Question

silicon forms a compound with chlorine in which 5.6 g of silicon is combined with 21.3g of chlorine . calculate the empirical formula of this compound

Answer 1

Answer:- The empirical formula of the compound formed is $SiCl_3$ .

Solution:- Mass of Si = 5.6 g

mass of Cl = 21.3 g

moles of Si = $5.6g(\frac{1mol}{28.08g})$

= 0.199 mol Si

moles of Cl = $21.3g Cl(\frac{1mol}{35.45g})$

= 0.601 mol Cl

Now we calculate the mol ratio of Si and Cl and for this we divide the moles of each by the least one of them. Here the least one is Si as we have less moles of it.

Si = $\frac{0.199}{0.199}$ = 1

Cl = $\frac{0.601}{0.199}$ = 3

So, the empirical formula is $SiCl_3$ .

Answer 2

Answer:

The empirical formula of the compound $Si_1Cl_3$

Explanation:

Mass od silcon = 5.6 g

Mass od chlorine = 21.3 g

Moles of silicon =$\frac{5.6 g}{28 g/mol}=0.2 mol$

Moles of chlorine = $\frac{21.3 g}{35.5 g/mol}=0.6 mol$

0.2 moles of silicon combines 0.3 moles of chlorine.

For empirical formula divide the smallest value of moles with the moles of each element coming in the compound.

Silicon = $\frac{0.2 mol}{0.2 mol}=1$

Chlorine = $\frac{0.6]{0.2}=3$

1 atom of silicon is associated with 3 chlorine atoms.

The empirical formula of the compound $Si_1Cl_3$