Chapter 5, Problem 170IP

Interpretation Introduction

Interpretation:

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

Concept introduction:

•  

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

• According to Thomas Graham the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

  Graham’s law of effusion,

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

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

• 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

Answer to Problem 170IP

$Molecularformulaofthecompound=C{H}_6{N}_2$

Explanation of Solution

To determine: The mass percent of elements in the unknown compound from the given data

Mass % of carbon

The combustion of compound produces $\text{33}\text{.5}\text{mg}$ of ${\text{CO}}_{\text{2}}$.

$\text{Number of}\text{moles}\text{of}\text{carbon dioxide =}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Therefore,

$\begin{array}{l}\text{Number of}\text{moles}\text{of}\text{carbon dioxide = }\frac{\text{33}\text{.5}\text{mg}{\text{CO}}_{\text{2}}}{\text{44}\text{.01}\text{mg}{\text{CO}}_{\text{2}}}\\ \text{=}\text{0}\text{.761}\text{mol}\end{array}$

Every moles of ${\text{CO}}_{\text{2}}$ contains 1 mole of carbon element.

Therefore,

$\text{Number}\text{of}\text{moles}\text{of}\text{carbon}\text{=0}\text{.761}\text{mol}$

$\begin{array}{l}\text{Mass}\text{of}\text{carbon}\text{inthe}\text{compound}\text{=}\text{Number of}\text{moles}\text{of}\text{carbon×atomic }\text{mass}\text{of}\text{carbon}\\ \text{Mass}\text{of}\text{carbon}\text{inthe}\text{compound=}\text{0}\text{.761}\text{mol×}\text{12}\text{.01}\text{mg}\text{C}\\ \text{=}\text{9}\text{.14}\text{mg}\text{C}\end{array}$

$\begin{array}{l}\text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}\\ \text{Since,35}\text{.0 mg}\text{of the}\text{compound}\text{is combusted to produce}\text{33}\text{.5}\text{g}{\text{CO}}_{\text{2}}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound}\text{=35}\text{.0}m\text{g}\\ \text{Mass}\text{\%}\text{of}\text{carbon}\text{=}\frac{\text{9}\text{.14}\text{mg}\text{C}}{\text{35}\text{.0}m\text{g}}\text{×100}\text{=}\text{26}\text{.1}\text{\%C}\end{array}$

Mass % of Hydrogen

The combustion of compound produces $\text{41}\text{.1}\text{mg}$ of ${\text{H}}_{\text{2}}\text{O}$.

$\text{Number of}\text{moles}\text{of}\text{water =}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Therefore,

$\begin{array}{l}\text{Number of}\text{moles}\text{of}\text{water = }\frac{\text{41}\text{.1mg}}{\text{18}\text{.02}\text{mg}}\\ \text{=}\text{2}\text{.27mol}\end{array}$

Every moles of ${\text{H}}_{\text{2}}\text{O}$ contains 2 moles of hydrogen elements.

Therefore,

$\text{Number}\text{of}\text{moles}\text{of}\text{Hydrogen}\text{= 2}\text{×}\text{2}\text{.28}\text{= }\text{4}\text{.54}\text{mol}$

$\begin{array}{l}\text{Mass}\text{of}\text{Hydrogen}\text{inthe}\text{compound}\text{=}\text{Number of}\text{moles}\text{of}\text{Hydrogen×Atomic}\text{mass}\text{of}\text{Hydrogen}\\ \text{Mass}\text{of}\text{Hydrogen}\text{in}\text{the}\text{compound=}\text{4}\text{.54}\text{mol×}\text{1}\text{.008}\text{mg}\text{H}\\ \text{=}\text{4}\text{.60mg}\text{H}\end{array}$

$\begin{array}{l}\text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}\\ \text{Since,35}\text{.0 mg}\text{of the}\text{compound}\text{is combusted to produce}\text{41}\text{.1}\text{mg}{\text{H}}_{\text{2}}\text{O}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound}\text{=35}\text{.0}\text{g}\\ \text{Mass}\text{\%}\text{of}\text{Hydrogen}\text{=}\frac{\text{4}\text{.60mg}\text{H}}{\text{35}\text{.0}\text{g}}\text{×100}\text{=}\text{13}\text{.1}\text{\%H}\end{array}$

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,

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

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

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

To convert: The mass percent to gram

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

Here the given gasses are carbon, hydrogen and nitrogen.

Hence,

The percentage composition of carbon, hydrogen and nitrogen out of 100 g of compoundare26.1g, 13.1g and 60.8g respectively.

To determine: the number of moles of each element

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

To divide: The mole value obtained by smallest mole value

From the mole values, smallest mole value is 2.17

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

To write: The empirical formula by mentioning the numbers after writing the symbols of respective elements

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 $C{H}_6{N}_2$

To determine: the molar mass of the given compound.

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

To determine: the empirical formula mass.

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

To divide: the molar mass by empirical formula.

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

To find: molecular formula of the given compound

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

The molecular formula of the compound is determined from the given data’s.