When 3.915 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 12.74 grams of CO2 and 3.913 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.

The empirical formula and the molecular formula of the hydrocarbon is and respectivley.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

where, 'x', 'y'  are the subscripts of Carbon, hydrogen

We are given:

Mass of = 12.74 g

Mass of = 3.913 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 12.74 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 3.913 g of water, = of hydrogen will be contained.  

Mass of C = 3.474 g

Mass of H = 0.435 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: 1.5

The whole number ratio will be = 2: 3

Hence the empirical formula is .

The empirical weight of = 2(12.01)+3(1.008)= 27.04 g.

The molecular weight = 54.09 g/mole

Now we have to calculate the molecular formula.

The molecular formula will be=

14 days is the half-life of phosphorus-32