# 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?

Jun 29, 2016

The empirical formula is ${C}_{2} {K}_{2} {O}_{3}$

#### Explanation:

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

$\text{ moles of C} :$ $\frac{15.8 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 1.32 \cdot m o l$

$\text{ moles of K} :$ $\frac{52.8 \cdot g}{39.10 \cdot g \cdot m o {l}^{-} 1} = 1.35 \cdot m o l$

$\text{ moles of O} :$ $\frac{32.1 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 2.00 \cdot m o l$

Now if we divide thru by the lowest molar quantity, we get $C K {O}_{1.5}$; if we multiply this preiminary formula by $2$ we get whole numbers:

${C}_{2} {K}_{2} {O}_{3}$ is the empirical formula.

But $\text{(empirical formula)} \times n$ $=$ $\text{molecular formula}$

Thus, solving for $n$:

$150.22 \cdot g \cdot m o {l}^{-} 1 = n \times \left(2 \times 12.011 + 2 \times 39.1 + 3 \times 15.999\right) \cdot g \cdot m o {l}^{-} 1$

Clearly, $n = 1$, and the molecular formula is ${C}_{2} {O}_{3} {K}_{2}$

This corresponds to no reasonable formula I know; ${C}_{2} {O}_{4} {K}_{2}$ would be reasonable. It is possible that you have been given duff values.