Rhodium has density of 12.41g/cc and crystallizes with the face-centered cubic unit cell.
Calculate the radius of a rhodium atom.
xpress your answer using four significant figures.
For this question I tried to calculate backwards from a previous question I had asked this past Friday. I somehow managed to not get it correct, however.
I figured to go:
(4 atoms)(atmic mass Rb)/(6.022*10^23 atoms mol^-1)
=5.677*10^-22g ??
then I went: d=m/v
& v=m/d
v=5.677*10^-22/12.41g/cc
v=4.57*10^23cc
Then I cube-rooted it & got 3.575*10^-8cm's
Then I multiplied that by 1*10^10 & got: 357.5pm's
Then I divided that by 2 to get the radius, or so I thought would be the right thing to do, & got 178.76pm's or: 178.8pm's for my final answer but it was wrong.
Could you pls tell me what I did wrong here?
Sorry you ran out of tries. I was working the problem to make sure it came out right.
First, you must look up Avogadro's number to four or five places since the problem asks for the radius to 4 s.f. I have used 6.022 x 10^23 but another place won't hurt to start.
You are ok on the first step in finding the mass EXCEPT that you apparently used Rb instead of Rh. Look up the atomic mass of Rh to at least 4 (and preferable 5 places). Using
4*102.9/6.022 x 10^23 = mass = 6.835 x 10^-22 g is what I have. Probably you have the same number if you substituted Rh for the Rb you apparently used.
Your second step is ok, also. with
v = m/d = 6.835 x 10^-22/12.41 = 5.5076 x 10^23 cc but check that, especially if you have new numbers for the atomic mass and Avogadro's number.
Cube root of that gives a = 3.8048 x 10^-8 cm = 380.5 pm
The next step is what I don't see in your work.
The diagonal of the fcc is df = a(sqrt 2) = 380.5 x sqrt 2 = 538.1 pm
r = df/4 = 538.1/4 = 134.5 pm radius.
Again, you must go through this onoe more time and substitute the "good" values to use. I hope this helps.
oh yes...the answer should look like this: r = _____pm
I think that I used Robidium's atomic mass instead of Rdodium's...I'll check now if that works better.
Does it look as though I am on the right track though??
I tried exactly what I had done with the above & got my answer to be: 190.2pm's & was still wrong!!
Could you please guide me as to what I should be doing? I only have 2more attempts left at the question.
I ran out of attempts but I still would like some help trying to figure out how to do this question properly, please.
I did manage to get as far as the 380.5pm's. I guess where I did go wrong is that I missed carrying it further than that. I missed the 380.5*sqrt 2=538.1pm's & then the remainder of the answer.
Thank you very much. It's good to know that I was getting the handle of it, aside from the last part, at least.
To calculate the radius of a rhodium atom, you need to follow the correct steps in the calculation. Let's break down the correct approach:
Step 1: Determine the volume of the unit cell.
For a face-centered cubic (FCC) structure, there are 4 atoms in the unit cell. The volume of an FCC unit cell can be calculated by dividing the molecular weight by the density and then multiplying by Avogadro's number:
Volume of the unit cell = (atomic mass of rhodium / density of rhodium) x Avogadro's number
Step 2: Calculate the volume occupied by each atom.
Since there are 4 atoms in the unit cell, the volume occupied by each atom is one-quarter of the total unit cell volume. Divide the volume of the unit cell by 4.
Volume occupied by each atom = Volume of the unit cell / 4
Step 3: Determine the radius of each atom.
The volume occupied by each atom in the unit cell is equal to the volume of a sphere with a radius, so you need to calculate the radius of each atom.
Radius = (3 * Volume occupied by each atom / (4 * π))^⅓
Now, let's apply these steps to the calculation:
Step 1: Calculate the volume of the unit cell:
Volume of the unit cell = (atomic mass of rhodium / density of rhodium) x Avogadro's number
= (102.91 g/mol / 12.41 g/cc) x (6.022 x 10^23 atoms/mol)
≈ 5.001 x 10^-23 cc
Step 2: Calculate the volume occupied by each atom:
Volume occupied by each atom = Volume of the unit cell / 4
= 5.001 x 10^-23 cc / 4
= 1.250 x 10^-23 cc
Step 3: Determine the radius of each atom:
Radius = (3 * Volume occupied by each atom / (4 * π))^⅓
= (3 * 1.250 x 10^-23 cc / (4 * π))^⅓
≈ 1.775 x 10^-8 cm
Finally, convert the radius to picometers:
Radius = 1.775 x 10^-8 cm * (1 x 10^10 pm / 1 cm)
≈ 177.5 pm
Therefore, the radius of a rhodium atom is approximately 177.5 picometers (pm).