An unknown compound has a percent composition of 52.10% potassium, 15.8% carbon, and 32.1% oxygen. The molar mass of the compound is 150.22 g/mol. What is the empirical formula and the molecular formula of this compound?

Question

6440 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-29T05:05:50

The empirical formula is #C_2K_2O_3#

As is typical with these questions, we assume #100*g# of unknown compound, and work out the MOLAR quantities of each element present:

#" moles of C":# #(15.8*g)/(12.011*g*mol^-1)=1.32*mol#

#" moles of K":# #(52.8*g)/(39.10*g*mol^-1)=1.35*mol#

#" moles of O":# #(32.1*g)/(15.999*g*mol^-1)=2.00*mol#

Now if we divide thru by the lowest molar quantity, we get #CKO_(1.5)#; if we multiply this preiminary formula by #2# we get whole numbers:

#C_2K_2O_3# is the empirical formula.

But #"(empirical formula)"xxn# #=# #"molecular formula"#

Thus, solving for #n#:

#150.22*g*mol^-1=nxx(2xx12.011+2xx39.1+3xx15.999)*g*mol^-1#

Clearly, #n=1#, and the molecular formula is #C_2O_3K_2#

This corresponds to no reasonable formula I know; #C_2O_4K_2# would be reasonable. It is possible that you have been given duff values.