The Empirical formula of compound is C₁₀H₂₀O₁. The Molecular Formula of compound is C₁₀H₂₀O₁.
Empirical formula is the simplest whole number ratio of atoms present in given compound.
[tex]\text{Mass of C} = 0.2829\ g\ CO_2 \times \frac{12\ g\ C}{44\ g\ CO_2}[/tex]
= 0.07720 g C
[tex]\text{Mass of H} = 0.1159\ g\ H_2O \times \frac{2\ g\ H}{18\ g\ H_2O}[/tex]
= 0.01297 g H
Mass of O = Mass of methanol - mass of C - mass of O
= 0.1005 g - 0.07720 g - 0.01297 g
= 0.01033 g
Element Mass/g Moles Ratio Integers
C 0.07720 0.006428 9.956 10
H 0.01297 0.01287 19.93 20
O 0.01033 0.0006456 1 1
The Empirical formula of compound is C₁₀H₂₀O₁.
Molecular formula = Empirical formula × n
[tex]n = \frac{\text{Molar Mass}}{\text{Empirical formula weight}}[/tex]
[tex]= \frac{156}{156}[/tex]
= 1
Molecular formula = n × Empirical formula
= 1 (C₁₀H₂₀O₁)
= C₁₀H₂₀O₁
Thus from the above conclusion we can say that The Empirical formula of compound is C₁₀H₂₀O₁. The Molecular Formula of compound is C₁₀H₂₀O₁.
Learn more about the Empirical Formula here: brainly.com/question/1603500
#SPJ1
