Question

Combine $0.2g$ atoms of silicon with $21.3g$ of chlorine. Find the empirical formula of the compound formed.

Answer 1

Step 1:

$0.2g$ atoms of silicon $\left(Si\right)$ means this is the mole of silicon $\left(Si\right)$.

Step 2:

$mole=\frac{weight}{molecularweight}$

Step 3 :

$0.2=\frac{weight}{28.8}$

$weight=5.76g$

Now, $molesofsilicon=\frac{5.76}{28.8}$

$=0.2$

$molesofchlorine=\frac{21.3g}{35.5g}$

$$\begin{gathered} =\frac{213}{355} \\ =0.6 \end{gathered}$$

Step 2:- Now take of both element

$Si:Cl=0.2:0.6$

$=1:3$

Here, we can see that one part of silicon reacts with $3$ part of chlorine.

So, $empiricalformula=SiC{l}_3.$