Calculate the atomic radius of iridium atom

Question

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.

Answer 1

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 $a = 2.83r$.

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

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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.

Answer 2

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

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