Answer
$a:{ \quad} H_2SO4 $ $ \qquad$ $b:{ \quad} AlCl_3 $
Work Step by Step
$a:$
Assuming 100g of the compound, in which,
$ $ $ $ $H = 2.1g, $ $ S = 32.6g, O = 65.3g.$
$ \qquad$First we need to convert grams of each element to moles.
The conversion factor needed is the molar mass of each element.
Let n represent the number of moles so that,
$ \qquad$$ \qquad$$ n_H = 2.1g $ $H $×$\frac {1 mol H}{1.008g H }= 2.08$ mol $H$ $
$ $ $ $ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$$ \qquad$ $n_S = 32.65g $ $S $×$ \frac{1 mol S}{1.008g S} = 1.01$ mol $S$
$ \qquad$$ \qquad$$ n_O= 2.1g$ $ O $× $\frac{1 mol O}{1.008g O} = 2.08$ mol $O$
Now divide each number of mole by the smaller, we get
$ \qquad$ $ H: \frac{2.08}{1.01} ≈ 2 $ $ \qquad$ $S: \frac{1.01}{1.01} = 1$ $ \qquad$ $O:\frac{ 4.08 }{1.01} ≈ 4$
So the required empirical formula of the compound is $: H_2SO_4 $.
$b:$
Assuming 100g of the compound, in which
$ \qquad$ $ Al = 20.2g $ $ \qquad$ $ Cl = 79.8g$
Let n represent the number of moles so that,
$ \qquad$$ nAl$ =$ 20.1g Al$ ×$ \frac{1 mol Al}{1.008g Al} = 0.74$ mol $Al$
$ \qquad$ $nCl $=$ 79.8g Cl$ ×$ \frac{1 mol Cl}{1.008g Cl} = 2.24$ mol $Cl$
Now divide each number of mole by the smaller, we get,
$ \qquad$ Al: $\frac {0.74}{0.74} =1 $ $ \qquad$ Cl: $\frac{ 2.24}{0.74} ≈3 $
So the required empirical formula of the compound is $: AlCl_3 $
