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0.265g of an organic compound produced an evaporation 102cm³ of vapour at 373k and 775mmHg percentage composition of the constituent element are 92.24% of C and

Sagot :

A. The molecular mass of the compound is 77.9 g/mol

B. The molecular formula of the compound is C₆H₆

Determination of the mole of the compound

We'll begin by calculating the number of mole of compound using the ideal gas equation as shown below:

  • Volume (V) = 102 cm³ = 102 / 1000 = 0.102 L
  • Temperature (T) = 373 K
  • Pressure (P) = 775 mmHg = 775 / 760 = 1.02 atm
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Number of mole (n) =?

n = PV / RT

n = (1.02 × 0.102) / (0.0821 × 373)

n = 0.0034 mole

A. Determination of the molecular mass of the compound.

  • Mass = 0.265 g
  • Number of mole = 0.0034 mole
  • Molecular mass =?

Molecular mass = mass / mole

Molecular mass = 0.265 / 0.0034

Molecular mass of compound = 77.9 g/mol

B. Determination of the molecular formula of the compound.

We'll begin by calculating the empirical formula of the compound.

  • Carbon (C) = 92.24%
  • Hydrogen (H) = 7.76%

Empirical formula =?

Divide by their molar mass

C = 92.24 / 12 = 7.69

H = 7.76 / 1 = 7.76

Divide by the smallest

C = 7.69 / 7.69 = 1

H = 7.76 / 7.69 = 1

Thus the empirical formula of the compound is CH

Finally, we shall determine the molecular formula.

  • Molecular mass = 77.9 g/mol
  • Empirical formula = CH
  • Molecular formula =?

Molecular formula = empirical × n = molecular mass

[CH]n = 77.9

[12 + 1]n = 77.9

13n = 77.9

Divide both side by 13

n = 77.9 / 13

n = 6

Molecular formula = [CH]n

Molecular formula = [CH]₆

Molecular formula = C₆H₆

Complete Question:

0.265g of an organic compound produced on evaporation 102cm cube of vapour at 373K and 775mmHg. Percentage composition of the constituent elements are 92.24% C and 7.76% H. Find the molecular mass and molecular formula of the composition.

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