A compound contains 54.52% carbon, 9.17% hydrogen, and 36.31% oxygen by mass. A sample weighing 3.023 grams has a volume of 2.00L at 25.0oC and 0.420 atm. a. Calculate the empirical formula of the compound.

Empirical formula is C₂H₄O

Explanation:

We assume, the compound is an ideal gas, so we apply the Ideal Gases law, to determine the moles:

P . V = n . R . T

We convert T° → 25°C + 273 = 298K

0.420 atm . 2L = n . 0.082 . 298K

n = (0.420 atm . 2L) / (0.082 . 298K) → 0.0343 moles

This moles are 3.023 g, so now, we can confirm the molar mass:

3.023 g / 0.0343 mol = 87.9 g/mol ≅ 88 g/m

Centesimal composition is: 54.52 % of C, 9.17 % of H, 36.31% of O.

100 g of compund are contained in: 100 g . 1mol / 87.9g = 1.14 moles

We state the moles of each element:

54.52 g / 12g/mol = 4.5 mol of C

9.17 g / 1g/mol = 9.17 mol of H

36.31 g / 16 g/mol = 2.3 moles of O

Then, we state:

In 1.14 moles of compound we have 4.5 moles of C

In 1 mol we may have (4.5 / 1.14) = 3.94 moles of C → 4C

In 1.14 moles of compound we have 9.17 moles of H

In 1 mol, we may have (9.17 / 1.14) = 8.04 moles of H → 8H

In 1.14 moles of compound we have 2.3 moles of O

In 1 mol we may have, (2.3 / 1.14) = 2 moles of O

Molecular formula is C₄H₈O₂

We confirm it by the molar mass: 12 g/m . 4 + 1g/m . 8 + 16g/m . 2 = 88g/m

Empirical formula will be (with the lowest subscripts) → C₂H₄O

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