Question

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

Short answer

The compound's empirical formula is \(C_3H_5Cl\), and its molecular formula is \(C_6H_{10}Cl_2\).

Step-by-step solution

Convert mass percentages to mass in grams

Assuming a \(100 \, g\) sample of the compound, we can convert the percentages directly to mass in grams: - \(47.08 \%\) carbon corresponds to \(47.08 \, g\) of carbon. - \(6.59 \%\) hydrogen corresponds to \(6.59 \, g\) of hydrogen. - \(46.33 \%\) chlorine corresponds to \(46.33 \, g\) of chlorine.

Convert mass to moles for each element

Now, we can convert the grams of each element into moles using their respective molar masses (C: \(12.01 \, g/mol\), H: \(1.01 \, g/mol\), Cl: \(35.45 \, g/mol\)): - \(47.08 \, g \, C \cdot \frac{1 \, mol \, C}{12.01 \, g \, C} \approx 3.92 \, mol \, C\) - \(6.59 \, g \, H \cdot \frac{1 \, mol \, H}{1.01 \, g \, H} \approx 6.52 \, mol \, H\) - \(46.33 \, g \, Cl \cdot \frac{1 \, mol \, Cl}{35.45 \, g \, Cl} \approx 1.31 \, mol \, Cl\)

Determine the mole ratio

Now, we divide each number of moles by the smallest of the three numbers (which is \(1.31 \, mol\)): - \(\frac{3.92 \, mol \, C}{1.31} \approx 3\) - \(\frac{6.52 \, mol \, H}{1.31} \approx 5\) - \(\frac{1.31 \, mol \, Cl}{1.31} = 1\) Thus, the mole ratio of the elements is \(C: H: Cl = 3: 5: 1\).

Determine the empirical formula

Using the mole ratio, we can determine the empirical formula to be \(C_3H_5Cl\).

Determine the empirical formula's molar mass and find the molecular formula

The molar mass of the empirical formula \(C_3H_5Cl\) is calculated as follows: - Carbon: \(3 \times (12.01 \, g/mol) = 36.03 \, g/mol\) - Hydrogen: \(5 \times (1.01 \, g/mol) = 5.05 \, g/mol\) - Chlorine: \(1 \times (35.45 \, g/mol) = 35.45 \, g/mol\) Adding all the individual molar masses, we get a total of approximately \(76.53 \, g/mol\). The problem states that the molar mass of the compound is \(153 \, g/mol\). To find the molecular formula, divide the molar mass of the compound by the empirical formula's molar mass and multiply the empirical formula by the result: - \(153 \, g/mol \div 76.53 \, g/mol \approx 2\) Hence, the molecular formula is \(2 \times (C_3H_5Cl) = C_6H_{10}Cl_2\).

Key concepts

Percent Composition

Understanding percent composition is crucial for determining the chemical makeup of a compound. It tells us the percentage, by mass, of each element within a compound. To calculate percent composition, we divide the mass of a specific element by the total mass of the compound and multiply by 100.For a compound with a given mass, finding percent composition requires converting mass percentages to actual masses, as seen in our exercise. The given percentages were presumed to be out of a total mass of 100 grams for simplicity. Knowing the percent composition enables us to transition towards the empirical formula of a compound, a steppingstone in identifying its molecular structure.

For students, a common improvement tip is to make sure the percent composition is correctly calculated and understood, as it's the foundation for the next steps in determining a compound's empirical and molecular formulas.

Molar Mass

Molar mass is the mass of one mole of a substance, generally expressed in grams per mole (g/mol). It's an intrinsic property of each element and compound, which allows us to convert between the mass of a substance and the number of moles. The molar mass of elements is found on the periodic table and is necessary for converting grams of elements to moles as shown in step 2 of our solution.

Relevance in Calculations

When calculating molar mass for a compound's empirical formula, we sum the molar masses of each individual element, multiplied by their quantity in the formula. This step is crucial to determine the mole-to-mole ratio and consequently helps in identifying the molecular formula, which involves comparing the empirical formula's molar mass to the given molar mass of the compound.

Mole Ratio

The mole ratio is the ratio of moles of one substance to the moles of another substance in a balanced chemical equation. It's a concept that lies at the heart of stoichiometry. In the context of our exercise, after converting the grams of each element to moles, the next step is to find the simplest whole number ratio of the moles of elements, which essentially gives us the empirical formula.

Application in Solving Problems

The smallest number of moles from step 2 is used as a divisor to determine the mole ratio. Students are often advised to ensure that they round off to the nearest whole number, unless the number is close to a simple fraction (like 1.5, which can be rounded to 3/2), to accurately reflect the proportion of elements in the empirical formula. This mole ratio guides us toward the actual chemical formula, linking the composition at the atomic level with the macroscopic amounts we began with.

Chemical Formula

A chemical formula represents the types and numbers of atoms in a substance. There are two main types of chemical formulas: empirical and molecular. The empirical formula is the simplest whole number ratio of atoms in a compound, while the molecular formula is the actual number of atoms of each element in a molecule of the compound.To find the molecular formula, relate the compound's molar mass to the empirical formula's molar mass. This comparison might result in a whole number, which is then used to multiply the empirical formula, just as done in step 5 of our exercise. Students often get confused between the empirical and molecular formulas; it's important to understand that while the empirical formula provides the simplest ratio, the molecular formula gives us the true number of atoms, which can be multiples of the empirical formula. Highlighting the relationship between these different formulas can significantly improve comprehension and application in solving chemistry problems.