Organic Chemistry: Elemental Analysis Part 2 Notes

Identification of Other Elements: Mineralization Process (Sodium Fusion Test)

• To identify elements other than carbon and hydrogen in an organic compound, a mineralization process called the sodium fusion test is used.
• Procedure:
  1. An organic substance is placed in a test tube with a fragment of sodium.
  2. The mixture is heated to redness while being mixed with an agitator.
  3. The hot tube is immersed in a beaker containing distilled water, causing it to break.
  4. The content of the beaker is heated to boiling point, and then filtered.

Identification of Sulfur

• Mineralization: The sample is treated with molten sodium metal at high temperature.
• If sulfur is present, it converts to sulfide ions ($S^{2-}$).
• Reaction: Lead ions ($Pb^{2+}$) are added to the filtrate, reacting with sulfide ions to form a black precipitate of lead sulfide ($PbS$).
• Equation: $Pb^{2+} + S^{2-} \rightarrow PbS(s)$
• Conclusion: A black precipitate of $PbS$ confirms the presence of sulfur.

Identification of Nitrogen

• Mineralization: The sample is treated with molten sodium metal at high temperature.
• If nitrogen is present, it converts to cyanide ions ($CN^{-}$).
• Reaction 1: Ferrous ions ($Fe^{2+}$) are added to the filtrate, reacting with cyanide ions to form ferrocyanide ions (colorless).
• Equation: $6CN^{-} + Fe^{2+} \rightarrow Fe(CN)_6^{4-}$
• Reaction 2: Ferric ions ($Fe^{3+}$) are then added, reacting with ferrocyanide ions to form a Prussian blue complex (ferric ferrocyanide).
• Conclusion: The formation of a Prussian blue complex confirms the presence of nitrogen.

Identification of Halogens (X)

• Mineralization: The sample is treated with molten sodium metal at high temperature.
• If a halogen is present, it converts to halide ions ($X^{-}$).
• Reaction: Silver ions ($Ag^{+}$) are added to the filtrate, reacting with halide ions to form a silver halide precipitate ($AgX$).
• Equation: $X^{-} + Ag^{+} \rightarrow AgX$
• Identification by Color: The halogen is identified based on the precipitate's color.
  • Chlorine (Cl): White precipitate of $AgCl$ that darkens in light: $Ag^{+} + Cl^{-} \rightarrow AgCl$
  • Bromine (Br): Cream-colored precipitate of $AgBr$: $Ag^{+} + Br^{-} \rightarrow AgBr$
  • Iodine (I): Yellow precipitate of $AgI$: $Ag^{+} + I^{-} \rightarrow AgI$

Quantitative Analysis

• Determines the quantity of elements in a compound to find its molecular formula.
• A common method is complete combustion of a known mass of the compound in a micro analyzer.

Experimental Determination of Masses

• The mass of $CO<em>2$ produced is determined by the increase in mass of potassium hydroxide ($KOH$) tubes, as $KOH$ absorbs $CO</em>2$.
• The mass of $H<em>2O$ produced is determined by the increase in mass of sulfuric acid ($H</em>2SO<em>4$) tubes, as $H</em>2SO<em>4$ absorbs $H</em>2O$.

Quantitative Determination of Carbon

• The quantity of carbon is calculated using the complete combustion reaction: $C + O<em>2 \rightarrow CO</em>2$
• Stoichiometric Relation: The number of moles of carbon equals the number of moles of carbon dioxide: $n(C) = n(CO_2)$
• Mass Percentage Calculation:
  • $$

Quantitative Determination of Hydrogen

• The quantity of hydrogen is calculated using the complete combustion reaction: $H<em>2 + \frac{1}{2} O</em>2 \rightarrow H_2O$
• Stoichiometric Relation: The number of moles of hydrogen equals the number of moles of water: $n(H<em>2) = n(H</em>2O)$
• Mass Percentage Calculation:
  • $$

Application Example

• Complete combustion of m = 0.6g of compound A (containing carbon, hydrogen, and oxygen) produces 0.88g of $CO<em>2$ and 0.36g of $H</em>2O$.
• The task is to calculate the masses and mass percentages of each element in compound A.