Assume 1mol of Al2O3:
mm(Al) = 26.98g/mol
x2(moles needed in 1mol of Al2O3)
= 53.96g
~54g
mm(O) = 16.00g/mol
x3(moles needed in 1mol of Al2O3)
=48g
mm(Al2O3) = m(Al) + m(O)
= 102g/mol
x1mol
=102g (in 1 mol)
<em>What</em><em> </em><em>percent</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>total</em><em> </em><em>does</em><em> </em><em>X</em><em> </em><em>or</em><em> </em><em>Y</em><em> </em><em>take</em><em> </em><em>up</em><em>?</em>
Just take the portion that it gets and divide by the total:
Al: 54/102 = 53%
O: 48/102 = 47% or 100%-53%=47%
<em>Notice that, with binary compounds, you only need to calculate one of the two elements: the remaining amount HAS to be taken up by the other element, or else there would be something else involved.</em>
