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

Short Answer

Expert verified
The molecular formula of Serotonin is \( C_{10}H_{12}N_{2}O \).

Step by step solution

01

Converting percentages to grams

Since we are given the percentage composition of Serotonin, we can assume that we have 100 g of Serotonin and convert the percentage to grams for each element. Then, we have: - 68.2 g of Carbon (C) - 6.86 g of Hydrogen (H) - 15.9 g of Nitrogen (N) - 9.08 g of Oxygen (O)
02

Converting grams to moles

Now, we will convert the mass of each element to moles using their respective molar masses: - Carbon - C: \( \frac{68.2 g}{12.01 \frac{g}{mol}} = 5.68 \space mol \) - Hydrogen - H: \( \frac{6.86 g}{1.01 \frac{g}{mol}} = 6.79 \space mol \) - Nitrogen - N: \( \frac{15.9 g}{14.01 \frac{g}{mol}} = 1.14 \space mol \) - Oxygen - O: \( \frac{9.08 g}{16.00 \frac{g}{mol}} = 0.57 \space mol \)
03

Determining the empirical formula

Now, we will divide the moles of each element by the lowest moles among them, which is for Oxygen (0.57 moles). By doing so, we have: - Carbon - C: \( \frac{5.68}{0.57} \approx 10 \) - Hydrogen - H: \( \frac{6.79}{0.57} \approx 12 \) - Nitrogen - N: \( \frac{1.14}{0.57} \approx 2 \) - Oxygen - O: \( \frac{0.57}{0.57} \approx 1 \) So, the empirical formula of Serotonin is \( C_{10}H_{12}N_{2}O \).
04

Calculating the molar mass of the empirical formula

Now, let's calculate the molar mass of the empirical formula: \( C_{10}H_{12}N_{2}O: = 10(12.01) + 12(1.01) + 2(14.01) + 16.00 = 176.23 g/mol \)
05

Determining the molecular formula

Now, we will use the molar mass of Serotonin (176 g/mol) and the molar mass of the empirical formula (176.23 g/mol) to find the molecular formula. Since the molar mass of Serotonin (176 g/mol) is approximately equal to the molar mass of the empirical formula (176.23 g/mol), the molecular formula is the same as the empirical formula. Therefore, the molecular formula of Serotonin is \( C_{10}H_{12}N_{2}O \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula Calculation
Understanding the empirical formula calculation is vital for students studying chemistry. It represents the simplest whole number ratio of atoms of each element in a compound. In our exercise involving Serotonin, calculating the empirical formula begins with converting the percentage composition into masses, assuming you have 100 g of the substance. This conversion is intuitive because percentage literally means 'per hundred'.

Once we have the mass of each element in grams, we need to determine how many moles of each element are present. This step utilizes the concept of molar mass, which is the mass of one mole of a substance. After finding the number of moles, we compare them by dividing each by the smallest number of moles present in the compound. The resulting ratios provide us with the simplest whole number ratio of atoms – our empirical formula, which in this case is the chemical blueprint for Serotonin, or more formally, C10H12N2O.
Molar Mass
The molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. Measured in grams per mole (g/mol), it's essentially the mass of one mole of a substance.

Understanding molar mass allows us to convert between the mass of a substance and the number of moles. This conversion is a foundation for stoichiometry and is often the stepping stone for solving complex chemical equations. In the Serotonin example, we use the molar mass of each element to convert the mass in grams we obtained from the percentage composition to moles, which helps us establish precise stoichiometric relationships between elements.
Percentage Composition
Percentage composition is the relative amount of each element within a compound, represented as a percentage of the total mass of the compound. This concept helps chemists understand the composition of substances and aids in calculating empirical and molecular formulas.

For instance, Serotonin's makeup was given in percentage by weight, so the calculation began by assuming you have a 100-gram sample, making it straightforward to convert the percentages directly to grams. This first step is critical because it allows you to determine the number of moles of each constituent element, which paves the way to finding the empirical formula. In practical situations, chemists use the percentage composition to deduce the purity of a compound or to identify its elemental constituents.
Stoichiometry
Stoichiometry can be likened to the recipe for a cake – it tells you how much of each ingredient you need. In chemistry, stoichiometry deals with the quantitative relationships, or ratios, that substances undergo in a chemical reaction. The mole concept is the cornerstone of stoichiometry, acting as a bridge to translate masses of substances to amounts of entities, such as atoms or molecules.

In determining the molecular formula of Serotonin, stoichiometry is used to compare the molar mass of Serotonin to that of the empirical formula. Since they match, it reveals that the empirical and molecular formulae are the same. Thus, stoichiometry isn't only about reactions—it also extends to understanding the relationship between empirical and molecular formulas, especially when dealing with unknown compounds or when synthesizing new ones.

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Most popular questions from this chapter

At least 25\(\mu \mathrm{g}\) of tetrahydrocannabinol \((\mathrm{THC}),\) the active ingredient in marijuana, is required to produce intoxication. The molecular formula of THC is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2} .\) How many moles of THC does this 25\(\mu \mathrm{g}\) represent? How many molecules?

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing numbers of atoms: 0.50 mol \(\mathrm{H}_{2} \mathrm{O}, 23 \mathrm{g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{N}_{2}\) molecules.

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right),\) the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) , but other products containing \(\mathrm{Cl},\) such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2},\) are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when 125 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with 255 \(\mathrm{g}\) of \(\mathrm{Cl}_{2},\) assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}\) and \(\mathrm{HCl} .(\mathbf{b})\) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces 206 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\)

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%) .\) In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called "\({NO}_{x}\)" gases. In \(2009,\) the United States emitted an estimated 19 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{x}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) into \(\mathrm{CO}_{2}\) and water. If 85\(\%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 g of octane.

(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 1.37 \(\mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\) . If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} . \mathrm{A}\) 0.1005 -gsample of mentholis combusted, producing 0.2829 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) and 0.1159 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of 156 \(\mathrm{g} / \mathrm{mol}\) what is its molecular formula?

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