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 analysis, a 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 CO2 + y 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 = 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 = 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 = 0.0167 mol C
moles of H = 0.08 mol H
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.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Combustion Analysis concept. You can view video lessons to learn Combustion Analysis. Or if you need more Combustion Analysis practice, you can also practice Combustion Analysis practice problems.
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Based on our data, we think this problem is relevant for Professor Romonosky's class at GENESEO.