Question

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.213g CO2 and 0.0310g of H2O. In another experiment, 0.103g of the compound produced 0.0230g of NH3. What is the emerical formula of the compound? In a seperate experiment, the molar mass is found to be 454.3 g/mol. what is the molecular formula of the compound? hint: assume that all carbon ends up in CO2 and all hydrogen ends up in H2O.Also assume that all nitrogen ends up in the NH3 in the second experiment

Answer 1

Solution :-

Lets calculate the mass of each element using the mass percent of the C, H and N in the CO2 , H2O and NH3

Mass of C= 0.231 g CO2 * 27.29 % C / 100 % = 0.06304 g C

Mass of H = 0.0310 g H2O * 11.18 % H / 100 % = 0.003466 g H

Mass of N = 0.0230 g NH3 * 82.87 % N / 100 % = 0.01906 g N

Mass of O = 0.157 g – (0.06304 g + 0.003466 g +0.01906 g)

                   = 0.071434 g O

Now lets calculate the moles of each element

Moles of C= 0.06304 g / 12.01 g per mol = 0.005249 mol C

Moles of H = 0.003466 g / 1.0079 g per mol = 0.003439 mol H

Moles of N = 0.01906 g / 14.007 g per mol = 0.001361 mol N

Moles of O = 0.071434 g / 16.0 g per mol = 0.004465 mol O

Now lets find the mole ratio of the each element by dividing the moles with smallest mole value

C= 0.005249 / 0.001361 = 3.89    

H= 0.003439 / 0.001361 = 2.5

N= 0.001361 / 0.001361 = 1

O = 0.004465 / 0.001361 = 3.28

To get the mole ratio in the integer number we need to multiply them with 4

So we get

C= 0.005249 / 0.001361 = 3.89 * 4 = 15

H= 0.003439 / 0.001361 = 2.5 * 4 = 10

N= 0.001361 / 0.001361 = 1 *4 = 4

O = 0.004465 / 0.001361 = 3.28 *4 = 13

So the empirical formula = C15H10N4O13

Lets find the empirical formula mass = C15H10N4O13 = 454.3 g per mol

Given molar mass is 454.3 g per mol

So the molecule formula is same as empirical formul

That is molecular formula = C15H10N4O13