Chapter 5, Problem 130AE

Interpretation Introduction

Interpretation:

From the given data, molecular formula of the compound should be determined.

Concept introduction:

Effusion is used to describe the passage of a gas through a tiny particle into an evacuated chamber.

The rate of effusion is the measure speed at which the gas is transferred to the chamber.  According to Thomas Graham the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

The relative rate of effusion of two gases at the same temperature and pressure are the inverse ratio of the square root of the masses of the gases particles. That is,

$\begin{array}{l}\frac{Rateofeffusionforgas1}{Rateofeffusionforgas2}=\frac{\sqrt{M_2}}{\sqrt{M_1}}\\ or\\ \frac{Rat{e}_1}{Rat{e}_2}={\left(\frac{M_2}{M_1}\right)}^{1/2}\end{array}$

${\text{M}}_{\text{1}}\text{and}{\text{M}}_{\text{2}}\text{are}\text{the}\text{molar}\text{masses}\text{of}\text{twogases}$

This equation is known as Graham’s law of effusion.

• Empirical formula can be determined from the mass per cent as given below

1. 1. 1. Convert the mass per cent to gram
   2. 2. Determine the number of moles of each elements
   3. 3. Divide the mole value obtained by smallest mole value
   4. 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements

• Molecular formula can be write by the following steps

1. 1. 1. Determine the empirical formula mass
   2. 2. Divide the molar mass by empirical formula mass

      $\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}}\text{=n}$

   3. 3. Multiply the empirical formula by n

$\text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}$

To find: The molecular formula of a compound from the given data

Answer to Problem 130AE

$\text{Molecular}\text{formula}\text{of}\text{the}\text{compound}\text{=}{\text{CH}}_{\text{6}}{\text{N}}_{\text{2}}$

Explanation of Solution

The mass percent from the given data is,

$\begin{array}{l}\text{26}\text{.1\%}\text{Carbon}\\ \text{13}\text{.1\%}\text{Hydrogen}\\ \text{61}\text{.0\%}\text{Nitrogen}\end{array}$

Mass % of carbon

$\begin{array}{l}\text{Since,}\text{complete}\text{combustion}\text{of}\text{35}\text{.0}\text{mg}\text{of}\text{the}\text{compound}\text{produced}\text{33}\text{.5}\text{mg}{\text{CO}}_{\text{2}}\text{and}\text{41}\text{.1}\text{mg}{\text{H}}_{\text{2}}\text{O}\\ \text{Number of}\text{moles}\text{of}\text{carbon dioxide =}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ \text{Since,35}\text{.0 mg}\text{of the}\text{compound}\text{is combusted to produce}\text{33}\text{.5}\text{g}{\text{CO}}_{\text{2}}\\ \text{Given}\text{mass}\text{=33}\text{.5}\text{mg}\\ \text{Molecular}\text{mass}\text{of}{\text{CO}}_{\text{2}}\text{=}\text{44}\text{.01}\text{mg}\\ \text{Number of}\text{moles}\text{of}\text{carbon dioxide = }\frac{\text{33}\text{.5}\text{mg}{\text{CO}}_{\text{2}}}{\text{44}\text{.01}\text{mg}{\text{CO}}_{\text{2}}}\\ \text{=}\text{0}\text{.761}\text{mol}\\ \text{Every}\text{mole}\text{of}{\text{CO}}_{\text{2}}\text{will}\text{contains}\text{1}\text{mole}\text{of}\text{carbon}\text{and2}\text{moles}\text{of}\text{oxygen}\text{.}\\ \text{Since,every}\text{mole}\text{of}{\text{CO}}_{\text{2}}\text{contains}\text{one}\text{mole}\text{ofcarbon}\\ \text{Number}\text{of}\text{moles}\text{of}\text{carbon}\text{=0}\text{.761}\text{mol}\\ \text{Mass}\text{of}\text{carbon}\text{inthe}\text{compound}\text{=}\text{Number of}\text{moles}\text{of}\text{carbon×atomic }\text{mass}\text{of}\text{carbon}\\ \text{Mass}\text{of}\text{carbon}\text{inthe}\text{compound=}\text{0}\text{.761}\text{mol×}\text{12}\text{.01}\text{mg}\text{C}\\ \text{=}\text{9}\text{.14}\text{mg}\text{C}\\ \text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}\\ \text{Since,35}\text{.0 mg}\text{of the}\text{compound}\text{is combusted to produce}\text{33}\text{.5}\text{g}{\text{CO}}_{\text{2}}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound}\text{=35}\text{.0}m\text{g}\\ \text{Mass}\text{\%}\text{of}\text{carbon}\text{=}\frac{\text{9}\text{.14}\text{mg}\text{C}}{\text{35}\text{.0}m\text{g}}\text{×100}\text{=}\text{26}\text{.1}\text{\%C}\end{array}$

Mass % of Hydrogen

$\begin{array}{l}\text{Since,}\text{complete}\text{combustion}\text{of}\text{35}\text{.0}\text{mg}\text{of}\text{the}\text{compound}\text{produced}\text{33}\text{.5}\text{mg}{\text{CO}}_{\text{2}}\text{and}\text{41}\text{.1}\text{mg}{\text{H}}_{\text{2}}\text{O}\\ \text{Number of}\text{moles}\text{of}\text{water =}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ \text{Since,35}\text{.0 mg}\text{of the}\text{compound}\text{is combusted to produce}\text{41}\text{.1}\text{mg}{\text{H}}_{\text{2}}\text{O}\\ \text{Given}\text{mass}\text{=41}\text{.1}\text{mg}\\ \text{Molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}\text{= 18}\text{.02 mg}\\ \text{Number of}\text{moles}\text{of}\text{water = }\frac{\text{41}\text{.1mg}}{\text{18}\text{.02}\text{mg}}\\ \text{=}\text{2}\text{.27mol}\\ \text{Every}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}\text{contains}\text{2}\text{moles}\text{of}\text{Hydrogen}\text{and1}\text{mole}\text{of}\text{oxygen}\text{.}\\ \text{Number}\text{of}\text{moles}\text{of}\text{Hydrogen}\text{= 2}\text{×}\text{2}\text{.28}\text{= }\text{4}\text{.54}\text{mol}\\ \text{Mass}\text{of}\text{Hydrogen}\text{inthe}\text{compound}\text{=}\text{Number of}\text{moles}\text{of}\text{Hydrogen×Atomic}\text{mass}\text{of}\text{Hydrogen}\\ \text{Mass}\text{of}\text{Hydrogen}\text{in}\text{the}\text{compound=}\text{4}\text{.54}\text{mol×}\text{1}\text{.008}\text{mg}\text{H}\\ \text{=}\text{4}\text{.60mg}\text{H}\\ \text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}\\ \text{Since,35}\text{.0 mg}\text{of the}\text{compound}\text{is combusted to produce}\text{41}\text{.1}\text{mg}{\text{H}}_{\text{2}}\text{O}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound}\text{=35}\text{.0}\text{g}\\ \text{Mass}\text{\%}\text{of}\text{Hydrogen}\text{=}\frac{\text{4}\text{.60mg}\text{H}}{\text{35}\text{.0}\text{g}}\text{×100}\text{=}\text{13}\text{.1}\text{\%H}\end{array}$

Mass % of Nitrogen

The Partial pressure of ${\text{N}}_{\text{2}}$ is $\text{740torr}\text{=}\text{740}\text{torr}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{torr}}\text{=}\text{0}\text{.973}\text{atm}$.

The volume of ${\text{N}}_{\text{2}}$ gas is given as $\text{35}\text{.6}\text{mL}\text{=}\text{35}\text{.6mL}\text{×}\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{=}35.6\times {10}^{-3}\text{L}$.

The temperature of ${\text{N}}_{\text{2}}$ gas is given as ${\text{25}}^{\text{o}}\text{C}\text{=}\text{(25+273)K}\text{=}\text{298}\text{K}$.

The equation for finding number of moles of a substance,

According to ideal gas equation,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{0}\text{.973}\text{atm}\text{×}35.6\times {10}^{-3}\text{L}}{\text{0}\text{.08206}\text{L}\text{atm/K}\text{mol}\text{×}298\text{K}}\\ =1.\text{42}\text{×}{\text{10}}^{\text{-3}}\text{mol}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of Nitroogen}\text{=}\text{1}{\text{.42×10}}^{\text{-3}}\text{mol}\\ \text{Mass}\text{of Nitrogen}\text{in the}\text{compound}\text{=}\text{Number of}\text{moles}\text{of}\text{Nitrogen×Molecular}\text{mass}\text{of}\text{nitrogen}\\ \text{Mass}\text{of}\text{Nitrogen}\text{in}\text{the}\text{compound=}\text{1}{\text{.42×10}}^{\text{-3}}\text{mol×}\text{28}\text{.02}\text{g}\\ \text{=}\text{3}\text{.98}\text{×}{\text{10}}^{\text{-2}}\text{=39}\text{.8}\text{mg}\\ \text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound=65}\text{.2}\text{mg}\\ \text{Mass}\text{\%}\text{of}\text{Nitrogen}\text{=}\frac{\text{39}\text{.8}\text{mg}}{\text{65}\text{.2}\text{mg}}\text{×100}\text{=}\text{61}\text{.0}\text{\%N}\\ \text{Mass}\text{\%}\text{of}\text{Nitrogen}\text{=}\text{100}\text{.0-}\left(\text{26}\text{.1+13}\text{.1}\right)\text{=60}\text{.8\%}\end{array}$

The conversion of mass percent to gram is,

$\begin{array}{l}\text{26}\text{.1\%}\text{Carbon}\text{= 26}\text{.1}\text{g}\\ \text{13}\text{.1\%}\text{Hydrogen }\text{=13}\text{.1 g}\\ \text{60}\text{.8\%}\text{Nitrogen = 60}\text{.8}\text{g}\end{array}$

First step to determine the empirical formula is to convert the mass percent to gram.

Here the given gasses are carbon, hydrogen

Percentage composition of carbon, hydrogen and nitrogen out of 100 g of compound.  Are 26.1%, 13.1% and 60.8% respectively.

The number of moles of each element is

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{carbon=2}\text{.17}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen=}13.01\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=4}\text{.34mol}\end{array}$

$\begin{array}{l}\text{ Number}\text{of}\text{moles from its given mass is,}\\ \text{Number}\text{of}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{in}\text{gram}}{\text{Molecular}\text{mass}}\\ \text{Masses}\text{of}\text{carbon,nitrgen,}\text{and}\text{hydrgrogen}\text{are}\text{given}\text{above}\text{.}\\ \text{Molecular}\text{masses}\text{of}\\ \text{Carbon}\text{=}\text{12}\text{.01g}\\ \text{Nitogen}\text{=}\text{14}\text{.01g}\\ \text{Hydrogen}\text{=1}\text{.008g}\\ \text{Number}\text{of}\text{moles}\text{of}\text{carbon}\text{=}\frac{\text{26}\text{.1}\text{g}}{\text{12}\text{.01}\text{g}}\\ \text{=}2.17\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}\text{=}\frac{\text{13}\text{.1g}}{\text{1}\text{.008g}}\\ \text{=}13.00\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}\text{=}\frac{\text{60}\text{.8g}}{\begin{array}{l}\text{14}\text{.01g}\\ \text{=}4.34\text{mol}\end{array}}\end{array}$

From the mole values, smallest mole value is 2.17

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{carbon=2}\text{.17}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen=}13.0\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=4}\text{.34}\text{mol}\\ \text{In}\text{the}\text{case}\text{of}\text{carbon,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value=}\frac{2.17}{\text{2}\text{.17}}\text{=}\text{1}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{hydrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{= }\frac{13.0}{2.17}\text{=6}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{nitrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{=}\frac{4.34}{2.17}\text{=2}\text{.00}\end{array}$

Empirical formula can be determined by dividing the mole value of each gas to smallest mole value gas.

$\begin{array}{l}\text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{of}\text{each}\text{gases}\text{is}\\ \text{C}\text{:}\text{H}\text{:}\text{N=}1\text{:}6\text{:2}\end{array}$

So, the empirical formula is $C{H}_6{N}_2$

$\begin{array}{l}\text{According}\text{to}\text{Thomas}\text{Graham,}\\ \frac{\text{Rate}\text{of}\text{effusion}\text{for}\text{gas}\text{1}}{\text{Rate}\text{of}\text{effusion}\text{for}\text{gas}\text{2}}\text{=}\frac{\sqrt{{\text{M}}_{\text{2}}}}{\sqrt{{\text{M}}_{\text{1}}}}\\ \text{or}\\ \frac{{\text{Rate}}_{\text{1}}}{{\text{Rate}}_{\text{2}}}\text{=}{\left(\frac{{\text{M}}_{\text{2}}}{{\text{M}}_{\text{1}}}\right)}^{\text{1/2}}\\ {\text{M}}_{\text{1}}\text{and}{\text{M}}_{\text{2}}\text{are}\text{the}\text{molar}\text{masses}\text{of}{\text{gas}}_{\text{1}}\text{and}{\text{gas}}_{\text{2}}\\ \text{Rate}\text{of}\text{effusion}\text{for}\text{gas}\text{1}\text{=26}\text{.4}\\ \text{Rate}\text{of}\text{effusion}\text{for}\text{gas}\text{2}\text{=}\text{24}\text{.6}\\ \text{Here,}\\ {\text{Gas}}_{\text{1}}\text{=}\text{Argon}\\ {\text{M}}_{\text{1}}\text{=39}\text{.95}\\ \frac{{\text{Rate}}_{\text{1}}}{{\text{Rate}}_{\text{2}}}\text{=}{\left(\frac{\text{M}}{\text{39}\text{.95}}\right)}^{\text{1/2}}\text{=}\frac{\text{26}\text{.4}}{\text{24}\text{.6}}\text{=}\text{1}\text{.07}\\ \text{M}\text{=}{\left(\text{1}\text{.07}\right)}^{\text{2}}\text{×39}\text{.95=45}\text{.7}\text{g/mol}\end{array}$

$\begin{array}{l}\text{Empirical}\text{formula}\text{is}{\text{CH}}_6{\text{N}}_2\text{.}\\ \text{The}\text{empirical}\text{formula}\text{mass}\text{of}{\text{CH}}_6{\text{N}}_2\approx \text{ 1}\times \text{12}\text{+ 1}\times \text{6}\text{+}2\text{×14 = 46}\text{.0}\text{g/mol}\end{array}$

$\begin{array}{l}\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}}\\ \text{Molar mass of the given compound is=}\text{45}\text{.7g/mol}\\ \text{Empirical formula mass=}\text{46}\text{.0g/mol}\\ \text{n}\text{=}\frac{\text{45}\text{.7g/mol}}{\text{46}\text{.0g/mol}}\text{=1}\end{array}$

By multiplying the empirical formula by n molecular formula of the compound should obtain.

$\begin{array}{l}\text{n}\text{=}\text{1}\text{and}\text{empirical}\text{of}\text{the}{\text{compound=CH}}_{\text{6}}{\text{N}}_{\text{2}}\\ {\left({\text{CH}}_{\text{6}}{\text{N}}_{\text{2}}\right)}_{}\text{×}\text{1}\text{=}{\text{CH}}_{\text{6}}{\text{N}}_{\text{2}}\\ \text{So,the}\text{molecular}\text{formula}\text{of}\text{the}\text{compound}\text{is}{\text{CH}}_{\text{6}}{\text{N}}_{\text{2}}\end{array}$

Conclusion

From the given data, molecular formula of the compound was determined.