What are the empirical and molecular formulae of an organic species, whose percentage composition is: #C:38.7%;H:9.7%;O,51.6%#? A #0.93*g# mass of this material occupies #0.336*L# at #"STP"#.

Question

What are the empirical and molecular formulae of an organic species, whose percentage composition is: #C:38.7%;H:9.7%;O,51.6%#?  A #0.93*g# mass of this material occupies #0.336*L# at #"STP"#.

Peyton Adams · Accepted Answer

We get, eventually, #"a molecular formula of"# #C_2H_6O_2#...........

As with all these problems, we assume a mass of #100*g#, and from this mass we calculate the molar quantity of each element. #"Moles of carbon"=(38.7*g)/(12.011*g*mol^-1)=3.22*mol#. #"Moles of hydrogen"=(9.7*g)/(1.00794*g*mol^-1)=9.62*mol#. #"Moles of oxygen"=(51.6*g)/(15.999*g*mol^-1)=3.22*mol#. (Note that an analyst normally does not give you #%O#, you are given #%C,H,N#, and if these percentages do not sum to #100%#, the balance is ASSUMED to be #%O#.) And now we divide thru by the smallest atomic quantity to get an empirical formula of #CH_3O#. And now we use the gas data to estimate a molecular mass. The molar volume an Ideal Gas at #STP# is #22.4*L#. And thus the molar mass of our vapour is #(0.93*g)/((0.336*L)/(22.4*L*mol^-1))=62.0*g*mol^-1#. The molecular formula is always a whole number multiple of the empirical formula, meaning that this is now FACT. #"Molecular formula"=nxx"Empirical formula"# Consequently,... #nxx(12.011+3xx1.00794+16.00)*g*mol^-1=62.0*g*mol^-1# We solves for #n#, to get #n=2#, and thus #"molecular formula"# #=# #C_2H_6O_2#. #"Phew!"#

Naomi Aronson · Answer

undefined The empirical formula of the organic species is CH2O, and the molecular formula is C6H12O6.