T là este của X,Y với Z nên X cũng là đơn chức

Muối E gồm XCOONa và YCOONa

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGybGaam4qaiaad+eacaWGpbGaamOtaiaadgga % aeqaaOGaeyypa0JaamOBamaaBaaaleaacaWGdbGaam4tamaaBaaame % aacaaIYaaabeaaaSqabaGccqGHsislcaWGUbWaaSbaaSqaaiaadIea % daWgaaadbaGaaGOmaaqabaWccaWGpbaabeaakiabg2da9iaaicdaca % GGSaGaaGimaiaaiAdaaaa!4B62! \to {n_{XCOONa}} = {n_{C{O_2}}} - {n_{{H_2}O}} = 0,06\)

Trong 6,9 gam M đặt: X là C_nH_{2n – 2}O₂ (u mol) và T là C_mH_{2m – 4}O₄ (v mol)

→ u + v = 0,06 (1)

m_M = u(14n + 30) + v(14m + 60) = 6,9 (2)

Trong phản ứng đốt cháy: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGybaabeaakiabgUcaRiaaikdacaWGUbWaaSbaaSqaaiaa % dsfaaeqaaOGaeyypa0JaamOBamaaBaaaleaacaWGdbGaam4tamaaBa % aameaacaaIYaaabeaaaSqabaGccqGHsislcaWGUbWaaSbaaSqaaiaa % dIeadaWgaaadbaGaaGOmaaqabaWccaWGpbaabeaakiabg2da9iaaic % dacaGGSaGaaGimaiaaiodaaaa!48E9! {n_X} + 2{n_T} = {n_{C{O_2}}} - {n_{{H_2}O}} = 0,03\)

→ n_O = 2n_X + 4n_T = 0,06

Bảo toàn O: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGpbWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabg2da9iaa % icdacaGGSaGaaGymaiaaicdacaaI1aaaaa!3D87! {n_{{O_2}}} = 0,105\)

Bảo toàn khối lượng → m_M = 2,3 trong phản ứng xà phòng hóa đã dùng lượng M nhiều gấp \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI2aGaaiilaiaaiMdaaeaacaaIYaGaaiilaiaaiodaaaGaeyypa0Ja % aG4maaaa!3C21! \frac{{6,9}}{{2,3}} = 3\) lần phản ứng cháy.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqa % baGccqGH9aqpdaWcaaqaaiaad6gadaWgaaWcbaGaamyDaaqabaGccq % GHRaWkcaWGUbWaaSbaaSqaaiaadAhaaeqaaaGcbaGaaG4maaaacqGH % 9aqpcaaIWaGaaiilaiaaigdadaqadaqaaiaaiodaaiaawIcacaGLPa % aaaaa!4805! \to {n_{C{O_2}}} = \frac{{{n_u} + {n_v}}}{3} = 0,1\left( 3 \right)\)

Giải hệ (1)(2)(3) ta có: u = 0,03; v = 0,03 → n + m =10

Do n ≥ 3 và m ≥ 6 và m ≥ n + 3 → n = 3; m = 7 là nghiệm duy nhất

X là CH₂=CH-COOH (0,03)

T là CH₂=CH-COO-CH₂-CH₂-OOC-CH₃ (0,03)

→ %T = 68,7%