Question

An unknown compound contains only C , H , and O . Combustion of 6.50 g of this compound produced 15.9 g CO2 and 6.50 g H2O . What is the empirical formula of the unknown compound? Insert subscripts as needed.

empirical formula:[CHO]

Answer 1

no of moles = mass / molar mass

no of moles of co2 = 15.9 / 44 = 0.3614 moles

so no of carbon also 0.3614 moles

mass of carbon = 0.3614 * 12 = 4.34 grams

no of moles of h2o = 6.50 / 18 = 0.3611 moles

for 1 mole h2o, 2 moles of hydrogens present

so moles of hydrogens => 0.3611*2 = 0.722 moles

mass of hydrogen = 1.083 grams

total mass = C mass + H mass + O mass

6.50 = 4.34 + 0.7222 + O mass

=> O mass = 1.4378 grams

mole ratio       =>     C                        H                       O

                        => 4.34/12 0.7222/1    1.4378/16

   => 0.3614    0.7222                0.089

                      => 0.3614/0.089         0.7222/0.089        0.089/0.089

                    => 4                                     8                             1

emperical formula => C₄H₈O