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 gof water and 0.374 g carbon dioxide. Is the unknown liquid propenoic acid? Support your reasoning with calculations.

Short answer

The percentage composition of elements in the unknown liquid (C: 49.60%, H: 5.56%, and O: 44.84%) is very close to the propenoic acid's percentage composition (C: 50.00%, H: 5.61%, and O: 44.38%). Therefore, it can be concluded that the unknown liquid is likely to be propenoic acid.

Step-by-step solution

Find the percentage of each element in the propenoic acid.

First, we must determine the molar mass of propenoic acid. Molar mass of C3H4O2: = (3 × 12.01 g/mol) + (4 × 1.008 g/mol) + (2 × 16.00 g/mol) = 72.03 g/mol Now, we'll find the percentage of each element in propenoic acid: Percentage of C = (mass of C / molar mass of propenoic acid) × 100 = (3 × 12.01) / 72.03 × 100 = 50.00 % Percentage of H = (mass of H / molar mass of propenoic acid) × 100 = (4 × 1.008) / 72.03 × 100 = 5.61 % Percentage of O = (mass of O / molar mass of propenoic acid) × 100 = (2 × 16) / 72.03 × 100 = 44.38 %

Determine the moles of products from the combustion reaction.

We are given the mass of the combustion products (water and carbon dioxide): mass of H2O = 0.102 g mass of CO2 = 0.374 g Now, we will determine the moles of each product using their respective molar mass. Moles of H2O (2 moles of H for every mole of H2O) = (0.102 g) / (18 g/mol) = 0.00567 mol (2 moles of H for every mole of H2O) = 0.01134 mol of H Moles of CO2 (1 mole of C for every mole of CO2) = (0.374 g) / (44 g/mol) = 0.00850 mol

Determine the percentage of each element in the unknown liquid.

Now, we'll find the percentage of each element in the unknown liquid using the calculated moles. Molar mass of H = 1.008 g/mol Molar mass of C = 12.01 g/mol Percentage of H in the sample = (moles of H × molar mass of H) / (mass of sample) × 100 = (0.01134 mol × 1.008 g/mol) / 0.275 g × 100 = 5.56 % Percentage of C in the sample = (moles of C × molar mass of C) / (mass of sample) × 100 = (0.00850 mol × 12.01 g/mol) / 0.275 g × 100 = 49.60 % To determine the percentage of oxygen in the sample, we'll subtract the sum of the percentages of carbon and hydrogen from 100. Percentage of O in the sample = 100% - (Percentage of C + Percentage of H) = 100 - (49.60 % + 5.56 %) = 44.84 %

Compare the percentage composition and make a conclusion.

Now, we'll compare the percentage composition of the elements in the unknown liquid with that of the propenoic acid. Percentage composition of elements in propenoic acid: C: 50.00 % H: 5.61 % O: 44.38 % Percentage composition of elements in the unknown liquid: C: 49.60 % H: 5.56 % O: 44.84 % The percentage composition of elements in the unknown liquid is very close to the propenoic acid's percentage composition. Therefore, it can be concluded that the unknown liquid is likely to be propenoic acid.

Key concepts

Stoichiometry

Understanding stoichiometry is like having a recipe for the perfect chemical reaction. It involves calculating the precise ratios of reactants and products involved in chemical reactions. This discipline of chemistry is essential when analyzing reactions such as the combustion of an organic compound, just like in the provided exercise where propenoic acid undergoes combustion.

In our recipe for stoichiometry, the first step is to balance the chemical equation for the reaction. Once balanced, we can relate the amounts of reactants to products through their coefficients. For example, the combustion of propenoic acid can be represented by the equation \( \text{C}_{3}\text{H}_{4}\text{O}_{2} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O} \). After balancing, we can identify the stoichiometric coefficients which allow us to determine the molar ratio between the reactants and products.

In the exercise, stoichiometry enables us to calculate the moles of water and carbon dioxide produced from the sample. It is pivotal to know that each mole of carbon dioxide reflects a mole of carbon from the compound, and each mole of water reflects two moles of hydrogen. By correctly applying stoichiometry, students can better understand the quantitative aspect of chemical reactions.

Molar Mass Calculation

When it comes to molar mass calculation, think of it as a way of finding out how much one mole (\(6.022 \times 10^{23}\) particles) of a substance weighs. Molar mass is the bridge between the mass of a substance and the number of moles, and it plays a crucial role in stoichiometry as well as in finding the empirical formula.

To calculate the molar mass of a compound, add up the atomic masses of each element multiplied by the number of atoms of that element in the compound's formula. For example, the molar mass of propenoic acid \( \text{C}_{3}\text{H}_{4}\text{O}_{2} \) is calculated by summing the products of the number of atoms of carbon, hydrogen, and oxygen with their respective atomic masses.

Students often confuse molar mass with molecular mass, but it's key to remember that molar mass is expressed in grams per mole (\text{g/mol}), while molecular mass is simply the sum of the atomic masses in the molecule's formula. Being able to accurately calculate molar mass is imperative for students to perform stoichiometric calculations and to assess the percentage composition of the elements in a compound.

Empirical Formula

The empirical formula is the simplest integer ratio of the elements present in a compound. It serves as a kind of 'identity card' for the molecule, telling you the proportions of each type of atom present without revealing the actual amount or structure.

Empirical formulas are derived from experimental data, such as the masses of the elements that combine to form the compound. In the context of the exercise, determining the empirical formula is the forensic tool we would use to identify an unknown substance by comparing its elemental composition with that of known substances.

To find the empirical formula, a student must first calculate the moles of each element in the sample. Then, these values are used to find the simplest whole number ratio of the elements. In the provided solution, we calculated the moles of hydrogen and carbon from the combustion products (water and carbon dioxide). Then, by taking the percentages of those elements and comparing them to the total mass of the sample, we found their ratios and confirmed the identity of the unknown liquid against the known empirical formula of propenoic acid. This process illustrates the practical application of empirical formulas in chemical analysis.