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

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

Step-by-step 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* 22.4L=2.0L`

`n=(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'* 22.4L=3.0L`

`n'=(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''* 22.4L=1.0L`

`n''=(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 =
`(0.08920 mol)/(0.04464 mol)=2`

Hydrogen =
`(0.2678 mol)/(0.04464 mol)=6`

Sulfur =
`(0.04464 mol)/(0.04464 mol)=1`

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