Initial info:
The atoms of crystalline solid pack together into a three-dimensional array of many small repeating units called unit cells. The simplest of the unit cells are those that have cubic symmetry, with atoms positioned at the corners of a cube. Atoms can also be found in the sides (faces) of the cube, or centered within the body of the cube. It is important to realize that a unit cell is surrounded by other unit cells in every direction. Therefore, face and corner atoms are shared with neighboring unit cells. The fraction varies with the type of atom as shown in the following table.
Type of atom / Fraction in unit cell
corner 1/8
face 1/2
body 1
The size of a unit cell in any given solid can be calculated by using its density. This and the reverse calculation are common test questions in general chemistry courses.
**I am having problems with the following 2 questions:
1. Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362pm. What is the radius of a gallium atom?
Express your answer numerically in picometers.
"radius = _____pm"
&
2. The face-centered gold crystal has an edge length of 407pm. Based on the unit cell, calculate the density of gold.
Express your answer numerically in grams per cubic centimeter.
density = ______g/cm^3
For Further Reading
CHEM - DrBob222, Thursday, November 29, 2007 at 10:49am
For a primitive unit cell, a = 2r where a is the length of the edge of the cell and r is the radius.
For the fcc gold atom, density = mass/volume.
volume calculation:
Volume = (edge)3. Don't forget to change the pm to cm if you want density in g/cc.
mass calculation:
There are 4 atoms per unit cell along the face of one cube.
mass = (4atoms)(atomic mass Au)/6.02 x 10^23 atoms mol-1
Then density = mass/volume
Does this mean for #1 then, when you say a=2r...
you just go: a*2=r therefore: 362*2 = r .... radius = 724pm's??
& for #2 (with the gold/Au):
Volume = (407pm)^3??? = 67,419,143 (or: 6.74 * 10^7)??
then does 6.74*10^-7pm's translate to: 6.74*10^+10 cm's??
Then go something like:
(4 atoms)(196.967g/mol Au)/6.02*10^23 atoms mol-1
= 1.30875*10^-21 g??
Then density = mass/volume:
d = 1.30875*10^-21 g / 6.74*10^+10 cm's
*****d = 1.9418*10^-32 g/cm^3????!!!***
PLEASE let me know if I have totally worked these two problems wrong!!! I am pretty sure I am not correct.
I am having problems with the following 2 questions:
1. Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362pm. What is the radius of a gallium atom?
Express your answer numerically in picometers.
"radius = _____pm"
&
2. The face-centered gold crystal has an edge length of 407pm. Based on the unit cell, calculate the density of gold.
Express your answer numerically in grams per cubic centimeter.
density = ______g/cm^3
For Further Reading
CHEM - DrBob222, Thursday, November 29, 2007 at 10:49am
For a primitive unit cell, a = 2r where a is the length of the edge of the cell and r is the radius.
For the fcc gold atom, density = mass/volume.
volume calculation:
Volume = (edge)3. Don't forget to change the pm to cm if you want density in g/cc.
mass calculation:
There are 4 atoms per unit cell along the face of one cube.
mass = (4atoms)(atomic mass Au)/6.02 x 10^23 atoms mol-1
Then density = mass/volume
Does this mean for #1 then, when you say a=2r...
you just go: a*2=r therefore: 362*2 = r .... radius = 724pm's??
No. a=2r. Why did you change it to a*2 = r? a*2=diameter but a=edge length=2*radius You take the edge length (a) and divide it by 2 to obtain the radius. So 362 pm/2 = ?? pm radius.
CHEM**** - K, Friday, November 30, 2007 at 1:53am
& for #2 (with the gold/Au):
Volume = (407pm)^3??? = 67,419,143 (or: 6.74 * 10^7)??
then does 6.74*10^-7pm's translate to: 6.74*10^+10 cm's??
No. Cubing the edge to obtain volume, using picometers, is ok BUT it makes the conversion from cubic picometers to cubic centimeters a little harder. I would convert 407 pm to cm FIRST, then cube.
407 pm x (10^-10 cm/pm) = 4.07 x 10^-8 cm. Then volume = (4.07 x 10^-8 cm)3 = 6.74 x 10^-23 cc. Check my arithmetic.
Then go something like:
(4 atoms)(196.967g/mol Au)/6.02*10^23 atoms mol-1
= 1.30875*10^-21 g??
The mass is ok (except you can't have that many significant figures. I would use 1.31 x -21 g.).
Then density = mass/volume:
d = 1.30875*10^-21 g / 6.74*10^+10 cm's
*****d = 1.9418*10^-32 g/cm^3????!!!***
I think this will be ok if you substitute the correct value for volume (in cc) in place of the 6.74 x 10^10 cc. I calculated a density of 19.3 g/cc. You can look up the value of gold. Gold is quite dense but not as dense as 10^10. :-).
PLEASE let me know if I have totally worked these two problems wrong!!! I am pretty sure I am not correct.
Let me know if you have additional questions on this.
this worked out perfectly!...thank you...that was the last question for the year that I had to finish through that program.
I really must not have been paying attention with "a*2", as opposed to dividing a/2 to solve for "r". Thank you again, & for having pointed that out & with the conversion of pm's to cm's. That looks much faster & correct!
Thank you for all of your help, & bobpursley's help this past semester!!! It all has guided me to understand what I am doing, for the most part, sooo tremendously!!!!!
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Yes, you are correct in solving the first question. In a primitive cubic unit cell, the length of one edge of the cube is equal to twice the radius of the atom. So, to find the radius of the gallium atom, you can use the equation a = 2r, where "a" is the length of the edge of the cube and "r" is the radius of the atom.
In this case, the length of the edge of the cube is given as 362pm. So, you can solve the equation: 362pm = 2r. Rearranging the equation, you get: r = 362pm / 2 = 181pm.
Therefore, the radius of the gallium atom is 181pm.
Now for the second question about calculating the density of gold in a face-centered cubic (fcc) unit cell:
First, you need to calculate the volume of the unit cell. The volume of a cube is given by the formula volume = (edge)³. The edge length of the gold crystal is given as 407pm. So, the volume of the unit cell would be (407pm)³.
Next, you need to calculate the mass of the unit cell. In an fcc unit cell, there are 4 atoms per unit cell along the face of one cube. So, the mass of the unit cell would be (4 atoms)(atomic mass Au) / (6.02 x 10^23 atoms/mol).
Finally, you can calculate the density by dividing the mass by the volume: density = mass / volume.
Remember to convert the edge length from picometers (pm) to centimeters (cm) before plugging it into the volume formula, and also to use the appropriate units for atomic mass and density (grams per cubic centimeter, g/cm³).
I hope this explanation helps you in solving the two questions.