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 empirical formula of the compound is \(C_3H_5Cl\), and the molecular formula is \(C_6H_{10}Cl_2\).

Step-by-step solution

Convert mass percentages to grams

Since the sample percentage gives the mass composition, we assume we have a 100 g sample of the compound. This way, the percentages can directly be considered as grams. Therefore, the composition of the compound is: - 47.08 g of carbon - 6.59 g of hydrogen - 46.33 g of chlorine

Convert grams into moles

The next step is to convert the masses of carbon, hydrogen, and chlorine into moles. Using their respective atomic masses: - Carbon: \(1 \mathrm{~mol ~C} = 12.01 \mathrm{~g}\) - Hydrogen: \(1 \mathrm{~mol ~H} = 1.008 \mathrm{~g}\) - Chlorine: \(1 \mathrm{~mol ~Cl} = 35.45 \mathrm{~g}\) Calculating the moles for each element: Moles of Carbon \(C = \frac{47.08 \mathrm{~g}}{12.01 \mathrm{~g/mol}} \approx 3.92 \mathrm{~mol}\) Moles of Hydrogen \(H = \frac{6.59 \mathrm{~g}}{1.008 \mathrm{~g/mol}} \approx 6.54 \mathrm{~mol}\) Moles of Chlorine \(Cl = \frac{46.33 \mathrm{~g}}{35.45 \mathrm{~g/mol}} \approx 1.31 \mathrm{~mol}\)

Determine the empirical formula

To find the empirical formula, we need to find the simplest whole number ratio of the elements present in the compound. Divide all the moles calculated above by the smallest mole value (1.31): Mole ratio of Carbon: \(\frac{3.92}{1.31} \approx 2.99 \approx 3\) Mole ratio of Hydrogen: \(\frac{6.54}{1.31} \approx 4.99 \approx 5\) Mole ratio of Chlorine: \(\frac{1.31}{1.31} = 1\) Thus, the empirical formula of the compound is \(C_3H_5Cl\)

Calculate the molar mass of the empirical formula

Next, find the molar mass of the empirical formula by multiplying the atomic masses of the elements by their mole ratio in the empirical formula: Molar mass of \(C_3H_5Cl: (3 \times 12.01) + (5\times 1.008) + (1 \times 35.45) \approx 76.5 \mathrm{~g/mol}\)

Determine the molecular formula

To find the molecular formula, we need to compare the molar mass of the compound (153 g/mol) to the molar mass of the empirical formula (76.5 g/mol). The ratio of the molar masses: \(\frac{153 \mathrm{~g/mol}}{76.5 \mathrm{~g/mol}} \approx 2\) The molecular formula is found by multiplying the empirical formula by this ratio: \(C_3H_5Cl \times 2 = C_6H_{10}Cl_2\) The molecular formula of the compound is \(C_6H_{10}Cl_2\).

Key concepts

Chemical Composition Analysis

Understanding the chemical composition of a substance is like deciphering a recipe for a special dish; it tells you what ingredients are present and in what amounts. In chemistry, knowing the percentage of each element in a compound is critical for identifying the substance and understanding its properties and potential reactions. To begin analyzing a compound's chemical composition, scientists often start with a given mass percentage, which conceptually represents a proportion out of a 100 gram sample. This approach simplifies calculations, as it directly equates the mass percentage to grams of each element.

The exercise provided reflects this method; it utilizes the mass percentages to express how much of each element—carbon, hydrogen, and chlorine—is contained in the sample. By stating the compound is composed of 47.08% carbon, 6.59% hydrogen, and 46.33% chlorine, we have a clear view of the 'chemical recipe' of this compound. This initial step sets the stage for further analysis and aids in deducing the empirical and molecular formulas, which are essentially the chemical IDs of the substance.

Molar Mass Calculation

To transition from grams to moles, a concept pivotal for understanding chemical quantities, we use molar mass, the mass of one mole of a substance, measured in grams per mole (g/mol). Think of it as converting a measurement from one unit (grams) to another (moles), using atomic mass as a conversion factor. Every element has its own unique molar mass, reflective of its atomic weight.

In our example, the conversion from grams to moles has been beautifully demonstrated. The molar mass of carbon is roughly 12.01 g/mol, hydrogen is about 1.008 g/mol, and chlorine comes in at around 35.45 g/mol. Dividing the given masses by these molar masses gives us moles of each element, revealing the amounts of each ingredient that make up the compound in a mole-based perspective, which is the language of chemistry.

Percent Composition by Mass

The percent composition by mass of a compound shows which portion of a compound's total mass is taken up by each component element. Put plainly, it indicates how much of the compound's weight comes from each type of atom within it. The concept is akin to understanding what share of a fruit salad's weight comes from apples, bananas, or grapes.

Using the earlier presented percentages, we observe how much 'weight' each element carries in the total mass of the compound. For accuracy, the mass percentages must add up to 100%, as they represent the entire mass of the compound. Calculating percent composition involves dividing the mass of each element in a sample by the total molar mass and then multiplying by 100 to obtain a percentage. This breakdown is key for tasks from verifying the purity of a sample to calculating empirical and molecular formulas.