Question

There are two oxides of chromium; one is 23.5% oxygen and the other is 31.6% oxygen by weight. Calculate the empirical formula and name each compound.

Answer 1

first one :

oxygen weight = 23.5

oxygen moles = 23.5 / 16 = 1.47

mass of Chromium = 100 - 23.5 = 76.5

moles of Cr = 76.5 / 51.99 = 1.47

moles ratio of Cr and O = 1.47 : 1.47 = 1:1

empirical formula = CrO

name : chromium(II)oxide

second compound

oxygen weight = 31.6

oxygen moles = 31.6 / 16 = 1.98

mass of Chromium = 100 - 31.6 = 68.4

moles of Cr = 68.4 / 51.99 = 1.32

moles ratio of Cr and O = 1.32 : 1.98 = 1:1.5

                                   = 2 : 3

empirical formula = Cr₂O₃

name : chromium (III) oxide