Question

1. A sample of a hydrocarbon weighing 4.99 grams was burned completely in excess oxygen, and 15.00 grams of carbon dioxide were obtained. What is the empirical formula of the compound? CHs CH4 О сн C3Hs

Answer 1

Answer:

The combustion reaction is

CxHy + (Excess)O2 ---> x CO2 +y/2 H2O

Here we need to find x and y inorder to get the empirical formula.

Given mass of CO2=15 g and molar mass CO2=44 g/mol

Therefore moles of CO2=mass/molar mass= 15.0 g / 44 g/mol = 0.3409 moles of CO2

In one moles of CO2 ------> 1 mol of C present.

Therefore moles of C=mol CO2=0.3409 mol and molar mass of C=12 g/mol.

Therefore the mass of C = 0.3409 mol x 12 g/mol
= 4.0909 g
Given mass of hydrocarbon=4.99 g
The mass of H =mass of hydrocarbon - mass of C= 4.99 g - 4.0909 g
=0.899 g of H
Molar mass of H=1 g/mol.

Therefore moles of H=mass / molar mass=0.899 g /1g/mol
Moles of H = 0.899 mol H.

The mole ratio between C:H=0.3409 : 0.899

Multiply the moles with 9, then C : H=(0.3409 x 9) : (0.899 x 9)=3:8.

Therefore empirical formula is C3H8.

Please let me know if you have any doubt. Thanks