When 1.666 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 5.422 grams of CO2 and 1.665 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Enter the elements in the order presented in the question.

empirical formula =

molecular formula =

To find the empirical formula of the hydrocarbon, first, determine the moles of carbon and hydrogen in the compound using the given masses of CO2 and H2O produced in the combustion reaction.

Moles of carbon in CO2 = mass of CO2 / molar mass of CO2

Moles of hydrogen in H2O = mass of H2O / molar mass of H2O

Then, calculate the ratio of moles of carbon to moles of hydrogen.

Next, determine the simplest whole-number ratio of carbon to hydrogen by dividing both by the smallest number of moles obtained.

Finally, use this ratio to determine the empirical formula.

Let's do the calculations:

Moles of carbon = 5.422 g / 44.01 g/mol (molar mass of CO2) ≈ 0.1233 mol

Moles of hydrogen = 1.665 g / 18.02 g/mol (molar mass of H2O) ≈ 0.0923 mol

Now, find the ratio of moles of carbon to moles of hydrogen:

Ratio = Moles of carbon / Moles of hydrogen ≈ 0.1233 / 0.0923 ≈ 1.335

The simplest whole-number ratio is approximately 1:1. Therefore, the empirical formula is CH.

Next, to find the molecular formula, you need the molar mass of the compound. The given molar mass is 54.09 g/mol.

Calculate the molar mass of the empirical formula (CH):

Molar mass of CH = 12.01 g/mol (molar mass of carbon) + 1.01 g/mol (molar mass of hydrogen) = 13.02 g/mol

Now, find the ratio of the given molar mass to the molar mass of the empirical formula:

Molar mass ratio = Given molar mass / Molar mass of empirical formula = 54.09 g/mol / 13.02 g/mol ≈ 4.15

This ratio suggests that the molecular formula is approximately 4 times the empirical formula.

So, the molecular formula is (CH)4, which simplifies to C4H4.