Answer : The emperical formula of hydrocarbon is [tex]C_{4} H_{9}[/tex].
Solution : Given,
Mass of [tex]CO_{2}[/tex] = 11.1 g
Mass of [tex]H_{2} O[/tex] = 5.11 g
Step 1 : convert given mass in moles.
Moles of [tex]CO_{2}[/tex] = [tex]\frac{\text{given mass}}{\text{molar mass}}\times 1 mole[/tex] = [tex]\frac{\text{11.1 g}}{\text{44 g/mole}}\times 1 mole CO_{2}[/tex] = 0.2522 moles
Moles of [tex]CO_{2}[/tex] = moles of C = 0.2522 moles
Moles of [tex]H_{2} O[/tex] = [tex]\frac{\text{given mass}}{\text{molar mass}}\times 1 mole[/tex] = [tex]\frac{\text{5.11 g}}{\text{18 g/mole}}\times 1 mole H_{2}O[/tex] = 0.2838 moles
Moles of [tex]H_{2} O[/tex] = 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 = [tex]\frac{1\times4}{2.25\times4}[/tex] = 4 : 9
The mole ratio of the element is repersented by subscripts in emperical formula.
Therefore, the Emperical formula = [tex]C_{4} H_{9}[/tex]