# Chemical Formulas of Compounds

The elements that make up the molecules of a compound can be described by the number of atoms of each element in a molecule or by the ratio of atoms of the elements in a molecule.
Molecular formulas show the number of atoms of each element in a molecule of a compound. For example, the molecular formula for glucose is C6H12O6. By convention, the numbers of atoms in a molecule are indicated by subscript numbers to the right of the element symbols. However, these formulas are not unique because they do not show the precise arrangements of these atoms. For example, fructose has the same molecular formula as glucose, C6H12O6. The difference between the compounds is in the way the atoms are arranged.
Empirical formulas are an even simpler form. They show only the ratio of atoms in the compound rather than the actual number of atoms. Thus, the empirical formula for glucose is CH2O, which is found by dividing each subscript in the molecular formula by the lowest common factor, 6. This formula is not unique because it also applies to acetic acid, which has the molecular formula C2H4O2.
When either the molecular formula or the empirical formula of a compound is known, the compound's percent composition can be calculated. Percent composition is the percentage by mass of each element present in a compound. For example, glucose has a molar mass of 180.18 g/mol. The molar masses of its constituent atoms are 12.01 g/mol for carbon (C), 1.01 g/mol for hydrogen (H), and 16.000 g/mol for oxygen (O). To calculate percent composition, consider one mole of glucose. Multiply the number of each type of atom by the element's molar mass, divide by the mass per mole of glucose, and multiply by 100.
$\%\;\rm C=\frac{(6)(12.01\;\rm {g})}{180.18\;\rm g}\times\;100=39.99\%\;\rm C$
$\%\;\rm H=\frac{(12)(1.01\;\rm g)}{180.18\;\rm g}\times\;100=6.727\%\;\rm H$
$\%\;\rm O=\frac{(6)(16.00\;\rm g)}{180.18\;\rm g}\times\;100=52.28\%\;\rm O$
Conversely, when the total mass and percent composition of a compound are known, its empirical formula can be derived.
Step-By-Step Example
Deriving the Empirical Formula of an Unknown Compound
An unknown substance is composed of 42.1% sodium (Na), 18.9% phosphorus (P), and 39.0% oxygen (O). Therefore, a 100-gram sample of the unknown substance contains 42.1 grams of sodium, 18.9 grams of phosphorus, and 39.0 grams of oxygen.
Step 1
Convert grams to moles by dividing the mass of each element by the element's atomic weight.
\begin{aligned}{\frac{42.1\;\rm g\;\rm{Na}}{22.99\;\rm g/\rm{mol}}=1.831\;\rm{mol}\;\rm{Na}}\\\;\\{\frac{18.9\;\rm g\;\rm P}{30.97\;\rm g/\rm{mol}}=0.6103\;\rm{mol}\;\rm P}\\\;\\{\frac{39.0\;\rm g\;\rm O}{16.00\;\rm g/\rm{mol}}=\;2.438\;\rm{mol}\;\rm O}\end{aligned}
Step 2
Use the mole amounts to find the ratio of moles of each element. At least one atom of each element must be present. So, convert the smallest quantity to 1 by dividing it by itself. Then calculate the other elements' quantities based on that quantity, rounding to the nearest whole number if needed.
\begin{aligned}{\rm{Na}\!:\;\frac{1.831\;\rm{mol}}{0.6103\;\rm{mol}}\approx3}\\\\{\rm{P}\!:\;\frac{0.6103\;\rm{mol}}{0.6103\;\rm{mol}}=1}\\\\{\rm{O}\!:\;\frac{2.44\;\rm{mol}}{0.610\;\rm{mol}}\approx4}\end{aligned}
Solution
The empirical formula for this compound is Na3PO4.

Various instruments and techniques, such as gas chromatography, an analytical method in which a compound is vaporized and separated into its components, can identify the percent composition for samples of unknown substances. The samples can then be identified or further analyzed for precise identification.

If the percent composition and molar mass of an unknown compound are known, the molecular formula can be calculated.

Step-By-Step Example
Calculate the Molecular Formula of an Unknown Compound
Suppose a substance has a molar mass of 62.07 g/mol. Its percent composition is 38.7% carbon, 9.70% hydrogen, and 51.6% oxygen. By convention, carbon atoms are usually listed first, followed by hydrogen atoms, and then all others in alphabetical order of their chemical symbols.
Step 1
To find its molecular formula, first find its empirical formula.
\begin{aligned}{\frac{38.7\;\rm g\;\rm C}{12.01\;\rm g/\rm{mol}}=3.22\;\rm{mol}\;\rm C}\\\;\\{\frac{9.70\;\rm g\;\rm H}{1.01\;\rm g/\rm{mol}}=9.60\;\rm{mol}\;\rm H}\\\;\\{\frac{51.6\;\rm g\;\rm O}{16.00\;\rm g/\rm{mol}}=3.23\;\rm{mol}\;\rm O}\end{aligned}
Step 2
Convert the lowest value to 1 and find the ratio of moles of each element in the compound, rounding to the nearest whole number.
\begin{aligned}\rm{C}\!:\frac{3.23\;\rm{mol}}{3.23\;\rm{mol}}=1\\\rm{H}\!:\frac{9.60\;\rm{mol}}{3.23\;\rm{mol}}\approx3\\\rm{O}\!:\frac{3.23\;\rm{mol}}{3.23\;\rm{mol}}=1\end{aligned}
Therefore, the empirical formula is CH3O.
Step 3
To determine the molecular formula, compute the mass of one mole of the substance.
$\begin{gathered}(1\;\rm{mol\;\rm C)(12.01\;\rm g/\rm{mol})\;+\;(3\;\rm{mol}\;\rm H)(1.01\;\rm g/\rm{mol})\;+\;(1\;\rm{mol}\;\rm O)(16.00\;\rm g/\rm{mol})}\\=31.04\;\rm g\end{gathered}$
Step 4
Then divide the molar mass of the unknown substance by the molar mass of the empirical formula to determine the multiplication factor, rounding to the nearest whole number.
$\frac{62.07\;\rm g}{31.04\;\rm g}=2$
Step 5
Multiply the subscripts of elements in the empirical formula by the multiplication factor to determine the subscripts in the molecular formula.
\begin{aligned}{\rm{C}\!:}\;1\times2&=2\;\\{\rm{H}\!:}\;3\times2&=6\;\\{\rm{O}\!:}\;1\times2&=2\end{aligned}
Solution
The molecular formula is C2H6O2. This substance is ethylene glycol.