Question

during the reduction of 16g of iron oxide 11.2g of metal were obtained.determine the composition of oxide and calculate its molar mass

Answer 1

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Answer 2

What is the oxidation state of iron? Usually its specified as Fe(II) or Fe(III). If its FeO, then I can only assume its Fe(II).Youre given the total weight of the compound and told that you retrieved 11.2 g Fe from a separation experiment. 16 total - 11.2 iron = 4.8 g oxide. Now we have the weight of each part, so in order to determine an empirical formula (which is the lowest whole number ratio of moles between elements) we use the molar mass of iron (55.845 g/mol) and the molar mass of oxygen (15.999g/mol), which are both found on the periodic table, to calculate the total molar composition of our compound. 11.2 g Fe x 55.845 g/mol = .2006 mol (rounding to an arbitrary number of places) and 4.8 g O x 15.999 g/mol = 0.3000 mol. So now we can take the lowest number of moles present (iron @ 0.2006 mol) and divide both values by it, which sets 0.2006 Iron to 1 mol and 0.3000 mol O to 1.5 mol. Oxide isnt a whole number, so we must multiply by 2 now on each value, thus making the lowest whole number ratio Fe2O3. From this, we can take the empirical formulas mass by multiplying the molar mass of each constituent element by the number of moles present in the empirical formula. So: (2 mol Fe x 55.845 g/mol)+(3 mol O x 15.999 g/mol) = 159.687 g/mol. If you want to know the percent composition (by mass) of Oxygen, remember, part/whole x 100 = percentage. So percent O weight / total formula weight x 100 = percent composition oxygen. 3 mol O x 15.999 g/mol = 47.997 g O / 159.687 g/mol total = 0.3005 x 100 = 30.05% O (you can round off on the percentage unless asked to do otherwise.) Just remember, in these types of problems, it usually comes down to determining the ratio of moles for each element, which can give you lots of information about a compound. Get good at that and if you can remember how each concept relates to it, you should do well on any test question that involves empirical/molecular formulas. Good luck and dont forget to rework equations! Its a process, not a conclusion!

Answer 3

thanks!! zeus87

Answer 4

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