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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?

Short Answer

Expert verified
The 25 渭g of THC represents 7.95 脳 10鈦烩伕 moles and approximately 4.79 脳 10鹿鈦 THC molecules.

Step by step solution

01

Find the molar mass of THC

To find the molar mass of THC, we will use the atomic masses of its constituent elements: Carbon (C), Hydrogen (H), and Oxygen (O). The molecular formula of THC is C鈧傗倎H鈧冣個O鈧, which means it contains 21 carbon atoms, 30 hydrogen atoms, and 2 oxygen atoms. The atomic mass of Carbon (C) = 12.01 g/mol The atomic mass of Hydrogen (H) = 1.01 g/mol The atomic mass of Oxygen (O) = 16.00 g/mol Now, we will calculate the molar mass of THC: Molar mass of THC = (21 脳 Atomic mass of C) + (30 脳 Atomic mass of H) + (2 脳 Atomic mass of O) Molar mass of THC = (21 脳 12.01 g/mol) + (30 脳 1.01 g/mol) + (2 脳 16.00 g/mol) Molar mass of THC = 252.21 g/mol + 30.30 g/mol + 32.00 g/mol Molar mass of THC = 314.51 g/mol
02

Calculate the number of moles of THC

Now, we will use the given mass of THC (25 渭g) and its molar mass (314.51 g/mol) to calculate the number of moles of THC. First, we need to convert the mass from micrograms to grams. Mass of THC = 25 渭g = 0.000025 g Next, we will use the formula: Number of moles = (Mass of the substance) / (Molar mass) Number of moles of THC = (0.000025 g) / (314.51 g/mol) Number of moles of THC = 7.95 脳 10鈦烩伕 mol
03

Calculate the number of THC molecules

Finally, we will use Avogadro's number to calculate the number of THC molecules. Avogadro's number is approximately 6.022 脳 10虏鲁 particles/mol. Number of THC molecules = (Number of moles of THC) 脳 (Avogadro's number) Number of THC molecules = (7.95 脳 10鈦烩伕 mol) 脳 (6.022 脳 10虏鲁 particles/mol) Number of THC molecules = 4.79 脳 10鹿鈦 particles So, 25 渭g of THC represents 7.95 脳 10鈦烩伕 moles and approximately 4.79 脳 10鹿鈦 THC molecules.

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Key Concepts

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

Molar Mass Calculation
Understanding molar mass calculation is essential for converting between mass and moles of a substance, which is a common practice in chemistry. Molar mass represents the mass of one mole of a substance and is measured in grams per mole (g/mol).

To calculate the molar mass of a compound, such as tetrahydrocannabinol (THC), you need to know the chemical formula and the atomic masses of each element within that compound. For THC, with the formula \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\), you calculate the molar mass by multiplying the number of atoms of each element by their respective atomic masses and adding these values together. You can find atomic masses on the periodic table, and the ones needed for THC are 12.01 g/mol for carbon, 1.01 g/mol for hydrogen, and 16.00 g/mol for oxygen.

To simplify, the molar mass calculation in our example would look like this:
  • Carbon: 21 atoms 脳 12.01 g/mol
  • Hydrogen: 30 atoms 脳 1.01 g/mol
  • Oxygen: 2 atoms 脳 16.00 g/mol
Then sum these values for the total molar mass of THC. Remember, precision in these numbers is critical as it directly affects the accuracy of all subsequent calculations.
Avogadro's Number
When dealing with atoms and molecules at the micro scale, Avogadro's number becomes a fundamental concept in chemistry. Avogadro's number, approximately \(6.022 \times 10^{23}\) entities per mole, represents the number of particles found in one mole of a substance. This constant enables chemists to count particles by weighing them.

In quantitative chemistry, to convert moles to the number of discrete particles - atoms, molecules, or ions - you multiply the moles by Avogadro鈥檚 number. In the example with THC, calculating the number of THC molecules from the moles determined involves this simple multiplication: \(\text{{Number of molecules}} = \text{{Number of moles}} \times \text{{Avogadro's number}}\). This provides an astonishingly large number, reflecting the incredibly small size and mass of individual molecules.
Chemical Formula Interpretation
The chemical formula interpretation is an invaluable skill for identifying the composition and proportions of atoms in a chemical compound. It provides an immediate understanding of the molecule's structure and is key for various chemical calculations, including determination of molar mass. The molecular formula, such as \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\) for THC, tells us that each molecule consists of 21 carbon atoms, 30 hydrogen atoms, and 2 oxygen atoms.

To interpret this effectively, think of the subscript numbers as a simple count for each type of atom in the molecule. If no subscript is present, it implies only one atom of that element. By knowing the atomic composition from the chemical formula, scientists can predict chemical properties, reaction behaviors, and how the molecule will interact with others. This information is foundational to tasks from balancing chemical equations to calculating reactions yields and is central to understanding chemistry.

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Most popular questions from this chapter

The thermite reaction, $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{Al} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{Fe} $$ produces so much heat that the Fe product melts. This reaction is used industrially to weld metal parts under water, where a torch cannot be employed. It is also a favorite chemical demonstration in the lecture hall (on a small scale). (a) Balance the chemical equation for the thermite reaction, and include the proper states of matter. (b) Calculate how many grams of aluminum are needed to completely react with 500.0 g of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in this reaction. (c) This reaction produces 852 kJ of heat per mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacted. How many grams of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are needed to produce \(1.00 \times 10^{4} \mathrm{kJ}\) of heat? (d) If you performed the reverse reaction aluminum-oxide plus iron makes iron oxide plus aluminum-would that reaction have heat as a reactant or a product?

The molecular formula of aspartame, the artificial sweetener marketed as NutraSweet, is \(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{N}_{2} \mathrm{O}_{5}\) . (a) What is the molar mass of aspartame? (b) How many moles of aspartame are present in 1.00 \(\mathrm{mg}\) of aspartame? (c) How many molecules of aspartame are present in 1.00 \(\mathrm{mg}\) of aspartame? (d) How many hydrogen atoms are present in 1.00 \(\mathrm{mg}\) of aspartame?

A piece of aluminum foil 1.00 \(\mathrm{cm}^{2}\) and 0.550 -mm thick is allowed to react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (The density of aluminum is 2.699 \(\mathrm{g} / \mathrm{cm}^{3} .\) ) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completely?

The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right) :\) $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) & \longrightarrow \\ & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment 1.00 g of sodium bicarbonate and 1.00 g of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when 1.85 mol \(\mathrm{NaOH}\) and 1.00 \(\mathrm{mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

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