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0.545 g of aluminum burns completely in oxygen, producing 1.030 g of aluminum oxide. find the empirical formula of the oxide​

Sagot :

Answer:

1.030-0.545=0.485 grams of oxygen in compound.

Convert grams of Al and grams of O into moles:

0.545g Al x 1 mol Al/26.98g Al = 0.0202 mol Al

0.485g O x 1 mol O/16.00g O = 0.0303 mol O

Next we divide the moles of both of them by the lowest number to get simplest ratio

0.0202/0.0202 = 1 Al

0.0303/0.0202 = 1.5 O

Next we multiply all numbers until we get a whole number for each one. In this case 2 works:

Al 1 x 2 : O 1.5 x 2

and we are left with Al2O3

Explanation:

Hope this helps :D