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I was delaying with this question as seen below.

Calculate the radius of iridium atom, given that Ir has an FFC crystal structure, a density of $\pu{22.4 g/cm^3}$ and atomic weight of $\pu{192.2 g/mol}$.

Let me show you how my teacher solved this question.

Givendata:

$$\text {Weight of crystal} = \pu{192.2 g/cm^3}$$ $$\text {Density} = \pu{22.4 g/cm^3}$$ $$\text {Volume of crystal} = V = \frac {192.2}{22.4}$$ $$V = \pu{8.58 cm^3/mol}$$

In FFC, there are 4 atoms in one unit cell.

$$\text {Volume of one cell} = 8.58 \times \frac {4}{6.023 \times 10^{23}}$$ $$ = 5.7 \times 10^{-23}$$ $$V = a^3$$

$$\therefore a = 2.83r$$ and $$V = (2.83r)^3$$ $$5.7 \times 10^{-23} = 2.83r^3$$ $$r = 1.34 \times 10^{-8}$$

Where did that $a = 2.83r$ come from? That's what I didn't get.

mhchem
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Cargobob
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2 Answers2

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Iridium has a face-center cubic (FCC) structure.

You assume that each iridium sphere has a radius $r$. But the edge of a unit cell has the dimension $a$. The volume of the unit cell, which is a cube, is $a^3$. The relationship between $a%$ and $r$ is that $a = 2.83r$.

FOLLOWING THE TEACHER"S(?) WORK...(or how you copied it?)

The following step is ok... $$V = (2.83r)^3\tag{1}$$

But the second step here is wrong. The 2.83 should be cubed too. $$5.7 \times 10^{-23} = 2.83r^3\tag{2}$$

(3) is totally wrong if we accept (2). $$r = 1.34 \times 10^{-8}\tag{3}$$

From (2) $$r = \sqrt[3]{\dfrac{5.7\times 10^{-23}}{2.83}} = 2.72 \times 10^{-8}\tag{4}$$

Starting over...

$$V = (2.83r)^3\tag{5}$$ $$5.7 \times 10^{-23} = (2.83r)^3\tag{6}$$ $$r = \dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = 1.36 \times 10^{-8}\tag{7}$$

After calculating the the volume of the unit cell, I'd solved for $a$ by taking the cube root, then divided by 2.83 to get $r$.

You have to be careful with significant figures too. It is not $5.7\times10^{-23}$ but rather $5.70\times10^{-23}$. Using a calculator I'd carried 5 significant figures in all the calculations then rounded to 3 at the end. In my day I used a slide rule so everything was to 3 significant figures.

MaxW
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  • So, is it constant? – Cargobob Dec 02 '17 at 07:35
  • Yes the 2.83 is a constant. – MaxW Dec 02 '17 at 14:45
  • That was what I was looking for! Also, shouldn't the answer be $1.36 \times 10^{-8}$? I mean this step $5.7 \times 10^{-23} = 2.83r^3$ – Cargobob Dec 02 '17 at 15:03
  • My teacher solved this two times. How did you get that answer? – Cargobob Dec 02 '17 at 16:01
  • You don't understand the difference between equations (2) and (6) ?!? – MaxW Dec 02 '17 at 16:05
  • My teachers has solved it and both got $1.34 \times 10^{-8}$ – Cargobob Dec 02 '17 at 16:09
  • Again, don't you see the difference between equations (2) and (6) ??? – MaxW Dec 02 '17 at 16:14
  • I see it. However, why did they get that answer? – Cargobob Dec 02 '17 at 16:15
  • The difference between 1.34 and 1.36 is some sort of rounding error. Is that what you're hung up on? – MaxW Dec 02 '17 at 16:18
  • So, isn't answer 1.34? – Cargobob Dec 02 '17 at 16:20
  • The critical part is that $(2.83r)^3 = 2.83^3r^3 = 22.67r^3$ so going from eq (1) to eq (2) is WRONG. – MaxW Dec 02 '17 at 16:23
  • What do you mean? – Cargobob Dec 02 '17 at 16:25
  • I showed my third teacher as well, He've found $1.34$. Is it really wrong? – Cargobob Dec 02 '17 at 21:38
  • As I said it is a rounding error. You are not supposed to trim significant figures until you get to the final result. So if you calculate $$\dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = \dfrac{3.8\times 10^{-8}}{2.83} = 1.34276\times10^{-8}= 1.34\times10^{-8}$$ but if you calculate $$\dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = \dfrac{3.8485\times 10^{-8}}{2.83} = 1.3598\times10^{-8} = 1.36\times 10^{-8}$$ So yes, they are wrong. – MaxW Dec 02 '17 at 21:51
  • PS - Your teacher also used the wrong number for the Avogadro constant. The accepted value is now $6.022140857(74)\times 10^{-23}\text{ mol}^{−1}$ and rounded to 3 decimals is $6.022\times 10^{-23}\text{ mol}^{−1}$ – MaxW Dec 02 '17 at 22:02
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This is how I would do the calculation...

Given data:

$$\text {Weight of crystal} = \pu{192.2 g/mole}$$ $$\text {Density} = \pu{22.4 g/cm^3}$$

Ok, the density is good to only 3 signifant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures throughout all the intermediate calculations to try avoid rounding errors within the multiple calculations. I'll round the final result to 3 significant figures.

$$\text {Volume of crystal} = V = \frac {192.2}{22.4}$$ $$V = \pu{8.5804 cm^3/mol}$$

In FFC, there are 4 atoms in one unit cell.

$$\text {Volume of one cell} = 8.5804 \times \frac {4}{6.0221 \times 10^{23}}$$ $$ = 5.6992 \times 10^{-23}\text{ cm}^3$$ A unit cell is a cube with each side being $a$ $$V = a^3$$ $$\therefore a = \sqrt[3]{5.6992 \times 10^{-23}} = 3.8483\times 10^{-8}\text{ cm}$$

But the FCC crystal, spheres of radius, $r$, packed inside a cube with side, $a$, the geometric relationship between $a$ and $r$ is: $$a = 2\sqrt{2}r = 2.8284r$$ so

$$r = \dfrac{3.8483\times 10^{-8}}{2.8284} = 1.3606\times 10^{-8}\text{ cm}$$

now rounding $r$ to 3 significant figures gives

$$r = 1.36\times 10^{-8}\text{ cm}$$

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