Question

determine the empirical formula of an oxide of iron which has 69.9 percent of iron and 30.1% dioxygen by mass ​

Answer 1

Answer:

atomic mass of Iron = 55.84

Mass percentage = 69.9%

Therefore atomic ratio = 1.25 = 1(approx)

atomic mass of Oxygen = 16

Mass percentage = 30.1%

therefore atomic ratio = 1.88 = 1.5 (approx)

therefore empirical formula will be:-

Fe2 03