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Difficulty: Very hardAlkanoic Acids, Esterification, Saponification, Fats, and Oils

A triglyceride derived from a single saturated alkanoic acid undergoes complete saponification with excess aqueous NaOH\text{NaOH}, yielding glycerol and a sodium soap salt. When the isolated soap salt is acidified with excess dilute HCl\text{HCl}, a pure saturated alkanoic acid XX is liberated. Neutralization of 10.24 g10.24\text{ g} of acid XX requires exactly 40.0 cm340.0\text{ cm}^3 of a 1.00 mol dm−31.00\text{ mol dm}^{-3} NaOH\text{NaOH} solution. What is the correct IUPAC name of the alkanoic acid XX? (Atomic masses: H=1\text{H} = 1, C=12\text{C} = 12, O=16\text{O} = 16)

  1. Hexadecanoic acidAnswer
  2. B
    Tetradecanoic acid
  3. C
    Pentadecanoic acid
  4. D
    Octadecanoic acid

Answer

Hexadecanoic acid
The acid XX reacts with NaOH\text{NaOH} in a 1:1 stoichiometry. 0.0400 mol0.0400\text{ mol} of NaOH\text{NaOH} neutralizes 0.0400 mol0.0400\text{ mol} of XX, giving a molar mass of 256 g mol−1256\text{ g mol}^{-1}. Setting the general formula for saturated alkanoic acids CnH2nO2\text{C}_n\text{H}_{2n}\text{O}_2 equal to 256256 yields 14n+32=25614n + 32 = 256, which solves to n=16n = 16. The 16-carbon saturated carboxylic acid is Hexadecanoic acid.

Step-by-Step Solution

1
Calculate the number of moles of NaOH used in the neutralization reaction
Moles of NaOH=Concentration×Volume in dm3=1.00 mol dm−3×40.01000 dm3=0.0400 mol\text{Moles of NaOH} = \text{Concentration} \times \text{Volume in dm}^3 = 1.00\text{ mol dm}^{-3} \times \frac{40.0}{1000}\text{ dm}^3 = 0.0400\text{ mol}
Neutralization uses volume and molarity to determine mole quantity.
2
Determine the molar mass of the monocarboxylic acid X
Since monocarboxylic acid reacts with NaOH in a 1:1 mole ratio (R-COOH+NaOH→R-COONa+H2O\text{R-COOH} + \text{NaOH} \rightarrow \text{R-COONa} + \text{H}_2\text{O}), moles of X=0.0400 molX = 0.0400\text{ mol}. Thus, Molar mass of X=10.24 g0.0400 mol=256 g mol−1\text{Molar mass of } X = \frac{10.24\text{ g}}{0.0400\text{ mol}} = 256\text{ g mol}^{-1}.
Molar mass is the mass divided by the amount in moles.
3
Use the general formula for a saturated alkanoic acid to find the number of carbon atoms (n)
The general formula for a saturated monocarboxylic acid is CnH2nO2\text{C}_n\text{H}_{2n}\text{O}_2. Molar mass =12n+2n+32=14n+32=256  ⟹  14n=224  ⟹  n=16= 12n + 2n + 32 = 14n + 32 = 256 \implies 14n = 224 \implies n = 16.
Determining nn gives the total number of carbon atoms in the IUPAC parent chain.
4
Assign the official IUPAC name for a 16-carbon saturated alkanoic acid
C16H32O2\text{C}_{16}\text{H}_{32}\text{O}_2 is named Hexadecanoic acid.
The IUPAC suffix for a 16-carbon alkanoic acid is hexadecanoic acid.

Key Concept

Determination of alkanoic acid stoichiometry from saponification and neutralization data
Estimated Time:3m 0s
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Topics

Alkanoic Acids, Esterification, Saponification, Fats, and Oils

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