determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass
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molar mass of iron is 56 and oxygen is 16
now we firstly have to convert mass element in moles As the weights are given in %age but we can write as in grams as 69.9gm and 30.1gm
mow firstly we convert mass of iron into moles by dividing given mass with molar mass
moles of iron 69.9/56=1.24
moles of oxygen 30.1/16=1.88
now we divide the moles with the lowest no of moles for iron=1.24/1.24=1
oxygen=1.88/1.24=1.5 now we multiply this 1.5 to 2 to get round off answer because cannot write fractions in formula so we get 1.5×2 =3
SO WE GET EMPIRICAL FORMULA OF IRON AND OXYGEN IS =FeO3
now we firstly have to convert mass element in moles As the weights are given in %age but we can write as in grams as 69.9gm and 30.1gm
mow firstly we convert mass of iron into moles by dividing given mass with molar mass
moles of iron 69.9/56=1.24
moles of oxygen 30.1/16=1.88
now we divide the moles with the lowest no of moles for iron=1.24/1.24=1
oxygen=1.88/1.24=1.5 now we multiply this 1.5 to 2 to get round off answer because cannot write fractions in formula so we get 1.5×2 =3
SO WE GET EMPIRICAL FORMULA OF IRON AND OXYGEN IS =FeO3
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