Problem: Combustion analysis of naphthalene, a hydrocarbon used in mothballs, produced 8.80 g CO2 and 1.44 g H2 O.Calculate the empirical formula for naphthalene.

FREE Expert Solution

We’re being asked to determine the empirical formula of a compound composed of C and H given the combustion analysis.

Recall that in combustion analysisa compound reacts with excess O2 to form products

For a compound composed of C, H, and O, the reaction looks like this:

CxHyOz + O2 (excess)  x CO2y H2O

This means we need to do the following steps:

Step 1: Calculate the mass of C and H in the compound.

Step 2: Calculate the mass of O in the compound.

Step 3: Determine the lowest whole number ratio of C, H, and O to get the empirical formula.

We can do a mole-to-mole comparison to determine the mass of C and H.

For C: The molar mass of CO2 is:

CO2     1 C × 12.01 g/mol C = 12.01 g/mol             

            2 O × 16.00 g/mol O = 32.00 g/mol     

                                     Sum = 44.01 g/mol

There is 1 mole of C in 1 mole of CO2. Finding the mass of C:

mass of C=8.80 g CO2×1 mol CO244.01 g CO2×1 mol C1 mol CO2

mass of C = 0.20 g C

For H: The molar mass of H2O is:

H2O     2 H × 1.01 g/mol H = 2.02 g/mol                  

            1 O × 16.00 g/mol O = 16.00 g/mol     

                                     Sum = 18.02 g/mol

There are 2 moles of H in 1 mole of H2O. Finding the moles of H:

mass of H=1.44 g H2O×1 mol H2O18.02 g H2O×1 mol H1 mol H2O

mass of H = 0.08 g H

We next need to calculate the moles of each element and determine the lowest whole number ratio. 

The moles of C and H are:

moles of C=0.20 g C×1 mol C12.01 g C

moles of C = 0.0167 mol C

moles of H=0.08 g H×1 mol H1 g H

moles of H = 0.08 mol H

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Problem Details

Combustion analysis of naphthalene, a hydrocarbon used in mothballs, produced 8.80 g CO2 and 1.44 g H2 O.

Calculate the empirical formula for naphthalene.

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What scientific concept do you need to know in order to solve this problem?

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