Question

Arrange the following compounds from left to right inorder of increasing percentage by mass of hydrogen: ${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O,}}\mathbf{}{\mathbf{\text{C}}}_{\mathbf{12}}{\mathbf{\text{H}}}_{\mathbf{26}}\mathbf{,}\mathbf{}{\mathbf{\text{N}}}_{\mathbf{4}}{\mathbf{\text{H}}}_{\mathbf{6}}\mathbf{,}\mathbf{}\mathbf{\text{LiH}}$

Short answer

The increasing order of mass percentage of hydrogen in each compound is${\text{N}}_4{\text{H}}_6<{\text{H}}_2\text{O}<\text{LiH}<{\text{C}}_{12}{\text{H}}_{26}$

Step-by-step solution

Mass percentage of hydrogen in H2O  .

The mass percentage can be calculated by dividing the molar mass of the element by the molar mass of the compound, multiplied by 100.

The elements present in the compound are hydrogen and oxygen.

The molar mass of hydrogen, ${\text{m}}_1\text{= 1.0079}{\text{gmol}}^{-1}$

The molar mass of oxygen, ${\text{m}}_2=15.999{\text{gmol}}^{-1}$

The molar mass of water,

$\begin{array}{rcl}{\text{M}}_1& =& \left(2\times 1.0079{\text{gmol}}^{-1}\right)+15.999{\text{gmol}}^{-1}\\ & =& 18.0148{\text{gmol}}^{-1}\end{array}$

The mass percentage ofhydrogen,

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_1}\times 100\\ & =& \frac{2\times 1.0079}{18.0148}\times 100\\ & =& 11.2\%\end{array}$

Hence, the mass percentage of hydrogen in water is 11.2%.

Step 2: Mass percentage of hydrogen in C12H26

The elements present in the compound are hydrogen and carbon.

Molar mass of hydrogen, ${\text{m}}_1\text{= 1.0079}{\text{gmol}}^{-1}$

The molar mass of carbon,${\text{m}}_3=12.011{\text{gmol}}^{-1}$

The molar mass of ${\text{C}}_{12}{\text{H}}_{26}$;

$\begin{array}{rcl}{\text{M}}_2& =& \left(26\times 1.0079{\text{gmol}}^{-1}\right)+\left(12\times 12.011{\text{gmol}}^{-1}\right)\\ & =& 170.3374{\text{gmol}}^{-1}\end{array}$

Mass percentage of hydrogen in ${\text{C}}_{12}{\text{H}}_{26}$,

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_2}\times 100\\ & =& \frac{26\times 1.0079}{170.3374}\times 100\\ & =& 15.4\%\end{array}$

Therefore, the mass percentage of hydrogen in${\text{C}}_{12}{\text{H}}_{26}$is 15.4%.

Mass percentage of hydrogen in  N4H6

The elements present in the compound are hydrogen and nitrogen.

The molar mass of hydrogen, ${\text{m}}_1\text{= 1.0079}{\text{gmol}}^{-1}$

The molar mass of nitrogen, ${\text{m}}_4=14.0067{\text{gmol}}^{-1}$

Themolar mass of ${\text{N}}_4{\text{H}}_6$

role="math" localid="1663410090404" $\begin{array}{rcl}{\text{M}}_3& =& \left(6\times 1.0079{\text{gmol}}^{-1}\right)+\left(4\times 14.0067\right){\text{gmol}}^{-1}\\ & =& 62.0742{\text{gmol}}^{-1}\end{array}$

Mass percentage of hydrogen,

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_3}\times 100\\ & =& \frac{6\times 1.0079}{62.0742}\times 100\\ & =& 9.7\%\end{array}$

So, the mass percentage of hydrogen in ${\text{N}}_4{\text{H}}_6$ is 9.7%.

Mass percentage of hydrogen in LiH

The elements present in the compound are hydrogen and lithium.

Molar mass of hydrogen,${\text{m}}_1\text{= 1.0079}{\text{gmol}}^{-1}$

Molar mass of lithium, ${\text{m}}_5=6.941{\text{gmol}}^{-1}$

Molar mass of LiH,

$\begin{array}{rcl}{\text{M}}_4& =& \left(1\times 1.0079{\text{gmol}}^{-1}\right)+6.941{\text{gmol}}^{-1}\\ & =& 7.9489{\text{gmol}}^{-1}\end{array}$

Mass percentage of hydrogen,

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_4}\times 100\\ & =& \frac{1\times 1.0079}{7.9489}\times 100\\ & =& 12.7\%\end{array}$

Hence, the mass percentage of hydrogen in LiH is 12.7%.

Comparison of mass percentages of hydrogen in each compound.

The increasing order of mass percentage of hydrogen in each compound is given below:

${\text{N}}_6{\text{H}}_6<{\text{H}}_2\text{O}<\text{LiH}<{\text{C}}_{12}{\text{H}}_{26}$