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.
The percent composition of Aluminum oxide (Al₂O₃) is 52.9 % of Al and 47.05 % of O.
The mass percentage of each type of element in a compound is called Percent composition.
%X = [tex]\frac{\text{Total atomic mass of X}}{\text{Molar mass which contains X}}[/tex]
Molar mass is also known as molecular weight is the sum total of the atomic mass of the atoms which is present in that compound or molecule.
The atomic weight of Aluminum (Al) is 27. The atomic weight of Oxygen (O) is 16.
Molar mass of Al₂O₃ = 2 (Atomic weight of Al) + 3 (Atomic weight of O)
= 2 (27) + 3 (16)
= 54 + 48
= 102 g/mol
Mass % of Al = [tex]\frac{54}{102} \times 100[/tex]
= 0.529 × 100
= 52.9
Mass % of O = [tex]\frac{48}{102} \times 100[/tex]
= 0.4705 × 100
= 47.05
Thus from the above conclusion we can say that The percent composition of Aluminum oxide (Al₂O₃) is 52.9 % of Al and 47.05 % of O.
Learn more about the Percent Composition here: https://brainly.com/question/17021926
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