Hỗn hợp X gồm glyxin, analin và axit glutamic (trong đó nguyên tố oxi chiếm 41,2% về khối lượng). Cho m gam X tác dụng với dung dịch NaOH dư, thu được 20,532 gam muối. Giá trị của m là

\(\eqalign{
& {m_O} = 0,412m \Rightarrow {n_O} = {{0,412m} \over {16}} \Rightarrow {n_{ - COOH}} = {1 \over 2}{n_O} = {{0,412m} \over {32}} \cr
& - COOH\,\,\, + \,\,NaOH\buildrel {} \over
\longrightarrow - COONa + \,\,{H_2}O \cr
& {{0,412m} \over {32}} \to {{0,412m} \over {32}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{0,412m} \over {32}} \cr
& \left\{ \matrix{
{H_2}N - C{H_2} - COOH \hfill \cr
{H_2}NCH(C{H_3}) - COOH \hfill \cr
{H_2}N{C_3}{H_5}{(COOH)_2} \hfill \cr} \right.\,\,\, + NaOH\,\,\,\,\,\,\,\buildrel {} \over
\longrightarrow \,\,\,\,\,\,\left\{ \matrix{
{H_2}N - C{H_2} - COONa \hfill \cr
{H_2}NCH(C{H_3}) - COONa \hfill \cr
{H_2}N{C_3}{H_5}{(COONa)_2} \hfill \cr} \right. + {H_2}O \cr
& \,\,\,\,\,\,\,m(gam)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{0,412m} \over {32}}(mol)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,20,532(gam)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{0,412m} \over {32}}(mol) \cr} \)

Bảo toàn khối lượng ta có:

\(\eqalign{
& m + {{0,412m} \over {32}}.40 = 20,532 + {{0,412m} \over {32}}.18 \cr
& \Rightarrow 1,28325m = 20,532 \cr
& \Rightarrow m = 16 \cr} \)