Question

An unknown liquid is composed of \(5.57 \% \mathrm{H}, 28.01 \% \mathrm{Cl}\), and \(66.42 \%\) C. The molecular weight found by mass spectrometry is \(126.58\). What is the molecular formula of the compound?

Short answer

The molecular formula is \(\text{C}_7\text{H}_7\text{Cl}\).

Step-by-step solution

Convert Percentages to Grams

Assume you have a 100g sample, which makes it easier to convert percentages to grams. Therefore, you have 5.57 grams of hydrogen (H), 28.01 grams of chlorine (Cl), and 66.42 grams of carbon (C).

Calculate Moles of Each Element

Convert the grams of each element to moles using their respective molar masses: the molar mass of H is approximately 1 g/mol, of Cl is approximately 35.45 g/mol, and of C is approximately 12 g/mol.- Moles of H = \( \frac{5.57\ ext{g}}{1\ ext{g/mol}} = 5.57\ ext{mol} \) - Moles of Cl = \( \frac{28.01\ ext{g}}{35.45\ ext{g/mol}} = 0.79\ ext{mol} \) - Moles of C = \( \frac{66.42\ ext{g}}{12\ ext{g/mol}} = 5.54\ ext{mol} \)

Find the Simplest Mole Ratio

To find the simplest whole number ratio, divide each of the mole amounts by the smallest number of moles calculated, which is 0.79 mol of Cl.- Ratio of H = \( \frac{5.57}{0.79} \approx 7.05 \) - Ratio of Cl = \( \frac{0.79}{0.79} = 1.00 \) - Ratio of C = \( \frac{5.54}{0.79} \approx 7.01 \)Round these ratios to the nearest whole number: H = 7, Cl = 1, C = 7.

Determine Empirical Formula

Based on the simplest mole ratios, the empirical formula of the compound is \(\text{C}_7\text{H}_7\text{Cl}\).

Calculate Empirical Formula Molar Mass

Calculate the molar mass of the empirical formula. - C: \(7 \times 12\ ext{g/mol} = 84 \ ext{g/mol}\) - H: \(7 \times 1\ ext{g/mol} = 7\ ext{g/mol}\) - Cl: \(1 \times 35.45\ ext{g/mol} = 35.45\ ext{g/mol}\) Total = 84 + 7 + 35.45 = 126.45 g/mol.

Determine Molecular Formula

Divide the given molecular weight by the empirical formula molar mass to find the multiplier.\[\frac{126.58\ ext{g/mol}}{126.45\ ext{g/mol}} \approx 1\]Since the ratio is approximately 1, the empirical formula is also the molecular formula.

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. It does not provide the exact number of atoms, but it gives a basic understanding of the composition.
To find it, you first need to determine the number of moles of each element in a sample. By using the percentages given, you convert them into grams, assuming a 100g total sample for simplicity:

• For instance, in the exercise, the compound is said to have 5.57% hydrogen, 28.01% chlorine, and 66.42% carbon by mass.
• This means out of 100g, you'll have 5.57g of hydrogen, 28.01g of chlorine, and 66.42g of carbon.

Understanding how to get these correct ratios helps break down the chemical composition to its simplest form easily.

Mass Spectrometry

Mass spectrometry is a technique used to determine the molecular weight and composition of a compound.
This process involves ionizing chemical species and sorting the ions based on their mass-to-charge ratio. It's especially useful for confirming the molecular formula.
The molecular weight obtained by mass spectrometry in this exercise is 126.58 g/mol. This value is crucial for determining if the empirical formula might differ from the molecular formula because you compare it with the empirical formula's molar mass. A close match means the empirical and molecular formulas are identical, as is the case here.

Mole Ratio Calculation

Mole ratio calculation helps in finding the relative number of moles of each element, leading to an empirical formula.
In the exercise, we've calculated moles by dividing each element's mass by its respective molar mass.

• Hydrogen: 5.57 grams converts to 5.57 moles;
• Chlorine: 28.01 grams converts to approximately 0.79 moles;
• Carbon: 66.42 grams converts to approximately 5.54 moles.

To find the simplest ratio, each number of moles is divided by the smallest value calculated, which is that of chlorine. This process simplifies the ratios into whole numbers, giving the empirical formula. Round off these results to the nearest whole number to complete the ratio analysis.

Molecule Composition Analysis

Molecule composition analysis involves understanding each element’s role and quantity in a specific molecule.
It starts with knowledge of the empirical formula but extends into evaluating the molecular structure. In our context, once we have the empirical formula \(\text{C}_7\text{H}_7\text{Cl}\), the molecular formula is determined by comparing the empirical molar mass to the mass given by mass spectrometry.
Since they match closely in this situation (126.58 g/mol), the empirical formula is confirmed as the molecular formula, meaning the ratios and amounts determined are correct for the actual molecular structure of the compound.

• This confirms that each unit of the compound contains exactly seven carbon atoms, seven hydrogen atoms, and one chlorine atom.

Such analysis is essential for chemists to ascertain not just what a compound consists of, but how the atoms are potentially arranged.