit cell (0.3615 nm)
We know that in the FCC structure, there are 4 atoms/cell
We know the mm Cu = 63.55 g Cu/mole Cu
And the density of the unit cell is 8.92 g/cm3
1.) Using the edge length, 0.3615 nm) we can calculate the volume of the cell. Remember the
volume of the box is L x W x H and since all sides are the same, you take 0.3615 nm x 0.3615 nm
x 0.3615 nm
This gives you the volume of the cell in nm3
2.) Is the entire unit cell – all the space in the unit cell – taken up by Cu atoms – or is there empty
space in the cell? The answer is, there is empty space in the cell. What number up above gives us
information about the amount of stuff in the cell? Which number up above relates the amount of
copper present in a cell? It’s the density! And notice that from number 1 – we calculated the
volume! So if we multiply the density x the volume of the cell – if our units are correct!! – we
will get the number of grams of Cu present in the FCC cell!
Turn the volume (nm3) into a volume with the units cm3.
Remember that if your units are cubed you MUST cube the numerical portion of the conversion
factor also!!
1 in 3
= 0.0610 in3
(2.54) 3 cm3
Remember that for the HW problem (and for your own sanity when you study your lab later in
preparation for the lab final!) you should show all your work!!
e.g. 1.00 cm3 x
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But, before we go any further, let’s check to make sure we are all on the same page: your cube
4.724 x 10 -23 cm3
volume =
cell
3.) We have the volume of the cube, we have the density of copper, if we multiply density x volume
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we get the number of grams of copper. BUT, we have to remember that there are FOUR copper
atoms per unit cell in the FCC structure. Set up your conversion factors as below
matom = 8.92
grams 4.724 x 10 -23 cm3
1 cell
??? grams Cu
x
x

3
cm
1 cell
4 Cu atoms
1 Cu atom
4.) We are sooooo close. Avogadro’s number has the units atoms Cu/mole Cu
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5.) So, the only conversion factor left to play with is the mm of Cu (63.55 grams/mole)
??? grams Cu
6.) Using your answer above
and the mm, 63.55 grams Cu/mole Cu
1 Cu atom
??? grams Cu
7.) I’ll let you all figure out how to arrange the
and the mm to give you the correct
1 Cu atom
units for Avogadro’s number!
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