Question

A compound contains \(47.08 \%\) carbon, \(6.59 \%\) hydrogen, and \(46.33 \%\) chlorine by mass; the molar mass of the compound is \(153 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Short answer

The compound contains carbon, hydrogen, and chlorine in the approximate mole ratios of 3:5:1, yielding the empirical formula C3H5Cl. Using the given molar mass of 153 g/mol, the molecular formula is determined to be C6H10Cl2.

Step-by-step solution

Calculate the moles of each element

To find the mole ratios, we will assume that we have exactly 100 g of the compound. This means that the given percentages can be directly used as masses. Then, we will convert the mass of each element to moles using their molar masses. - Carbon: \(47.08 g\) C (\(\frac{1 mol}{12.01 g}\)) = \(3.92\) mol C - Hydrogen: \(6.59 g\) H (\(\frac{1 mol}{1.008 g}\)) = \(6.54\) mol H - Chlorine: \(46.33 g\) Cl (\(\frac{1 mol}{35.45 g}\)) = \(1.31\) mol Cl

Determine the empirical formula

First, divide each mole value by the smallest mole value to find the mole ratio of each element. - Carbon: \(\frac{3.92}{1.31}\) ≈ \(3\) - Hydrogen: \(\frac{6.54}{1.31}\) ≈ \(5\) - Chlorine: \(\frac{1.31}{1.31}\) = \(1\) The empirical formula is C3H5Cl.

Calculate the molar mass of the empirical formula

Now we need to find the molar mass of the empirical formula to determine how many times the empirical formula repeats in the molecular formula. Molar mass of C3H5Cl = (3 × 12.01) + (5 × 1.008) + (1 × 35.45) ≈ \(77.5 g/mol\)

Find the molecular formula

We can find the number of times the empirical formula repeats in the molecular formula by dividing the given molar mass by the empirical formula's molar mass and rounding to the nearest whole number: \(\frac{153 g/mol}{77.5 g/mol}\) ≈ \(2\) The molecular formula is thus the empirical formula repeated 2 times: (C3H5Cl)2 = C6H10Cl2. The empirical formula is C3H5Cl, and the molecular formula is C6H10Cl2.

Key concepts

Empirical Formula

The empirical formula of a compound provides the simplest whole-number ratio of elements within the compound. Think of it as a recipe that uses the fewest ingredients possible. To find the empirical formula, we start by determining the percentage composition of each element. We often assume a 100 g sample, which directly translates the percentage values to grams. From there, we convert these masses to moles using each element's molar mass.
For example, in a compound with 47.08% carbon, we have 47.08 g of carbon in our 100 g sample. We convert this mass using the molar mass of carbon, 12.01 g/mol, resulting in approximately 3.92 moles of carbon. This process is repeated for each element, giving us the mole ratios.
By dividing each mole amount by the smallest number of moles present among the elements, we ensure that the results reflect the simplest whole-number ratio, crucial for defining the compound's empirical formula.

Mole Ratio

The concept of mole ratio is essential in chemistry as it translates raw data into a simplified, usable format. Once we have calculated the moles for each element in a compound, the next step is to compare them relative to one another. To do this, each mole value is divided by the smallest mole quantity obtained among the elements.
For instance, if a compound has calculated moles of 3.92, 6.54, and 1.31 for carbon, hydrogen, and chlorine respectively, the smallest value here is 1.31 (chlorine). Dividing all mole numbers by 1.31 aligns these values to a simple ratio: 3:5:1. These numbers are crucial because they become the indices of the elements in the empirical formula, showing their relative amounts. This systematic approach helps chemists understand the internal structure of a compound.

Molar Mass

Understanding molar mass is vital because it bridges the gap between the empirical formula and the molecular formula. The molar mass tells us the mass of one mole of a substance. For example, if we have one mole of carbon atoms, it weighs approximately 12.01 grams.
To find the full molar mass of a compound, we sum up the molar masses of each atom in its empirical formula. For the empirical formula C3H5Cl, the molar mass is calculated by adding the weights of all its constituent atoms:

• 3 carbon atoms, each weighing 12.01 grams
• 5 hydrogen atoms, each weighing 1.008 grams
• 1 chlorine atom weighing 35.45 grams

The total gives us a molar mass of about 77.5 g/mol for the empirical formula. This figure helps us determine the number of times the empirical formula is present in the actual molecular formula by comparing it to the compound's given molar mass.

Chemical Composition

Chemical composition refers to the types and amounts of elements present in a compound, expressed in terms of their relative abundance. It provides a deeper understanding of what makes up a substance, much like a list of ingredients in a recipe.
Knowing the chemical composition is key to figuring out both empirical and molecular formulas. This is because it reveals the percentage, mass, and mole ratio of each element within the substance. For example, in our compound, 47.08% was carbon, 6.59% was hydrogen, and 46.33% was chlorine. These percentages allow us to determine the gram amounts in a 100 g sample, setting the stage for further analysis.
This deep dive into chemical composition also assists in knowing how each element contributes to the compound's total mass and reacts during chemical changes. It forms the foundation of understanding chemical phenomena and real-world applications of chemistry.