Answer to Question #204764 in Chemistry for Master j

Question #204764

An organic compound, in liquid form, comprises of 59.2% C, 13.6% H and 27.2% O, by mass; and has a molar mass of 60 gmol-1 . (RAM: C = 12; H = 1; O = 16) Determine the:


i. Empirical formula of the compound, (4 marks)

ii. Molecular formula of the compound. (2 marks)

iii. Balanced chemical equation, with the correct physical states, when the molecule from C ii. above is reacted with oxygen to give carbon dioxide and water


1
Expert's answer
2021-06-11T22:49:57-0400

Solution:

Assume that the sample has a mass of 100 g.

Convert the %values to grams:

Mass of C = w(C) × Mass of sample = 0.592 × 100 g = 59.2 g

Mass of H = w(H) × Mass of sample = 0.136 × 100 g = 13.6 g

Mass of O = w(O) × Mass of sample = 0.272 × 100 g = 27.2 g


Convert to moles:

(59.2 g C) × (1 mol C / 12 g C) = 4.93 mol C

(13.6 g H) × (1 mol H / 1 g H) = 13.60 mol H

(27.2 g O) × (1 mol O / 16 g O) = 1.70 mol O


Divide all moles by the smallest of the results:

C: 4.93 / 1.70 = 2.9 = 3

H: 13.60 / 1.70 = 8

O: 1.70 / 1.70 = 1


The empirical formula of the compound is C3H8O


Empirical formula mass = 3×Ar(C) + 8×Ar(H) + Ar(O) = 3×12 + 8×1 + 16 = 60 (g/mol)

Molar mass of the compound = 60 g/mol

Thus:

molar mass / empirical formula mass = n formula units/molecule

(60 g/mol) / (60 g/mol) = 1 formula units/molecule

Therefore,

The molecular formula of the compound is C3H8O


Balanced chemical equation:

2C3H8O(l) + 9O2(g) → 6CO2(g) + 8H2O(l)


Answers:

i: The empirical formula of the compound is C3H8O

ii: The molecular formula of the compound is C3H8O

iii: 2C3H8O(l) + 9O2(g) → 6CO2(g) + 8H2O(l)

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