Answer:
The empirical formula of the organic compound is = ![C_2H_6S_1](https://tex.z-dn.net/?f=C_2H_6S_1)
Explanation:
At STP, 1 mole of gas occupies 22.4 L of volume.
Moles of
gas at STP occupying 2.0 L = n
![n\times 22.4L=2.0L](https://tex.z-dn.net/?f=n%5Ctimes%2022.4L%3D2.0L)
![n=\frac{2.0 L}{22.4 L}=0.08929 mol](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B2.0%20L%7D%7B22.4%20L%7D%3D0.08929%20mol)
Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol
Moles of
gas at STP occupying 3.0 L = n'
![n'\times 22.4L=3.0L](https://tex.z-dn.net/?f=n%27%5Ctimes%2022.4L%3D3.0L)
![n'=\frac{3.0 L}{22.4 L}=0.1339 mol](https://tex.z-dn.net/?f=n%27%3D%5Cfrac%7B3.0%20L%7D%7B22.4%20L%7D%3D0.1339%20mol)
Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol
Moles of
gas at STP occupying 1.0 L = n''
![n''\times 22.4L=1.0L](https://tex.z-dn.net/?f=n%27%27%5Ctimes%2022.4L%3D1.0L)
![n''=\frac{1.0 L}{22.4 L}=0.04464 mol](https://tex.z-dn.net/?f=n%27%27%3D%5Cfrac%7B1.0%20L%7D%7B22.4%20L%7D%3D0.04464%20mol)
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](https://tex.z-dn.net/?f=%5Cfrac%7B0.08920%20mol%7D%7B0.04464%20mol%7D%3D2)
Hydrogen = ![\frac{0.2678 mol}{0.04464 mol}=6](https://tex.z-dn.net/?f=%5Cfrac%7B0.2678%20mol%7D%7B0.04464%20mol%7D%3D6)
Sulfur = ![\frac{0.04464 mol}{0.04464 mol}=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.04464%20mol%7D%7B0.04464%20mol%7D%3D1)
The empirical formula of the organic compound is = ![C_2H_6S_1](https://tex.z-dn.net/?f=C_2H_6S_1)