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

6878 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.