College Chemistry : Combustion Reactions

Study concepts, example questions & explanations for College Chemistry

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Example Questions

Example Question #1 : Reactions

Suppose that the only elements a compound contains are carbon, hydrogen, and oxygen. If complete combustion of \(\displaystyle 70.86\:g\) of this compound produces \(\displaystyle 103.9\:g\) of \(\displaystyle CO_{2}\) and \(\displaystyle 42.52\:g\) of \(\displaystyle H_{2}O\), what is this compound's molecular formula?

Note: The molecular weight of this compound is \(\displaystyle 120.104\:\frac{g}{mol}\).

Possible Answers:

\(\displaystyle C_{5}H_{12}O_{3}\)

\(\displaystyle C_{3}H_{4}O_{5}\)

\(\displaystyle C_{4}H_{8}O_{4}\)

\(\displaystyle C_{7}H_{4}O_{2}\)

Correct answer:

\(\displaystyle C_{4}H_{8}O_{4}\)

Explanation:

To answer this question, it's important to remember the details of combustion reactions. In the presence of oxygen and a sufficient amount of energy to initiate the process, combustion reactions that go to completion will result in the complete oxidation and breakdown of the initial reactant.

The \(\displaystyle CO_{2}\) produced from the reaction contains all of the carbon in the original reactant. Likewise, the \(\displaystyle H_{2}O\) produced contains all the hydrogen in the original compound. The oxygen content in the original compound is everything left over.

The first step then is to use the mass of each product to find the mass of each element in the original compound. To do this, we'll need to use the molar mass for each of the products as well as the individual elements.

\(\displaystyle 103.9\:g\:CO_{2}\cdot \frac{12\:\frac{g}{mol}\:C}{44\:\frac{g}{mol}\:CO_{2}}=28.34\:g\:C\)

\(\displaystyle 42.52\:g\:H_{2}O\cdot \frac{2\:\frac{g}{mol}\:H}{18\:\frac{g}{mol}\:H_{2}O}=4.72\:g\:H\)

The combined mass of carbon and hydrogen in the original compound is \(\displaystyle 28.34\:g+4.72\:g=33.06\:g\).

The remaining mass will come from oxygen. Thus, the original compound will contain \(\displaystyle 70.86\:g-33.06\:g=37.8\:g\) of oxygen.

Once we have the mass of each element, we can use that information to calculate the number of moles of each element in the original compound.

\(\displaystyle 28.34\:g\:C\cdot\frac{1\:mol\:C}{12\:g\:C}=2.36\:mol\:C\)

\(\displaystyle 4.72\:g\:H\cdot\frac{1\:mol\:H}{1\:g\:H}=4.72\:mol\:H\)

\(\displaystyle 37.8\:g\:O\cdot \frac{1\:mol\:O}{16\:g\:O}=2.36\:mol\:O\)

Now that we have the relative molar amounts of each element in the original compound, we can find the empirical formula by finding the smallest whole number ratios.

\(\displaystyle C_{2.36}H_{4.72}O_{2.36}=CH_{2}O\)

From this, we can conclude that the empirical mass of the compound is \(\displaystyle 30\:g\).

Finally, since we are given the compound's molecular weight, we can use that information to find the molecular formula for the compound.

\(\displaystyle \frac{120.104}{30}\approx4\)

\(\displaystyle 4(CH_{2}O)=C_{4}H_{8}O_{4}\)

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