Question

A 8.024 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 20.75 grams of CO₂ and 4.248 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 136.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Enter the elements in the order C, H, O

empirical formula =

molecular formula =

Answer 1

Let the moleculear formula of the compound be C_xH_yO_z

Molar mass of the compound = molar mass of C*x + molar maas of H*y + molar mass of O = 12g/mole*x + 1g/mole*y + 16g/mole*z

The molar mass of the compound is given to us as 136.2 g/mol

So, 12x+y+16z = 136.2 ---- Eqn1

and number of moles of this compound in 8.024 g sample = 8.024g / 136.2g/mole = 0.05891 moles

moles of H2O produced by the combustion = weight/molar mass = 4.248/18 g/mole = 0.2355 moles

Each mole of H2O has 2 moles of H atoms so, 0.2355 moles of H2O has 0.2355*2 = 0.471 moles of H atoms

Now, all the H atoms must've come from the organic compund so,

0.471 moles= y*0.05891 moles (As each mole of compound contains y atoms of H)

So, y = 8

moles of CO2 produced by the combustion = weight/molar mass = 20.75g/44 g/mole = 0.471 moles

Each mole of CO2 has 1 mole of C atoms so, 0.471 moles of CO2 has 0.471 moles of C atoms

Also, Every C atom in CO2 must've come from the compound

So, equating the mole of C we have

0.471 moles = x*0.05891 moles (As each mole of compound contains x atoms of C)

so, x = 8

Also,

from Eqn1 we have

12x+y+16z = 136.2

so, by putting the values of x,y we have

12*8 + 8 + 16z = 136.2

so, z = 2

so, the molecular formula will be C₈H₈O₂

Empirical formula can be derived from the molecular formula by dividing all the number of atoms in molecule by the lowest number of atoms (in our case its 2 from O) so, we get 8/2 = 4

So, empirical formula will be C₄H₄O