Problem 42 - Chemistry: The Central Science

Question

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol (THC), the active ingredient in marijuana, is required to produce intoxication. The molecular formula of \(\mathrm{THC}\) 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?

Step-by-Step Solution

Answer

In 25 µg of tetrahydrocannabinol (THC), with a molecular formula of C₂₁H₃₀O₂, there are approximately 7.95 x 10⁻⁸ moles and 4.79 x 10¹⁶ molecules.

Step 1: Find the molar mass of THC

To find the molar mass of THC (C₂₁H₃₀O₂), we can add up the molar mass of each of the individual elements, multiplied by the number of atoms of that element in the molecule: Molar mass of THC = (21 x Molar mass of carbon) + (30 x Molar mass of hydrogen) + (2 x Molar mass of oxygen) Using the periodic table, the molar mass of carbon is 12.01 g/mol, hydrogen is 1.008 g/mol, and oxygen is 16 g/mol: Molar mass of THC = (21 x 12.01 g/mol) + (30 x 1.008 g/mol) + (2 x 16 g/mol) = 314.47 g/mol.

Step 2: Convert micrograms to grams

Since the given mass is in micrograms, we need to convert it to grams, as the molar mass is in grams per mole. 25 µg = 25 x 10⁻⁶ grams

Step 3: Find the number of moles in 25 µg of THC

Now that we have the molar mass of THC and the mass in grams, we can find the number of moles using the formula: moles = mass / molar mass moles of THC = (25 x 10⁻⁶ g) / (314.47 g/mol) = 7.95 x 10⁻⁸ mol

Step 4: Find the number of molecules in the given number of moles

Now, we can use Avogadro's number (6.022 x 10²³ molecules/mole) to find the number of molecules in these moles: Number of molecules = moles x Avogadro's number Number of molecules of THC = (7.95 x 10⁻⁸ mol) x (6.022 x 10²³ molecules/mol) = 4.79 x 10¹⁶ molecules So, in 25 µg of THC, there are approximately 4.79 x 10¹⁶ molecules of tetrahydrocannabinol.

Key Concepts

Chemical Formulas

Chemical formulas are crucial in chemistry as they provide a shorthand way of representing the composition of molecules and compounds. These formulas indicate the type and number of atoms in a molecule. For example, in the original exercise, the chemical formula of tetrahydrocannabinol (THC) is \(\text{C}_{21}\text{H}_{30}\text{O}_{2}\). This illustrates that each molecule of THC consists of 21 carbon atoms, 30 hydrogen atoms, and 2 oxygen atoms.
The subscripts in a chemical formula convey the moles of each element in one mole of the compound. By understanding chemical formulas, scientists can easily determine the proportions of different elements in a substance and calculate properties like molar mass. Recognizing these formulas is essential for performing further calculations based on the mole concept or other chemical principles.

Mole Concept

The mole concept is a fundamental principle in chemistry that allows chemists to count particles, such as atoms and molecules, using a measurable quantity. One mole of any substance contains the same number of particles, which facilitates calculations between the mass and number of atoms in a sample.
The mole concept ties the macroscopic world of grams and liters to the microscopic world of atoms and molecules. By using the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\), we can determine the number of moles of a substance when given the mass and the molar mass. In the exercise, converting 25 µg of THC into moles helps us transition from a tiny mass into an understandable quantity on the mole scale. This allows us to further compute how many molecules we are considering.

Avogadro's Number

Avogadro's number is a key constant in chemistry, defined as \(6.022 \times 10^{23}\) particles per mole. It provides the link between the macroscopic scale (that we can measure) and the atomic scale (which we cannot directly observe).
For instance, once we know the number of moles of THC, Avogadro's number allows us to calculate the exact number of molecules in that amount. By multiplying the moles obtained from the mass by Avogadro's number \((\text{moles} \times 6.022 \times 10^{23})\), students can transition from grams to individual molecules. This tool is invaluable for understanding reactions as it bridges the gap between the scale of laboratory measurements and molecular interactions.