3,248 (g) C_xH_y →  CO2(a)H2O(b)→ddCa(OH)218,62(g)↓Δmdd giam=3,724(g)

m_X = 12a + 2b = 3,248   (1)

∆m_{dd giảm} = m_↓ - ( )

→ = 18,62 – 3,724 = 14,896 (g)

→ 44a + 18b = 14,896 (2)

(1,2) →a=0,224=nCb=0,28⇒nH=0,56

x: y = 0,224 : 0,56 = 2 : 5 → (C₂H₅)_n

y ≤ 2x + 2 → 5n ≤ 4n + 2 → n ≤ 2 → n = 2 → C₄H₁₀

CH₃ – CH₂ – CH₂ – CH₃

CH₃ – CH(CH₃) – CH₃