Calculate the percent composition of aluminum oxide (al2o3). *show how you got the answer ! *

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)

What percent of the total does X or Y take up?

Just take the portion that it gets and divide by the total:

Al: 54/102 = 53%

O: 48/102 = 47% or 100%-53%=47%

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.

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