$4.40\;g$ of a branched aliphatic carboxylic acid are dissolved in $400\;mL$ of water. $20\;mL$ of this solution consume $25.0\;mL\;NaOH\;0.1\frac{mol}{L}$ by titration.
Calculate the molecular formula of the acid.
Find the condensed formula and the name of this acid.
1)
$20\; mL$ contain $\frac{20\cdot 4,40}{400}$ = $0,220\; g$ acid
At the equivalence point of the titration:
$n_{acid}$ $= $ $n_{NaOH}$ $=$ $0,1\cdot 0,025$ $=$ $0,0025$
$M_{acid}$ $=$ $\frac{0,220}{0,0025}$ $ =$ $ 88 \frac{g}{mol}$
General formula:
$C_nH_{2n}O_2$
$12n$ $+$ $2n$ $+$ $32$ $=$ $88$
$n=4$
Molecular formula:
$C_4H_8O_2$
2)
$CH_3CH(CH_3)COOH$
methylpropanoic acid
