\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGobGaamyyamaaBaaameaacaaIYaaabeaaliaadoeacaWG % pbWaaSbaaWqaaiaaiodaaeqaaaWcbeaakiabg2da9maalaaabaGaam % OBamaaBaaaleaacaWGobGaamyyaiaad+eacaWGibaabeaaaOqaaiaa % ikdaaaGaeyypa0JaaGimaiaacYcacaaIYaGaaGPaVlaad2gacaWGVb % GaamiBaiaacUdacaWGUbWaaSbaaSqaaiaad+eacaaMe8Uaaiikaiaa % dAeacaGGPaaabeaakiabg2da9iaaikdacaWGUbWaaSbaaSqaaiaad6 % eacaWGHbGaam4taiaadIeaaeqaaOGaeyypa0JaaGimaiaacYcacaaI % 4aGaaGjbVlaad2gacaWGVbGaamiBamaaoqcaleaacaWGcbGaamivai % aacQdacaaMc8UaaGjbVlaad+eaaeqakiaawkziaiaad6gadaWgaaWc % baGaamisamaaBaaameaacaaIYaaabeaaliaad+eaaeqaaOGaeyypa0 % JaaGimaiaacYcacaaIZaGaaGjbVlaad2gacaWGVbGaamiBaaaa!71B3! {n_{N{a_2}C{O_3}}} = \frac{{{n_{NaOH}}}}{2} = 0,2\,mol;{n_{O\;(F)}} = 2{n_{NaOH}} = 0,8\;mol\xrightarrow{{BT:\,\;O}}{n_{{H_2}O}} = 0,3\;mol\)

Muối gồm C_nH_mO₂Na (0,1 mol) và C_n’H_m’O₂Na (0,3 mol)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaa4ajaSqaai % aadkeacaWGubGaaiOoaiaaykW7caWGdbaabeGccaGLsgcacaaIWaGa % aiilaiaaigdacaWGUbGaey4kaSIaaGimaiaacYcacaaIZaGaamOBai % aacEcacqGH9aqpcaWGUbWaaSbaaSqaaiaad6eacaWGHbWaaSbaaWqa % aiaaikdaaeqaaSGaam4qaiaad+eadaWgaaadbaGaaG4maaqabaaale % qaaOGaey4kaSIaamOBamaaBaaaleaacaWGdbGaam4tamaaBaaameaa % caaIYaaabeaaaSqabaGccqGHshI3caWGUbGaey4kaSIaaG4maiaad6 % gacaGGNaGaeyypa0JaaGOnaiabgkDiElaad6gacqGH9aqpcaaIZaGa % ai4oaiaad6gacaGGNaGaeyypa0JaaGymaaaa!6184! \xrightarrow{{BT:\,C}}0,1n + 0,3n' = {n_{N{a_2}C{O_3}}} + {n_{C{O_2}}} \Rightarrow n + 3n' = 6 \Rightarrow n = 3;n' = 1\) là nghiệm duy nhất → m’ = 1

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaa4ajaSqaai % aadkeacaWGubGaaiOoaiaaykW7caWGibaabeGccaGLsgcacaWGUbWa % aSbaaSqaaiaadIeaaeqaaOGaeyypa0JaaGimaiaacYcacaaIXaGaam % yBaiabgUcaRiaaicdacaGGSaGaaG4maiaad2gacaGGNaGaeyypa0Ja % aGimaiaacYcacaaIZaGaeyO0H4TaamyBaiabg2da9iaaiodaaaa!4F36! \xrightarrow{{BT:\,H}}{n_H} = 0,1m + 0,3m' = 0,3 \Rightarrow m = 3\)→ CH₂=CH-COONa: 0,1 mol và HCOONa: 0,3 mol

Quy đổi E thành: HCOOH (0,3 mol), CH₂=CH-COOH (0,1 mol), C₃H₅(OH)₃ (0,04 mol), H₂O (-a mol)

Với m_E = 23,06 → a = 0,09 mol → n_T =  = 0,03 mol mà n_X = 8n_T = 0,24 mol → n_{X (T)} = 0,06 mol

Ta có: n_{X (T)} = 2n_T nên phân tử T có 2 gốc X và 1 gốc Y.

Vậy T là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGibGaam4qaiaad+eacaWGpbaacaGLOaGaayzkaaWaaSbaaSqaaiaa % ikdaaeqaaOWaaeWaaeaacaWGdbWaaSbaaSqaaiaaikdaaeqaaOGaam % isamaaBaaaleaacaaIZaaabeaakiaadoeacaWGpbGaam4taaGaayjk % aiaawMcaaiaadoeadaWgaaWcbaGaaG4maaqabaGccaWGibWaaSbaaS % qaaiaaiwdaaeqaaOGaaiOoaiaaicdacaGGSaGaaGimaiaaiodacaaM % e8UaamyBaiaad+gacaWGSbGaeyO0H4Taaiyjaiaac2gadaWgaaWcba % GaamivaaqabaGccqGH9aqpcaaIYaGaaGOnaiaacYcacaaIYaGaaGio % aiaacwcaaaa!58FA! {\left( {HCOO} \right)_2}\left( {{C_2}{H_3}COO} \right){C_3}{H_5}:0,03\;mol \Rightarrow \% {m_T} = 26,28\% \)