Respuesta :
Answer:
Empirical formula :
[tex]C_{3}H_{3}O_{1}[/tex]
Molecular formula :
[tex]C_{6}H_{6}O_{2}[/tex]
Explanation:
STEP 1: Calculate the moles of each species i.e C , H , O
Molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen means
Mass of Carbon = 65.5 gram
Mass of Hydrogen = 5.5 gram
Mass of oxygen = 29.0 gram
[tex]moles =\frac{Given\ mass}{Molar\ mass}[/tex]
Moles of C:
Molar mass of C = 12.0 gram
[tex]C =\frac{65.5}{12}[/tex]
Moles of C = 5.45
Moles of H:
Molar mass = 1.0 gram
[tex]H=\frac{5.5}{1}[/tex]
Moles of H = 5.5 moles
Moles of O :
Molar mass of O = 16.00 gram
[tex]O=\frac{29.0}{16}[/tex]
Moles of O = 1.81 moles
STEP2 : Calculate the simple ratio of C ,H,O by dividing with 1.81 (Because it is the minimum moles present)
Always divide by minimum number of moles
For C
[tex]C = \frac{5.45}{1.81}[/tex]
= 3.00
For H
[tex]H = \frac{5.5}{1.81}[/tex]
= 3.03
= 3.00 (round off the answer to nearest whole number)
For O
[tex]O = \frac{1.81}{1.81}[/tex]
= 1.00
So we get the empirical formula:
[tex]C_{3}H_{3}O_{1}[/tex]
STEP 3: Calculate the "n" value to obtain the molecular formula using:
[tex]n=\frac{Molar\ mass}{Empirical\ mass}[/tex]
Empirical mass =
[tex]C_{3}H_{3}O_{1}[/tex]
= 3(mass of C)+3(mass of H)+ mass of O
= 3x12 + 3x1 + 16
= 55 grams
Molar mass = 110 g/mole
[tex]n=\frac{Molar\ mass}{Empirical\ mass}[/tex]
[tex]n=\frac{110}{55}[/tex]
n = 2
STEP 4: Multiply the empirical formula with n
[tex](C_{3}H_{3}O_{1})\times 2[/tex]
[tex]C_{6}H_{6}O_{2}[/tex]
