A compound of oxygen and fluorine has a relative molecular mass which is twice that of its empirical formula mass. 105g of the compound contains 57g of fluorine.
What is the molecular formula of the compound? (Ar values: O=16; F=19)
for this question, i did- 57/19 and 58/16, i got 3 and 3.6, so i thought it was F3O4, but it is not correct
hi
what is the correct answer??
57g of fluorine so there is 105-57=48g oxygen
O : 48/16 = 3
F : 57/19 = 3
so the ratio is = 3 : 3 = 1 : 1
the molecular mass is twice the empirical mass so it has a ratio 2 : 2, we find F2O2
yes, its O2F2, i did not subtract properly, thank you for bringing it to my attention, it is 48 not 58, dont know what i was thinking. thanks again juliette!