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Find the empirical formula of chromium oxide containing 68.4% of chromium. (Cr=52, O=16) |
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Answer» Solution :Let the weight of chromium oxide be 100g. `THEREFORE` Weight of chromium= 68.4gand weight of oxygen =31.6g `{:(therefore "moles of chromium"=(68.4)/(52)=1.32),("and moles of oxygen"=(31.6)/(16)=1.98):}}` Relative NUMBER of moles of Cr and O atoms `= ("moles of Cr")/("moles of O")= (1.32)/(1.98)= (132)/(198) = (2 xx 66)/(3 xx 66)= (2)/(3)` (by inspection only). The empirical formula is `Cr_(2)O_(3)`. |
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