Question

The empirical formula of a
compound whose molecular
formula is C8H604 & empirical
formula weight is 83:​

Answer 1

Answer:

ANSWER

We have,

An organic compound C=57.8%

H=3.6%

Thus the oxygen is 38.6% Since the total %is100

So, 

Molecular weight i=2×vapordensity

=2×83=166

So, Mass of C=\dfrac{57.8}{100}\times166$$

=95.948g

No pof carbon atoms =1295.948=7.99∼8atoms Since 

12=mass of the carbon

Mass of H=1003.6×166=5.976∼6 atoms

Mass of O=10038.6×166=64.076g

No of atoms =1664.076=4atoms

Thus. The formulaa will be C8H6O4

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Answer 2

Given: Empirical formula weight = 83

Molecular formula of a compound = $C_{8} H_{6} O_{4}$

To Find: The empirical formula of the compound.

Solution:

• The number of atoms in the molecular formula  $C_{8} H_{6} O_{4}$ are 8 atoms of carbon, 6 atoms of hydrogen and 4 atoms of oxygen respectively.
• Firstly, determine a number that can be divisible completely by the number of atoms of the compound with a molecular formula $C_{8} H_{6} O_{4}$ to find the empirical formula of a com und.
• By taking 2 as the number which can be divisible completely with the atoms of the compounds and give 4atoms of carbon,3 atoms of hydrogen and 2 atoms of oxygen respectively.
• The number of atoms calculated above are depicted in the empirical formula.
• So, the empirical formula concluded is  $C_{4} H_{3} O_{2}$₂.
• Given that the molecular weight of the empirical formula is 83.
• To check whether the calculated empirical formula is correct or not, equalize the atoms of the empirical formula depicted with the molecular weight of the empirical formula.
• $(4 \times 12) \: + (3 \times 2) \: + (2 \times 16) \: = 83$
• As the molecular weight from the depicted empirical formula is also 83. This signifies that the empirical formula of the compound is $C_{4} H_{3} O_{2}$.

Hence, the empirical formula of the compound is $C_{4} H_{3} O_{2}$.