Info: The carboxylic groups in a fatty acid can be neutralized by the addition o
ID: 1016723 • Letter: I
Question
Info:
The carboxylic groups in a fatty acid can be neutralized by the addition of potassium hydroxide (KOH). The neutralization reaction is the following:
CH3-(CH2)n-COOH + KOH --> CH3-(CH2)n-COO- K+ + H2O
The mass Index (Ia) is defined as the ratio between MKOH, mass (expressed in mg) of KOH needed to neutralize all the fatty acids, and MFA, the mass of fatty acids (expressed in mg) that was neutralized. Ia = 103 x (MKOH/MFA)
Ignoring the fact that the fatty acid may be unsaturated, you can calculate the molecular weight of a fatty acid (expressed in g.mol-1 ) by applying the following formula: MWFA = 14n + 32 with n the total number of carbon atoms contained in the fatty acid. Indeed, the chemical composition of a saturated fatty acid is CnH2nO2
Based on the numerical value of Ia calculated above, calculate how many carbon atoms are contained in this unknown fatty acid. Show your calculations. Additional information: the molecular weight of KOH is 56 g.mol-1 .
Explanation / Answer
As KOH and Fatty acid react in equimolar ratio ,1:1
so mass (KOH)/MW (KOH)=x = moles of KOH reacted and
mass (FA)/MW (FA)=Y =moles of FA reacted
X:Y=1:1 or X=Y gives mass (KOH)/MW (KOH) =mass (FA)/MW (FA)
or, mass (KOH)/mass (FA) =MW (KOH) /MW (FA)
or,mass (KOH)/mass (FA)=MW (KOH)/MW(FA)=Ia
As Ia value is not clearly reported in the question i would proceed with an arbitrary value ,Ia=1 say
If Ia=1000*MW (KOH)/MW(FA)
22.775=1000*MW (KOH)/MW(FA)
22.775*10^-3=MW (KOH)/MW(FA)
22.775*10^-3=56/MW(FA)
MW(FA)=56/22.775*10^-3=2458.84=14n+32
14n=2458.84-32=2426.84
n=2426.84/14=173
n=173