Problem 80 - Chemistry The Molecular Science

Question

L-Dopa is a drug used for the treatment of Parkinson's disease. Elemental analysis shows it to be \(54.82 \%\) carbon, \(7.10 \%\) nitrogen, \(32.46 \%\) oxygen, and the remainder hydrogen. (a) Determine \(\mathrm{L}\) -dopa's empirical formula. (b) The molar mass of \(\mathrm{L}\) -dopa is \(197.19 \mathrm{~g} / \mathrm{mol}\). Determine its molecular formula.

Step-by-Step Solution

Answer

(a) Empirical formula is \( C_9H_{11}NO_4 \). (b) Molecular formula is \( C_9H_{11}NO_4 \).

Step 1: Determine percent hydrogen

Given the percentages for carbon, nitrogen, and oxygen, calculate the percentage of hydrogen. Since the sum of percentages must be 100%, subtract the given percentages from 100%:\[\text{Percent hydrogen} = 100\% - (54.82\% + 7.10\% + 32.46\%) = 5.62\%\]

Step 2: Convert percentages to grams

Assume you have 100 grams of L-Dopa, which makes the calculations straightforward because the percentage for each element directly converts to grams: - Carbon (C): 54.82 grams - Nitrogen (N): 7.10 grams - Oxygen (O): 32.46 grams - Hydrogen (H): 5.62 grams.

Step 3: Convert grams to moles

Use the molar mass of each element to convert grams to moles:- Carbon: \( \frac{54.82 \text{g}}{12.01 \text{g/mol}} = 4.566 \text{ moles} \)- Nitrogen: \( \frac{7.10 \text{g}}{14.01 \text{g/mol}} = 0.507 \text{ moles} \)- Oxygen: \( \frac{32.46 \text{g}}{16.00 \text{g/mol}} = 2.029 \text{ moles} \)- Hydrogen: \( \frac{5.62 \text{g}}{1.008 \text{g/mol}} = 5.579 \text{ moles} \)

Step 4: Determine empirical formula ratios

Divide each element's mole amount by the smallest number of moles, which is for nitrogen (0.507), to find the ratio:- C: \( \frac{4.566}{0.507} = 9.005 \approx 9 \)- N: \( \frac{0.507}{0.507} = 1 \)- O: \( \frac{2.029}{0.507} = 4.001 \approx 4 \)- H: \( \frac{5.579}{0.507} = 11.001 \approx 11 \)Thus, the empirical formula is \( C_9H_{11}NO_4 \).

Step 5: Calculate empirical formula mass

Calculate the molar mass of the empirical formula \( C_9H_{11}NO_4 \):- Carbon: \( 9 \times 12.01 = 108.09 \)- Hydrogen: \( 11 \times 1.008 = 11.088 \)- Nitrogen: \( 1 \times 14.01 = 14.01 \)- Oxygen: \( 4 \times 16.00 = 64.00 \)Adding these gives the empirical formula mass:\[ \text{Empirical mass} = 108.09 + 11.088 + 14.01 + 64.00 = 197.188 \text{ g/mol} \]

Step 6: Determine molecular formula

The empirical formula mass is approximately the same as the given molecular mass of L-Dopa (197.19 g/mol), which means the empirical formula is the molecular formula. Therefore, the molecular formula is also \( C_9H_{11}NO_4 \).

Key Concepts

Molecular Formula Determination

To determine the molecular formula of a compound, we start with its empirical formula. This is the simplest whole number ratio of elements in the compound. First, we calculate the empirical formula mass by adding up the molar masses of all the atoms in the empirical formula. Once we have the empirical formula mass, it's used alongside the actual molar mass of the compound to identify the molecular formula. If the empirical formula mass equals the given molar mass, as it does for L-Dopa in this example, then the empirical formula and molecular formula are identical. In cases where the empirical mass is a fraction of the molar mass, we multiply the subscripts in the empirical formula by the factor that converts the empirical mass to the actual molar mass. This helps us find the correct number of atoms of each element in a molecule.

Elemental Analysis

Elemental analysis is a technique used to determine the percentage composition of each element in a compound. This involves breaking down a compound into its basic elements and measuring the amount of each. In the case of L-Dopa, we calculated the percentages of carbon, nitrogen, and oxygen from the elemental analysis given, and by subtracting these from 100%, we find the percentage of hydrogen. This comprehensive method provides the foundation for the empirical formula determination as it quantifies each element's contribution to the compound, simplifying molecular calculations later on.

Percent Composition

Percent composition is the percentage by mass of each element in a compound. To get these percentages, you assume a certain mass of the compound, typically 100 grams, which makes the math straightforward. For example, in L-Dopa, the percent composition was given as 54.82% carbon, 7.10% nitrogen, and 32.46% oxygen, with the rest being hydrogen. Using percent composition allows students to easily convert these percentages into grams, and subsequently into moles, which are critical steps in deriving the empirical formula. Understanding percent composition is vital for identifying how the compound's mass is distributed among different elements.

Molar Mass Conversion

Converting molar mass is a key step in solving chemical formula problems. First, convert element percentages to grams assuming 100 grams of the compound. Then, use the molar mass of each element to convert these grams into moles. For instance, for carbon in L-Dopa, the computation was \( \frac{54.82 \text{ g}}{12.01 \text{ g/mol}} = 4.566 \text{ moles} \). Repeat this for all elements. This conversion is essential for establishing the ratio of atoms in the empirical formula. Finally, we ensure these elemental moles are accurately scaled to find both the empirical and molecular formulas, aligning with the compound's actual molar mass.