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What is the molecular weight of a monoprotic carboxylic acid if 11.20 mL of 0.9635 M sodium hydroxide is required to titrate 1.253 g of this acid? The reaction is represented by following equation. NOTE: In the equation, R represents an unspecified carbon containing structure RCO-H (aq)NaOH (aq) RCO,Na (aq) HOH (U) Separate experiments suggest the unknown acid is likely pentanoic acid, C4H,CO2H. Is the unknown pentanoic acid? Why or why not?

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dsdrajlin

Answer:

116.1 a.m.u.

It is not likely that RCOOH is the pentanoic acid

Explanation:

Let's consider the generic neutralization between NaOH and a monoprotic carboxylic acid.

RCOOH(aq) + NaOH(aq) ⇒ RCOONa(aq) + H₂O(l)

The molar ratio of RCOOH to NaOH is 1:1. The moles of RCOOH are:

[tex]11.20 \times 10^{-3}L.\frac{0.9635molNaOH}{1L}. \frac{1molRCOOH}{1molNaOH} =1.079 \times 10^{-2}molRCOOH[/tex]

The molar mass of RCOOH is:

[tex]\frac{1.253g }{1.079 \times 10^{-2}mol } =116,1g/mol[/tex]

Thus, the molecular weight is 116.1 a.m.u.

Pentanoic acid has the formula C₅H₁₀O₂ with a molecular weight of 102.1 a.m.u. So, it is not likely that RCOOH is the pentanoic acid.