12.4: Empirical and Molecular Formulas
Lesson Objectives
The student will:
 reduce molecular formulas to empirical formulas.
 determine the empirical formula of a compound given either percent composition or masses.
 determine the molecular formula of a compound given either percent composition and molar mass or masses.
Vocabulary
 molecular formula
Introduction
The empirical formula is the simplest ratio of atoms in a compound. Formulas for ionic compounds are always empirical formulas, but for covalent compounds, the empirical formula is not always the actual formula for the molecule. Molecules such as benzene, \begin{align*}\text{C}_6\text{H}_6\end{align*}
Finding Empirical Formula from Experimental Data
Empirical formulas can be determined from experimental data or from percent composition. Consider the following example.
Example:
We find that a \begin{align*}2.50 \ \mathrm{gram}\end{align*}
Solution:
First, we convert the mass of each element into moles.



\begin{align*}\text{moles of Ca} = \frac {0.900\ \text{g}} {40.1 \ \text{g/mol}} = 0.0224 \ \text{mol Ca}\end{align*}
moles of Ca=0.900 g40.1 g/mol=0.0224 mol Ca

\begin{align*}\text{moles of Ca} = \frac {0.900\ \text{g}} {40.1 \ \text{g/mol}} = 0.0224 \ \text{mol Ca}\end{align*}




\begin{align*}\text{moles of Cl atoms} = \frac {1.60\ \text{g}} {35.5\ \text{g/mol}} = 0.0451 \ \text{mol Cl}\end{align*}
moles of Cl atoms=1.60 g35.5 g/mol=0.0451 mol Cl

\begin{align*}\text{moles of Cl atoms} = \frac {1.60\ \text{g}} {35.5\ \text{g/mol}} = 0.0451 \ \text{mol Cl}\end{align*}

At this point, we have the correct ratio for the atoms in the compound. The formula \begin{align*}\text{Ca}_{0.0224}\text{Cl}_{0.0451}\end{align*}
\begin{align*}\text{Ca} = \frac {0.0224} {0.0224} = 1.00 && \text{Cl} = \frac {0.0451} {0.0224} = 2.01\end{align*}
Now, we can see the correct empirical formula for this compound is \begin{align*}\text{CaCl}_2\end{align*}
Finding Empirical Formula from Percent Composition
When finding the empirical formula from percent composition, the first thing to do is to convert the percentages into masses. For example, suppose we are given that the percent composition of a compound as \begin{align*}40.0\%\end{align*} carbon, \begin{align*}6.71\%\end{align*} hydrogen, and \begin{align*}53.3\%\end{align*} oxygen. Since every sample of this compound will have the same composition in terms of the ratio of atoms, we could choose a sample of any size. Suppose we choose a sample size of \begin{align*}100. \ \mathrm{grams}\end{align*}. The masses of each of the elements in this sample will be \begin{align*}40.0 \ \mathrm{grams}\end{align*} of carbon, \begin{align*}6.71 \ \mathrm{grams}\end{align*} of hydrogen, and \begin{align*}53.3 \ \mathrm{grams}\end{align*} of oxygen. These masses can then be used to find the empirical formula. You could use any size sample, but choosing a sample size of \begin{align*}100. \ \mathrm{grams}\end{align*} is usually most convenient because it makes the arithmetic simple.
Example:
Find the empirical formula of a compound whose percent composition is \begin{align*}40.0 \%\end{align*} carbon, \begin{align*}6.71 \%\end{align*} hydrogen, and \begin{align*}53.3 \%\end{align*} oxygen.
Solution:
We choose a sample size of \begin{align*}100. \ \mathrm{grams}\end{align*} and multiply this \begin{align*}100. \ \mathrm{gram}\end{align*} sample by each of the percentages to get masses for each element. This would yield \begin{align*}40.0 \ \mathrm{grams}\end{align*} of carbon, \begin{align*}6.71 \ \mathrm{grams}\end{align*} of hydrogen, and \begin{align*}53.3 \ \mathrm{grams}\end{align*} of oxygen. The next step is to convert the mass of each element into moles.


 \begin{align*}\text{moles of C} = \frac {40.0\ \text{g}} {12.0\ \text{g/mol}} = 3.33 \ \text{moles C}\end{align*}



 \begin{align*}\text{moles of H} = \frac {6.71\ \text{g}} {1.01\ \text{g/mol}} = 6.64 \ \text{moles H}\end{align*}



 \begin{align*}\text{moles of O} = \frac {53.3\ \text{g}} {16.0\ \text{g/mol}} = 3.33 \ \text{mole Ca}\end{align*}

Then, we divide all three numbers by the smallest one to get simple whole number ratios:
\begin{align*}\text{C} = \frac {3.33} {3.33} = 1 && \text{H} = \frac {6.64} {3.33} = 2 && \text{O} = \frac {3.33} {3.33} = 1\end{align*}
Finally, we can write the empirical formula \begin{align*}\text{CH}_2\text{O}\end{align*}.
Sometimes, this technique of dividing each of the moles by the smallest number does not yield whole numbers. Whenever the subscript for any element in the empirical formula is 1, dividing each of the moles by the smallest will yield a simple whole number ratio, but if none of the elements in the empirical formula has a subscript of 1, then this technique will not yield a simple whole number ratio. In those cases, a little more work is required.
Example:
Determine the empirical formula for a compound that is \begin{align*}66.0 \%\end{align*} calcium and \begin{align*}34.0 \%\end{align*} phosphorus.
Solution:
We choose a sample size of \begin{align*}100. \ \mathrm{grams}\end{align*} and multiply the \begin{align*}100. \ \mathrm{grams}\end{align*} by the percentage of each element to get masses. This yields \begin{align*}66.0 \ \mathrm{grams}\end{align*} of calcium and \begin{align*}34.0 \ \mathrm{grams}\end{align*} of phosphorus. We then divide each of these masses by their molar mass to convert the masses into moles:


 \begin{align*}\text{moles of Ca} = \frac {66.0\ \text{g}} {40.1\ \text{g/mol}} = 1.65 \ \text{moles Ca}\end{align*}



 \begin{align*}\text{moles of P} = \frac {34.0\ \text{g}} {31.0\ \text{g/mol}} = 1.10 \ \text{moles P}\end{align*}

We then divide each of these moles by the smallest.
\begin{align*}\text{Ca} = \frac {1.65} {1.10} = 1.50 && \text{P} = \frac {1.10} {1.10} = 1.00\end{align*}
In this case, dividing each of the numbers by the smallest one does not yield a simple whole number ratio. In such a case, we must multiply both numbers by some factor that will produce a whole number ratio. If we multiply each of these by \begin{align*}2\end{align*}, we get a whole number ratio of \begin{align*}3\end{align*} Ca to \begin{align*}2\end{align*} P. Therefore, the empirical formula is \begin{align*}\text{Ca}_3\text{P}_2\end{align*}.
Finding Molecular Formulas
Empirical formulas show the simplest whole number ratio of the atoms in a compound. Molecular formulas show the actual number of atoms of each element in a compound. When you find a empirical formula from either masses of elements or from percent composition, you are finding the empirical formula. For the compound \begin{align*}\text{N}_2\text{H}_4\end{align*}, you will get an empirical formula of \begin{align*}\text{NH}_2\end{align*}, and for \begin{align*}\text{C}_3\text{H}_6\end{align*}, you will get \begin{align*}\text{CH}_2\end{align*}. If we want to determine the actual molecular formula, we need one additional piece of information. The molecular formula is always a whole number multiple of the empirical formula. In order to get the molecular formula for \begin{align*}\text{N}_2\text{H}_4\end{align*}, you must double each of the subscripts in the empirical formula. Since the molecular formula is a whole number multiple of the empirical formula, the molecular mass will be the same whole number multiple of the formula mass. The formula mass for \begin{align*}\text{NH}_2\end{align*} is \begin{align*}16 \ \mathrm{g/mol}\end{align*}, and the molecular mass for \begin{align*}\text{N}_2\text{H}_4\end{align*} is \begin{align*}32 \ \mathrm{g/mol}\end{align*}. When we have the empirical formula and the molecular mass for a compound, we can divide the formula mass into the molecular mass and find the whole number that we need to multiply each of the subscripts in the empirical formula.
Example:
Find the molecular formula for a compound with percent composition of \begin{align*}40.0\%\end{align*} carbon, \begin{align*}67.1\%\end{align*} hydrogen, and \begin{align*}53.3\%\end{align*} oxygen. The molecular mass of the compound is \begin{align*}180 \ \mathrm{g/mol}\end{align*}.
Solution:
This is the same as an earlier example, except now we also have the molecular mass of the compound. Earlier, we determined the empirical formula of this compound to be \begin{align*}\text{CH}_2\text{O}\end{align*}. The empirical formula has a formula mass of \begin{align*}30.0 \ \mathrm{g/mol}\end{align*}. In order to find the molecular formula for this compound, we divide the formula mass into the molecular mass (\begin{align*}180\end{align*} divided by \begin{align*}30\end{align*}) and find the multiplier for the empirical formula to be \begin{align*}6\end{align*}. As a result, the molecular formula for this compound will be \begin{align*}\text{C}_6\text{H}_{12}\text{O}_6\end{align*}.
Example:
Find the molecular formula for a compound with percent composition of \begin{align*}85.6\%\end{align*} carbon and \begin{align*}14.5 \%\end{align*} hydrogen. The molecular mass of the compound is \begin{align*}42.1 \ \mathrm{g/mol}\end{align*}.
Solution:
We choose a sample size of \begin{align*}100. \ \mathrm{g}\end{align*} and multiply each element percentage to get masses for the elements in this sample. This yields \begin{align*}85.6 \ \mathrm{g}\end{align*} of C and \begin{align*}14.5 \ \mathrm{g}\end{align*} of H. Dividing each of these by their atomic mass yields \begin{align*}7.13 \ \mathrm{moles}\end{align*} of C and \begin{align*}14.4 \ \mathrm{moles}\end{align*} of H. Dividing each of these by the smallest yields a whole number ratio of \begin{align*}1\end{align*} carbon to \begin{align*}2\end{align*} hydrogen. Thus, the empirical formula will be \begin{align*}\text{CH}_2\end{align*}.
The formula mass of \begin{align*}\text{CH}_2\end{align*} is \begin{align*}14 \ \mathrm{g/mol}\end{align*}. Dividing \begin{align*}14 \ \mathrm{g/mol}\end{align*} into the molecular mass of \begin{align*}42.1 \ \mathrm{g/mol}\end{align*} yields a multiplier of 3. The molecular formula will be \begin{align*}\text{C}_3\text{H}_6\end{align*}.
Lesson Summary
 The empirical formula of a compound indicates the simplest whole number ratio of atoms present in the compound.
 The empirical formula of a compound can be calculate from the masses of the elements in the compound or from the percent composition.
 The molecular formula of a compound is some whole number multiple of the empirical formula.
Further Reading / Supplemental Links
This website has solved example problems for a number of topics covered in this lesson, including the determination of empirical and molecular formulas.
Review Questions
 What is the empirical formula for \begin{align*}\text{C}_8\text{H}_{18}\end{align*}?
 What is the empirical formula for \begin{align*}\text{C}_6\text{H}_6\end{align*}?
 What is the empirical formula for \begin{align*}\text{WO}_2\end{align*}?
 A compound has the empirical formula \begin{align*}\text{C}_2\text{H}_8\text{N}\end{align*} and a molar mass of \begin{align*}46 \ \mathrm{g/mol}\end{align*}. What is the molecular formula of this compound?
 A compound has the empirical formula \begin{align*}\text{C}_2\text{H}_4\text{NO}\end{align*}. If its molar mass is \begin{align*}116.1 \ \mathrm{g/mol}\end{align*}, what is the molecular formula of the compound?
 A sample of pure indium chloride with a mass of \begin{align*}0.5000 \ \mathrm{grams}\end{align*} is found to contain \begin{align*}0.2404 \ \mathrm{grams}\end{align*} of chlorine. What is the empirical formula of this compound?
 Determine the empirical formula of a compound that contains \begin{align*}63.0 \ \mathrm{grams}\end{align*} of rubidium and \begin{align*}5.90 \ \mathrm{grams}\end{align*} of oxygen.
 Determine the empirical formula of a compound that contains \begin{align*}58.0 \%\end{align*} \begin{align*}\text{Rb}\end{align*}, \begin{align*}9.50 \%\end{align*} \begin{align*}\text{N}\end{align*}, and \begin{align*}32.5 \%\end{align*} \begin{align*}\text{O}\end{align*}.
 Determine the empirical formula of a compound that contains \begin{align*}33.3 \%\end{align*} \begin{align*}\text{Ca}\end{align*}, \begin{align*}40.0 \%\end{align*} \begin{align*}\text{O}\end{align*}, and \begin{align*}26.7 \%\end{align*} \begin{align*}\text{S}\end{align*}.
 Find the molecular formula of a compound with percent composition \begin{align*}26.7\%\end{align*} \begin{align*}\text{P}\end{align*}, \begin{align*}12.1\%\end{align*} \begin{align*}\text{N}\end{align*}, and \begin{align*}61.2\%\end{align*} \begin{align*}\text{Cl}\end{align*} and with a molecular mass of \begin{align*}695 \ \mathrm{g/mol}\end{align*}.
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