When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 56.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question.

The empirical formula is and molecular formula is

Explanation:

We are given:

Mass of = 18.95 g

Mass of = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =

Moles of H=

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =

For H =

The ratio of C : H = 1: 2

Hence the empirical formula is .

The empirical weight of = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

The molecular formula will be=