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Answer :
To determine the empirical formula of the compound containing 10.0g aluminum and 39.4g chlorine, we need to first calculate the moles of each element.
Moles of Aluminum = 10.0g / 26.98 g/mol (molar mass of aluminum) = 0.370 mol
Moles of Chlorine = 39.4g / 35.45 g/mol (molar mass of chlorine) = 1.11 mol
We then need to find the simplest whole number ratio between the moles of the two elements. To do this, we divide both values by the smallest value:
0.370 mol / 0.370 mol = 1
1.11 mol / 0.370 mol = 3
This means that the empirical formula of the compound is AlCl3, which indicates that there are 1 mole of aluminum and 3 moles of chlorine in the compound.
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