Formulas from masses.

Step 1 of 6

A share of the mass

How much of water’s mass is oxygen? A molecule has two hydrogens and one oxygen, but oxygen is $16$⁠ times heavier than hydrogen. So the question is about mass, not the number of atoms:

$\%\,\mathrm{O} = \frac{A_r(\mathrm{O}) \times 1}{M_r(\mathrm{H_2O})} \times 100\%$$\%\,\mathrm{O} =$$\frac{A_r(\mathrm{O}) \times 1}{M_r(\mathrm{H_2O})} \times 100\%$⁠

The readings beside the Scene have a row for one element of the formula, here oxygen, and for now it is hidden.

Step 2 of 6

An element in a sample

Once you know an element’s share of the mass, you can work out how many grams of it any sample holds. The balance holds 50 g of chalk, CaCO₃. How many grams of carbon are in it?

Step 3 of 6

Atoms of one element

A mole of CaCO₃ is one mole of formula units, but three moles of oxygen atoms. The Scene shows 25 g of chalk, and Look inside shows the ions: calcium ions and carbonate ions, CO₃²⁻, each one carbon and three oxygens.

Step 4 of 6

The simplest ratio

Now the other way round: measured masses give a formula. To analyse an unknown compound, chemists measure how many grams of each element it holds. To turn that into numbers of atoms, divide each mass by the element’s $A_r$⁠ to get moles. The ratio of the moles is the ratio of the atoms. Then divide by the smallest and round to whole numbers. The result is the empirical formula: the simplest ratio of the atoms.

Step 5 of 6

Empirical and molecular

The empirical formula is not necessarily the molecule itself. C₆H₁₂O₆, glucose, and C₂H₄O₂, the acid in vinegar, have the same simplest ratio. The formula that describes the molecule itself is the molecular formula, and it is a whole-number multiple of the empirical one.

Step 6 of 6

Back to the molecule

To go from the empirical formula to the molecular formula you need one more fact: the molecule’s formula mass, measured separately. Divide it by the mass of the empirical formula, and you get how many times bigger the molecule is. The Scene shows glucose, with the readings showing its formula mass.