Which sample of fluorine gas, F2, contains the greatest number of fluorine molecules? (multiple choice)

A. 1.9 x 10^231023 F2 molecules.
B. 1.9 dm^3dm3 of F2 at 273 K and 1.00 x 10^5105 Pa pressure.
C. 1.9 g F2
D. 1.9 mol F2

The answer is D. 1.9 mol F2, why? How do you calculate it?

1 Answer
Feb 16, 2016

Let's use the symbol N_ANA, where N_ANA == 6.022xx10^236.022×1023.

Explanation:

A. there are approx. 1/3N_A13NA, INDIVIDUAL F_2F2 molecules.

B. there are approx 1/12N_A112NA, INDIVIDUAL F_2F2 molecules.

C. there are approx. 1/19N_A119NA, INDIVIDUAL F_2F2 molecules.

D. there are approx. 1.9xxN_A1.9×NA, INDIVIDUAL F_2F2 molecules.

By using the symbol N_ANA (where N_ANA is Avogadro's number) it is clear that option D is BY FAR THE MOST NUMEROUS.

Option A. was simply a number. To assess B, I know that N_ANA F_2F2 molecules has a volume of 22.4*L22.4L. To assess C, I know that N_ANA F_2F2 molecules has a mass of 3838 gg, so in 1.91.9 gg F_2F2 there are approx. 1/19xxN_A119×NA such molecules. But D. specifies approx. 2N_A2NA F_2F2 miolecules.

So, when we use Avogadro's number as a NUMBER, and indeed it is a number, option D CLEARLY contains the most molecules.

I could have replaced the "mole" term with a dozen (because that's the number the mole is equivalent to!) and got nearly the same answer. If you are unconvinced with this treatment, state your objections, and we will try again. The point is that N_ANA is simply a number, that has the property that N_ANA F_2F2 molecules has a mass of 38.038.0 gg.