Quantitative Analysis and Formula Determination of Organic Compounds

Overview of Quantitative Analysis of Organic Compounds

• Quantitative analysis of an organic compound determines the exact amount, specifically the masses, of each element that constitutes the compound.
• Determining the masses of various elements allows for the derivation of the empirical and molecular formulas of the organic compound.

Test for Carbon and Hydrogen

• Process: A known mass of an organic compound is heated with Copper (II) oxide ($CuO$) in the presence of oxygen, typically at temperatures between $600^{\circ}C$ and $800^{\circ}C$.
• Oxidation Reaction: Both Carbon ($C$) and Hydrogen ($H$) atoms within the organic sample are oxidized to Carbon dioxide ($CO_2$) and Water ($H_2O$) respectively.
• Detection and Collection:
  • Solutions of Potassium hydroxide ($KOH$) and Calcium chloride ($CaCl_2$) are utilized to detect the presence and quantify the amounts of $C$ and $H$.
  • The increase in the mass of the $KOH$ solution represents the mass of $CO_2$ absorbed.
  • The increase in the mass of the $CaCl_2$ solution represents the mass of $H_2O$ absorbed.

Calculations for Hydrogen and Carbon Content

• Mass of Hydrogen:
  • The mass of hydrogen is calculated using the formula:   $\text{mass of hydrogen} = \frac{\text{Atomic mass of H}_2\text{ molecule}}{\text{Molar mass of H}_2\text{O}} \times \text{increase in mass of CaCl}_2\text{ solution}$
• Percentage Mass:
  • $\%\text{ mass of element} = \frac{\text{mass of element}}{\text{mass of starting compound}} \times 100$

Worked Example: Carbon and Hydrogen Analysis

• Problem: During a quantitative analysis of a $1.2\,g$ organic sample, the mass of the $CaCl_2$ solution increased from $2\,g$ to $3.8\,g$, while the $KOH$ solution increased by $2.6\,g$. Calculate the masses and percentages of hydrogen and carbon.
• Solution:
  • (a) Increase in mass of $CaCl_2$ solution: $3.8\,g - 2\,g = 1.8\,g$.
  • Mass of Hydrogen: $\frac{2}{18} \times 1.8\,g = 0.2\,g\text{ of Hydrogen}$.
  • (b) Increase in mass of $KOH$ solution: $2.6\,g$.
  • Mass of Carbon: $\frac{\text{Atomic mass of Carbon}}{\text{Molecular mass of CO}_2} \times \text{increase in mass of KOH}$
  • $\text{Mass of Carbon} = \frac{12}{44} \times 2.6\,g = 0.71\,g\text{ of Carbon}$.
  • (c) Percentage mass of hydrogen: $\frac{0.2}{1.2} \times 100 = 16.7\%\text{ of Hydrogen}$.
  • (d) Percentage mass of carbon: $\frac{0.71}{1.2} \times 100 = 59.2\%\text{ of Carbon}$.

Analysis of Nitrogen: Kjeldahl's Method

• Method: In Kjeldahl's method, an organic compound of known mass is heated with concentrated Sulphuric acid ($H_2SO_4$).
• Reaction: If nitrogen is present in the sample, it is converted into Ammonium sulphate ($(NH_4)_2SO_4$).
• Calculations: The mass and percentage of nitrogen are calculated from the mass of $(NH_4)_2SO_4$ formed.
• Limitations: This method is not applicable to organic compounds containing nitrogen in rings (such as pyridine), nitro groups ($-NO_2$), or diazo groups ($-N=N-$), as these cannot be converted to ammonium sulphate under these specific conditions.

Worked Example: Kjeldahl's Method

• Problem: The starting mass of an organic sample is $5.3\,g$. Treatment with concentrated $H_2SO_4$ yielded $3.4\,g$ of Ammonium sulphate. Calculate the mass of nitrogen present and its percentage composition.
• Solution:
  • Mass of Nitrogen: $\frac{\text{MM of N}_2}{\text{MM of (NH}_4\text{)}_2\text{SO}_4} \times \text{mass of (NH}_4\text{)}_2\text{SO}_4$
  • $\text{Mass of Nitrogen} = \frac{28}{132} \times 3.4\,g = 0.72\,g$
  • Percentage of Nitrogen: $\frac{\text{mass of nitrogen}}{\text{mass of org. cpd}} \times 100$
  • $\%\text{ Nitrogen} = \frac{0.72\,g}{5.3\,g} \times 100 = 13.6\%$

Analysis of Halogens: Carius Method

• Method: The Carius method is used to analyze the amount of halogens and/or sulphur in an organic compound.
• Halogen Procedure: A known mass of the organic compound is heated with fuming trioxonitrate (V) acid (concentrated $HNO_3$) in the presence of excess Silver nitrate ($AgNO_3$) solution inside a sealed Carius tube.
• Reaction: Any halogen present reacts with the silver nitrate to form a silver halide precipitate ($AgX$).
• Process: The silver halide is filtered, washed, dried, and weighed.
• Calculations:
  • $\text{Mass of halogen (X)} = \frac{\text{Atomic mass of X}}{\text{MM of AgX}} \times \text{Mass of AgX obtained}$
  • $\%\text{ of halogen} = \frac{\text{mass of X}}{\text{mass of org. sample}} \times 100$

Worked Example: Carius Method

• Problem: A $5.2\,g$ organic sample was heated with concentrated $HNO_3$ and $AgNO_3$. A white precipitate weighing $1.8\,g$ was formed. (a) Identify the halogen. (b) Determine the percentage composition ($\text{MM of AgCl} = 143.32\,g/mol$).
• Solution:
  • (a) Identification: Since the precipitate is white, it is a chloride ($AgCl$) precipitate.
  • (b) Mass of Halogen: $\frac{\text{Atomic mass of Cl}_2}{\text{MM of AgCl}} \times \text{mass of AgCl obtained}$
  • (Note: Using $71$ for $Cl_2$ atomic mass as per transcript context)
  • $\text{Mass of Cl} = \frac{71}{143.3} \times 1.8 = 0.89\,g\text{ of Cl}$ (approx. $0.9\,g$).
  • Percentage Composition: $\frac{0.9\,g}{5.2\,g} \times 100 = 17.3\%$

Analysis of Sulphur

• Process: A known mass of the organic compound is heated with fuming $HNO_3$ acid or Sodium peroxide in a Carius tube.
• Reaction: Sulphur is oxidized to Sulphuric acid.
• Precipitation: After cooling, Barium chloride ($BaCl_2$) is added to precipitate the sulphur as Barium sulphate ($BaSO_4$).
• Calculations:
  • $\text{Mass of S} = \frac{\text{Atomic mass of S}}{\text{MM of BaSO}_4} \times \text{mass of BaSO}_4\text{ formed}$
  • $\%\text{ of S} = \frac{\text{mass of S}}{\text{mass of org. sample}} \times 100$

Analysis of Oxygen

• Method: The percentage of oxygen is determined indirectly by subtracting the sum of the percentages of all other analyzed elements from $100\%$.
• Formula: $\%\text{ Oxygen} = 100 - (\%\text{ Element A} + \%\text{ Element B} + \dots)$

Worked Example: Multi-Element Analysis

• Problem: $0.3\,g$ of an organic compound gives $0.733\,g$ of $CO_2$ and $0.30\,g$ of water. Calculate the percentage composition of all elements.
• Solution:
  • Mass of C: $\frac{12}{44} \times 0.733\,g = 0.1999\,g$
  • Percentage of C: $\frac{0.1999}{0.30\,g} \times 100 = 66.3\%$
  • Mass of H: $\frac{2}{18} \times 0.30\,g = 0.0333\,g$ (approx. $0.034\,g$ in transcript)
  • Percentage of H: $\frac{0.034\,g}{0.30\,g} \times 100 = 11.3\%$
  • Determination of Oxygen: The sum $66.3\% + 11.3\% = 77.6\%$, which is less than $100\%$, indicating oxygen is present.
  • Percentage of O: $100 - (66.3 + 11.3) = 22.4\%$

Determination of Empirical Formula

• Definition: The empirical formula shows the simplest whole number ratio of atoms present in a compound.
• Steps for Determination:
  1. Start with the mass (in grams) of each element. If percentages are given, assume a $100\,g$ sample or check if they sum to $100\%$. If the sum is less than $100$, the difference is usually oxygen.
  2. Convert the mass of each element to moles using the respective molar masses ($n = m/M$).
  3. Divide each mole value by the smallest calculated mole value in the set.
  4. Approximate the resulting mole ratios to the nearest whole number.

Worked Examples: Empirical Formula

• Example 1: An organic compound contains $0.32\,g$ of $H$ and $0.96\,g$ of $C$. ($C=12, H=1$).

  • Moles of C: $\frac{0.96}{12.01} = 0.08\,moles$
  • Moles of H: $\frac{0.32}{1.008} = 0.32\,moles$
  • Relative number of atoms:
  • $C: \frac{0.08}{0.08} = 1$
  • $H: \frac{0.32}{0.08} = 4$
  • Empirical Formula: $CH_4$

• Example 2: A sample contains $0.430\,g$ of $C$, $0.161\,g$ of $H$, $0.640\,g$ of $O$, and $1.121\,g$ of $N$.

  • Calculate Moles:
  • $C: 0.430 / 12.01 = 0.0358$
  • $H: 0.161 / 1.008 = 0.160$
  • $O: 0.640 / 16 = 0.04$
  • $N: 1.121 / 14 = 0.08$
  • Relative Ratios (Divide by 0.036):
  • $C: 0.036 / 0.036 = 1$
  • $H: 0.160 / 0.036 = 4.44 \approx 4$
  • $O: 0.04 / 0.036 = 1.11 \approx 1$
  • $N: 0.08 / 0.036 = 2.22 \approx 2$
  • Empirical Formula: $C_1H_4O_1N_2$ or $CH_4ON_2$

Determination of Molecular Formula

• Definition: The molecular formula can be the same as the empirical formula or a whole-number multiple of it.
• Calculation: The ratio ($n$) between the molecular formula and the empirical formula is determined by dividing the molecular mass of the compound by the total mass of the elements in the empirical formula ($\text{Empirical Mass}$).
• Formula: $(\text{Empirical Formula})_n = \text{Molecular Mass}$

Worked Example: Molecular Formula Calculation

• Problem: A substance contains $42.8\%\,C$, $2.38\%\,H$, and $16.67\%\,N$. It has a molecular mass of $168$. Calculate the molecular formula.
• Solution:
  • Calculate Oxygen %: $42.8 + 2.38 + 16.67 = 61.85\%$. Oxygen $= 100 - 61.85 = 38.15\%$.
  • Calculate Moles (per 100g):
  • $C: 42.8 / 12.01 = 3.56\,mol$
  • $H: 2.38 / 1.008 = 2.36\,mol$
  • $N: 16.67 / 14.0 = 1.19\,mol$
  • $O: 38.15 / 16.0 = 2.38\,mol$
  • Relative Ratios (Divide by 1.19):
  • $C: 2.99 \approx 3$
  • $H: 1.98 \approx 2$
  • $N: 1$
  • $O: 2$
  • Empirical Formula: $C_3H_2NO_2$
  • Determine Multiplier ($n$):
  • $\text{Empirical Mass} = (3 \times 12) + (2 \times 1) + (14) + (2 \times 16) = 36 + 2 + 14 + 32 = 84 \, g/mol$
  • $84n = 168$
  • $n = 168 / 84 = 2$
  • Molecular Formula: $(C_3H_2NO_2)_2 = C_6H_4N_2O_4$