If a compound is made in a lab or found in nature, the chemical formula must be determined experimentally. First, the types of elements in the compound must be identified as well as the amounts of each element in the compound. The percent composition of a compound can be determined from its chemical formula. For example, a compound with the chemical formula C₆H₁₄ has a molar mass of 86.18 g/mol. This means that 1 mole of C₆H₁₄ has a mass of 86.18 grams. We can use this information to find the percent composition of the compound. Recall, percent is parts per hundred. We take the amount of one component and divide by the mass of the whole and then multiply by 100. For C₆H₁₄ we will find the percent of carbon in 1 mole (86.18 g) of the compound. One mole of the compound has a mass of 86.18 g and we have 6 moles of carbon in one mole of C₆H₁₄ which needs to be multiplied by its molecular mass of 12.011 g/mol.

6 x 12.011 g = 72.066 g carbon

\(\displaystyle \frac{72.066\;g\;C}{86.18\;g\;C_6H_{14}}\times\;{100}\;=\;\mathbf{83.62\%\;carbon}\)

The same calculation can be done for hydrogen. There are 14 moles of hydrogen in C₆H₁₄. This is 14 x 1.00794 g H = 14.111 g H.

\(\displaystyle \frac{14.111\;g\;H}{86.18\;g\;C_6H_{14}}\times\;{100}\;=\;\mathbf{16.37\%\;hydrogen}\)

We could have easily subtracted 83.62% from 100% because there are only two elements in C₆H₁₄ and the percentages of each element must add to 100%. For this compound we have 83.62% carbon + 16.37% hydrogen = 99.99% ~ 100%. There can be a small error introduced due to rounding.

Our compound, C₆H₁₄, whether we have one mole or a truckload of C₆H₁₄, will have a percent composition of 83.62% carbon and 16.37% hydrogen.

Another example is a hydrocarbon (a substance composed of carbon and hydrogen) that is 84.6148% carbon and 15.3852% hydrogen. This information will allow us to determine the molecular formula of the compound. Assuming 100 g of compound, 84.6148 g is carbon and 15.3852 g is hydrogen. First we find the molar mass of carbon and hydrogen in the compound. One mole of carbon is 12.011 g, and one mole of hydrogen is 1.00794 g.

\(\displaystyle moles\;carbon\;=\;84.6148\;g\;C\times\frac{1\;mol\;C}{12.011\;g\;C}\;=\;\mathbf{7.0448\;mol\;C}\)

\(\displaystyle moles\;hydrogen\;=\;15.3852\;g\;H\times\frac{1\;mol\;H}{1.00794\;g\;H}\;=\;\mathbf{15.2640\;mol\;H}\)

The formula can be written as:

C_7.0448H_15.2640

We can then divide both subscripts by the smallest subscript which is 7.0448.

\(\displaystyle C_{\frac{7.0448}{7.0448}}H_{\frac{15.2640}{7.0448}}\;=\;C_1H_{2.1667}\)

The 2.1667 means the subscripts must be multiplied by a small integer that will result in whole numbers. This will give us the empirical formula of the compound–the smallest whole number ratios of atoms. Let’s start multiplying the subscripts by 2 and then go on until we get whole number subscripts. Any discrepancy must be small:

C_{(1 x 2)}H_{(2 x 2.1667)} = C₂H_4.3334
C_{(1 x 3)}H_{(3 x 2.1667)} = C₃H_6.5001
C_{(1 x 4)}H_{(4 x 2.1667)} = C₄H_8.6668
C_{(1 x 5)}H_{(5 x 2.1667)} = C₅H_10.8335
C_{(1 x 6)}H_{(6 x 2.1667)} = C₆H_13.0002

Multiplying the subscripts by 6 gives us C₆H₁₃ as the empirical formula. The empirical formula only tells us the ratios of atoms in a compound, but the molecular formula tells us the actual number of atoms in the chemical formula. The compound could have the empirical formula or it could be a multiple of the empirical formula. To determine the molecular formula, the molecular mass of the compound is needed. The molecular mass of this compound was found using mass spectrometry and is 170.335 g/mol. To find the whole number multiple divide the molecular mass by the empirical formula mass.

\(\displaystyle\frac{molecular\;mass}{empirical\;formula\;mass}\;=\;\frac{170.335\;g/mol}{85.169\;g/mol}\;=\;{1.99996}\;=\;2\)

Multiply the subscripts in the empirical formula, C₆H₁₃, by 2. The molecular formula of the compound is C₁₂H₂₆.

It is possible for an empirical formula to be the same as the molecular formula, but the molecular mass of the compound is required. Recall, ionic compounds are always represented with an empirical formula. Manganese(IV) oxide is an example, MgO₂. We would not write it as Mg₂O₄. There is a 1:2 ratio of manganese to oxygen. In NaCl there is a 1:1 ratio of sodium to chlorine.

One analytical technique for determining an empirical formula for compounds that contain carbon and hydrogen is combustion analysis. A sample of the compound is first weighed and then combusted (burned in oxygen). The apparatus has two absorbers: one for H₂O another for the CO₂ that is produced. The absorbers are weighed before the compound is combusted and again after combustion. The difference in the final and initial masses of the absorbers corresponds to the amounts of carbon and hydrogen produced from the sample compound. The amount of hydrogen is determined from the H₂O and the amount of carbon produced is determined from the CO₂. If there is a third element in the compound, its mass is determined by subtracting the masses of carbon and hydrogen produced, from the mass of the compound.

Here, we work an example. If 0.255 g of an unknown compound that contains carbon, hydrogen, and oxygen was placed in a combustion apparatus and 0.560 g of CO₂ and 0.305 g of H₂O was produced, what is the empirical formula for the compound?

First, we need to calculate the amounts, in both grams and moles, of C and H that are formed. The amount of C is determined from CO₂ and the amount of hydrogen is determined from H₂O.

\(\displaystyle mols\;C\;=\;0.560\;g\;CO_2\times\frac{1\;mol\;CO_2}{44.01\;g\;CO_2}\times\frac{1\;mol\;C}{1\;mol\;CO_2}\;=\;\mathbf{0.0127\;mol\;C}\)

\(\displaystyle g\;C\;=\;0.0127\;mol\;C\times\frac{12.011\;g\;C}{1\;mol\;C}\;=\;\mathbf{0.15\underline{2}5\;g\;C}\)

\(\displaystyle mols\;H\;=\;0.305\;g\;H_2O\times\frac{1\;mol\;H_2O}{18.015\;g\;H_2O}\times\frac{2\;mol\;H}{1\;mol\;H_2O}\;=\;\mathbf{0.033\underline{8}6\;mol\;H}\)

\(\displaystyle g\;H\;=\;0.033\underline{8}6\;mol\;H\times\frac{1.00794\;g\;H}{1\;mol\;H}\;=\;\mathbf{0.034\underline{1}3\;g\;H}\)

Now that we have found the number of moles and grams of C and H, we can calculate the number of grams and moles of oxygen. We take the grams of H and C and subtract from the mass of the compound which is 0.255 g.

0.255 g Compound – (0.03413 g H + 0.1525 g C) = 0.06837 g O

\(\displaystyle moles\;O\;=\;0.068\underline{3}7\;g\;\times\frac{1\;mol\;O}{15.9994\;g\;O}\;=\;\mathbf{0.0042\underline{7}3\;mol\;O}\)

We can now write the formula that contains C, H, and O, and divide the subscripts by the smallest subscript which is 0.004273.

\(\displaystyle C_{\frac{0.0127}{0.004273}}H_{\frac{0.03386}{0.004273}}O_{\frac{0.004273}{0.004273}}\;=\;C_3H_8O\)

The empirical formula is C₃H₈O. To determine the molecular formula, we would need the molar mass of the compound. Again, the empirical formula could be the molecular formula, but there is no way of knowing without the molecular mass. For this compound the molecular mass was determined, by mass spectrometry, to be 60.096 amu. The molecular mass of the empirical formula is 60.0959 amu.

\(\displaystyle\frac{molecular\;mass}{empirical\;formula\;mass}\;=\;\frac{60.096\;g/mol}{60.0959\;g/mol}\;=\;{1.001}\;=\;1\)

In this case the empirical formula, C₃H₈O, is our molecular formula–we multiply the subscripts by the whole number multiple of 1.

Worksheet: Percent Mass, Empirical Formula, and Combustion Analysis
 
Please watch the following videos before attempting the exercises.

Determine Empirical and Molecular Formulas from Combustion Analysis

EXERCISES

Exercise 1 What is the mass percent of carbon and hydrogen in C₁₀H₂₂?

Check Answer/Solution to Exercise 1

Exercise 2 What is the mass percent of nitrogen in 365 grams (NH₄)₂CO₃

Check Answer/Solution to Exercise 2

Exercise 3 A compound is 55.3% K, 14.6% P and 30.1% O. What is the empirical formula?

Check Answer/Solution to Exercise 3

Exercise 4 A substance contains by mass 7.1% H, 59.0% C, 26.2% O and 7.7% N. The molecular mass of the substance was determined by mass spectrometry to be 180. amu. Determine both the empirical and molecular formula of the compound.

Check Answer/Solution to Exercise 4

Exercise 5 Combustion analysis of 0.960 g of a hormone that contains hydrogen, carbon, and oxygen yielded 2.650 g of CO₂ and 0.818 g of H₂. Which of the following substances could be the unknown hormone?

 

Check Answer/Solution to Exercise 5

Exercise 6 There are several oxides of tungsten, W. If 7.56 g of the compound contains 5.99 g of W, what is the empirical formula of the compound?

Check Answer/Solution to Exercise 6

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