Question

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?

Answer 1

116.1 a.m.u.

It is not likely that RCOOH is the pentanoic acid

Step-by-step 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:

`11.20 * 10^(-3)L.(0.9635molNaOH)/(1L). (1molRCOOH)/(1molNaOH) =1.079 * 10^(-2)molRCOOH`

The molar mass of RCOOH is:

`(1.253g )/(1.079 * 10^(-2)mol ) =116,1g/mol`

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.