Question

Serotonin is a compound that conducts nerve impulses in the brain. It contains \(68.2\) mass percent \(C, 6.86\) mass percent \(H\), \(15.9\) mass percent \(\mathrm{N}\), and \(9.08\) mass percent \(\mathrm{O}\). Its molar mass is \(176 \mathrm{~g} / \mathrm{mol}\). Determine its molecular formula.

Short answer

The molecular formula of Serotonin is C10HNO.

Step-by-step solution

Convert mass percentages to grams

Since we have the mass percentages of each element in Serotonin, we can convert these percentages into grams. By assuming we have 100g of Serotonin, the mass of each element in grams will be equal to its mass percentage. C: 68.2 g H: 6.86 g N: 15.9 g O: 9.08 g

Calculate moles of each element

Now, we will convert the mass of each element in grams to moles using their molar masses. Molar masses: C: 12.01 g/mol H: 1.008 g/mol N: 14.01 g/mol O: 16.00 g/mol Moles of each element: C: \( \frac{68.2}{12.01} \) H: \( \frac{6.86}{1.008} \) N: \( \frac{15.9}{14.01} \) O: \( \frac{9.08}{16} \)

Find the simplest whole number ratio of moles

To find the empirical formula, we need to find the simplest whole number ratio of the moles of each element. We will do this by dividing all the mole values by the smallest mole value, then rounding to the nearest whole number: Smallest mole value = Moles of H C: \(\frac{68.2 / 12.01}{6.86 / 1.008} \) H: \(\frac{6.86 / 1.008}{6.86 / 1.008} \) N: \(\frac{15.9 / 14.01}{6.86 / 1.008} \) O: \(\frac{9.08 / 16}{6.86 / 1.008} \) Rounded to the nearest whole number, we get: C: 10 H: 1 N: 1 O: 1 So the empirical formula is C10H1N1O1 or C10HNO.

Determine the molecular formula

Now, we will find the molar mass of the empirical formula and use it to determine the molecular formula by comparing it with the given molar mass of Serotonin (176 g/mol). Empirical formula molar mass: C: 10 x 12.01 = 120.1 g/mol H: 1 x 1.008 = 1.008 g/mol N: 1 x 14.01 = 14.01 g/mol O: 1 x 16 = 16 g/mol Total: 120.1 + 1.008 + 14.01 + 16 = 151.118 g/mol Next, we will divide the given molar mass of Serotonin by the empirical formula's molar mass: \(\frac{176}{151.118} \) ≈ 1.16 Since the ratio is close to 1, this indicates that the molecular formula is roughly the same as the empirical formula. Given that we are limited by the number of significant figures used in the initial mass percentages, we consider C10HNO the molecular formula of Serotonin.

Key concepts

Elemental Composition

Understanding the elemental composition of a compound is key to determining its molecular structure. In the case of serotonin, a neurotransmitter in the brain, we started with percentages:

• 68.2% Carbon (C)
• 6.86% Hydrogen (H)
• 15.9% Nitrogen (N)
• 9.08% Oxygen (O)

These percentages reflect how much of each element is present. To simplify calculations, it is often helpful to imagine we have a 100g sample. This way, the mass percentages translate directly into grams, making the next steps in the process intuitive and straightforward. These values lay the groundwork for determining the empirical formula, which will eventually assist us in finding the molecular formula.

Molar Mass Calculation

The calculation of molar mass is a crucial step in converting mass percentages into practical mole values. To achieve this, you must know the molar mass of each element in a compound:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.008 g/mol
• Nitrogen (N): 14.01 g/mol
• Oxygen (O): 16.00 g/mol

By using these values, you can compute the number of moles for each element using the formula: \[\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass}}\]This step converts the mass of each element in the 100g sample into moles, which are then used to find the empirical formula. The conversion is essential because it allows for comparison on a mole basis, reflecting how substances react in real chemical processes.

Empirical Formula

An empirical formula represents the simplest whole-number ratio of elements in a compound. To find this for serotonin, we take the mole values from the previous step. The aim is to divide all by the smallest number of moles among the elements, which was Hydrogen (H) in this case. Doing so gives:

• Carbon: \( \frac{68.2 / 12.01}{6.86 / 1.008} \) ≈ 10
• Hydrogen: \( \frac{6.86 / 1.008}{6.86 / 1.008} \) = 1
• Nitrogen: \( \frac{15.9 / 14.01}{6.86 / 1.008} \) ≈ 1
• Oxygen: \( \frac{9.08 / 16}{6.86 / 1.008} \) ≈ 1

This results in the empirical formula \(\text{C}_{10}\text{H}_1\text{N}_1\text{O}_1\)\, or simply \(\text{C}_{10}\text{HNO}\). The empirical formula provides a basic picture of the compound's composition, though it may not always match the actual molecular formula.

Stoichiometry

Stoichiometry bridges the gap between the empirical and molecular formulas, using the compound's molar mass. For serotonin, its molar mass is 176 g/mol. First, we calculate the empirical formula’s molar mass:

• \(10 \times 12.01 = 120.1\) g/mol for Carbon
• \(1 \times 1.008 = 1.008\) g/mol for Hydrogen
• \(1 \times 14.01 = 14.01\) g/mol for Nitrogen
• \(1 \times 16 = 16\) g/mol for Oxygen

Adding these gives an empirical molar mass of 151.118 g/mol. To find the relationship between the empirical and molecular formulas, divide the given molar mass by the empirical molar mass:\[\frac{176}{151.118} \approx 1.16\]Though this result is close to 1, indicating that the empirical formula \(\text{C}_{10}\text{HNO}\) is also the molecular formula, rounding and significant figures play a role in determining precision. The empirical formula is therefore adjusted to fit the molecular needs, here reaffirmed by stoichiometry to correctly reflect serotonin's structure.