Question

In: Chemistry

A scientist is studying a monoprotic carboxylic acid that consists of only carbon, hydrogen, and oxygen....

A scientist is studying a monoprotic carboxylic acid that consists of only carbon, hydrogen, and oxygen. From a titration he finds that 13.4 mL of a 0.250 M sodium hydroxide solution are required to neutralize 0.5303 grams of the carboxylic acid. From a combustion experiment he finds that 2.982 grams of the carboxylic acid yields 0.9835 grams of carbon dioxide and 4.80 grams of water.

(a).What is the molecular weight of the carboxylic acid? g/mol

(b.)What is the empirical formula of the carboxylic acid?

(c.) What is the molecular formula of the carboxylic acid?

Expert Solution

NaOH + HA -------------> NaA + H2O

no of moles of NaOH = molarity * volume in L

= 0.25*0.0134 = 0.00335 moles

from balanced equation

1 mole of NaOH react with 1 mole of HA

0.00335 moles of NaOH react with 0.00335 moles of HA

molecular weight of acid = weight of acid/no of moles

= 0.5303/0.00335 = 158.3g/mole

C% = (12/44) * weight of CO2*100/weight of organic compound

= (12/44) *0.9835*100/2.982

= 12*0.9835*100/44*2.982 = 8.99%

H% = (2/18) * weight of H2O*100/weight of organic compound

= (2/18) *4.8*100/2.982

= 2*4.8*100/18*2.982 = 17.9%

O% = 100-(C%+ H%)

= 100-(8.99+17.9) = 73.11%

Element % A.Wt relative number simple ratio

C 8.99 12 8.99/12 = 0.75 0.75/0.75 = 1

H 17.9 1 17.9/1 = 17.9 17.9/0.75 = 24

O 73.11 16 73.11/16 = 4.57 4.57/0.75 = 6

Empirical formula = CH24O6

molecular formula = (empirical formula)n

n = M.Wt/E.F.Wt

= 158.3/132 = 1

molecular formula = CH24O6

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