Find the empirical formula from a reaction

Silicon reacts with $H_2$, to form a compound with formula $Si_xH_y$. From the combustion of 6.22gr of $Si_xH_y$ we obtain 11.64gr of $SiO_2$ and 6.98gr of $H_2O$. What is the empirical formula of the compound?

SOLUTION:
First we can calculate the number of mole of $SiO_2$ and $H_2O$:
$$MW_{SiO_2}=60gr/mol$$
$$MW_{H_2O}=18gr/mol$$
$$n_{SiO_2}=\frac{11.64gr}{60gr/mol}=0.194mol$$
$$n_{H_2O}=\frac{6.98gr}{18gr/mol}=0.388mol$$

Every mole of $SiO_2$ comes from one mole of $Si$; every mole of $H_2O$ comes from two mole of $H$:
$$n_{Si}=n_{SiO_2}=0.194mol$$
$$n_H=2\cdot n_{H_2O}=0.776mol$$

Now we can calculate the mass (m) of Si and H in that number of moles:
$$m_{Si}=n_{Si}\cdot Si=0.194mol \cdot 28.09gr/mol=5.45gr$$
$$m_H=n_H \cdot H=0.776mol \cdot 1.01gr/mol=0.784gr$$

The percentage composition of $Si_xH_y$ is:
$$\%Si=\frac{5.45gr}{6.22gr}\cdot 100=89.6\%$$
$$\%H=\frac{0.784gr}{6.22gr}\cdot 100=12.6\%$$

In 100gr of $Si_xH_y$ there are 87.6gr of Si and 12.6gr of H; the relative numbers of mole are:
$$n_{Si}=\frac{87.6}{28.09}=3.12$$
$$n_H=\frac{12.6}{1.01}=12.47$$

Now we can calculate the empirical formula of $Si_xH_y$:
$$x:y=3.12:12.47=\frac{3.12}{3.12}:\frac{12.47}{3.12}=1:3.99$$

The empirical formula is $$SiH_4$$.

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PROBLEM 6:
0.520 grams of $NiSO_4 \cdot xH_2O$ are heated to drive off the water. The reaction gave a residue of 0.306gr. Find the formula of the hydrate.