An oxide of iron contains $72.4%$ Fe by mass (Fe $= 56$ , O $= 16$ ). Find its simplest formula and then the AVERAGE oxidation number of iron in it (O is $- 2$ and there is no peroxo link). Give the answer to two decimals.

Numerical answer type, tolerance 0.01.

Answer: 2.67

Moles: Fe

= 72.4/56 = 1.293

and O

= 27.6/16 = 1.725

. The ratio Fe : O

= 0.75

, so the formula is

Fe X_{3} O X_{4}

. Then

3 x + 4 (- 2) = 0

gives

x = 8/3 = 2.67

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