Chapter 5, Problem 145AE

Interpretation Introduction

Interpretation: From the given data, empirical formula and molecular formula of the compound should be determined.

Concept introduction:

• 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 inverse ratio of the square root of the masses of the gases particles. That is,

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

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

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

Answer to Problem 145AE

Answer

$\text{Empirical}\text{formula}\text{and}\text{molecular}\text{formula}\text{of}\text{the}\text{compound=}{C}_2{H}_{3}N$

Explanation of Solution

Explanation

• To find: the empirical formula of the compound

Convert the mass percent to gram

$\begin{array}{l}\text{58}\text{.51\%}\text{Carbon=}\text{58}\text{.51}\text{g}\text{C}\\ \text{7}\text{.37}\text{\%}\text{Hydrogen=7}\text{.37}\text{g}\text{H}\\ \text{34}\text{.12\%}\text{Nitrogen}\text{=34}\text{.12g}\text{N}\end{array}$

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

Here the given gasses are carbon, hydrogen, nitrogen

Percentage composition of carbon and hydrogen are, 58.51% and 7.37% respectively.

Out of 100 g of compound, percent composition of Nitrogen is 34.12%

• To determine: the number of moles of each element

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{carbon=4}\text{.872}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen=}\text{7}\text{.31}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=2}\text{.435}\text{mol}\end{array}$

$\begin{array}{l}\text{ Number}\text{of}\text{moles from its given mass is,}\\ \text{Number}\text{of}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{in}\text{gram}}{\text{Molecular}\text{mass}}\\ \text{Masses}\text{of}\text{carbon,nitrgen,}\text{and}\text{hydrgrogen}\text{are}\text{given}\text{above}\text{.}\\ \text{Molecular}\text{masses}\text{of}\\ \text{Carbon}\text{=}\text{12}\text{.01g}\\ \text{Nitogen}\text{=}\text{14}\text{.01g}\\ \text{Hydrogen}\text{=1}\text{.008g}\\ \text{Number}\text{of}\text{moles}\text{of}\text{carbon}\text{=}\frac{\text{58}\text{.51}\text{g}}{\text{12}\text{.01}\text{g}}\\ \text{=}\text{4}\text{.872}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}\text{=}\frac{\text{7}\text{.37g}}{\text{1}\text{.008g}}\\ \text{=}\text{7}\text{.31}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=}\frac{\text{34}\text{.12g}}{\begin{array}{l}\text{14}\text{.01g}\\ \end{array}}\\ \text{=}\text{2}\text{.435}\text{mol}\\ \end{array}$

• To divide:  the mole value obtained by smallest mole value

$\begin{array}{l}\text{In}\text{the}\text{case}\text{of}\text{carbon,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value=}\text{2}\text{.001}\\ \text{In}\text{the}\text{case}\text{of}\text{hydrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value=}\text{3}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{nitrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{=}\text{1}\text{.00}\end{array}$

From the mole values, smallest mole value is 2.435

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{carbon=4}\text{.872}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen=}\text{7}\text{.31}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=2}\text{.435}\text{mol}\\ \text{In}\text{the}\text{case}\text{of}\text{carbon,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value=}\frac{\text{4}\text{.872}}{\text{2}\text{.435}}\text{=}\text{2}\text{.001}\\ \text{In}\text{the}\text{case}\text{of}\text{hydrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{= }\frac{\text{7}\text{.31}}{\text{2}\text{.435}}\text{=3}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{nitrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{=}\frac{\text{2}\text{.435}}{\text{2}\text{.435}}\text{=1}\text{.00}\end{array}$

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

The empirical formula is $C_2{H}_{3}N$

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

$\begin{array}{l}\text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{of}\text{each}\text{gases}\text{is}\\ \text{C}\text{:}\text{H}\text{:}\text{N=}\text{2}\text{:}\text{3}\text{:1}\end{array}$

So, the empirical formula is $C_2{H}_{3}N$

• To find: molecular formula of the given compound

Determine the molar mass of the given compound.

Molar mass of the given compound is $\text{=}\text{41}\text{.0g/mol}$

$\begin{array}{l}\text{According}\text{to}\text{Thomas}\text{Graham,}\\ \frac{\text{Rate}\text{of}\text{effusion}\text{for}\text{gas}\text{1}}{\text{Rate}\text{of}\text{effusion}\text{for}\text{gas}\text{2}}\text{=}\frac{\sqrt{{\text{M}}_{\text{2}}}}{\sqrt{{\text{M}}_{\text{1}}}}\\ \text{or}\\ \frac{{\text{Rate}}_{\text{1}}}{{\text{Rate}}_{\text{2}}}\text{=}{\left(\frac{{\text{M}}_{\text{2}}}{{\text{M}}_{\text{1}}}\right)}^{\text{1/2}}\\ {\text{M}}_{\text{1}}\text{and}{\text{M}}_{\text{2}}\text{are}\text{the}\text{molar}\text{masses}\text{of}{\text{gas}}_{\text{1}}\text{and}{\text{gas}}_{\text{2}}\\ \text{Let}{\text{gas}}_{\text{1}}\text{=}\text{Helium}\\ {\text{M}}_{\text{1}}\text{=}\text{4}\text{.003}\\ \text{Helium}\text{effuses}\text{through}\text{a}\text{porus}\text{3}\text{.20}\text{times}\text{as}\text{fast}\text{as}\text{the}\text{compound}\text{does}\\ \text{So,}\\ \text{3}\text{.20}\text{=}{\left(\frac{{\text{M}}_{\text{2}}}{\text{4}\text{.003}}\right)}^{\text{1/2}}\\ {\text{M}}_{\text{2}}\text{=}{\left(\text{3}\text{.20}\right)}^{\text{2}}\text{×4}\text{.003}\\ \text{=}\text{10}\text{.24}\text{×}\text{4}\text{.003}\\ \text{=}\text{41}\text{.0}\text{g/mol}\\ \end{array}$

• To determine: the empirical formula mass

Empirical formula mass $\text{=}\text{41}\text{.0g/mol}$

$\begin{array}{l}\text{Empirical}\text{formula}\text{is}{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{N}\text{.}\\ \text{The}\text{empirical}\text{formula}\text{mass}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{N }\approx \text{ 2×12}\text{+ 3×1}\text{+}\text{1×14 = 41}\text{.0}\text{g/mol}\end{array}$

• To divide: the molar mass by empirical formula

$\text{n}\text{= }\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}}=1$

$\begin{array}{l}\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}}\\ \text{Molar mass of the given compound is=}\text{41}\text{.0g/mol}\\ \text{Empirical formula mass=}\text{41}\text{.0g/mol}\\ \text{n}\text{=}\frac{\text{41}\text{.0g/mol}}{\text{41}\text{.0g/mol}}\text{=1}\end{array}$

• To multiply: the empirical formula by n

By multiplying the empirical formula by n,

$\text{molecular}\text{formula}\text{of}\text{the}\text{compound=}{C}_2{H}_{3}N$

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

$\begin{array}{l}\text{n}\text{=}\text{1}\text{and}\text{empirical}\text{of}\text{the}{\text{compound=C}}_{\text{2}}{\text{H}}_{\text{3}}\text{N}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{N}\text{×}\text{1}\text{=}{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{N}\\ \text{So,the}\text{molecular}\text{formula}\text{of}\text{the}\text{compound}\text{is}{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{N}\end{array}$

Conclusion

Conclusion

Empirical formula of the compound is determined from the mass per cent as given below

1. Converted the mass per cent to gram

2. Determined the number of moles of each element

3. Divided the mole value obtained by smallest mole value

4. Wrote empirical formula by mentioning the numbers after writing the symbols of respective elements

Molecular formula of the given compound is determined by the following steps

1. Determined the empirical formula mass

2. Determined the molar mass from rate effusion

3. Divided the molar mass by empirical formula mass

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

1. 4. Multiplied the empirical formula by n