Question

Arrange the following compounds from left to right in order of increasing percentage by mass of fluorine: ${\mathbf{\text{HF,C}}}_{\mathbf{6}}{\mathbf{\text{HF}}}_{\mathbf{5}}\mathbf{,}\mathbf{\text{BrF}}\mathbf{,}{\mathbf{\text{UF}}}_{\mathbf{6}}$

Short answer

The increasing order of mass percentage of fluorine in each compounds is${\text{UF}}_6<{\text{C}}_6{\text{HF}}_5<\text{BrF}<\text{HF}$

Step-by-step solution

Mass percentage of fluorine in HF

Mass percentage of any element is the fraction of molar mass of the element and the compound, multiplied with 100.

The compound contains one hydrogen and fluorine

Molar mass of fluorine, ${\text{m}}_1=18.9984{\text{gmol}}^{-1}$

Molar mass of hydrogen, ${\text{m}}_2=1.0079{\text{gmol}}^{-1}$

Molar mass of the compound $\text{HF}$

$\begin{array}{rcl}{\text{M}}_1& =& 1.0079+18.9984\\ & =& 20.0063{\text{gmol}}^{-1}\end{array}$

Mass percentage of fluorine in $\text{HF}$

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_1}\times 100\\ & =& \frac{18.9984}{20.0063}\times 100\\ & =& 94.96\%\end{array}$

So, the mass percentage of fluorine in $\text{HF}$ is 94.96%.

Step 2: Mass percentage of fluorine in C6HF5

The compound contains 6 carbon, one hydrogen and 5 fluorine atoms.

Molar mass of fluorine, ${\text{m}}_1=18.9984{\text{gmol}}^{-1}$

Molar mass of hydrogen, ${\text{m}}_3=1.0079{\text{gmol}}^{-1}$

Molar mass of carbon, ${\text{m}}_4=12.011{\text{gmol}}^{-1}$

Molar mass of the compound ${\text{C}}_6{\text{HF}}_5$,

$\begin{array}{rcl}{\text{M}}_2& =& \left(6\times 12.011\right)+1.0079+\left(5\times 18.9984\right)\\ & =& 168.0659{\text{gmol}}^{-1}\end{array}$

Mass percentage of fluorine in ${\text{C}}_6{\text{HF}}_5$

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_2}\times 100\\ & =& \frac{18.9984}{168.0659}\times 100\\ & =& 11.3\%\end{array}$

So, the mass percentage of fluorine in ${\text{C}}_6{\text{HF}}_5$ is 11.3%.

Mass percentage of fluorine in BrF

The compound contains one bromine and fluorine

Molar mass of fluorine,${\text{m}}_1=18.9984{\text{gmol}}^{-1}$

Molar mass of hydrogen, localid="1663416955235" ${\text{m}}_5=79.904{\text{gmol}}^{-1}$

Molar mass of the compound $\text{BrF}$

$\begin{array}{rcl}{\text{M}}_3& =& 79.904+18.9984\\ & =& 98.9024{\text{gmol}}^{-1}\end{array}$

Mass percentage of fluorine in $\text{BrF}$

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_3}\times 100\\ & =& \frac{18.9984}{98.9024}\times 100\\ & =& 19.21\%\end{array}$

So, the mass percentage of fluorine in $\text{BrF}$is 19.21%.

Mass percentage of fluorine in  UF6

The compound contains one uranium and fluorine

Molar mass of fluorine,${\text{m}}_1=18.9984{\text{gmol}}^{-1}$

Molar mass of hydrogen, ${\text{m}}_6=238.02891{\text{gmol}}^{-1}$

Molar mass of the compound ${\text{UF}}_6$

$\begin{array}{rcl}{\text{M}}_4& =& 238.0289+18.9984\\ & =& 257.0273{\text{gmol}}^{-1}\end{array}$

Mass percentage of fluorine in role="math" localid="1663416507766" ${\text{UF}}_6$

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{{\text{m}}_1}{{\text{M}}_4}\times 100\\ & =& \frac{18.9984}{257.0273}\times 100\\ & =& 7.39\%\end{array}$

So, the mass percentage of fluorine in${\text{UF}}_6$ is 7.39%.

Comparison of the mass percentage of fluorine in each compounds

The increasing order of mass percentage of fluorine in each compounds is

${\text{UF}}_6<{\text{C}}_6{\text{HF}}_5<\text{BrF}<\text{HF}$