# An organic compound has a percentage composition of 40.1% C, and 6.6% H. What is its probable empirical formula?

Sep 6, 2017

I makes it $C {H}_{2} O$

#### Explanation:

The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.

We assumes that we got $100 \cdot g$ of some compound....

And thus $\text{moles of carbon} \equiv \frac{40.1 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 3.34 \cdot m o l$.

And $\text{moles of hydrogen} \equiv \frac{6.6 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 6.55 \cdot m o l$.

But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....

And $\text{moles of oxygen} \equiv \frac{100 \cdot g - 40.1 \cdot g - 6.6 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 3.33 \cdot m o l$.

And please note that here WE CAN MAKE NO OTHER ASSUMPTION. %O is usually UNREPORTED because there are few analytical methods for measurement of oxygen in microanalysis, and it is assumed to be the missing percentage. This is a standard practice in analysis.

And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....

..........................$C {H}_{2} O$.

We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.