Problem 58 - Chemistry The Central Science In Si Units

Question

Propenoic acid, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2},\) is a reactive organic liquid that is used in the manufacturing of plastics, coatings, and adhesives. An unlabeled container is thought to contain this liquid. A 0.275 -g sample of the liquid is combusted to produce \(0.102 \mathrm{~g}\) of water and \(0.374 \mathrm{~g}\) carbon dioxide. Is the unknown liquid propenoic acid? Support your reasoning with calculations.

Step-by-Step Solution

Answer

Yes, the empirical formula matches that of propenoic acid.

Step 1: Calculate Moles of Water

Start by determining the moles of water produced. The molecular weight of water (\(\mathrm{H}_2\mathrm{O}\)) is 18.015 g/mol. Thus, the moles of water are calculated by \(\frac{0.102 \text{ g}}{18.015 \text{ g/mol}} \approx 0.00566 \text{ moles}.\)

Step 2: Calculate Moles of Carbon Dioxide

Calculate the moles of carbon dioxide produced. The molecular weight of carbon dioxide (\(\mathrm{CO}_2\)) is 44.01 g/mol. Thus, the moles of carbon dioxide are \(\frac{0.374 \text{ g}}{44.01 \text{ g/mol}} \approx 0.00850 \text{ moles}.\)

Step 3: Determine the Amount of Carbon and Hydrogen in the Sample

Each mole of \(\mathrm{H}_2\mathrm{O}\) contains 2 moles of hydrogen. Therefore, the moles of hydrogen are \(2 \times 0.00566 \approx 0.01132\text{ moles of } \mathrm{H}.\)Each mole of \(\mathrm{CO}_2\) contains 1 mole of carbon. Therefore, the moles of carbon are \(0.00850 \text{ moles of } \mathrm{C}.\)

Step 4: Calculate Grams of Each Element

The mass of hydrogen is \(0.01132 \text{ moles} \times 1.008 \text{ g/mol} \approx 0.01141 \text{ g of } \mathrm{H}.\)The mass of carbon is \(0.00850 \text{ moles} \times 12.01 \text{ g/mol} \approx 0.102 \text{ g of } \mathrm{C}.\)

Step 5: Calculate Mass of Oxygen in the Compound

The total mass of carbon and hydrogen is \(0.01141 \text{ g} + 0.102 \text{ g} = 0.11341 \text{ g}.\) Subtract this from the original mass of the sample to find the mass of oxygen:\(0.275 \text{ g} - 0.11341 \text{ g} \approx 0.1616 \text{ g of } \mathrm{O}.\)

Step 6: Calculate Moles of Oxygen

The molecular weight of oxygen (\(\mathrm{O}\)) is 16.00 g/mol. Therefore, the moles of oxygen are \(\frac{0.1616 \text{ g}}{16.00 \text{ g/mol}} \approx 0.0101 \text{ moles of } \mathrm{O}.\)

Step 7: Determine Empirical Formula

Convert the moles of elements to the simplest whole number ratio:For \(\mathrm{C}\): \(0.00850\)For \(\mathrm{H}\): \(0.01132\)For \(\mathrm{O}\): \(0.0101\)Divide each by the smallest number, \(0.00850\):\(\mathrm{C}: \frac{0.00850}{0.00850} = 1\); \(\mathrm{H}: \frac{0.01132}{0.00850} \approx 1.33\); \(\mathrm{O}: \frac{0.0101}{0.00850} \approx 1.19\).The approximate empirical formula is \(\mathrm{C}_3\mathrm{H}_4\mathrm{O}_2\), consistent with propenoic acid.

Key Concepts

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of a compound. This is especially useful in organic chemistry, where compounds consist mainly of carbon, hydrogen, and oxygen. In the combustion process, a sample is burned and the products, typically carbon dioxide and water, are analyzed. By measuring the amount of these products, the amounts of carbon and hydrogen in the original substance can be calculated.

• For instance, the carbon from the sample ends up in carbon dioxide. By calculating the moles of carbon dioxide, you can also determine the moles of carbon in the sample.
• Similarly, the hydrogen from the sample contributes to the production of water. From the moles of water, you derive the moles of hydrogen.

This information can then be used to deduce the empirical formula of a compound, which is the simplest ratio of atoms in the compound. This process is critical for identifying unknown compounds and confirming their identity against known substances.

Molecular Weight Calculation

Understanding how to calculate molecular weight is essential in many aspects of chemistry, especially when confirming the identity of unknown compounds. Molecular weight, or molar mass, is the weight of one mole of a substance and is typically expressed in grams per mole (g/mol). To find the molecular weight of a compound, you sum the atomic weights of all atoms in its formula:

• For water ( ext{H}_2 ext{O}), you sum the atomic weights of 2 hydrogen atoms (1.008 g/mol each) and 1 oxygen atom (16.00 g/mol), resulting in 18.015 g/mol.
• Carbon dioxide ( ext{CO}_2) involves adding the atomic weight of 1 carbon atom (12.01 g/mol) and 2 oxygen atoms (16.00 g/mol each) to yield 44.01 g/mol.

These calculations are vital in determining how much of each molecule is present in a sample. Knowing the molecular weight allows you to convert grams to moles, setting the stage for determining the empirical formula of a compound from combustion data.

Elemental Composition

Elemental composition refers to the percentage and ratios of different elements within a compound. This is crucial for deriving empirical formulas through combustion analysis. The elemental composition is derived from the mass of each element present, calculated through data from burning a sample:

• The weight of carbon can be calculated from measuring produced carbon dioxide.
• The weight of hydrogen is derived from the amount of water formed.
• Oxygen's weight can often be deduced by subtracting known weights of carbon and hydrogen from the total mass.

This step enables translation from mass to moles using the known atomic masses of the elements, progressing to the simplest whole number ratio or the empirical formula. Knowledge of elemental composition is fundamental to both qualitative and quantitative chemical analysis.

Organic Chemistry

Organic chemistry is a vast field focused on the study of carbon-containing compounds. These compounds often also include hydrogen, oxygen, nitrogen, sulfur, phosphorus, and other elements. Organic chemistry underpins many scientific, pharmaceutical, and industrial processes due to the versatility and prevalence of carbon-based compounds.
In the realm of organic chemistry, combustion analysis serves as a powerful tool to investigate the presence and ratios of elements within organic compounds. The field covers numerous classes of compounds, each playing essential roles in various applications:

• Acids like propenoic acid are crucial in manufacturing polymers and other chemical products.
• Alkanes, alkenes, and alkynes form the foundation for fuels and chemical building blocks.
• Complex molecules and macromolecules derived from simpler organic substances are vital for life itself, including DNA, proteins, and carbohydrates.

Harnessing the insights of combustion analysis enables chemists to classify, verify, and synthesize these organic compounds, contributing broadly to scientific and industrial advancements.