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 empirical formula of the compound is C3H5Cl, and the molecular formula is C6H10Cl2.

Step-by-step solution

Calculate moles of each element

Assume there are 100 grams of the compound. The mass of each element is given as a percentage of the total mass. Therefore, - Carbon: 47.08 g - Hydrogen: 6.59 g - Chlorine: 46.33 g Now, we will convert the mass of each element to moles using their respective molar masses: - Carbon: 12.01 g/mol (C) - Hydrogen: 1.01 g/mol (H) - Chlorine: 35.45 g/mol (Cl) Moles of each element: - Carbon: \( \frac{47.08}{12.01} \approx 3.92 \) moles - Hydrogen: \( \frac{6.59}{1.01} \approx 6.52 \) moles - Chlorine: \( \frac{46.33}{35.45} \approx 1.31 \) moles

Find the empirical formula

To find the empirical formula, we need to determine the simplest whole number ratio of the moles of each element: - Carbon: \( \frac{3.92}{1.31} \approx 3 \) - Hydrogen: \( \frac{6.52}{1.31} \approx 5 \) - Chlorine: \( \frac{1.31}{1.31} \approx 1 \) Now, we can see that the empirical formula is C3H5Cl.

Find the molecular formula

Using the molar mass of the empirical formula, we can find the molecular formula. The molar mass of the empirical formula is as follows: - Carbon: 3 x 12.01 g/mol (C) = 36.03 g/mol - Hydrogen: 5 x 1.01 g/mol (H) = 5.05 g/mol - Chlorine: 1 x 35.45 g/mol (Cl) = 35.45 g/mol Total molar mass of the empirical formula: \(36.03 + 5.05 + 35.45 \approx 76.53 \: g/mol \) Now, we will find the ratio of the molar mass of the compound to the molar mass of the empirical formula: Ratio: \( \frac{153}{76.53} \approx 2 \) Since the ratio is approximately 2, this means that the molecular formula is twice the empirical formula. Therefore: Molecular formula: (C3H5Cl)2 = C6H10Cl2 The empirical formula of the compound is C3H5Cl, and the molecular formula is C6H10Cl2.

Key concepts

Molar Mass

The molar mass is a fundamental concept in chemistry and is crucial in finding both empirical and molecular formulas. It refers to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To determine the molar mass of the compound from a formula, you simply add up the molar masses of all the atoms present. You can find the molar mass of each element on the periodic table. For example:

• Carbon (C) has a molar mass of 12.01 g/mol.
• Hydrogen (H) has a molar mass of 1.01 g/mol.
• Chlorine (Cl) has a molar mass of 35.45 g/mol.

When given the molar mass of a compound, like the 153 g/mol in our exercise, you can use it to determine the molecular formula once you've established the empirical formula. By dividing the given molar mass of the compound by the molar mass of the empirical formula, you can find out how many empirical formula units are in the molecular formula.

Percentage Composition

Percentage composition is a way of describing a compound in terms of the percentage by mass of each element present in it. This concept is essential for converting mass percentages into a pattern that quickly helps in determining the empirical formula. In our exercise:

• 47.08% carbon refers to the mass of carbon relative to the total mass of the compound.
• 6.59% hydrogen signifies the mass of hydrogen relative to the total.
• 46.33% chlorine reflects the mass of chlorine in the compound.

To use percentage composition in calculations, you assume a certain mass of the compound, commonly 100 g, for ease of converting percentages to grams. Once you have the grams, you then convert them to moles using the respective molar masses. This step bridges the mass composition to a mole ratio needed for empirical formula calculation.

Chemical Formula Calculation

Calculating the chemical formula of a compound begins with deriving the empirical formula. First, using the percentage composition to determine the mass of each element, you convert these masses into moles. This gives you the mole ratio, which is then simplified to the smallest whole number ratio to find the empirical formula. In our given compound:

• The empirical formula, after simplification, is C\(_3\)H\(_5\)Cl.

Once the empirical formula is known, calculate its molar mass. Using the molar mass of the entire compound, as provided, divide this by the empirical formula's molar mass to obtain a ratio. This ratio signifies how many times the empirical formula needs to be multiplied to match the molecular formula. In our exercise: - We find the ratio by dividing 153 g/mol by the molar mass of C\(_3\)H\(_5\)Cl.- The resulting ratio (approximately 2) means the molecular formula is twice the empirical formula, leading us to C\(_6\)H\(_10\)Cl\(_2\).This approach ensures the molecular formula fully reflects the structure of the compound, aligning its mass and elemental composition.