An empirical formula gives a ratio, a molecular formula gives a count

1. The two formulae describe the same compound in different ways.
2. Glucose has the molecular formula $C_6{H}_{12}{O}_6$ and the empirical formula ${CH}_{2}O$.
3. Dividing every subscript by their highest common factor turns a molecular formula into an empirical one.
4. Ionic compounds are always written as empirical formulae, because a lattice contains no molecules.
5. Some molecular formulae are already in their simplest ratio, such as $H_{2}O$.

From reacting masses to an empirical formula

1. Write down the mass of each element present in the sample.
2. Divide each mass by that element's relative atomic mass: $$ratio\; number=\frac{mass}{A_r}$$
3. Divide every ratio number by the smallest of them.
4. Turn the results into whole numbers, multiplying them all by the same factor if one comes out as $1.5$ or $1.33$.
5. Use those whole numbers as the subscripts.
6. $2.40g$ of magnesium gives $2.40÷24=0.100$, and $1.60g$ of oxygen gives $1.60÷16=0.100$.
7. Dividing both by $0.100$ gives $1$ and $1$, so the empirical formula is $MgO$.

From percentage composition to an empirical formula

1. Treat the percentages as masses in a $100g$ sample, so $40.0\%$ carbon becomes $40.0g$ of carbon.
2. That works because only the ratio between the masses matters.
3. The rest of the method is identical: divide by $A_r$, divide by the smallest, then simplify.
4. For $40.0\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen the ratio numbers are $3.33$, $6.7$ and $3.33$.
5. Dividing by $3.33$ gives $1$, $2$ and $1$, so the empirical formula is ${CH}_{2}O$.

From an empirical formula to a molecular formula

1. Add the $A_r$ values in the empirical formula to get its empirical formula mass.
2. Divide the compound's $M_r$ by that empirical formula mass.
3. The answer is a whole number, because a molecule contains whole atoms.
4. Multiply every subscript in the empirical formula by that number.
5. For glucose, ${CH}_{2}O$ has an empirical formula mass of $12+2+16=30$.
6. With $M_r=180$, the multiplier is $180÷30=6$.
7. Multiplying every subscript by $6$ gives $C_6{H}_{12}{O}_6$.

Finding the formula of magnesium oxide by experiment

1. Wear eye protection throughout, and move the hot crucible only with tongs.
2. Clean a length of magnesium ribbon with emery paper, which removes the oxide layer already on it.
3. Weigh an empty crucible with its lid, then weigh it again with the magnesium inside.
4. Heat strongly, lifting the lid a little at intervals so that air reaches the magnesium while the solid oxide stays inside.
5. Let the crucible cool, then weigh it again.
6. Reheat, cool and reweigh until the mass stops changing, which shows the reaction has reached constant mass.
7. The mass of magnesium is the second weighing minus the first.
8. The mass of oxygen is the final weighing minus the second, because that increase is the oxygen taken in.
9. With weighings of $20.00g$, $20.24g$ and $20.40g$, the masses are $0.24g$ of magnesium and $0.16g$ of oxygen.
10. Dividing by $24$ and by $16$ gives $0.010$ and $0.010$, a $1:1$ ratio, so the formula is $MgO$.