If 41.6g of chromium is fully reacted with oxygen to produce 60.8g of chromium oxide, what is the empirical formula of the chromium oxide
Mass of Cr2O3 =151.99 g/mol
We know atomic mass of oxygen =16g
Mass of oxygen in chromium oxide =(16×3/151.99 )×60.8
=19.20g
Percentage of Chromium= (41.6/41.6+19.20)×100%
=68.42%
Percentage of oxygen = (19.2/41.6+19.2)× 100%
=31.57%
| Element | Percentage of element | atomic mass | moles of atom | mole ratio |
| Chromium | 68.42 | 52 | 1.315 | 1 |
| Oxygen | 31.57 | 16 | 1.973 | 1.5 |
Now rounding off to simplest whole no ratio
Chromium =2
Oxygen= 3
Therefore empirical formula = Cr2O3
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