Question

how many grams of fluorine are contained in 5 molecules of BF3

Answer 1

ok lets suppose that by fluorine you mean F^-
then:

FIRST METHOD:

n=N/L
n(BF3) = 5 / 6,022 * 10^23 mol-1
n(BF3) = 8,3 * 10^-24 mol
to continue:
n= m / M -> m= n*M
n(F-) / n(BF3)= 3/1 = 3
n(F-) = 3 * n(BF3)
n(F-) = 2,5 * 10^-23 mol

and now:
m(F-) = n(F-) * M(F-)
m(F-) = 2,5*10^-23 mol * 67,8 gmol-1
m(F-) = 4,7 *10^-22 g

SECOND METHOD:

n=N/L
n(BF3) = 5 / 6,022 * 10^23 mol-1
n(BF3) = 8,3 * 10^-24 mol

m(F-) = n(BF3) * M(BF3)
m(F-) = 4,7*10^-24 mol * [3*Ar(F) / M(BF3)]
m(F-) = 4,7*10^-24 mol * [(3*19)/67,8] gmol-1
m(F-) = 4,7 * 10^-22 g

here was used [3*Ar(F) / M(BF3)] which is sort of "equivalence unit"

THIRD METHOD:

task says that you have 5 molecules of BF3 and you need mass of F- so you just use common sense and multiply 5*3 = 15 ions of F- and now:

n(F-) = N(F-) / L
n(F-) = 15 / 6,022 * 10^23 mol-1
n(F-) = 2,5 * 10^-23 mol

m(F-) = n(F-) * Ar(F-)
m(F-) = 2,5 * 10^-23 mol * 19 gmol-1
m(F-) = 4,7 * 10^-22 g