Obtaining Empirical and Molecular Formulas from Combustion Data 

Empirical and molecular formulas for compounds that contain only carbon and hydrogen (C_aH_b) or carbon, hydrogen, and oxygen (C_aH_bO_c) can be determined with a process called combustion analysis. The steps for this procedure are:

• Weigh a sample of the compound to be analyzed and place it in the apparatus shown in the image below.

• Burn the compound completely. The only products of the combustion of a compound that contains only carbon and hydrogen (C_aH_b) or carbon, hydrogen, and oxygen (C_aH_bO_c) are carbon dioxide and water.

• The H₂O and CO₂ are drawn through two tubes. One tube contains a substance that absorbs water, and the other contains a substance that absorbs carbon dioxide. Weigh each of these tubes before and after the combustion. The increase in mass in the first tube is the mass of H₂O that formed in the combustion, and the increase in mass for the second tube is the mass of CO₂ formed.

• Assume that all the carbon in the compound has been converted to CO₂ and trapped in the second tube. Calculate the mass of carbon in the compound from the mass of carbon in the measured mass of CO₂ formed.

• Assume that all of the hydrogen in the compound has been converted to H₂O and trapped in the first tube. Calculate the mass of hydrogen in the compound from the mass of hydrogen in the measured mass of water.

• If the compound contains oxygen as well as carbon and hydrogen, calculate the mass of the oxygen by subtracting the mass of carbon and hydrogen from the total mass of the original sample of compound.

• Use this data to determine the empirical and molecular formulas in the usual way. (See the text for a reminder of how this is done.) 

Apparatus for Combustion Analysis     A compound containing carbon and hydrogen (C_aH_b) or carbon, hydrogen, and oxygen (C_aH_bO_c) is burned completely to form H₂O and CO₂. The products are drawn through two tubes. The first tube absorbs water, and the second tube absorbs carbon dioxide.

To illustrate how empirical and molecular formulas can be determined from data derived from combustion analysis, lets consider a substance called trioxane. Formaldehyde, CH₂O, is unstable as a pure gas, readily forming a mixture of a substance called trioxane and a polymer called paraformaldehyde. That is why formaldehyde is dissolved in a solvent, like water, before it is sold and used. The molecular formula of trioxane, which contains carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 17.471 g of trioxane is burned in the apparatus shown above, and 10.477 g H₂O and 25.612 g CO₂ are formed. In the second experiment, the molecular mass of trioxane is found to be 90.079. 

We can get the molecular formula of a compound from its empirical formula and its molecular mass. (See the text for a reminder of how this is done.) To get the empirical formula, we need to determine the mass in grams of the carbon, hydrogen, and oxygen in 17.471 g of trioxane. Thus, we need to perform these general steps.

• First, convert from the data given to grams of carbon, hydrogen, and oxygen.

• Second, determine the empirical formula from the grams of carbon, hydrogen, and oxygen.

• Third, determine the molecular formula from the empirical formula and the given molecular mass.

Because we assume that all the carbon in trioxane has reacted to form in CO₂, we can find the mass of carbon in 17.471 g trioxane by calculating the mass of carbon in 25.612 g CO₂.

Because we assume that all of the hydrogen in trioxane has reacted to form H₂O, we can find the mass of hydrogen in 17.471 g trioxane by calculating the mass of hydrogen in 10.477 g H₂O.

Because trioxane contains only carbon, hydrogen, and oxygen, we can calculate the mass of oxygen by subtracting the masses of carbon and hydrogen from the total mass of trioxane.

? g O  =  17.471 g trioxane  - 6.9899 g C  -  1.1724 g H  =  9.309 g O

We now calculate the empirical formula.

The empirical formula is CH₂O, which can be used to calculate the molecular formula.

Empirical formula mass  =  1(12.011)  +  2(1.00794)  +  1(15.9994)

                                     =  30.026

molecular formula        C₃H₆O₃

Click here to see a Sample Study Sheet that summarizes this procedure.

Click here to see an example. 

Click here to see an exercise. 

  bishopmark@comcast.net

[ Home ] [ Up ] [ Study Sheet ] [ Example ] [ Exercise ]