In an organic compound, 40%Carbon,6.66% Hydrogen and rest Oxygen is present. Its vapour density is 30 .Calculate empirical formula and molecular formula of compound.
Answers
Let me write the actual data.
Percentage amount of C = 40%
Percentage amount of H₂ = 6.67%
Percentage amount of O₂ = 100 - (40 + 6.67)% = 53.33%
Vapour density of the compound = 30
Since vapour density = 30, molecular mass = 30 × 2g = 60g
Now find the relative no. of each element, i.e., percentage amount / atomic mass of one element.
Relative no. of C = 40 / 12 = 3.33
Relative no. of H₂ = 6.67 / 1 = 6.67
Relative no. of O₂ = 53.33 / 16 = 3.33
Here the relative no. of C and O₂ each are same, 3.33, and is the least among the three.
3.33 < 6.67
Now let's find the simplest whole number ratios of each element, i.e., relative no. of each element / the least relative no. among them.
Simplest ratio of C = 3.33 / 3.33 = 1
Simplest ratio of H₂ = 6.67 / 3.33 = 2
Simplest ratio of O₂ = 3.33 / 3.33 = 1
So we get the empirical formula of the compound - CH₂O.
Empirical formula mass = 12 + 2 + 16 = 30g.
We know Molecular formula = n · Empirical formula
Here n = Molecular mass / Empirical formula mass
∴ n = 60g / 30g = 2
∴ Molecular formula = 2(CH₂O) = C₂H₄O₂
Answer:
The Molecular Formula of the compound is C2H4O2
Explanation:
Percentage oxygen = 100- ( 40+ 6.66)= 100 - 46.6 = 53.4%
Hence, empirical formula = CH2O
Now, empirical formula mass
=12+(2×1)+16=12+2+16=30
Molecular mass of given organic compound
= vapor density x 2
= 30 x 2 = 60
n
= Molecular mass / Empirical formula mass
= 30/60
=2
Thus , Molecular formula
= (Empirical formula) ×n
=(CH2O)×2
=C2H4O2
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