7.77 g of rush (an oxide of iron) is heated until all the O2 atoms combine with charcoal. When this is done, a pellet of pure iron with a mass of 5.44 g remains. What is the empirical forumla for rust?

Answer 1

5.44/55.847 = 0.0974 moles Fe.
g oxygen = 7.77-5.44 = 2.33
moles oxygen = 2.33/16 = 0.145 moles oxygen.
Now you want to find the small whole number ratio of Fe to O. The easiest way to do this is to divide the smaller number by itself (0.0974) and divide the other number by the same small number. that should give you a whole number ratio or one that can be converted to a small whole number ratio.
By the way, rust is not quite this formula. It is much more complicated than that.

Answer 2

Thank you!
I don't get one part though,
to find the mass of the oxygen, why did you minus 5.44 and 7.77?

Answer 3

7.77 is the mass of the iron oxide in the problem and it produces 5.44 g iron. Ergo, the mass oxygen must be 7.77 - 5.44. Said another way, the iron + oxygen has a mass of 7.77 grams. The oxygen is driven off and the amount of iron remaining is 5.44, so.......

Answer 4

OH, i understand now, 7.77 is the combination of iron and oxygen.. whereas 5.44 is just iron, so therefore to find the mass of the oxygen, you minus the two..
thank you so much!

Answer 5

You have it.