A compound contains only carbon hydrogen nitrogen and oxygen combustion of 0.157g of the compound produced 0.213g CO2 and 0.0310g H2O In another experiment it is found that 0.103g of the compound?

Answer 1

Molar masses:

C = 12 g/mol

O = 16 g/mol

H = 1 g/mol

N = 14 g/mol

CO2 = 44 g/mol

H2O = 18 g/mol

NH3 = 17 g/mol

* 44 g CO2 is produced from 12 g C

0.213 g CO2 is produced from = (12)(0.213) / 44 = 0.0581 g C present in 0.157 g compound.

* 18 g H2O is produced from 2 g H

0.0310 g H2O is produced from = (2)(0.0310) / 18 = 0.0034 g H present in 0.157 g compound.

* 17 g NH3 is produced from 14 g N

0.0230 g NH3 is produced from = (14)(0.0230) / 17 = 0.0189 g N present in 0.103 g compound.(this is different)

If in 0.103 g compound there is 0.0189 g N

in 0.157 g compound there is ;

(0.157)(0.0189) / 0.103 = 0.0288 g N

* Mass of C, H and N = 0.0581 + 0.0034 + 0.0288 = 0.0903 g

The rest is mass of O = 0.157 - 0.0903 = 0.0667 g

Moles of elements in the compound sample:

C = 0.0581 g / 12 g/mol = 0.00484 mol

H = 0.0034 g / 1 g/mol = 0.0034 mol

N = 0.0288 g / 14 g/mol = 0.00205 mol

O = 0.0667 g / 16 g/mol = 0.00417 mol

Smallest mole ratio:

N = 0.00205 / 0.00205 = 1.00

C = 0.00484 / 0.00205 = 2.36

H = 0.0034 / 0.00205 = 1.66

O = 0.00417 / 0.00205 = 2.00

Multiplying by 3 we get;

N = 3

C = 7

H = 5

O = 6

Simplest formula:

C7H5O6N3