Question

On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of SO2 at STP. What is the empirical formula of the compound?

Answer 1

Answer:

The empirical formula of the organic compound is  = $C_2H_6S_1$

Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

Moles of $CO_2$ gas at STP occupying 2.0 L = n

$n\times 22.4L=2.0L$

$n=\frac{2.0 L}{22.4 L}=0.08929 mol$

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of $H_2O$ gas at STP occupying 3.0 L = n'

$n'\times 22.4L=3.0L$

$n'=\frac{3.0 L}{22.4 L}=0.1339 mol$

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of $SO_2$ gas at STP occupying 1.0 L = n''

$n''\times 22.4L=1.0L$

$n''=\frac{1.0 L}{22.4 L}=0.04464 mol$

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = $\frac{0.08920 mol}{0.04464 mol}=2$

Hydrogen =  $\frac{0.2678 mol}{0.04464 mol}=6$

Sulfur = $\frac{0.04464 mol}{0.04464 mol}=1$

The empirical formula of the organic compound is  = $C_2H_6S_1$