How many atoms are contained in 4.5 ng of flourine?

Answer 1

This is simple, but try not to make any errors! The first thing we need to do is convert 4.5 ng to grams. Remember nanograms=10^-9 (n for nine is how I remember it), so it becomes 4.5*10^-9 grams. Now we must convert from grams to moles, making sure to remember that naturally occuring fluorine is diatomic (we will need the molar mass of F2 rather than just F). From a periodic table, we see the molar mass of fluorine to be 19 g/mol. That means that the molar mass of F2 is 38 g/mol.

Set up the conversion:
(4.5*10^-9 grams)*(1 mol/38 grams)=1.18*10^-10 mol

To get atoms, we must use Avogadro's number (6.022*10^23 atoms/mol)

We just have to multiply the amount of moles we have by this number.

We get 7.13*10^13 atoms of fluorine.

Answer 2

There are 7.1 x 10^13 F atoms.