Question

The combustion of \(20.63 \mathrm{mg}\) of compound \(\mathrm{Y}\), which contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O},\) with excess oxygen gave \(57.94 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(11.85 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). (a) Calculate how many milligrams of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) were present in the original sample of \(Y .(b)\) Derive the empirical formula of Y. (c) Suggest a plausible structure for \(Y\) if the empirical formula is the same as the molecular formula.

Short answer

Mass of C: 15.82 mg, H: 1.33 mg, O: 3.48 mg; Empirical formula is C₆H₆O, likely phenol.

Step-by-step solution

Convert mass of CO2 to moles of C

First, calculate the number of moles of carbon in the carbon dioxide produced. The molar mass of CO2 is calculated as follows: \( \text{Molar mass of } \mathrm{CO}_{2} = 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} \).\Now, convert the mass of CO2 to moles: \[ \text{Moles of } \mathrm{CO}_{2} = \frac{57.94 \, \text{mg}}{44.01 \, \text{g/mol}} \times 10^{-3} = 1.317 \times 10^{-3} \, \text{mol} \]\Since each mole of CO2 contains exactly 1 mole of carbon, we have the same number of moles of C: \[ \text{Moles of } \mathrm{C} = 1.317 \times 10^{-3} \, \text{mol} \]

Convert mass of H2O to moles of H

Calculate the moles of hydrogen in the water produced. The molar mass of H2O is given by \( 2.02 + 16.00 = 18.02 \, \text{g/mol} \).\Convert the mass of H2O to moles: \[ \text{Moles of } \mathrm{H}_{2}\mathrm{O} = \frac{11.85 \, \text{mg}}{18.02 \, \text{g/mol}} \times 10^{-3} = 0.658 \times 10^{-3} \, \text{mol} \] \Since each mole of H2O contains 2 moles of hydrogen, calculate moles of hydrogen: \[ \text{Moles of } \mathrm{H} = 2 \times 0.658 \times 10^{-3} = 1.316 \times 10^{-3} \, \text{mol} \]

Determine mass of C, H, O in compound Y

Determine the mass of carbon and hydrogen in compound Y:\Mass of carbon: \[ \text{Mass of } \mathrm{C} = 1.317 \times 10^{-3} \, \text{mol} \times 12.01 \, \text{g/mol} = 15.82 \, \text{mg} \] \Mass of hydrogen: \[ \text{Mass of } \mathrm{H} = 1.316 \times 10^{-3} \, \text{mol} \times 1.01 \, \text{g/mol} = 1.33 \, \text{mg} \] \Total mass of compound Y = 20.63 mg, so the mass of oxygen = Total mass - Mass of C - Mass of H: \[ \text{Mass of } \mathrm{O} = 20.63 \, \text{mg} - 15.82 \, \text{mg} - 1.33 \, \text{mg} = 3.48 \, \text{mg} \]

Calculate moles of O in compound Y

Calculate the moles of oxygen present in compound Y, using its mass: \[ \text{Moles of } \mathrm{O} = \frac{3.48 \, \text{mg}}{16.00 \, \text{g/mol}} \times 10^{-3} = 0.217 \times 10^{-3} \, \text{mol} \]

Determine empirical formula

Convert the moles to the simplest whole number ratio for the empirical formula. Divide the number of moles of C, H, and O by the smallest number of moles to get the empirical formula.\C: \( \frac{1.317 \times 10^{-3} \, \text{mol}}{0.217 \times 10^{-3} \, \text{mol}} \approx 6.07 \) \H: \( \frac{1.316 \times 10^{-3} \, \text{mol}}{0.217 \times 10^{-3} \, \text{mol}} \approx 6.06 \) \O: \( \frac{0.217 \times 10^{-3} \, \text{mol}}{0.217 \times 10^{-3} \, \text{mol}} = 1 \)\The empirical formula is approximately C₆H₆O.

Suggest molecular structure

The empirical formula matches the molecular formula, suggesting the compound is aromatic. Therefore, a plausible structure for Y could be phenol (C₆H₅OH), where one hydroxyl group is attached to a benzene ring.

Key concepts

Empirical Formula

An empirical formula represents the simplest whole number ratio of atoms of each element present in a compound. To find it, you need to determine the number of moles of each element in the compound and then simplify these numbers to their smallest ratio.
In our example, combustion analysis allowed us to calculate the moles of carbon, hydrogen, and oxygen in compound Y. By using the given masses of CO₂ and H₂O produced, we converted them into moles of C and H respectively. We found:

• 1.317 mmol of C
• 1.316 mmol of H
• 0.217 mmol of O

To get the empirical formula, we divided each number of moles by the smallest number, which was for oxygen, resulting in an empirical formula of \(C_6H_6O\) for the compound Y.
This step ensures the correct stoichiometric proportions and empowers chemists to predict how atoms combine at a basic level.

Molecular Formula

While an empirical formula gives the simplest ratio, the molecular formula specifies the actual number of atoms of each element in a molecule. Simply put, it is a multiple of the empirical formula that matches the molar mass of the compound.
For compound Y, since the calculated empirical formula is \(C_6H_6O\), and it was suggested that this is also the true molecular formula, the molecular mass would have to match experimental data through advanced analysis like mass spectrometry.
If the molecular formula isn't immediately obvious, multiply the empirical formula by an integer factor that equates to the experimentally determined molecular mass. For example, if a compound has an empirical formula \(CH_2\) and an actual molar mass of about 56 g/mol, then the molecular formula is \(C_4H_8\), indicating butene.

Chemical Composition

Chemical composition refers to the specific arrangement of different types and quantities of atoms in a compound. It provides insight into the compound's properties and behavior during chemical reactions. To deduce the chemical composition from combustion data, as done for Y, one must carefully calculate the percentage by mass, moles, and actual atoms from the combustion results.
Here, the chemical composition reflects in the masses calculated:

• 15.82 mg of Carbon
• 1.33 mg of Hydrogen
• 3.48 mg of Oxygen

This breakdown allows for understanding of how each element contributes to the overall structure and potential reactions of the compound.
Accurate composition calculations ensure that what you theoretically analyze matches the experimental results tightly, which is crucial in scientific research and industrial applications.

Organic Chemistry

Organic chemistry is the study of carbon-containing compounds and their properties. Compounds like Y fall under this branch due to their carbon backbone typically bound to hydrogen, oxygen, nitrogen, and other elements. Organic chemistry shines in explaining the behavior, synthesis, and transformations of these molecules.
Combustion analysis is a critical tool within organic chemistry to determine stoichiometry. Organic compounds are combusted, and the gaseous products are analyzed to deduce the original compound’s composition.
The compound Y, with an empirical formula of \(C_6H_6O\), aligns with common organic structures like phenol, highlighting aromatic compounds characterized by their stable ring structures. Understanding the nuances of such aromaticity is essential for fields like pharmaceuticals, where these compounds serve as core building blocks for drug design and synthesis.