Empirical & Molecular Formulas

The subscripts in a formula count atoms, not grams. In water, the formula says two hydrogen atoms for every oxygen atom, yet by mass water is only $11.21\%$ hydrogen, because an oxygen atom is about 16 times heavier than a hydrogen atom. So a mass ratio can never be read directly as an atom ratio. Think of a jar of nuts and bolts sold by weight: to know how many of each you have, you divide each total mass by the mass of one piece. Chemistry does the same thing with $n=m/M$. Once every element is expressed in moles, the numbers are proportional to atom counts (one mole is always $6.02\times 10^{23}$ atoms, whatever the element), and their ratio is the ratio of subscripts.

1. Assume a $100\text{ g}$ sample, so each percentage converts directly into grams of that element.
2. Convert each element's mass to moles using $n=m/M$.
3. Divide every mole value by the smallest mole value in the set — this gives a ratio close to whole numbers.
4. If any ratio ends in a recognizable fraction like $.5$ or $.33/.67$, multiply every subscript by the smallest number that clears the fraction (by 2 for $.5$, by 3 for $.33$/$.67$).

Each step has a reason. The $100\text{ g}$ assumption is legal because percent composition doesn't depend on sample size (law of definite proportions), and $100\text{ g}$ turns "$40.0\%$" into "$40.0\text{ g}$" with no arithmetic. Dividing by the smallest mole value sets the scarcest element to exactly 1, so every other element is expressed as "atoms per atom of the scarcest element"; that is guaranteed to give numbers of 1 or more, never awkward decimals below 1. Keep at least 3 or 4 significant figures in the mole values until the ratio step: rounding $1.333$ to $1.3$ too early hides the fact that the true ratio is $4:3$. Watch especially for ".5": a mole ratio like $\text{Fe}:\text{O}=1:1.5$ is not "close enough" to $1:2$ — it's exactly $2:3$ once you double both numbers, and rounding $1.5$ to $2$ gives the wrong formula entirely.

If you're also told the compound's actual molar mass, compare it to the empirical formula's mass to see how many empirical units make up one real molecule. A real molecule must be a whole number of empirical units, so $n$ should come out very close to an integer (1, 2, 3, 6 …). If you get something like $2.5$, recheck the empirical formula or its mass. The actual molar mass comes from a separate measurement, such as the density of the gas or a mass spectrometer, which is why a question must give it to you.

Real labs rarely hand you percentages. For compounds of carbon and hydrogen (and often oxygen), a weighed sample is burned in excess oxygen. Every carbon atom ends up in ${\text{CO}}_2$ and every hydrogen atom in ${\text{H}}_2\text{O}$, and both products are trapped and weighed. The mass of carbon is the fraction of the ${\text{CO}}_2$ that is carbon, $m(\text{C})=m({\text{CO}}_2)\times \frac{12.01}{44.01}$, and the mass of hydrogen is $m(\text{H})=m({\text{H}}_2\text{O})\times \frac{2.02}{18.02}$. Oxygen can't be found this way, because extra oxygen was added to burn the sample, so any oxygen in the compound is found by difference: sample mass minus the C and H masses. From there, the usual moles-and-ratio steps give the empirical formula.