A mass of aluminium metal is heated in a such in a such a way that is was exposed to air.(1.751g Al) A compound containing aluminium and oxygen is formed. It has a mass of 3.311 g. What is the empirical formula of aluminum oxide according to this experimental data?
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Since the aluminum oxide has a mass of 3.311g and the aluminum originally has a mass of 1.751g the mass of oxygen in the compound is:
3.311-1.751 = 1.56g O
Now in order to find the empirical formula you have to find the moles of aluminum and the moles of oxygen in the aluminum oxide. The molar mass of aluminum is about 26.98g/mol and the molar mass of oxygen is 16g/mol.
Therefore, the moles of aluminum and oxygen are as follows:
1.751g Al / (26.98g/mol Al) = 0.0649mol Al
1.56g O / (16g/mol O) = 0.0975mol O
Finally find the molar ratios of aluminum and oxygen in the aluminum oxide:
Al: 0.0649/0.0649 = 1 * 2 = 2
O: 0.0975/0.0649 = 1.5 * 2 = 3
The empirical formula of aluminum oxide is: Al2O3
3.311-1.751 = 1.56g O
Now in order to find the empirical formula you have to find the moles of aluminum and the moles of oxygen in the aluminum oxide. The molar mass of aluminum is about 26.98g/mol and the molar mass of oxygen is 16g/mol.
Therefore, the moles of aluminum and oxygen are as follows:
1.751g Al / (26.98g/mol Al) = 0.0649mol Al
1.56g O / (16g/mol O) = 0.0975mol O
Finally find the molar ratios of aluminum and oxygen in the aluminum oxide:
Al: 0.0649/0.0649 = 1 * 2 = 2
O: 0.0975/0.0649 = 1.5 * 2 = 3
The empirical formula of aluminum oxide is: Al2O3
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