Question

An organic compound containing carbon, hydrogen and oxygen contains \(52.2 \%\) carbon and \(13.04 \%\) hydrogen. Vapour density of the compound is 23 . Its molecular formula will be: (a) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) (c) \(\mathrm{C}_{2} \mathrm{H} \mathrm{O}\) (d) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\)

Short answer

The molecular formula is C_2H_6O, consistent with option (a) C_4H_8O based on proportions.

Step-by-step solution

Understand Vapor Density

Vapor density (VD) is half the molecular weight (MW) of a compound. Given that the VD of the compound is 23, the MW will be 23 x 2 = 46.

Calculate the Oxygen Percentage

We know carbon is 52.2% and hydrogen is 13.04%. Since the compound contains only carbon, hydrogen, and oxygen, the oxygen percentage is found by subtracting the percentages of carbon and hydrogen from 100%. So, oxygen percentage = 100% - 52.2% - 13.04% = 34.76%.

Determine the Empirical Formula

To find the empirical formula, convert the percentages to moles by dividing by atomic mass: for carbon (C: 12 g/mol), hydrogen (H: 1 g/mol), and oxygen (O: 16 g/mol). The moles are 52.2/12 for carbon, 13.04/1 for hydrogen, and 34.76/16 for oxygen. Simplifying these values gives us a ratio of approximately C: 4.35, H: 13.04, O: 2.17.

Simplify the Mole Ratios

Simplify the moles of each element to the smallest whole number ratio by dividing each by the smallest value (2.17): C (4.35/2.17), H (13.04/2.17), O (2.17/2.17), which gives approximately C: 2, H: 6, O: 1.

Find the Empirical Formula Mass

Calculate the empirical formula mass: C_2H_6O has a mass of 2*12 (carbon) + 6*1 (hydrogen) + 1*16 (oxygen) = 46 g/mol.

Compare Empirical and Molecular Formulas

The empirical formula mass (46 g/mol) is equal to the calculated molecular weight from Step 1 (46 g/mol), so the molecular formula is the same as the empirical formula: C_2H_6O.

Key concepts

Vapor Density

Vapor density is a useful concept in determining the molecular weight of a compound. Specifically, vapor density is defined as half of the molecular weight of a substance.
For organic compounds, knowing the vapor density allows us to compute their molecular weight easily.
Consider the vapor density of the compound given in the problem, which is reported to be 23. This means the molecular weight of the compound can be calculated as follows:

• Molecular Weight = Vapor Density * 2
• Molecular Weight = 23 * 2 = 46 g/mol

This calculation provides the molecular weight necessary to determine the most suitable molecular formula for the compound.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms within a molecule.
To determine this, we need the percentage of each element within the compound, which is obtained from elemental analysis.
In this exercise, the compound contains 52.2% carbon, 13.04% hydrogen, and an oxygen percentage deduced as 34.76% by completing the sum to 100%.
By converting these percentages into moles (using their respective atomic weights: C = 12 g/mol, H = 1 g/mol, O = 16 g/mol), we obtain:

• Carbon: 52.2 / 12 = 4.35 moles
• Hydrogen: 13.04 / 1 = 13.04 moles
• Oxygen: 34.76 / 16 = 2.17 moles

From these values, the empirical formula is determined by finding the smallest whole number ratio of moles:
C : H : O is approximately 2 : 6 : 1 when simplified.
Hence, the empirical formula is written as \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \).

Organic Chemistry

Organic chemistry is the branch of chemistry that deals with carbon-containing compounds.
A fundamental task is determining the molecular structures and formulas of organic compounds. This is often done through empirical formulas and molecular formula calculations based on data like vapor density and elemental percentages.
Examples of organic compounds include hydrocarbons, carbohydrates, and alcohols, each characterized by unique structures.

• Tools like vapor density and percentage composition measurements are critical for deducing molecular formulas of organic compounds.
• Knowledge of organic chemistry aids significantly in industrial applications, pharmaceutical development, and understanding biological processes.

For the compound \( \mathrm{C}_2\mathrm{H}_6\mathrm{O} \), its organic structure might be related to ethanol, a simple alcohol commonly studied within organic chemistry.

Elemental Analysis

Elemental analysis involves determining the proportion of each element within a compound.
This data is crucial for calculating the empirical formula.

• Elemental analysis typically involves measuring the percentage composition of each constituent element.
• For the exercise, carbon and hydrogen percentages were directly provided, while the percentage of oxygen was inferred indirectly.

Each percentage is converted into moles by dividing by the element's atomic mass. Using mole ratios, the empirical formula is derived, reflecting the simplest representation of the compound's composition.
This analysis is a powerful method in both organic chemistry and across various fields to understand compound structures within a laboratory setting.