A 3.556g sample of a pure aluminum oxide decomposes under heat to produce 1.674g of oxygen in addition to pure aluminum metal. What is the empirical formula of the aluminum oxide?
we have mass of each elements as:
Al: 3.556 - 1.674 = 1.882 g
O: 1.674 g
Divide by molar mass to get number of moles of each:
Al: 1.882/26.98 = 6.976*10^-2
O: 1.674/16.0 = 0.1046
Divide by smallest:
Al: 6.976*10^-2/6.976*10^-2 = 1
O: 0.1046/6.976*10^-2 = 1.5
Multiply by 2 to get simplest whole number ratio:
Al: 1*2 = 2
O: 1.5*2 = 3
So empirical formula is:Al2O3
Answer: Al2O3
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