Question

24. Rust occurs when iron metal reacts with the oxygen in air. Chemical analysis shows that dry rust is 69.9% iron and 30.1% oxygen, by mass. Determine its empirical formula and its empirical formula mass.

Answer 1

Empirical formula = Fe₂O₃

Amount given is in percentage (%). Therefore given amount is out of 100. mass of iron = 69.9% = 69.9 grams t mass o gygen = 30.1% - 30.1 grams ② moles = masse mw 6 Iron = 69.9 g 55.8hgim = 1.25 mole 6 sygen = 30.19 = 1.89 more. 42 T6 g/m therefore we livicle 1.25 is smallest number, both the values by 1.25 @m= 1.25). 6 sygen - 1.89/1.25 = 1.5 Hex, 1.5 is bruction number, therefore, we multiply by two to both to get whole number @ 100 = $x2 = 2 © osygens = 1.5*2 = 3, That is dry rust empirical fermula contains 2 mote Iron & 3 mole asygen. Empirical formula - Fez O₂)