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?

1 Answer
Jun 29, 2016

The empirical formula is C_2K_2O_3C2K2O3

Explanation:

As is typical with these questions, we assume 100*g100g of unknown compound, and work out the MOLAR quantities of each element present:

" moles of C": moles of C: (15.8*g)/(12.011*g*mol^-1)=1.32*mol15.8g12.011gmol1=1.32mol

" moles of K": moles of K: (52.8*g)/(39.10*g*mol^-1)=1.35*mol52.8g39.10gmol1=1.35mol

" moles of O": moles of O: (32.1*g)/(15.999*g*mol^-1)=2.00*mol32.1g15.999gmol1=2.00mol

Now if we divide thru by the lowest molar quantity, we get CKO_(1.5)CKO1.5; if we multiply this preiminary formula by 22 we get whole numbers:

C_2K_2O_3C2K2O3 is the empirical formula.

But "(empirical formula)"xxn(empirical formula)×n == "molecular formula"molecular formula

Thus, solving for nn:

150.22*g*mol^-1=nxx(2xx12.011+2xx39.1+3xx15.999)*g*mol^-1150.22gmol1=n×(2×12.011+2×39.1+3×15.999)gmol1

Clearly, n=1n=1, and the molecular formula is C_2O_3K_2C2O3K2

This corresponds to no reasonable formula I know; C_2O_4K_2C2O4K2 would be reasonable. It is possible that you have been given duff values.