Question

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.

Answer 1

To determine the empirical formula of a compound with a given percent composition, the mass percentages are converted to moles, and then the mole ratio of elements is calculated to find the simplest whole number ratio, which in this example is C2H4O.

Step-by-step explanation:

To calculate the empirical formula of the compound with given percent composition, we start by assuming a 100 g sample, which makes it easy to convert percent composition to grams. For carbon (C), we have 54.52 g, for hydrogen (H), we have 9.17 g, and for oxygen (O), we have 36.31 g. To find the empirical formula, we need to convert these amounts to moles.

Using the atomic masses (C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol), we calculate the moles of each element:

• Carbon: 54.52 g C × (1 mol C/12.01 g) = 4.54 mol C
• Hydrogen: 9.17 g H × (1 mol H/1.01 g) = 9.07 mol H
• Oxygen: 36.31 g O × (1 mol O/16.00 g) = 2.27 mol O

Divide each by the smallest number of moles to find the simplest whole number ratio:

• Carbon: 4.54 mol C ÷ 2.27 = 2.00, approximately 2
• Hygen: 9.07 mol H ÷ 2.27 = 4.00, approximately 4
• Oxygen: 2.27 mol O ÷ 2.27 = 1.00, approximately 1

The empirical formula is C2H4O.

Answer 2

Empirical formula is C₂H₄O

Step-by-step 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