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.

Short answer

The unknown liquid is not propenoic acid. The empirical formula calculated from the combustion results is \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_4\), which is different from the given propenoic acid formula \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_2\).

Step-by-step solution

Calculate the moles of carbon, hydrogen, and oxygen in the sample

First, we need to determine the moles of each element present in the sample. We are given that combustion produces \(0.374 \mathrm{~g}\) of carbon dioxide and \(0.102 \mathrm{~g}\) of water. Let's start by finding the moles of carbon and hydrogen. Moles of carbon (in 0.374 g CO₂)= \(\frac {0.374 g}{44.01 g/mol} =0.0085 mol\) Moles of hydrogen (in 0.102 g H₂O)= \(\frac{2×0.102 g}{18.02 g/mol} = 0.0113 mol\) Now, let's find the mass of oxygen in the sample. To do this, we subtract the masses of carbon and hydrogen from the total mass of the sample (\(0.275 g\)): Mass of oxygen = \(0.275 g - (0.0085 mol × 12.01 g/mol) - (0.0113 mol × 1.008 g/mol)\) = \(0.191 g\) Now, we can calculate the moles of oxygen: Moles of oxygen = \(\frac{0.191 g}{16.00 g/mol} = 0.0119\) We have the moles of each element in the sample: Carbon: \(0.0085 \mathrm{mol}\) Hydrogen: \(0.0113 \mathrm{~mol}\) Oxygen: \(0.0119 \mathrm{~mol}\)

Compare the ratio of elements in the sample

To find the ratio of the elements in the sample, we must determine the smallest whole number ratio between moles of carbon, hydrogen, and oxygen. We start by finding the smallest mole value among the three elements, which is 0.0085 in this case (carbon). Next, divide all the mole values by this smallest value: Carbon (\(\mathrm{C}\)) ratio = \(\frac{0.0085}{0.0085} = 1\) Hydrogen (\(\mathrm{H}\)) ratio = \(\frac{0.0113}{0.0085} \approx 1.33\) Oxygen (\(\mathrm{O}\)) ratio = \(\frac{0.0119}{0.0085} \approx 1.4\) Since the ratios aren't close to whole numbers, we can try multiplying all ratios by the smallest ratio (in this case, 3) to obtain whole number ratios: Carbon (\(\mathrm{C}\)) ratio = \(1 × 3 = 3\) Hydrogen (\(\mathrm{H}\)) ratio = \(1.33 × 3 \approx 4\) Oxygen (\(\mathrm{O}\)) ratio = \(1.4 × 3 \approx 4\) This gives us the approximate empirical formula: \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_4\).

Compare the empirical formula with propenoic acid formula

Now that we have determined the empirical formula of the unknown liquid sample, we can compare it to the given formula for propenoic acid, which is \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_2\). Our empirical formula is \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_4\), which is different from the given propenoic acid formula \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_2\). Based on this comparison, we can conclude that the unknown liquid is not propenoic acid.

Key concepts

Combustion Analysis

Combustion analysis is a process used to determine the empirical formula of a compound by analyzing the products of its combustion. When an organic compound undergoes complete combustion, it typically produces carbon dioxide (CO₂) and water (H₂O). By measuring the amounts of these products, you can find out how much carbon and hydrogen were present in the original compound.

In the given exercise, a sample of an unknown liquid was combusted to produce 0.374 g of CO₂ and 0.102 g of H₂O. To uncover the identity of the liquid, the moles of carbon and hydrogen were calculated from these products. This is possible because:

• Each mole of CO₂ contains one mole of carbon.
• Each mole of H₂O contains two moles of hydrogen.

Understanding these relationships allows us to reverse-engineer the original elemental composition of the sample, setting the stage for determining its empirical formula.

Elemental Composition

Determining the elemental composition involves calculating the individual amounts of carbon, hydrogen, and oxygen in the sample. By knowing the mass of CO₂ and H₂O produced in the combustion, we can uncover how much of each element was in the unknown compound.

This process starts with "back-calculating" using the formulas:

• For carbon: Moles of carbon = \(\frac{\text{mass of CO}_2}{44.01 \text{ g/mol}}\)
• For hydrogen: Moles of hydrogen = \(\frac{2 \times \text{mass of H}_2\text{O}}{18.02 \text{ g/mol}}\)

After calculating these, any remaining mass in the sample can be attributed to oxygen, which we calculate by subtracting the mass of carbon and hydrogen from the total mass.

This step is crucial for understanding the elemental make-up of the sample, which is necessary for determining the empirical formula. It provides insight into the exact amounts of each element that contributed to the combustion products.

Mole Calculations

Mole calculations are fundamental in chemical analysis, especially when determining empirical formulas. Once the amount of each element in moles is known, these values are used to find the simplest whole-number ratio among the elements.

In this scenario, the moles calculated were:

• Carbon: 0.0085 mol
• Hydrogen: 0.0113 mol
• Oxygen: 0.0119 mol

To find the empirical formula, you divide each of these by the smallest mole value (carbon, in this case). If these ratios aren't whole numbers, you might need to multiply by an integer to reach a more accurate, simple ratio.

The goal is to express the moles of each element as a ratio close to small integers, revealing the empirical formula of the substance. By this method, the closest empirical formula derived from this set of data was different from the known formula of propenoic acid, leading to the conclusion on the identity of the liquid.