Chapter 5, Problem 139AE

An organic compound contains C, H, N, and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g CO₂ and 0.0991 g H₂O. A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129). At STP, 27.6 mL of dry N₂ was obtained. In a third experiment, the density of the compound as a gas was found to be 4.02 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound?

Interpretation Introduction

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

Concept introduction:

• 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

• $\text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}$
• According to ideal gas equation,

  $\text{Pressure}\text{×}\text{Volume}\text{=}\frac{\text{Mass}\text{×}\text{R}\text{×}\text{Temperature}}{\text{Molecular}\text{mass}}$

• Ideal gas equation in terms of Density,

$\text{Density}\text{=}\frac{\text{Pressure}\text{×}\text{Molecular mass}}{\text{R}\text{×}\text{Temperature}}$

Answer to Problem 139AE

Answer

$Molecularformulaofthecompound=C_{24}{H}_{42}{N}_2{O}_2$

Explanation of Solution

Explanation

• To determine: The mass percent from the given data

$\begin{array}{l}\text{73}\text{.78\%}\text{Carbon}\\ \text{10}\text{.9\%}\text{Hydrogen}\\ \text{7}\text{.14\%}\text{Nitrogen}\\ \text{8}\text{.2\% Oxygen}\end{array}$

Mass % of carbon

$\begin{array}{l}\text{Since,}\text{complete}\text{combustion}\text{of}\text{0}\text{.1023}\text{g}\text{of}\text{the}\text{compound}\text{produced}\text{0}\text{.2766mg}{\text{CO}}_{\text{2}}\text{and}\text{0}\text{.0991g}{\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,0}\text{.1023g}\text{of the}\text{compound}\text{is combusted to produce}\text{0}\text{.2766g}{\text{CO}}_{\text{2}}\\ \text{Given}\text{mass}\text{=0}\text{.2766}\text{g}\\ \text{Molecular}\text{mass}\text{of}{\text{CO}}_{\text{2}}\text{=}\text{44}\text{.01}\text{g}\\ \text{Number of}\text{moles}\text{of}\text{carbon dioxide = }\frac{\text{0}\text{.2766g}{\text{CO}}_{\text{2}}}{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}\\ \text{=}\text{0}\text{.00628}\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{.00628mol}\\ \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{.00628}\text{mol×}\text{12}\text{.01}\text{g}\text{C}\\ \text{=}\text{0}\text{.0757mg}\text{C =}\text{7}{\text{.548 ×10}}^{\text{-2}}\text{gC}\\ \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,0}\text{.1023 g}\text{of the}\text{compound}\text{is combusted to produce}\text{0}\text{.2766g}{\text{CO}}_{\text{2}}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound}\text{=0}\text{.1023}\text{g}\\ \text{Mass}\text{\%}\text{of}\text{carbon}\text{=}\frac{\text{7}{\text{.548 ×10}}^{\text{-2}}\text{gC}}{\text{0}\text{.1023}\text{g}}\text{×100}\text{=}\text{73}\text{.78}\text{\%C}\end{array}$

Mass % of Hydrogen

$\begin{array}{l}\text{Since,}\text{complete}\text{combustion}\text{of}\text{0}\text{.1023mg}\text{of}\text{the}\text{compound}\text{produced}\text{0}\text{.2766g}{\text{CO}}_{\text{2}}\text{and}\text{0}\text{.0991g}{\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,0}\text{.1023g}\text{of the}\text{compound}\text{is combusted to produce}\text{0}\text{.0991g}{\text{H}}_{\text{2}}\text{O}\\ \text{Given}\text{mass}\text{=0}\text{.0991g}\\ \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{0}\text{.0991g}}{\text{18}\text{.02}\text{g}}\\ \text{=}\text{0}\text{.00550mol}\\ \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{0}\text{.00550}\text{= }\text{0}\text{.011}\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{0}\text{.011mol×}\text{1}\text{.008}\text{mg}\text{H}\\ \text{=}\text{0}\text{.011g}\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,0}\text{.1023g}\text{of the}\text{compound}\text{is combusted to produce}\text{0}\text{.0991g}{\text{H}}_{\text{2}}\text{O}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound}\text{=0}\text{.1023}\text{g}\\ \text{Mass}\text{\%}\text{of}\text{Hydrogen}\text{=}\frac{\text{0}\text{.011g}\text{H}}{\text{0}\text{.1023}\text{g}}\text{×100}\text{=}\text{10}\text{.9\%H}\end{array}$

Mass % of Nitrogen

The pressure of ${\text{N}}_{\text{2}}$ is1.0 atm

The volume of ${\text{N}}_{\text{2}}$ gas is $\text{27}\text{.6}\text{mL}\text{=27}\text{.6}\text{×}{\text{10}}^{\text{-3}}\text{L}$

The temperature of ${\text{N}}_{\text{2}}$ gas is 273K.

The equation for finding number of moles of a substance,

According to ideal gas equation,

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

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{1}\text{.0}\text{atm}\text{×}\text{27}\text{.6}{\text{×10}}^{\text{-3}}\text{L}}{\text{0}\text{.08206}\text{L}\text{atm/K}\text{mol}\text{×}\text{273}\text{K}}\\ \text{=}\text{1}\text{.23}\text{×}{\text{10}}^{\text{-3}}\text{mol}\end{array}$

$\begin{array}{l}\\ \text{Number}\text{of}\text{moles}\text{of Nitroogen}\text{=}\text{1}{\text{.23×10}}^{\text{-3}}\text{mol}\\ \text{Mass}\text{of Nitrogen}\text{in the}\text{compound}\text{=}\text{Number of}\text{moles}\text{of}\text{Nitrogen×Molecular}\text{mass}\text{of}\text{nitrogen}\\ \text{Mass}\text{of}\text{Nitrogen}\text{in}\text{the}\text{compound=}\text{1}{\text{.23×10}}^{\text{-3}}\text{mol×}\text{28}\text{.02}\text{g}\\ \text{=}\text{3}\text{.45}\text{×}{\text{10}}^{\text{-2}}\text{g}\\ \text{Mass}\text{\%}\text{of}\text{an}\text{element}\text{=}\frac{\text{Mass}\text{of}\text{that}\text{element}\text{in}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}\text{×100}\\ \text{Molar}\text{mass}\text{of}\text{the}\text{compound=0}\text{.4831}\text{g}\\ \text{Mass}\text{\%}\text{of}\text{Nitrogen}\text{=}\frac{\text{3}\text{.45}\text{×}{\text{10}}^{\text{-2}}\text{g}}{\text{0}\text{.4831}\text{g}}\text{×100}\text{=}\text{7}\text{.14\%N}\\ Mass\%ofoxgen\text{=}\text{100}\text{.0-}\left(\text{73}\text{.78+10}\text{.9+7}\text{.14}\right)\text{=8}\text{.2\%O}\end{array}$

• To convert: The mass percent to gram

$\begin{array}{l}\text{73}\text{.78}\text{\%}\text{Carbon}\text{= 73}\text{.78}\text{g}\\ \text{10}\text{.9\%}\text{Hydrogen }\text{=10}\text{.9 g}\\ \text{7}\text{.14\%}\text{Nitrogen = 7}\text{.14}\text{g}\\ \text{8}\text{.2\% Oxygen}\text{=}\text{8}\text{.2 }\text{g}\end{array}$

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, nitrogen and oxygen out of 100 g of compound

Are73.78%, 10.9%, 7.14%and 8.2%   respectively.

• To determine: the number of moles of each element

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{carbon=6}\text{.143}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen=}\text{10}\text{.8}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=0}\text{.510mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{oxygen=0}\text{.51}\text{mol}\end{array}$

$\begin{array}{l}\text{ Number}\text{of}\text{moles from its given mass is,}\\ \text{Number}\text{of}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{in}\text{gram}}{\text{Molecular}\text{mass}}\\ \text{Masses}\text{of}\text{carbon,nitrgen,}\text{hydrgrogen and oxygen}\text{are}\text{given}\text{above}\text{.}\\ \text{Molecular}\text{masses}\text{of}\\ \text{Carbon}\text{=}\text{73}\text{.78g}\\ \text{Nitogen}\text{=}\text{7}\text{.14g}\\ \text{Hydrogen}\text{=10}\text{.9g}\\ \text{Oxygen =}\text{8}\text{.2 g}\\ \text{Number}\text{of}\text{moles}\text{of}\text{carbon}\text{=}\frac{\text{73}\text{.78}\text{g}}{\text{12}\text{.01}\text{g}}\\ \text{=}\text{6}\text{.143}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}\text{=}\frac{\text{10}\text{.9g}}{\text{1}\text{.008g}}\\ \text{=}\text{10}\text{.8mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=}\frac{\text{7}\text{.14g}}{\begin{array}{l}\text{14}\text{.01g}\\ \end{array}}\\ \text{=}\text{0}\text{.510mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{oxygen}\text{=}\frac{\text{8}\text{.2g}}{\text{16}\text{.00}}\\ \text{=}\text{0}\text{.51}\text{mol}\\ \end{array}$

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

$\begin{array}{l}\text{In}\text{the}\text{case}\text{of}\text{carbon,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value=}\text{12}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{hydrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value=}\text{21}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{nitrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{=}\text{1}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{oxygen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{=1}\text{.00}\end{array}$

From the mole values, smallest mole value is 0.51

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{carbon=6}\text{.143}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{hydrogen=}\text{10}\text{.8}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{nitrogen=0}\text{.510}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{oxygen}\text{=0}\text{.51}\text{mol}\\ \text{In}\text{the}\text{case}\text{of}\text{carbon,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value=}\frac{\text{6}\text{.143}}{\text{0}\text{.51}}\text{=}\text{12}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{hydrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{= }\frac{\text{10}\text{.8}}{\text{0}\text{.51}}\text{=21}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{nitrogen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{=}\frac{\text{0}\text{.51}}{\text{0}\text{.51}}\text{=1}\text{.00}\\ \text{In}\text{the}\text{case}\text{of}\text{oxygen,}\\ \text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{=}\frac{\text{0}\text{.51}}{\text{0}\text{.51}}\text{=1}\text{.00}\end{array}$

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

The empirical formula is ${\text{C}}_{\text{12}}{\text{H}}_{\text{21}}\text{NO}$

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

$\begin{array}{l}\text{Ratio}\text{of}\text{mole}\text{value}\text{to}\text{smallest}\text{mole}\text{value}\text{of}\text{each}\text{gases}\text{is}\\ \text{C}\text{:}\text{H}\text{:}\text{N:O=}\text{12}\text{:}\text{21:1:1}\end{array}$

So, the empirical formula is ${\text{C}}_{\text{12}}{\text{H}}_{\text{21}}\text{NO}$

• To find: molecular formula of the given compound

Determine the molar mass of the given compound.

Molar mass of the given compound is $\text{=}392\text{g/mol}$

$\begin{array}{l}\text{Ideal gasequation in terms of Density, }\\ \text{Density}\text{=}\frac{\text{Pressure}\text{×}\text{Molecular mass}}{\text{R}\text{×}\text{Temperature}}\\ \text{Molecular mass}\text{=}\frac{\text{Density×}\text{R}\text{×}\text{Temperature}}{\text{Pressure}}\\ \text{Here,}\\ \text{Density=}\text{4}\text{.02g/L}\\ \text{R}\text{=}\frac{\text{0}\text{.08206L}\text{atm}}{\text{K}\text{mol}}\\ \text{At}\text{STP}\\ \text{T}\text{=127°C}\text{= 400K}\text{Since,°C+273}\text{=K,127°C+273=400K}\\ \text{P}\text{=}\text{256}\text{torr=0}\text{.33}\text{atm}\text{Since,1}\text{atm=}\text{760}\text{torr}\\ \text{256}\text{torr×}\frac{\text{1}\text{atm}}{\text{760}\text{torr}}\text{=}\text{0}\text{.33}\text{atm}\\ \text{By}\text{adding}\text{the}\text{given}\text{values}\text{tothe}\text{equation,}\\ \text{Molecular mass}\text{=}\frac{\text{4}\text{.02g/L×}\frac{\text{0}\text{.08206L}\text{atm}}{\text{K}\text{mol}}\text{×400}\text{K}}{\text{0}\text{.33}\text{atm}}\\ \text{=}\text{392}\text{g/}\text{mol}\end{array}$

• To determine: the empirical formula mass

Empirical formula mass $\text{=}195\text{g/mol}$

$\begin{array}{l}\text{Empirical}\text{formula}\text{is}{\text{C}}_{12}{\text{H}}_{21}\text{NO}\text{.}\\ \text{The}\text{empirical}\text{formula}\text{mass}\text{of}{\text{C}}_{12}{\text{H}}_{21}\text{NO }\approx \text{ 1}2\times \text{12}\text{+ 1}\times 21\text{+}1\text{×14+1}\times \text{16 = 195}\text{g/mol}\end{array}$

• To divide: the molar mass by empirical formula

$\text{n}\text{= }\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}}=2$

$\begin{array}{l}\text{n}\text{=}\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}}\\ \text{Molar mass of the given compound is=}\text{392g/mol}\\ \text{Empirical formula mass=}\text{195g/mol}\\ \text{n}\text{=}\frac{\text{392g/mol}}{\text{195g/mol}}\text{=2}\end{array}$

• To multiply: the empirical formula by n

By multiplying the empirical formula by n,

$\text{molecular}\text{formula}\text{of}\text{the}{\text{compound=C}}_{\text{24}}{\text{H}}_{\text{42}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}$

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

$\begin{array}{l}\text{n}\text{=}\text{2}\text{and}\text{empirical}\text{of}\text{the}{\text{compound=C}}_{\text{12}}{\text{H}}_{\text{21}}\text{NO}\\ {\left({\text{C}}_{\text{12}}{\text{H}}_{\text{21}}\text{NO}\right)}_{}\text{×}\text{2}\text{=}{\text{C}}_{\text{24}}{\text{H}}_{\text{42}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\\ \text{So,the}\text{molecular}\text{formula}\text{of}\text{the}\text{compound}\text{is}{\text{C}}_{\text{24}}{\text{H}}_{\text{42}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\end{array}$

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