Question

Serotonin is a compound that conducts nerve impulses in the brain. It contains 68.2 mass percent C, 6.86 mass percent \(\mathrm{H}\), 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 for serotonin is \(C_{10}H_{12}N_{2}O\).

Step-by-step solution

Calculate Moles of Each Element

Given the mass percentages and assuming a 100g sample, we can calculate the mass of each element in the sample. Then, we'll use their molar masses to convert these values to moles: For Carbon (C): Mass of C = 68.2 g Molar Mass of C = 12.01 g/mol Moles of C = (Mass of C)/(Molar Mass of C) = 68.2/12.01 For Hydrogen (H): Mass of H = 6.86 g Molar Mass of H = 1.008 g/mol Moles of H = (Mass of H)/(Molar Mass of H) = 6.86/1.008 For Nitrogen (N): Mass of N = 15.9 g Molar Mass of N = 14.01 g/mol Moles of N = (Mass of N)/(Molar Mass of N) = 15.9/14.01 For Oxygen (O): Mass of O = 9.08 g Molar Mass of O = 16.00 g/mol Moles of O = (Mass of O)/(Molar Mass of O) = 9.08/16.00

Find the Simplest Ratio

Next, we'll find the simplest whole-number ratio of the moles. To do this, divide all the calculated moles by the smallest number of moles: moles of C: 68.2/12.01 = 5.68 moles of H: 6.86/1.008 = 6.81 moles of N: 15.9/14.01 = 1.14 moles of O: 9.08/16.00 = 0.568 Divide each mole value by 0.568 (the smallest value) to get the simplest ratio: moles of C: 5.68/0.568 = 10.0 moles of H: 6.81/0.568 = 12.0 moles of N: 1.14/0.568 = 2.00 moles of O: 1.00 The simplest whole-number ratio is C: 10, H: 12, N: 2, O: 1.

Calculate the Empirical Formula Molar Mass

Now we will calculate the molar mass of the empirical formula (simplest whole-number ratio): Empirical formula molar mass (EFMM) = (Moles of C)(Molar Mass of C) + (Moles of H)(Molar Mass of H) + (Moles of N)(Molar Mass of N) + (Moles of O)(Molar Mass of O) EFMM = (10)(12.01) + (12)(1.008) + (2)(14.01) + (1)(16.00)

Determine the Molecular Formula

To get the molecular formula, we'll divide the given molar mass by the empirical formula molar mass and multiply the whole-number ratio of elements by this factor: Molecular formula multiplier = (Given molar mass)/(Empirical formula molar mass) Molecular formula multiplier = 176/EFMM Multiply the empirical formula by the molecular formula multiplier: C: (10)(molecular formula multiplier) H: (12)(molecular formula multiplier) N: (2)(molecular formula multiplier) O: (1)(molecular formula multiplier) Calculating the values and rounding to the nearest whole number, we'll get the molecular formula for serotonin.

Key concepts

Elemental Analysis

Elemental analysis is a process that tells us what elements are present in a compound and in what proportion. Typically, this analysis provides the percentage by mass of each element in a sample. For serotonin, the given data shows it contains 68.2% carbon (C), 6.86% hydrogen (H), 15.9% nitrogen (N), and 9.08% oxygen (O). These percentages are crucial as they provide the starting point for determining the compound's molecular formula.

This analysis often assumes a 100g sample for ease of calculation, allowing us to straightforwardly convert percentages into grams. Therefore, each percentage directly correlates with mass: 68.2g of carbon, 6.86g of hydrogen, 15.9g of nitrogen, and 9.08g of oxygen. From here, we can calculate the number of moles for each element, which is an essential next step in figuring out how these elements fit together to form the molecular formula.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the atoms of each element present in the compound. It is derived from the moles of each element calculated from the elemental analysis.

For serotonin, once we know the number of moles for each element (from the molecular weights given: C = 12.01 g/mol, H = 1.008 g/mol, N = 14.01 g/mol, O = 16.00 g/mol), the simplest ratio can be determined. By dividing each element's moles by the smallest mule value in the set, we ultimately achieve a straightforward whole-number ratio.

This step is vital because the empirical formula often leads directly to understanding the compound's chemical properties and helps in calculating the compound's molar mass, which is necessary for determining the molecular formula.

Moles Calculation

Calculating the moles of each element is essential for deriving both the empirical and molecular formulas of a compound. The mole, a fundamental unit in chemistry, measures the amount of substance and is analogous to counting parts or particles on a much larger scale. The number of moles for an element can be found using the formula:

\[\text{Moles} = \frac{\text{Mass of element (g)}}{\text{Molar mass of element (g/mol)}}\]

In our serotonin example, the calculation for carbon is 68.2 g divided by 12.01 g/mol, which equals approximately 5.68 moles. This process is repeated for hydrogen, nitrogen, and oxygen. Estimating moles for each element assists in determining the simplest ratio of atoms for the empirical formula.

This calculation is critical because the values derived provide a foundational understanding of the compound's composition and serve as a precursor to further calculations in chemical analysis.

Molar Mass

Molar mass is the mass of one mole of a given substance and is usually expressed in g/mol. It's a fundamental concept in calculating the molecular formula of any compound. For determination purposes, molar mass is often needed to convert mass percentages to moles and later to validate the molecular formula.

The known molar mass of serotonin is 176 g/mol. After calculating the empirical formula mass, which arises from the sum of the molar masses of the elements in their simplest ratio, we can establish the molecular formula. By dividing the compound's molar mass by the empirical formula's mass, we'll obtain a multiplier factor, which scales up the empirical formula to the actual molecular formula of serotonin.

Recognizing the molar mass is therefore crucial as it bridges the gap between the empirical formula and the complete understanding of the substance's actual molecular structure, helping chemists design useful applications and further analyses.