Complete combustion of 3.60 g of a hydrocarbon produced 11.1 g of co2 and 5.11 g of h2o. what is the empirical formula for the hydrocarbon?

Answer :  The emperical formula of hydrocarbon is .

Solution :  Given,

                 Mass of = 11.1 g

                 Mass of = 5.11 g

Step 1 : convert given mass in moles.

Moles of = = = 0.2522 moles

Moles of = moles of C = 0.2522 moles

Moles of =   = = 0.2838 moles

Moles of = moles of H = 0.2838 × 2 = 0.5677 moles

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

For C = 0.2522/0.2522 = 1

For H = 0.5677/0.2522 = 2.25

C : H =  1 : 2.25

To make the ratio as a whole number multiply numerator and denominator by 4.

Ratio of  C : H =   = 4 : 9

The mole ratio of the element is repersented by subscripts in emperical formula.

Therefore, the Emperical formula =

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