Question

omposition: C = 24.78%; H = 2.08 en and chlorine had the following composition: C = 24.78% % and Cl = 73.14%. A 1.26 g sample of this gas occupies a volume of 292 mL at STP. a) What is the empirical formula of the unknown gas (3 points) show all setup & work Answer: empirical formula = b) What is the molar mass of the unknow gas (3 points) show all setup & work answer: molar mass = What is the molecular formula of the unknown gas? (4 points show all setup & work

Answer 1

a)

First calculate moles of C,H & O.

Moles of C = % of C / Atomic mass of C

Moles of C = 24.78 / 12.01 = 2.063

Moles of H = % of H / Atomic mass of H

Moles of H = 2.08 / 1.0079 = 2.064

Moles of Cl = % of Cl / Atomic mass of Cl

Moles of Cl  = 73.14 / 35.45 = 2.063

Calculate ratio of number of moles of C:H:Cl

2.063 / 2.063 = 1.000 mol C

2.063 / 2.063 = 1.000 mol H

2.063 / 2.063 = 1.000 mol Cl

Hence, empirical formula of compound is CHCl.

ANSWER : Empirical formula of compound : CHCl

b)

We know that at STP , 22.4 L gas $\equiv$ 1 mol gas

$\therefore$ 0.292 L gas = 1 $\times$ 0.292 / 22.4 mol gas

0.292 L gas = 0.0130 mol gas.

We have , no. of moles = Mass / Molar mass

$\therefore$ Molar mass of gas = Mass / No. of moles of gas

$\therefore$ Molar mass of gas = 1.26 g / 0.0130 mol = 96.92 g / mol

ANSWER : Molar mass of gas = 96.92 g /mol

C)

We have, Molecular formula = Y  $\times$ empirical formula

Where Y is the ratio of molar mass of substance to the empirical formula mass.

Empirical formula mass = 12.01 + 1.0079 + 35.45 = 48.47 g / mol

$\therefore$ Y = 96.92 g /mol / 48.47 g /mol = 2

$\therefore$  Molecular formula =Y $\times$ empirical formula = 2 ( CHCl ) = C₂H₂Cl₂

ANSWER :  Molecular formula of gas = C₂H₂Cl₂