Question

an unknown compound contains only C,H,and O. combustion of 5.60g of compound produced 11.2 g of CO2 and 4.58g of H2O.    what is the emprical formula of the unknown formula?

Answer 1

Answer – We are given, mass of compound = 5.60 g

Mass of CO₂ = 11.2 g , mass of H₂O = 4.58 g

Now, moles of CO₂ and H₂O from the given mass

Moles of CO₂ = 11.2 g / 44.0 g.mol^-1 = 0.255 mole

Moles of H₂O = 4.58 g / 18.015 g.mol^-1 = 0.255 moles

Moles of C from the moles of CO₂

1 moles of CO₂ = 1 moles of C

So, 0.255 moles of CO₂ = ?

= 0.255 moles of C

Moles of H from moles of H₂O

1 moles of H₂O = 2 moles of H

So, 0.255 moles of H₂O = ?

= 0.508 moles of H

Mass of C = 0.255 moles * 12.011 g/mol

                  =3.06 g of C

Mass of H = 0.508 moles * 1.0079 g/mol

                  = 0.512 g of H

Total mass of compound = mass of C + mass of H + mass of O

5.60 g = 3.06 g + 0.512 g + mass of O

Mass of O = 5.60 g – 3.06 g – 0.512 g

                  = 2.03 g of O

Moles of O = 2.03 g / 15.998 g.mol^-1

                   = 0.127 moles of O

we need to divided each mole by this mole of O, since moles of O is smallest

So , C = 0.255 /0.127 = 2

      H = 0.508 / 0.127 = 4

      O = 0.127 /0.127 = 1

So empirical formula is C₂H₄O