Problem 114 - Chemistry Matter and Change

Question

Terephthalic acid is an organic compound used in the formation of polyesters. It contains 57.8\(\% \mathrm{C}\) , \(3.64 \% \mathrm{H},\) and 38.5\(\%\) O. The molar mass is approximately 166 \(\mathrm{g} / \mathrm{mol} .\) What is the molecular formula of terephthalic acid? (Chapter 10)

Step-by-Step Solution

Answer

The molecular formula of terephthalic acid is \(\text{C}_8\text{H}_6\text{O}_4\).

Step 1: Calculate Moles of Each Element

Assume you have 100 grams of terephthalic acid. Calculate the moles of carbon (C), hydrogen (H), and oxygen (O) in the compound using the given percent composition and atomic masses: - Moles of C = \( \frac{57.8\, \text{g}}{12.01\, \text{g/mol}} \approx 4.81\, \text{mol} \)- Moles of H = \( \frac{3.64\, \text{g}}{1.008\, \text{g/mol}} \approx 3.61\, \text{mol} \)- Moles of O = \( \frac{38.5\, \text{g}}{16.00\, \text{g/mol}} \approx 2.41\, \text{mol} \)

Step 2: Write the Empirical Formula

Divide each mole value by the smallest number of moles to find the simplest whole-number ratio. Here, divide by the smallest moles, which is for oxygen:- Moles ratio for C: \( \frac{4.81}{2.41} \approx 2 \)- Moles ratio for H: \( \frac{3.61}{2.41} \approx 1.5 \)- Moles ratio for O: \( \frac{2.41}{2.41} = 1 \)The empirical formula isn't yet in whole numbers: \(\text{C}_2\text{H}_{1.5}\text{O}_1\). We need whole numbers, so multiply all by 2 to simplify: \(\text{C}_4\text{H}_3\text{O}_2\).

Step 3: Determine Molecular Formula

Calculate the empirical formula mass of \(\text{C}_4\text{H}_3\text{O}_2\):- \(4 \times 12.01\, \text{g/mol} + 3 \times 1.008\, \text{g/mol} + 2 \times 16.00\, \text{g/mol} \approx 83.072\, \text{g/mol}\).Compare this with the given molar mass of 166 \(\text{g/mol}\). Find the multiplier \(n\) by dividing the molar mass by the empirical formula mass:\[ n = \frac{166}{83.072} \approx 2 \]Hence, the molecular formula is \(\text{C}_8\text{H}_6\text{O}_4\).

Key Concepts

Empirical Formula Calculation

An empirical formula represents the simplest whole-number ratio of different types of atoms in a compound. To calculate it, the first step involves determining the number of moles of each element within a given sample. Think of it as laying the foundation for understanding a compound's basic composition.

For example, consider a compound with a mass percentage of each element provided. The first method to determine moles is by assuming you have a 100 grams sample of the compound. This way, the percentage of each element translates directly to grams.

• Convert grams to moles using each element's atomic mass. For carbon (C), hydrogen (H), and oxygen (O), this might look like: \( \text{moles of C} = \frac{\text{mass of C}}{12.01 \, \text{g/mol}} \),\( \text{moles of H} = \frac{\text{mass of H}}{1.008 \, \text{g/mol}} \), and\( \text{moles of O} = \frac{\text{mass of O}}{16.00 \, \text{g/mol}} \).

Then by dividing all mole values by the smallest number of moles calculated, you determine the smallest ratio.This mathematical action yields a preliminary empirical formula, which might sometimes feature fractional subscripts.

To resolve any fractional subscripts, multiply all elements by an appropriate integer so the subscripts are whole numbers. This gives you the empirical formula preferred for chemical calculations.

Molecular Formula Determination

The molecular formula provides the actual number of each type of atom in a molecule, unlike the empirical formula which only gives a ratio. To find the molecular formula, you often need the compound's empirical formula and its molar mass.

Start with the empirical formula mass, which is the combined atomic mass of all atoms present in the empirical formula:

• Use the empirical formula to sum up the atomic masses of each atom. For example, \( C_4H_3O_2: (4 \times 12.01) + (3 \times 1.008) + (2 \times 16.00) = 83.072 \, \text{g/mol} \).

Next, compare the empirical formula mass to the compound's known molar mass. The molecular formula is determined by finding the factor or multiplying number \( n \) that scales up the empirical formula mass to the molar mass:

• Calculate \( n = \frac{\text{molar mass}}{\text{empirical formula mass}} \).

Multiply each subscript in the empirical formula by \( n \) to find the molecular formula. Thus, the empirical formula \( C_4H_3O_2 \) becomes the molecular formula \( C_8H_6O_4 \) when scaled by \( n \approx 2 \). Hence, this reveals the true number of each atom type within a molecule.

Moles and Percent Composition

Understanding moles and percent composition is crucial in chemistry, especially when deciphering a compound's formula. The concept of moles serves as a bridge between the macroscopic world we can measure and the microscopic world of atoms and molecules.

Each element has a characteristic atomic mass in grams per mole. By knowing the number of moles present, you essentially know how much of the substance is present on an atomic level. Consider a measurement example:

• If a compound is made of 60% carbon by mass, in a 100g sample, you'd have 60 grams of carbon, which can be converted to moles for detailed study.

Percent composition, on the other hand, reveals the relative weight percentages of each element within a compound. This unlocks a straightforward analysis of the primary components of any chemical compound.

• It aids in calculating the weight of elements like \( \text{percent }_{\text{C}} = \frac{\text{mass of C}}{\text{total mass}} \times 100 \% \), alongside hydrogen and oxygen for composition purposes.

These calculations form a foundation for further chemical analysis. They transform measurable mass into a vivid understanding of molecular makeup, supporting empirical and molecular formula determinations.