A given metal M has two oxides,,one with 22.53% of oxygen and the other with 30.38% of oxygen. if the formula of the second compound is M2O3,then find the formula of the first compound.
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Answers
Answer:For MO, assuming you have 100 g of material, then the mass composition is 77.47 g of M and 22.53 g of O. This means you have 22.53 / 16 = ~1.41 moles of O. Because the number of moles of M and O are the same, that also means you have ~1.41 moles of M. The molar mass of M can then be determined from MM = mass / number of moles = 77.47 / 1.41 = ~55 g/mol (which incidentally makes the metal likely to be manganese).
Then for the second oxide, again assume 100 g of material, so you have 69.62 g of M and 30.38 g of O. This means you have 69.62 / 55 = 1.27 moles of M and 30.38 / 16 = 1.9 moles of O. So that’s a mole ratio of 1.27:1.9 M:O, respectively. Divide both numbers by 1.27 to make one of them 1 and we get a mole ratio of 1:1.5. Now multiply both of these by 2 to get rid of the fraction and we get a ratio of 2:3. Therefore the second oxide is M2O3.
Explanation: