# Determining Empirical Formula (Example)

The outcome of this problem is to understand how to find the empirical formula of the unknown reactant in this combustion reaction.

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Given Problem: The combustion reaction X+O2→ CO2 + H2O contains 1.29 grams of X reactant, where X is an unknown compound containing C,H, & O. The reaction yields 1.13 grams of H2O and 2.086 grams of CO2. Oxygen (O2) is in excess. What is the empirical formula for X?

1. Everything in grams needs to be converted to moles. You can do this by using the compound’s molar mass. The identity of the X reactant is unknown, so its amount in moles can’t be calculated.

2. The question asks for the X compound’s empirical formula. We can assume that X fully combusts in this reaction, so it yields all of its carbon and hydrogen to the products. This is why the equation goes on further to find the amount of moles carbon yielded in the CO2 product. From this, we can hopefully backtrack from moles carbon in the CO2 product to the moles carbon found in the X compound. It is given that the molar mass of CO2 is ~44.01 grams/mol CO2.

3. The conversion step of grams to moles is repeated for hydrogen and the H2O product yielded from this reaction. It’s given that the molar mass of H2O is ~18.0 grams/mol H2O.

4. The mass of carbon and hydrogen yielded from the combustion reaction was found in grams and then subtracted from the total mass of the unknown compound. This equation conceptually looks like:

Total mass CHO= mass C+ mass H+ mass O (all in grams)

5. So far we have calculated:

• Moles CO2
• Moles C
• Moles H2O
• Moles H
• Mass O

Now we can convert the mass of oxygen found in compound X to moles of oxygen. This step is included because now we can look at whole compound in terms of moles, and we can compare the different components more easily to each other.

6. Now that all of the components of the unknown compound have been found, we compare them. The one that is smallest is the one that everything else gets divided by. This is to get the molar ratio for the compound. In this case, oxygen had the least amount of moles, so the moles of carbon and hydrogen were divided through by that value.

7. These values were rounded. This case yielded an uncommon value of 1.5, but that can be fixed in this case because we are dealing with the empirical formula and not the molecular formula. Moles carbon is rounded to 1.5 and moles hydrogen is rounded to 4.

8. One might think that this is our final answer: C1.5H4O. But that isn’t the case. The formula does have to be multiplied by 2 in order to get rid of all decimals in the subscript.

9. Finally, we multiply our initial compound by 2 and get C3H8O2 as our answer. That is the empirical formula for compound X.