Ustal skład wagowy tlenku żelaza.
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2*56:3*16=112:48=28:12=14:6=7:2 w takim stosunku wagowym
mFeO = 56u+16u = 72u
mFe/mO = 56/16=7/2
tlenek żelaza (III)
mFe₂O₃ =2*56u + 3*16u = 160u
2*mFe/3*mO = 2*56u/3*16u = 112/48=7/3