Question

Propenoic acid, as shown here, is a reactive organic liquid used in the manufacture of plastics, coatings, and adhesives. An unlabeled container is thought to contain this acid. A 0.2033 -g sample is combusted in an apparatus such as that shown in Figure \(3.14 .\) The gain in mass of the \(\mathrm{H}_{2} \mathrm{O}\) absorber is \(0.102 \mathrm{~g},\) whereas that of the \(\mathrm{CO}_{2}\) absorber is \(0.374 \mathrm{~g}\). Is this analysis consistent with the contents of the container being propenoic acid?

Short answer

After calculating the moles of H, C, and O, we find their simplest ratio to determine the empirical formula. If the obtained empirical formula is the same as the formula of propenoic acid (C₃H₄O₂), then the unknown sample is indeed an unlabeled container of propenoic acid; otherwise, it's not propenoic acid.

Step-by-step solution

Determine moles of H, C and O

Propenoic acid has the molecular formula C₃H₄O₂. In order to get the empirical formula of the unknown substance, we need to know the number of moles of each element in it. Let's first find the moles of hydrogen (H), carbon (C), and oxygen (O) from the masses of the byproducts of combustion H2O and CO2. The molar mass of H2O is \(18.02 \mathrm{~g/mol}\) and the molar mass of CO2 is \(44.01 \mathrm{~g/mol}\). Since the gaseous product continues 2 moles of H atoms per mole of H2O produced, we can find the moles of H present in the original sample by the following equation: Moles of \(H = \frac{0.102\mathrm{~g}}{18.02\mathrm{~g/mol}}\times 2\) Similarly, we have only 1 mole of C atoms in each mole of CO2. Moles of \(C = \frac{0.374\mathrm{~g}}{44.01\mathrm{~g/mol}}\) Since propenoic acid has 3 atoms of carbon, 4 atoms of hydrogen, and 2 atoms of oxygen, we can now calculate for the moles of oxygen: Let moles of O be x Mass of unknown sample = Mass of C + Mass of H + Mass of O \(0.2033\mathrm{~g} = (0.374\mathrm{~g}) + (0.102\mathrm{~g}) + x\) x = \(0.2033\mathrm{~g} – (0.374\mathrm{~g} + 0.102\mathrm{~g})\) Moles of \(O = \frac{x}{16.00\mathrm{~g/mol}}\)

Get the simplest ratio of H, C and O

Now we have the moles of H, C, and O. In order to find the empirical formula, we need to get the simplest ratio of these moles, which can be done by dividing each value by the smallest value among them. Let's divide the moles of each element by the smallest number of moles obtained: Moles of C : H : O = \( \frac{\mathrm{moles \thinspace of \thinspace C}}{\mathrm{min(moles \thinspace of \thinspace H, \thinspace moles \thinspace of \thinspace C, \thinspace moles \thinspace of \thinspace O)}} : \frac{\mathrm{moles \thinspace of \thinspace H}}{\mathrm{min(moles \thinspace of \thinspace H, \thinspace moles \thinspace of \thinspace C, \thinspace moles \thinspace of \thinspace O)}} : \frac{\mathrm{moles \thinspace of \thinspace O}}{\mathrm{min(moles \thinspace of \thinspace H, \thinspace moles \thinspace of \thinspace C, \thinspace moles \thinspace of \thinspace O)}} \)

Compare the empirical formula

Now that we have the empirical formula of the unknown substance, let's compare it with the known formula of propenoic acid. If the empirical formula is the same as the formula of propenoic acid (C₃H₄O₂), then the unknown sample is indeed an unlabeled container of propenoic acid. If not, then the unknown sample is not propenoic acid. From the step-by-step solution process, you will now be able to calculate the moles of each of the elements and find out if the empirical formula matches the formula of propenoic acid.

Key concepts

Empirical Formula Calculation

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It's calculated from the relative number of moles of each element in the compound. To calculate it, chemists use a clear procedure. Firstly, they determine the number of moles of each element from experimental data, which involves converting the mass of each element obtained from combustion to moles using the molar mass of the element (e.g., hydrogen's molar mass is approximately 1.01 g/mol, carbon's is 12.01 g/mol, and oxygen's is 16.00 g/mol).

In the context of the solved problem, the analysis of the combustion products (water for hydrogen content, and carbon dioxide for carbon content) allows for the calculation of the moles of carbon and hydrogen. For oxygen—which isn’t directly measured because it's part of both the water and carbon dioxide—its amount is determined by the difference in the mass of the original compound and the sum of the masses of carbon and hydrogen.
After obtaining the moles of each element, the next step is to divide each by the smallest amount of moles present to get the simplest ratio. This ratio translates directly into the empirical formula, which might not be the actual molecular formula – it’s essentially the 'reduced version.' For instance, a compound with an empirical formula of CH2 could have a molecular formula of C2H4, C3H6, and so on, depending on the actual molecular mass.

Molecular Formula Determination

Once the empirical formula is established, the molecular formula – the actual number of atoms of each element within a molecule – can be found if the compound's molar mass is known. It’s identified by comparing the empirical formula mass to the molar mass of the compound. The molecular formula is a whole number multiple of the empirical formula.

The molecular formula is calculated by dividing the molar mass of the compound by the mass of its empirical formula. For example, if the empirical formula of a compound is CH2O and the molar mass of the compound is found to be 60 g/mol, then the molecular formula is found by dividing 60 g/mol by the mass of the empirical formula, which is 30 g/mol (12 for carbon, plus 2 for hydrogen, plus 16 for oxygen), resulting in a factor of 2. Therefore, the molecular formula would be C2H4O2.It is this molecular formula that can provide insight into the actual structure and composition of a substance. In the case of the propenoic acid analysis in the exercise, confirming that the empirical formula matches the molecular formula of known propenoic acid (C3H4O2) would solidify the identification of the compound in the unknown sample.

Stoichiometry

Stoichiometry is the section of chemistry that involves using balanced chemical equations to calculate the relative quantities of reactants and products involved in a chemical reaction. It’s pivotally used for calculating the quantities in a chemical reaction to ensure optimal usage of reactants without wastage, and it helps predict the amounts of products that will be formed.

In the context of combustion analysis, stoichiometry allows for the prediction of the amount of carbon dioxide and water that will be produced given a certain amount of a substance. For example, according to the stoichiometry of the combustion of a hydrocarbon, one mole of a hydrocarbon will typically produce one mole of carbon dioxide for each carbon atom, and one mole of water for every two hydrogen atoms.Understanding stoichiometry is essential to determine the empirical formula because it aids in relating the masses of reactants and products to their mole ratios in the chemical equation. Thus, it connects the mass of the substances consumed to the mass of the products formed, providing a foundation for empirical and molecular formula calculations.