Q3.115CP - Chemistry: Molecular Nature Of Matter And Change

Question

Hydrogen-containing fuels have a “fuel value” based on their mass % H. Rank the following compounds from highest mass % H to lowest: ethane, propane, benzene, ethanol, cetyl palmitate (whale oil, C₃₂H₆₄O₂).

Step-by-Step Solution

Answer

The compound from the highest mass percent to the lowest is:

$C{}_2{H}_6>C_3{H}_8>C_{32}{H}_{64}{O}_2>C_2{H}_{5}OH>C_6{H}_6.$

Step 1: Formula for the mass percentage of H

To get the mass percent of an element, use the formula below.

$Mass \% of H=\frac{moles of H in formula\times molar mass of H}{mass of 1 mol of compound}\times 100$

Step 2: Finding the mass percentage

The mass percentage can be calculated as:

$\begin{array}{rcl}Mass \% of H& =& \frac{6 mol H\times \frac{1.01 g H}{1 mol H}}{30.08 g C_2{H}_6}\times 100\\ & =& 20.1\% H.\\ Mass \% of H& =& \frac{8 mol H\times \frac{1.01 g H}{1 mol H}}{44.11 g C_3{H}_8}\times 100\\ & =& 18.3\% H.\\ Mass \% of H & =& \frac{6 mol H\times \frac{1.01 g H}{1 mol H}}{78.12 g C_6{H}_6}\times 100\\ & =& 7.8\% H.\\ Mass \% of H& =& \frac{6 mol H\times \frac{1.01 g H}{1 mol H}}{46.08 g C_2{H}_{5}OH}\times 100\\ & =& 13.2\% H.\\ Mass \% of H& =& \frac{64 mol H\times \frac{1.01 g H}{1 mol H}}{480.96 g C_{32}{H}_{64}{O}_2}\times 100\\ & =& 13.4\% H.\end{array}$

The mass percent from the highest to the lowest based on the calculations are:

$C{}_2{H}_6>C_3{H}_8>C_{32}{H}_{64}{O}_2>C_2{H}_{5}OH>C_6{H}_6.$