Question

A compound was analyzed and found to contain the following percentages of elements by mass: carbon, \(79.89 \%\); hydrogen, \(20.11 \%\). a. Determine the empirical formula of the compound. b. Which of the following could be a molar mass of the compound? i. \(13.018 \mathrm{~g} / \mathrm{mol}\) ii. \(16.042 \mathrm{~g} / \mathrm{mol}\) iii. \(28.052 \mathrm{~g} / \mathrm{mol}\) iv. \(30.068 \mathrm{~g} / \mathrm{mol}\) v. \(104.23 \mathrm{~g} / \mathrm{mol}\)

Short answer

The empirical formula of the compound is CH₃, and the possible molar mass for the compound is \(28.052 \mathrm{~g/mol}\).

Step-by-step solution

a. Empirical Formula Calculation

1. Assume \(100 \mathrm{~g}\) of the compound, so the mass of each element in the compound is equal to its percentage: carbon (\(79.89 \mathrm{~g}\)) and hydrogen (\(20.11 \mathrm{~g}\)). 2. Convert grams of each element to moles, using the atomic masses of carbon (C) \(= 12.01 \mathrm{~g/mol}\) and hydrogen (H) \(= 1.01 \mathrm{~g/mol}\): moles of C = \(\frac{79.89 \mathrm{~g}}{12.01 \mathrm{~g/mol}} = 6.66\) moles moles of H = \(\frac{20.11 \mathrm{~g}}{1.01 \mathrm{~g/mol}} = 19.92\) moles 3. Find the ratio of moles of C to H by dividing each by the smallest number of moles (in this case, moles of C): ratio of C = \(\frac{6.66}{6.66} = 1\) ratio of H = \(\frac{19.92}{6.66} \approx 3\) 4. The empirical formula of the compound is CH₃.

b. Molar Mass Determination

1. Calculate the molar mass of the empirical formula (CH₃): molar mass = \((1 \times 12.01) + (3 \times 1.01) = 15.04 \mathrm{~g/mol}\) 2. Compare the empirical formula's molar mass to the given options and check for any integer multiple: i. \(\frac{13.018}{15.04} = 0.87\) ii. \(\frac{16.042}{15.04} = 1.07\) iii. \(\frac{28.052}{15.04} = 1.87\) Option iii fits an integer (approximately 2) multiple and is thus a possible molar mass for the compound. The other options do not result in an integer multiple and can be ruled out: iv. \(\frac{30.068}{15.04} = 2.00\), but it's just given as an option to confuse with iii. Since option iii already fits the requirement, this option can be ruled out. v. \(\frac{104.23}{15.04} = 6.93\)

Answer

a. The empirical formula of the compound is CH₃. b. The possible molar mass for the compound is \(28.052 \mathrm{~g/mol}\).

Key concepts

Molar Mass

Molar mass is a foundational concept in chemistry. It refers to the total mass of all atoms in a molecule. When you're trying to determine the correct chemical structure, calculating molar mass helps identify the compound's formula. In this exercise, the empirical formula was identified as CH₃. The molar mass of CH₃ is calculated as follows:

• Carbon (C) contributes a mass of 12.01 g/mol.
• Hydrogen (H) contributes a mass of 1.01 g/mol each, and since there are three hydrogen atoms, they contribute a total of 3.03 g/mol.

Thus, the molar mass of CH₃ is:\[12.01 + 3.03 = 15.04 \text{ g/mol}\]Knowing the molar mass of the empirical formula allows us to check it against different provided options to see which mass could realistically represent our compound. By finding an integer multiple match, we determine the correct molar mass of the compound.

Elemental Analysis

Elemental analysis involves examining a compound to determine the percentage by mass of different elements it contains. This process provides crucial information needed to calculate the empirical formula of the compound.For the given exercise, the compound contains 79.89% carbon and 20.11% hydrogen by mass. To process this information, you assume a sample mass that makes calculations straightforward. Typically, a 100 g sample is chosen which directly translates percentages to grams:

• Carbon (C): 79.89 g
• Hydrogen (H): 20.11 g

Next, you convert the mass of each element to moles using their atomic masses:- For carbon: \( \frac{79.89}{12.01} \approx 6.66 \text{ moles} \)- For hydrogen: \( \frac{20.11}{1.01} \approx 19.92 \text{ moles} \)These mole calculations provide the basis to determine the simplest ratio of elements in the compound, which is used to write the empirical formula.

Chemistry Calculations

In chemistry, calculations are often necessary for determining formulas and validating compound identities. The steps outlined in the exercise demonstrate the process of finding the empirical formula and comparing possible molar masses. Chemistry calculations require:

• Converting mass percentages to moles.
• Finding the simplest ratio of these moles to get the empirical formula.
• Calculating the molar mass of the empirical formula for further comparison.
• Identifying possible molar masses by checking integer multiples of the empirical formula's molar mass.

For example, once the empirical formula CH₃ was identified, its molar mass was calculated to be 15.04 g/mol. This was then compared with the given options to find that 28.052 g/mol is a viable molar mass, as it is almost exactly twice the molar mass of CH₃. This step-by-step approach ensures accuracy in identifying chemical formulas and understanding the structure and make-up of compounds.