How many GRAMS of fluorine are present in 5.35×1022 molecules of boron trifluoride?

I don’t know if I did this right, but I hope it helps.

Molar mass of BF3
M(BF3) = (1x10.8) + (3x18.91)
= 67.53 grams per mol

Molar mass of Boron in BF3
M(B) = 1x10.8 = 10.8 grams per mol

5.35x10^22 molecules of BF3 into mol
n = N/NA

n(BF3) = 5.35x10^22/6.02x10^23
= 8.89x10^44 mol of BF3

Mass of B in BF3
m = n x M
m(b) = 8.89x10^44 x 10.8
= 9.6x10^45g

The answer is a i think