Question

What is the empirical formula of a compound that contains 27.0% s, 13.4% o, and 59.6% cl by mass?

Answer 1

Answer : The empirical formula of a compound is, $SOCl_2$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of S = 27.0 g

Mass of O = 13.4 g

Mass of Cl = 59.6 g

Molar mass of S = 32 g/mole

Molar mass of O = 16 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{27.0g}{32g/mole}=0.844moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{13.4g}{16g/mole}=0.837moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{59.6g}{35.5g/mole}=1.68moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For S = $\frac{0.844}{0.837}=1.00\aprrox 1$

For O = $\frac{0.837}{0.837}=1$

For Cl = $\frac{1.68}{0.837}=2.00\approx 2$

The ratio of S : O : Cl = 1 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $S_1O_1Cl_2=SOCl_2$

Therefore, the empirical of the compound is, $SOCl_2$

Answer 2

27g of 's × 1 mole/32.065g equals 0.84
13.4g of o × 1 mole/16g equals 0.83
Divide all these by the smallest number then round to the nearest number.