An alcohol is oxidized to a carboxylic acid, and .2003 g of the acid is tritrate

Question

An alcohol is oxidized to a carboxylic acid, and .2003 g of the acid is tritrated with 45.25 mL of .03811 M NaOH. What is the formula of the alcohol? Please explain how you would approach this problem and the problem soving techniques used but keep in mind that we have only really covered naming the various functional groups and organic compounds. And that we haven't covered mechanisms either. I'd really appreciate help on this we haven't really been given any instruction on how to solve stuff like this but it's still in our homework. Thanks

Explanation / Answer

Let the acid be HA

HA + NaOH => NaA + H2O

Moles of HA = moles of NaOH = volume x concentration of NaOH

= 45.25/1000 x 0.03811 = 0.0017245 mol


Molar mass of HA = mass/moles of HA

= 0.2003/0.0017245 = 116.15 g/mol


Carboxylic acids have general formula C(n)H(2n)O2

Molar mass = 12.01 x n + 1.008 x 2n + 16.00 x 2

= (14.026n + 32.00) = 116.15

n = 6

Thus the acid has formula C6H12O2


The corresponding alcohol has formula C(n)H(2n+2)O or C(n)H(2n+1)OH

=> C6H14O or C6H13OH

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