Answer to Question #100616 in General Chemistry for Jackson Halvorsen

Question

Calculate the molar mass of a metal that forms an oxide having the empirical formula M₂O₃ and contains 68.04% of the metal by mass. identify the metal.

Answer 1

 We know that 68.4% of the mass is the unknown metal. Since this is a binary compound, the only other element present is oxygen. That means that oxygen makes up 31.6% of the mass. There are three oxygen atoms in the molecule and 2 molecules of metal. Assume there is 100 g of substance; therefore there is 68.04 g of metal and 31.96 g of oxygen. Use the molar mass of oxygen to calculate how many moles of oxygen you have in this sample.1 mole =16.0 g. Therefore, 68.4g=4.275 moles O. Now, oxygen forms 2− anions; this should tell you that the chemical formula of the metal oxide will be 2M3+3O2−→M2O3.Here you need 3 oxygen anions to balance the positive charge coming from 2 metal cations. Consequently, you can say that this sample will contain 4.275moles O2 =3. 2=2*4.275/3 =2.85 moles M. Since you know that you have 31.6 g of metal in this sample, you can say that the molar mass of the metal will be.MM metal=31.6 g /2.85 moles=11.1 g mol−1.This is equivalent to an atomic mass of 11.1u.The closest match to this value is actually boron, B, which has an atomic mass of ≈10.8 u. Despite the fact that boron does form oxides, it can't really be classified as a metal. Boron is actually a metalloid, an element that shares properties with both metals and non-metals. B2O3→ boron trioxide