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Chapter 15: Problem 75

Lactic acid (found in sour milk) has an empirical formula of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\). A 1.0-g sample of lactic acid required \(17.0 \mathrm{~mL}\) of \(0.65 \mathrm{M}\) \(\mathrm{NaOH}\) to reach the end point of a titration. What is the molecular formula for lactic acid?

Short Answer

Expert verified
The molecular formula for lactic acid is \(\text{HC}_3 \text{H}_5 \text{O}_3\).

Step by step solution

01

Determine the moles of \(\text{NaOH}\)

First, calculate the moles of \(\text{NaOH}\) used in the titration. Use the formula: \(\text{Moles} = \text{Molarity} \times \text{Volume (L)}\). Here, the volume should be converted to liters: \(17.0 \mathrm{mL} = 0.017 \mathrm{L}\). The molarity of \(\text{NaOH}\) is 0.65 M. Thus, moles of \(\text{NaOH} = 0.65 \mathrm{M} \times 0.017 \mathrm{L} = 0.01105 \text{moles}\).
02

Determine the moles of lactic acid

Lactic acid reacts with \(\text{NaOH}\) in a 1:1 molar ratio in the titration process. Therefore, the moles of lactic acid will be the same as the moles of \(\text{NaOH}\), which is \(0.01105 \text{moles}\).
03

Calculate the molar mass of lactic acid

Use the mass of the sample (1.0 g) and the moles of lactic acid to find the molar mass. The formula is \( \text{Molar mass} = \frac{ \text{Mass} }{ \text{Moles} } \). So, \( \text{Molar mass} = \frac{1.0 \mathrm{g}}{0.01105 \text{moles}} = 90.5 \mathrm{g/mol} \).
04

Determine the molecular formula

The empirical formula of lactic acid is \(\text{HC}_3 \text{H}_5 \text{O}_3\). The empirical formula mass is: \(1 (\text{H}) + 3 (\text{C}) + 5 (\text{H}) + 3 (\text{O}) = 1 \times 1 + 3 \times 12 + 5 \times 1 + 3 \times 16 = 90 \mathrm{g/mol}\). Since the empirical formula mass is roughly equal to the molar mass, the molecular formula is the same as the empirical formula: \(\text{HC}_3 \text{H}_5 \text{O}_3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

titration
Titration is a method used to determine the concentration of a substance in a solution by reacting it with a solution of known concentration. In this exercise, we used NaOH (sodium hydroxide) to titrate lactic acid.

The endpoint of a titration is the point at which the reaction between the titrant and the analyte is complete. An indicator is often used to signal this point, usually by a color change.

It's important to convert the volume of titrant used to liters and use the molarity to find the moles of titrant. In our case, we converted 17.0 mL NaOH to 0.017 L, and used its molarity (0.65 M) to find the moles of NaOH involved in the reaction.
empirical formula
The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element in the compound. For lactic acid, the empirical formula is \(\text{HC}_3\text{H}_5\text{O}_3\).

To find the molecular formula, we often need additional information, such as the molar mass. In this exercise, it was given that the empirical formula is known and needs to be confirmed as the molecular formula.
molar mass calculation
Calculating the molar mass of a substance involves dividing the mass of the sample by the number of moles present in that sample.

In this exercise, we had a 1.0 g sample of lactic acid. After completing the titration, we found there were 0.01105 moles of lactic acid present.

The formula to find molar mass is: \(\text{Molar mass} = \frac{ \text{Mass} }{ \text{Moles} }\). Therefore, the molar mass was calculated as \(\frac{1.0 \text{g}}{0.01105 \text{moles}} = 90.5 \text{g/mol}\).
stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on balanced chemical equations.

In this exercise, the reaction between lactic acid and NaOH can be simplified as a 1:1 molar ratio. This means one mole of lactic acid reacts with one mole of NaOH.

Therefore, the moles of NaOH calculated during titration (0.01105 moles) also represent the moles of lactic acid present in the sample. This stoichiometric relationship helped us find the molar mass and confirmed the empirical formula.
chemical reactions in acid-base titration
In an acid-base titration, an acid reacts with a base to form water and a salt. The reaction is usually represented as: \(\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}\).

For lactic acid reacting with NaOH, the balanced chemical equation would be: \(\text{HC}_3\text{H}_5\text{O}_3 + \text{NaOH} \rightarrow \text{NaC}_3\text{H}_5\text{O}_3 + \text{H}_2\text{O}\).

This reaction confirms the stoichiometric relationship and allows us to use the moles of NaOH to determine the moles of lactic acid, thereby helping in determining the molecular formula of lactic acid.

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Most popular questions from this chapter

Write a reaction between the sulfide ion \(\left(\mathrm{S}^{2-}\right)\) and water showing how it acts as a Brønsted-Lowry base.

The bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\), has the ability to act as both an acid and a base. (a) What is the conjugate acid of \(\mathrm{HCO}_{3}^{-}\)? (b) What is the conjugate base of \(\mathrm{HCO}_{3}^{-}\)?

Determine whether each of the following is a strong acid, weak acid, strong base, or weak base. Then write an equation describing the process that occurs when the substance is dissolved in water. (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{HCl}\) (c) \(\mathrm{KOH}\) (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Write an equation to illustrate the acid-base reaction that will take place between perchloric acid, \(\mathrm{HClO}_{4}\), and a hydroxide ion, \(\mathrm{OH}^{-}\). Identify the acids, bases, and their conjugates.

Calculate the \(\mathrm{pH}\) of (a) orange juice, \(3.7 \times 10^{-4} \mathrm{M} \mathrm{H}^{+}\) (b) vinegar, \(2.8 \times 10^{-3} \mathrm{M} \mathrm{H}^{+}\) (c) shampoo, \(2.4 \times 10^{-6} \mathrm{M} \mathrm{H}^{+}\) (d) dishwashing detergent, \(3.6 \times 10^{-8} \mathrm{M} \mathrm{H}^{+}\)
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