Question

Propionic acid, an organic acid, contains only \(\mathrm{C}, \mathrm{H},\) and O. When \(0.236 \mathrm{~g}\) of the acid burns completely in \(\mathrm{O}_{2}\) it gives \(0.421 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.172 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) Determine the empirical formula of the acid.

Short answer

The empirical formula of the acid is \(C_3H_6O_2\).

Step-by-step solution

Find moles of carbon in CO₂

Calculate the moles of carbon by using the mass of CO₂ produced. The molar mass of CO₂ is approximately 44.01 g/mol. Since CO₂ contains one carbon atom, the moles of C in CO₂ is given by: \[\text{moles of } C = \frac{0.421 \, \text{g CO}_2}{44.01 \, \text{g/mol}} \approx 0.00957 \, \text{mol of } C.\]

Find moles of hydrogen in H₂O

Calculate the moles of hydrogen using the mass of H₂O produced. The molar mass of H₂O is approximately 18.02 g/mol. Since each molecule of H₂O contains two hydrogen atoms, the moles of H is: \[\text{moles of } H = \frac{2 \times 0.172 \, \text{g H}_2\text{O}}{18.02 \, \text{g/mol}} \approx 0.0191 \, \text{mol of } H.\]

Calculate moles of oxygen in propionic acid

The total mass of the propionic acid is 0.236 g, consisting of C, H, and O. First, find the mass of carbon using the moles calculated: \[\text{mass of C} = 0.00957 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.115 \, \text{g.} \]Find the mass of hydrogen: \[\text{mass of H} = 0.0191 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.0192 \, \text{g.} \]Subtract these masses from the total to get the mass of oxygen:\[\text{mass of O} = 0.236 \, \text{g} - (0.115 \, \text{g} + 0.0192 \, \text{g}) = 0.1018 \, \text{g.}\]Calculate the moles of oxygen:\[\text{moles of } O = \frac{0.1018 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.00636 \, \text{mol of } O.\]

Determine the empirical formula

Now, use the moles of each element to find the simplest whole-number ratio. Divide each by the smallest number of moles calculated:\[\text{C: } \frac{0.00957}{0.00636} \approx 1.50, \quad \text{H: } \frac{0.0191}{0.00636} \approx 3.0, \quad \text{O: } \frac{0.00636}{0.00636} = 1.0.\]These ratios correspond to approximately C:3/2, H:3, O:1. Multiply through by 2 to rid the fractional component:\[C_3H_6O_2.\]The empirical formula is \(C_3H_6O_2\).

Key concepts

Stoichiometry

Stoichiometry is the study of measuring chemical quantities, such as mass and moles, in chemical reactions. It is essentially about finding a balanced relationship between reactants and products in a chemical equation.
In stoichiometry, the key is to use the mole concept and the law of conservation of mass to solve problems. Moles allow chemists to count particles by weighing them, as atoms and molecules are too small to count directly. Based on this, it’s critical to understand that whether conducting analysis or synthesis, stoichiometry connects the theoretical world with real-world calculations through:

• Balanced equations: Every element in a chemical equation has the same number of atoms before and after the reaction.
• Mole ratios: Allow conversion between grams and moles using the molar mass of substances.
• Conservation laws: Mass and atoms are conserved across chemical reactions.

In the case of propionic acid, stoichiometry helps calculate the quantity of each element produced during combustion. By using the known masses of carbon dioxide and water, one can deduce the number of moles of carbon and hydrogen involved, which, in turn, helps find the empirical formula of the compound.

Combustion Analysis

Combustion analysis is a method used to determine the elemental composition of a compound, especially when it contains organic elements like carbon, hydrogen, and oxygen. During combustion, the organic compound burns in oxygen, producing carbon dioxide, water, and sometimes other gases like nitrogen.
Understanding how combustion analysis works involves a few important steps:

• Complete combustion converts all carbon to carbon dioxide and all hydrogen to water.
• The masses of carbon dioxide and water produced are measured to calculate the moles of carbon and hydrogen.
• For oxygen, the element in the product not directly measured, its moles are determined by subtracting the calculated masses of carbon and hydrogen from the total mass of the original substance.

This technique is particularly practical in organic chemistry for verifying empirical formulas. In the exercise, combustion analysis of propionic acid determined the moles of carbon, hydrogen, and then oxygen, supporting the calculation of its empirical formula. The careful measurement and conversion of these reaction products give insight into the makeup of the initial organic compound.

Organic Chemistry

Organic chemistry is the study of compounds primarily composed of carbon, often in conjunction with hydrogen, oxygen, nitrogen, and other elements. This field investigates structures, properties, compositions, reactions, and synthesis of organic compounds.
In organic chemistry, understanding the basic structure and formula of organic acids is crucial. Empirical formulas provide the simplest integer ratios of elements in a compound, which is fundamental when working with complex molecules.
Organic acids, like propionic acid, demonstrate characteristic behaviours and structures:

• They contain a carboxyl group (-COOH), giving them acidic properties.
• They typically follow standard formula derivations, initial stoichiometric calculations helping to define their empirical formulas.
• Empirical formulas align with stoichiometric ratios deduced from reactions like combustion.

The discipline of organic chemistry requires a blend of theoretical insight and practical measurement techniques, such as combustion analysis, to elucidate the structure and composition of organic compounds. This allows chemists to create a bridge between molecular descriptions and experiments, underscoring the principles of organic chemistry and facilitating the discovery of new compounds and reactions.