Answer to Question #123313 in General Chemistry for Izzy

Question

INFO: 5.25 g of an unknown organic compound that is known to contain only carbon, hydrogen, nitrogen, and oxygen are inserted into a combustion analysis device.
The following data is obtained:
7.955 g CO2 3.803 g of H2O 5.544 g of NO2
A second sample of the same unknown is put into a molecule counting device. The 2nd sample of unknown had a mass of 0.500 g and contained 1.15X1021 molecules.

TASK:
Find the empirical and molecular formula of the unknown compound.

Answer 1

In 44 g CO2, C present = 12 g

So in 7.995 g of CO2, C present = 2.18g

In 18g H2O, H present = 2g

So in 3.803 g of H2O, H present = 0.422 g

In 46 g of NO2, N present = 14 g

So in 5.544 g of NO2, N present = 1.68 g

Oxygen present = 5.25 - (2.18+0.422+1.68)

=0.968 g

Emperical formula is

=C(2.18/ 12)H(0.422/2)N(1.68/14)O(0.968/16)

=C3H4N2O

2nd part

Mass of 1.15×1021 molecules =0.500g

Mass of 6.023×1023 molecules = 262 g

Molecular weight of the sample = 262g

Mass of the emperical formula

= 84

The molecular formula

= C9H12N6O3