Question

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 1

Answer :  The emperical formula of hydrocarbon is $C_{4} H_{9}$.

Solution :  Given,

                 Mass of $CO_{2}$ = 11.1 g

                 Mass of $H_{2} O$ = 5.11 g

Step 1 : convert given mass in moles.

Moles of $CO_{2}$ = $\frac{\text{given mass}}{\text{molar mass}}\times 1 mole$ = $\frac{\text{11.1 g}}{\text{44 g/mole}}\times 1 mole CO_{2}$ = 0.2522 moles

Moles of $CO_{2}$ = moles of C = 0.2522 moles

Moles of $H_{2} O$ =  $\frac{\text{given mass}}{\text{molar mass}}\times 1 mole$ = $\frac{\text{5.11 g}}{\text{18 g/mole}}\times 1 mole H_{2}O$ = 0.2838 moles

Moles of $H_{2} O$ = 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 =  $\frac{1\times4}{2.25\times4}$ = 4 : 9

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

Therefore, the Emperical formula = $C_{4} H_{9}$