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}}* 1 mole` =
`\frac{\text{11.1 g}}{\text{44 g/mole}}* 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}}* 1 mole` =
`\frac{\text{5.11 g}}{\text{18 g/mole}}* 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 =
`(1*4)/(2.25*4)` = 4 : 9

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

Therefore, the Emperical formula =
`C_(4) H_(9)`