The empirical formula : C₁₁O₁₄O₃
The assumption of the compound consists of C, H, and O
mass of C in CO₂ =
[tex]\tt \dfrac{12}{44}\times 2.492=0.680~g[/tex]
mass of H in H₂O =
[tex]\tt \dfrac{2.1}{18}\times 0.6495=0.072~g[/tex]
mass of O :
mass sample-(mass C + mass H)
[tex]\tt 1-(0.68+0.072)=0.248`g[/tex]
mol of C :
[tex]\tt \dfrac{0.68}{12}=0.056[/tex]
mol of H :
[tex]\tt \dfrac{0.072}{1}=0.072[/tex]
mol of O :
[tex]\tt \dfrac{0.248}{16}=0.0155[/tex]
divide by 0.0155(the lowest ratio)
C : H : O ⇒
[tex]\tt \dfrac{0.056}{0.0155}\div \dfrac{0.072}{0.0155}\div \dfrac{0.0155}{0.0155}=3.6\div 4.6\div 1\\\\\dfrac{11}{3}\div \dfrac{14}{3}\div \dfrac{3}{3}=11:14:3[/tex]
