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At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol (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?

Short Answer

Expert verified
25 µg of THC represents approximately \(7.95 \times 10^{-11}\) moles and \(4.79 \times 10^{13}\) molecules of THC.

Step by step solution

01

Calculate the molar mass of THC

First, we need to find the molar mass of THC. To do this, we'll use the molecular formula \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\) and the atomic masses of each element. We know the atomic masses of carbon (C), hydrogen (H), and oxygen (O) are approximately 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively. So, the molar mass of THC is: Molar mass of THC = (21 × 12.01 g/mol) + (30 × 1.01 g/mol) + (2 × 16.00 g/mol) = 314.47 g/mol
02

Convert the given mass of THC to moles

Next, we will convert the given mass of THC, \(25 \mu \mathrm{g}\), to moles. To do this, we will use the molar mass of THC (314.47 g/mol) and the conversion factor: 1 g = 1,000,000 µg. Moles of THC = \(\frac{25 \mu \mathrm{g}}{314.47 \frac{\mathrm{g}}{\mathrm{mol}} \times 1,000,000 \frac{\mu \mathrm{g}}{\mathrm{g}}}\) = \(\frac{25}{314.47 \times 10^6}\) mol ≈ 7.95 × 10⁻¹¹ mol
03

Convert the number of moles to the number of molecules

Finally, we will convert the number of moles of THC (7.95 × 10⁻¹¹ mol) to the number of molecules. To do this, we will use Avogadro's number (6.022 × 10²³ molecules/mol): Number of molecules = (7.95 × 10⁻¹¹ mol) × (6.022 × 10²³ molecules/mol) ≈ 4.79 × 10¹³ molecules So, 25 µg of THC represents approximately 7.95 × 10⁻¹¹ moles and 4.79 × 10¹³ molecules of THC.

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Key Concepts

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

Molar Mass
Molar mass is a fundamental concept in chemistry that helps us understand how to relate the mass of a substance to the amount of that substance in moles. It is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). To calculate the molar mass, you need the molecular formula of the compound and the atomic masses of its elements.

In the example of tetrahydrocannabinol (THC), which has the molecular formula \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\), we find its molar mass by adding up the molar masses of each atom in the formula:
  • Carbon (C): 21 atoms × 12.01 g/mol = 252.21 g/mol
  • Hydrogen (H): 30 atoms × 1.01 g/mol = 30.30 g/mol
  • Oxygen (O): 2 atoms × 16.00 g/mol = 32.00 g/mol
When we sum these values, the molar mass of THC comes out to be 314.47 g/mol. This means that one mole of THC weighs 314.47 grams.
Avogadro's Number
Avogadro's number, which is 6.022 × 10²³, is a constant that denotes the number of particles in one mole of a substance. These particles can be atoms, molecules, ions, or other entities, depending on the context.

This number is immensely useful when converting between moles and the actual number of molecules or atoms. For instance, in our THC example, we calculated that the 25 micrograms of THC corresponds to 7.95 × 10⁻¹¹ moles. Using Avogadro's number, we can further determine the number of THC molecules:
  • Number of molecules = 7.95 × 10⁻¹¹ moles × 6.022 × 10²³ molecules/mole
  • This calculation results in approximately 4.79 × 10¹³ THC molecules
Understanding Avogadro's number is critical because it allows chemists to work on measurable scales, bridging the gap between atomic-scale and macro-scale quantities.
Molecular Formula
The molecular formula provides specific information about the number and types of atoms in a molecule, and it allows us to determine the molecular composition of a compound. The formula shows the exact number of each type of atom present in a molecule.

For THC, the molecular formula is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\). This tells us that each molecule of THC is composed of:
  • 21 carbon atoms
  • 30 hydrogen atoms
  • 2 oxygen atoms
The specific composition in the molecular formula is crucial for determining both the chemical properties and the molar mass of the compound. By knowing the molecular formula, chemists can predict reactions and understand the structure and function of the molecule in various contexts. For THC, this understanding is essential to grasp its biological activity and how it interacts with the body.

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Most popular questions from this chapter

(a) If an automobile travels \(225 \mathrm{mi}\) with a gas mileage of \(20.5 \mathrm{mi} / \mathrm{gal}\), how many kilograms of \(\mathrm{CO}_{2}\) are produced? Assume that the gasoline is composed of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) whose density is \(0.69 \mathrm{~g} / \mathrm{mL}\). (b) Repeat the calculation for a truck that has a gas mileage of \(5 \mathrm{mi} / \mathrm{gal}\).

What is the mass, in kilograms, of an Avogadro's number of people, if the average mass of a person is \(160 \mathrm{lb}\) ? How does this compare with the mass of Earth, \(5.98 \times 10^{24} \mathrm{~kg}\) ?

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to NO: \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) In a certain experiment, \(2.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Why is it essential to use balanced chemical equations when determining the quantity of a product formed from a given quantity of a reactant?

Write a balanced chemical equation for the reaction that occurs when (a) \(\mathrm{Mg}(s)\) reacts with \(\mathrm{Cl}_{2}(g) ;\) (b) barium carbonate decomposes into barium oxide and carbon dioxide gas when heated; \((\mathbf{c})\) the hydrocarbon styrene, \(\mathrm{C}_{8} \mathrm{H}_{8}(l)\), is combusted in air; (d) dimethylether, \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(g)\), is combusted in air.

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