Question

Caproic acid is composed of C, H, and O atoms. The combustion of a 0.225g sample yields 0.512g of CO2 and 0.209g of H2O. What is the empirical formula for the compound?

Answer 1

Let the empirical formula be C_xH_yO_z

Now,

C_xH_yO_z + {x+(y/4) - z}O₂   -------------> xCO₂ + (y/2)H₂O

Now, molar mass of CO₂ = 44 g/mole

Thus, moles of CO₂ released = mass/molar mass = 0.512/44 = 0.0116

Thus, moles of C in the given mass of acid = moles of CO₂ = 0.0116

Molar mass of C = 12 g/mole

Thus, mass of C in given mass of the acid = moles*molar mass of C = 0.0116*12 = 0.1392 g...........(1)

Molar mass of H₂O = 18 g/mole

Thus, moles of H₂O formed = mass/molar mass = 0.209/18 = 0.0116

Thus, moles of H in the given mass of the acid = 2*moles of H₂O = 0.0232

Molar mass of H = 1g/mole

Thus, mass of H in given mass of the acid = molar mass*moles = 0.0232*1 = 0.0232 g............(2)

Now, mass of O in the acid = 0.512 - mass of C - mass of H = 0.3496 g

Now, molar mass of O = 16 g/mole

Thus, moles of O in the acid = mass/molar mass of O = 0.022.............(3)

Now, molar ratio of C:H:O in the acid = 0.0116:0.0232:0.022

or, on simplification, molar ratio of C:H:O in the acid = 1:2:2

Hence the empirical formula is C₁H₂O₂