Problem 106 - Chemistry The Central Science

Question

(a) By titration, \(15.0 \mathrm{~mL}\) of \(0.1008 \mathrm{M}\) sodium hydroxide is needed to neutralize a \(0.2053-\mathrm{g}\) sample of an organic acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that it is composed of \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C},\) and \(23.5 \% \mathrm{O}\) by mass. What is its molecular formula?

Step-by-Step Solution

Answer

The molar mass of the monoprotic organic acid is 135.6 g/mol, and its molecular formula is C₈H₁₆O₂.

Step 1: Calculate the moles of sodium hydroxide

To calculate the moles of sodium hydroxide (NaOH) used in the titration, we'll use the formula: moles = volume × molarity where: - volume = 15.0 mL of NaOH (convert to liters: \(\frac{15.0}{1000}\)) - molarity = 0.1008 M NaOH moles = \(\frac{15.0}{1000}\) L × 0.1008 mol/L = 0.001512 mol of NaOH

Step 2: Deduce the moles of the organic acid

Since the organic acid is monoprotic, it reacts with NaOH in a 1:1 ratio. That means the moles of the organic acid are equal to the moles of NaOH used in the titration: moles of organic acid = 0.001512 mol

Step 3: Calculate the molar mass of the organic acid

To find the molar mass of the organic acid, we can use the formula: molar mass = \(\frac{mass}{moles}\) where: - mass = 0.2053 g - moles = 0.001512 mol (from step 2) molar mass = \(\frac{0.2053}{0.001512}\) g/mol = 135.6 g/mol

Step 4: Find the empirical formula of the organic acid

Given the percentages of each element, first, convert them to moles by dividing the mass percentage by the molar mass of each element, then find the ratio between elements by dividing each by the smallest value obtained: For H: grams H = 0.0589 × 135.6 g/mol = 7.98 g moles H = 7.98 g / 1.0079 g/mol = 7.92 mol For C: grams C = 0.706 × 135.6 g/mol = 95.75 g moles C = 95.75 g / 12.011 g/mol = 7.97 mol For O: grams O = 0.235 × 135.6 g/mol = 31.86 g moles O = 31.86 g / 15.999 g/mol = 1.99 mol Divide each by the smallest value: H ratio = 7.92 / 1.99 ≈ 4 C ratio = 7.97 / 1.99 ≈ 4 O ratio = 1.99 / 1.99 ≈ 1 Empirical formula = C₄H₈O

Step 5: Determine the molecular formula

To find the molecular formula of the organic acid, we'll first calculate the molar mass of the empirical formula and then compare it with the molar mass of the organic acid: Molar mass of empirical formula: (4 × 12.011) + (8 × 1.0079) + (1 × 15.999) = 72.08 g/mol Compare empirical formula mass to organic acid molar mass: \(\frac{135.6}{72.08} \approx 1.88\) Considering significant figures, we can round the ratio to 2. That means the molecular formula is twice the empirical formula: Molecular formula = C₈H₁₆O₂

Key Concepts

Molar Mass Calculation

Understanding how to calculate the molar mass of a compound is a fundamental skill in chemistry, which ties into many different areas such as stoichiometry, analytical chemistry, and physical chemistry.

Let's look at the process as applied in our exercise. The molar mass is defined as the mass of one mole of a substance. In the context of a titration, once we know the moles of the titrant (in this case, sodium hydroxide), we can find the moles of the substance of interest (the organic acid), assuming we know the stoichiometry of the reaction.

In this scenario, because our organic acid is monoprotic – which means it can donate only one proton (H⁺) – it will react with NaOH at a 1:1 ratio. Given this information and the mass of the acid used in the titration, we use the formula:
\[\text{molar mass} = \frac{\text{mass}}{\text{moles}}\]
Therefore, if we divide the mass of the organic acid sample by the number of moles of NaOH reacted, we arrive at the molar mass. It's important to keep track of your units here – we need to convert milliliters to liters and grams to moles – to ensure that our final molar mass is in the standard units of grams per mole (g/mol).

Empirical and Molecular Formula Determination

The empirical formula represents the simplest whole-number ratio of elements in a compound, while the molecular formula is the actual ratio of atoms in a molecule of that compound, which could be the same as or a multiple of the empirical formula.

By determining the percentage composition of the elements (as given in the exercise), and converting these to moles, we can deduce the simplest ratio. Each element's moles are divided by the smallest moles calculated to find the whole-number ratio for each element. This leads to the empirical formula, which is a crucial stepping stone in finding the molecular formula.

Finding the Empirical Formula

For each element present in the compound, we convert the mass percent to absolute mass (assuming a 100g sample, or in our case, using the previously calculated molar mass), and then to moles. Next, each mole value is divided by the smallest mole value to find the empirical formula ratios.

Once the empirical formula is established, we determine its molar mass and then compare it with the already-known molar mass of the compound derived from the titration experiment. If the molar mass of the empirical formula is less than the molar mass of the compound from titration, the molecular formula will be a multiple of the empirical formula. The multiplier is found by dividing the molar mass of the compound by the molar mass of the empirical formula.

Stoichiometry of Titration

Stoichiometry is the part of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In a titration, stoichiometry helps us to determine the concentration of a solution by reacting it with a standard solution of known concentration.

During an acid-base titration, the unknown concentration of an acid can be found by carefully measuring the amount of a base of known concentration required to neutralize it. The point at which neutralization occurs is often indicated by a color change with an indicator or by reaching a specific pH value when using a pH meter.

To solve stoichiometry problems in titrations, it's essential to write down the balanced chemical equation for the reaction. This will tell you the stoichiometry, or the mole-to-mole ratios, which you apply to convert between moles of the titrant and moles of the analyte. In our exercise, the reaction between a monoprotic acid and NaOH is at a 1:1 mole ratio.

It is important to remember that stoichiometry is not only about the amounts of reactants and products but also about the concept of conservation of mass. This principle states that in a closed system, matter cannot be created or destroyed in a chemical reaction. Therefore, the total mass of the reactants in a reaction will equal the total mass of the products.