Chemistry: Find Atomic Mass, Radius from Unit Cell

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Chemists like to think of atoms as hard spheres, so here is, for example, a hard sphere of silver. Obviously, it's not really silver, it's Styrofoam, but suppose this is a silver atom. Silver, it turns out, crystallizes in a face-centered cubic lattice. So here's the face-centered cubic lattice. What we have is silver atoms in each of the 8 corners, so 4 up here and 4 down here, and then additional silver atoms in the face center. So 1 here, 1 on top, 1 in back, 1 in the bottom and 1 on each of the 2 sides. Now this, of course, this model, represents more than 1 unit cell. So, if you take a look at the graphic, there will be a cutaway picture, which illustrates how many atoms are really in the unit cell. Remember that only 1/8 of each of the corners is in the unit cell, and only ½ of each of the atom on the face center is inside the unit cell. And so, there are actually 4 complete atoms inside the unit cell, 1/8 of each of the corners, of which there are 8, so that's 8 times 1/8, plus 6 times ½, so 6 face center, each of which contain ½ an atom that's inside the unit cell. So, given that silver crystallizes in a face-centered cubic lattice and the idea that these are hard spheres, if we know what the length of the unit cell is - and remember, it's a cubic unit cell, so we only need to describe it by 1 length - if we know the length of the unit cell and we know something macroscopic, like the density of this macroscopic piece of silver, what can we learn? Well, the answer is that we can learn the mass of a silver atom, and also, knowing that the unit cell has a certain length and knowing that it's face-centered cubic, we can learn something about the radius of a silver atom, again, assuming that it's just a hard sphere.
So let's look at a problem. Silver crystallizes, again, in this face-centered cubic lattice, and the length of the unit cell is determined to be 408.6 picometers, where a picometer is 10^-12 meters. The density of silver is 10.05 grams per cubic centimeter. How could you measure that? Well, you could take a graduated cylinder and put some water in it and measure the volume, and then take a known mass of silver and drop it into the grad cylinder and the water would rise, and the volume change is the volume of the silver that you added. And, since you knew the mass of the silver, the density is just the mass of the silver divided by the volume of the silver. Now, given these 2 pieces of information, it's possible to calculate the mass of a silver atom, and it's also possible to calculate the radius of a silver atom. Actually, these 3 pieces of information: the face-centered cubic lattice, the length of the edge of the unit cell and the macroscopic density. So let's look at the mass of the silver atom first.
First, what we need to do is calculate the volume of the unit cell. So, if we express 408.6 picometers in terms of scientific notation, it's 4.086 times 10^-10 meters. And it's a cubic unit, so we cube the length of the side of the unit cell, and that's 6.822 times 10^-29 cubic meters. Now, we also need to convert the density into units of cubic meters, so we take the density, which is in grams per cubic centimeter, and multiply by conversion factor that relates centimeters to meters. And we have to take that to the power of 3, because we have 3 units of centimeters in the denominator here. And so, we multiply through here and, again, express it in scientific notation, 1.050 times 10^7 grams per cubic meters. Now, we're going to use these 2 pieces of information, because the mass of the unit cell is going to be just the volume of the unit cell times the macroscopic density. And the density is 1.050 times 10^7 grams per cubic meter, and the volume is 6.822 times 10^-29 cubic meters, and so the mass of a unit cell is 7.163 times 10^-22 grams per unit cell. Again, this is specifically for silver, because we use the macroscopic of density and we use the size of the unit cell of silver.
Now, inside each unit cell, remember, there are 4 silver atoms. You have bits and pieces of them as they're in the cutaway view, but they sum up to 4 silver atoms. So, if we know the mass of a unit cell, we also know the mass of 1 silver atom, because each unit cell contains 4 silver atoms. So the mass of 1 silver atom is the mass of the unit cell times 1 unit cell divided by 4 silver atoms. In other words, this is a conversion factor that relates the number of silver atoms to the unit cell, and so the mass of a silver atom is ¼ of this, which is 1.791 times 10^-22 grams. Now, this is obviously a very tiny number, but we expect that the mass of an atom is going to be a very tiny number.
And we can also calculate the radius of a silver atom based on the information that it's face-centered cubic and we know the length of the unit cell. And the reason is, if we look at the unit cell again, or this extrapolation of the unit cell, the balls on the corners are exactly touching the ball in the face center. So here's the ball in the one corner, here's the ball in the opposite diagonal corner and, along the face diagonal, we have that the balls are exactly touching and they're exactly in align. And because they're exactly in align, we can relate the radius of a hard sphere to a length that is related to the length of the unit cell. So let's go from this 3-dimensional picture to a 2-dimensional picture, and that 2-dimensional picture is a projection of the unit cell onto the picture of one of the front faces, or one of the faces. And again, let me repeat what I just said. If the red box is the size of the unit cell in 2 dimensions and d is the length of the unit cell, which was 408.6 picometers across a face diagonal, we have that the radius - the sum of 4 radii, so to get from the center of the ball in the corner to this point, that's 1 radius of a silver atom. And then from here to here, that's another radius, and from here to here is another radius, and from here to here is another radius. So, in other words, if we add up the radius 4 times, what do we get? Well, we get a length that is the diagonal of the face. So it's a face diagonal. And the way to say that algebraically is to say that 4 times the radius - remember, the radius is r-4 times the radius is equal to - and let me add a step here - is equal to the face diagonal. So, ignore this right now. 4 times the radius is exactly equal to the face diagonal. Now, we can do a little bit of trigonometry, and the trigonometry says this is a right triangle, where this side is d and this side is d. And the Pythagorean theorem tells us that, if this side is d, and this side is d and this is a right triangle, then the length of the hypotenuse, which is the hypotenuse of this right triangle, is equal to the square root of 2 times d. And so, we now have an equation that relates the radius of a hard sphere of a silver atom to the length of the unit cell. 4 times the radius of the silver atom is equal to the square root of 2 times the length of the unit cell, of 1 edge of the unit cell. We can solve this for r, r is equal to the square root of 2 over 4 times d, and we can plug in for d. And we get that the radius of a silver atom is 144.5 picometers. So, by knowing just the length of the unit cell - and then remember, this is only for a face-centered cubic lattice, we can calculate what the radius of a hard sphere atom ought to be.
Now, we can do related calculations for our other 2 cubic structures, and the other 2 cubic structures are simple cubic and body-centered cubic, and let me try to walk you through the relationships that we're going to derive here on this piece of paper. So here is a simple cubic lattice. And, if you think about it, where these balls are touching is they're touching along a line that is exactly the edge of the cubic unit cell. And so, we go from the center of this ball to the center of this ball, and it's going to be 2 radii, so 1 radius from the center of this ball to where they touch, and then another radius from where they touch to the center of this ball, that's exactly equal to the length of 1 side of the unit cell. And so, for simple cubic, we have the relationship d is equal to 2r, instead of the relationship that the square root of 2 times d is equal to 4r. But, given this, if you know what d is, you can solve for r. Again, for a simple cubic lattice.
Now, for body-centered cubic - and to remind you, body-centered cubic is 1 atom in each of the 4 corners plus 1 more in the very center - it's a little trickier. But, if you'll take this and take a look at it really closely, what you'll realize is that where balls are touching exactly is along the body diagonal; that is, from this front left corner to the back right corner, that is a series of 4 radii. So, if we go from this atom over here and if we go to 1 radius, it's just touching the atom that's in the middle. If we go another radius, it's to the very middle of the cube. Another radius is going to get us to just where it's touching the outside corner cube, and then 1 more will get us to the corner of the cube. So the length will be body diagonal, using a little bit of trigonometry, is the square root of 3 times the length of the edge of the cube. So the length of the edge of the cube in a body-centered cubic structure is the square root of 3 times d, and that's equal to 4r again, because we have a sum of 4 radii to go from this corner through the ball that's in the middle and out to the corner that's opposite on the body diagonal. So here again, we have a relationship between r and d and we can use this to calculate things like the radius of an atom, if it happens to be in a body-centered cubic structure.
So what have we talked about? Well, we've shown that if you know something about the crystal structure - and these are, obviously really simple. They're all in the cubic system. But, if you know something about the crystal structure and something about, for instance, the macroscopic density, you can calculate the mass of an atom. And this, actually, is one of the ways that they've calculated Avogadro's number rather accurately. And then the other thing is, if you know something about the size of the unit cell, you can calculate the radius of the atom.
Condensed Phases: Liquids and Solids
Solid State: Structure and Bonding
Calculating Atomic Mass and Radius from a Unit Cell Page [2 of 2]