Problem 138 - Chemistry The Molecular Science

Question

A chemical company was interested in characterizing a competitor's organic acid (it consists of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O})\). After determining that it was a diacid, \(\mathrm{H}_{2} \mathrm{X},\) a \(0.1235-\mathrm{g}\) sample was neutralized with \(15.55 \mathrm{~mL}\) of \(0.1087-\mathrm{M}\) \(\mathrm{NaOH}\). Next, a \(0.3469-\mathrm{g}\) sample was burned completely in pure oxygen, producing \(0.6268 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.2138 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). (a) Calculate the molar mass of \(\mathrm{H}_{2} \mathrm{X}\). (b) Calculate the empirical formula for the diacid. (c) Calculate the molecular formula for the diacid.

Step-by-Step Solution

Answer

(a) Molar mass: 146 g/mol; (b) Empirical formula: C₃H₅O₂; (c) Molecular formula: C₆H₁₀O₄.

Step 1: Determine Moles of NaOH

The moles of NaOH can be found using the concentration and volume:\[ \text{Moles of NaOH} = M \times L = 0.1087 \text{ M} \times 0.01555 \text{ L} = 0.001691 \text{ moles} \]

Step 2: Calculate Moles of Diacid H₂X

Since the diacid H₂X reacts with NaOH in a 1:2 ratio, the moles of H₂X are:\[ \text{Moles of H}_2\text{X} = \frac{1}{2} \times 0.001691 \text{ moles} = 0.0008455 \text{ moles} \]

Step 3: Calculate Molar Mass of Diacid H₂X

The molar mass is calculated using mass and moles:\[ \text{Molar Mass} = \frac{0.1235 \text{ g}}{0.0008455 \text{ moles}} \approx 146.0 \text{ g/mol} \]

Step 4: Determine Moles of C and H from Combustion Data

Use the masses of CO₂ and H₂O to find moles of C and H:\[ \text{Moles of C} = \frac{0.6268 \text{ g of CO}_2}{44.01 \text{ g/mol}} = 0.01425 \text{ moles} \]\[ \text{Moles of H} = \frac{0.2138 \text{ g of H}_2\text{O}}{18.02 \text{ g/mol}} \times 2 = 0.02373 \text{ moles} \]

Step 5: Calculate Mass of O in Diacid H₂X

Find the mass of C and H and subtract from the total mass of the sample to find O:\[ \text{Mass of C} = 0.01425 \text{ moles} \times 12.01 \text{ g/mol} = 0.1710 \text{ g} \]\[ \text{Mass of H} = 0.02373 \text{ moles} \times 1.008 \text{ g/mol} = 0.0239 \text{ g} \]\[ \text{Mass of O} = 0.3469 \text{ g} - (0.1710 \text{ g} + 0.0239 \text{ g}) = 0.1520 \text{ g} \]

Step 6: Determine Moles of O

Calculate moles of O from its mass:\[ \text{Moles of O} = \frac{0.1520 \text{ g}}{16.00 \text{ g/mol}} = 0.00950 \text{ moles} \]

Step 7: Calculate Empirical Formula

Divide the moles by the smallest value to find the ratio:\[ \text{C:} \frac{0.01425}{0.00950} \approx 1.5 \Rightarrow 3 \]\[ \text{H:} \frac{0.02373}{0.00950} \approx 2.5 \Rightarrow 5 \]\[ \text{O:} \frac{0.00950}{0.00950} = 1 \Rightarrow 2 \]Empirical formula is \( \text{C}_3\text{H}_5\text{O}_2 \).

Step 8: Calculate Molecular Formula

Use the molar mass to find the multiple of the empirical formula:\[ \text{Empirical Mass} = (3 \times 12.01) + (5 \times 1.008) + (2 \times 16.00) = 73.08 \text{ g/mol} \]\[ \text{Factor} = \frac{146.0}{73.08} \approx 2 \]Molecular formula is \( \text{C}_6\text{H}_{10}\text{O}_4 \).

Key Concepts

Molar Mass Calculation

Understanding the concept of molar mass calculation is fundamental in organic chemistry. It allows us to determine the amount of a substance by comparing its mass to the number of moles present. The molar mass is essentially the mass of one mole of a compound, measured in grams per mole (g/mol).
To calculate the molar mass, we use the formula:

• Divide the mass of the sample by the number of moles of the compound in the sample.
• For instance, in the exercise, the diacid \(\mathrm{H}_{2}\mathrm{X}\) was given as having a mass of 0.1235 g and 0.0008455 moles.
• Using these values, we calculate the molar mass as:\[\text{Molar Mass} = \frac{0.1235 \, \text{g}}{0.0008455 \, \text{moles}} \approx 146.0 \, \text{g/mol}\]

This calculation is crucial when determining other chemical properties, like the empirical or molecular formula.

Empirical Formula Determination

The empirical formula determination is all about finding the simplest whole number ratio of atoms in a compound. This is crucial for identifying unknown compounds in organic chemistry.
The steps to determine the empirical formula include:

• Calculate moles of each element present in the compound.
• Use the combustion data, as shown in the exercise, to find moles of carbon (C) and hydrogen (H).
• For oxygen (O), subtract the masses of C and H from the total mass to find the remaining mass, which is attributed to O.
• Then, divide the moles of each element by the smallest number of moles to find the ratio:\[\text{C:} \frac{0.01425}{0.00950} \approx 1.5 \Rightarrow 3\]\[\text{H:} \frac{0.02373}{0.00950} \approx 2.5 \Rightarrow 5\]\[\text{O:} \frac{0.00950}{0.00950} = 1 \Rightarrow 2\]

This results in the empirical formula \( \text{C}_{3}\text{H}_{5}\text{O}_{2} \), which represents the simplest ratio of the elements in the compound.

Molecular Formula Analysis

Once we have the empirical formula, the next logical step is molecular formula analysis. This analysis gives us the actual number of each type of atom in a molecule, which may be a multiple of the empirical formula.
To find the molecular formula, follow these steps:

• Calculate the empirical formula mass by adding the atomic masses of all atoms in the empirical formula.
• In the example, this mass is:\[(3 \times 12.01) + (5 \times 1.008) + (2 \times 16.00) = 73.08 \, \text{g/mol}\]
• Determine how many times the empirical formula mass fits into the molar mass (determined previously as 146.0 g/mol).
• The factor is:\[\frac{146.0}{73.08} \approx 2\]
• Multiply the subscripts in the empirical formula by this factor to get the molecular formula, which in this case is \( \text{C}_{6}\text{H}_{10}\text{O}_{4} \).

The molecular formula gives us the complete picture of the composition of the compound.