# Volume of Iron (dealing with the units)

1. Aug 10, 2008

### blipped

1. The problem statement, all variables and given/known data
What is the volume of one mol of iron where iron is 55.8 g/mol and the density of iron is 7.86 x 10^3 kg/m^3 ?

2. Relevant equations
p = m/v

3. The attempt at a solution

I can get the numerical answer (checked back of book), but I'm not sure how my units worked out?

v = m/p = (55.8 g/mol) / (7.86 x 10^3 kg/m^3)

I just knock off the 10^3 and divie 55.8 by 7.86 and I get the answer of 7.10 and my units would be cm, but why? How did this work out?

Thanks.

2. Aug 10, 2008

### alphysicist

Hi blipped,

Your mass can't be 55.8 g/mol, can it? That's not the right units for mass. That's the factor to convert moles of iron to grams of iron. So what is the mass with the right units, and how do you get it?

(And the units you get at the final step is not just centimeters; that would be for a length.)

3. Aug 10, 2008

### blipped

Thanks for help!

I suppose then it's just 55.8 g. The problem stated that there are 55.8 grams or iron per mol, so that's what I put..

I figured out what I'm supposed to do I think.. Just take the cube root of 7.86 x 10^3 kg/m^3 in order to get them to be the same units (ratio of mass to volume) then divide

55.8 g / 7.86 g /m^3 which is 7.1 m^3 for the volume..

The answer is correct according to the book but is that the right way to do it?

4. Aug 10, 2008

### alphysicist

That's right; you convert the 1 mole of iron to grams of iron, and so your expression will not have units of moles in it at all.
I'm not sure what you are referring to with the cube root. I don't see where you took a cube root. (I don't see where it is needed, either.)
I think you went the wrong way in the unit conversion.

$$7860 \mbox{ kg/m}^3 \neq 7.86 \mbox{ g/m}^3$$

But didn't you want the answer to be in cm$^3$? So if you convert both kg $\to$ g and m$^3\to\mbox{cm}^3$, what do you get?