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Question:
An unknown organic compound weighing 0.486 g was dissolved in water and titrated with 0.100 N NaOH solution. At the end point, 58.55 mL of base had been added. A second neutralization experiment, on the same mass of compound, required 57.25 mL of base. What is the neutralization equivalent for this unknown organic compound?
Answer:
We need to use the equation for neutralization equivalent:
Neutralization equivalent (NE) = (weight of sample * 1000) / (ml of base * normality)
Solve the equation for NE:
NE = (0.486 g * 1000) / (58.55 mL * 0.100 mol equivalents / L)
NE = 83.0 g / mol equivalents
NE = (0.486 g * 1000) / (57.25 mL * 0.100 mol equivalents / L)
NE = 84.9 g / mol equivalents
The average NE is 84.0 g / mol equivalents. The error associated with this experiment is ± 1. Therefore, the neutralization equivalent is reported as 84 ± 1 g / mol equivalents.
Remember, that the molecular weight of the unknown compound is some multiple of the neutralization equivalent. For example, if the unknown had only one acidic functionality, the molecular weight would be 84 ± 1 g / mol. If the unknown had 2 acidic functionalities, then the MW would be 168 ± 2 g / mol.
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