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Manufacturers of headache remedies routinely claim that their own brands are more potent pain relievers than the competing brands. Their way of making the comparison is to compare the number of molecules in the standard dosage. Tylenol uses \(325 \mathrm{mg}\) of acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2}\right)\) as the standard dose, whereas Advil uses \(2.00 \times 10^{2} \mathrm{mg}\) of ibuprofen \(\left(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\right) .\) Find the number of molecules of pain reliever in the standard doses of (a) Tylenol and (b) Advil.

Short Answer

Expert verified
Tylenol: \(1.29 \times 10^{21}\) molecules; Advil: \(5.84 \times 10^{20}\) molecules.

Step by step solution

01

Determine Molar Mass of Acetaminophen

Find the molar mass of acetaminophen (C8H9NO2) by adding the atomic masses of each element: Carbon (C) is 12.01 g/mol, Hydrogen (H) is 1.008 g/mol, Nitrogen (N) is 14.01 g/mol, and Oxygen (O) is 16.00 g/mol. Thus, \[\text{Molar mass of C}_8\text{H}_9\text{NO}_2 = 8\times 12.01 + 9\times 1.008 + 1\times 14.01 + 2\times 16.00 = 151.17 \text{ g/mol}\]
02

Calculate Moles of Acetaminophen in Tylenol Dose

Convert the mass of acetaminophen in the Tylenol dose (325 mg) to grams (325 \text{ mg} = 0.325 \text{ g}) and then find the number of moles using the molar mass.\[\text{Moles of acetaminophen} = \frac{0.325 \text{ g}}{151.17 \text{ g/mol}} = 2.15 \times 10^{-3} \text{ mol}\]
03

Find Number of Molecules in Tylenol Dose

Use Avogadro's number (6.022 \times 10^{23} \text{ molecules/mol}) to find the number of molecules in the moles of acetaminophen.\[\text{Molecules} = 2.15 \times 10^{-3} \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.29 \times 10^{21} \text{ molecules}\]
04

Determine Molar Mass of Ibuprofen

Calculate the molar mass of ibuprofen (C13H18O2) using the atomic masses: \ Carbon (C) is 12.01 g/mol, Hydrogen (H) is 1.008 g/mol, and Oxygen (O) is 16.00 g/mol. \[\text{Molar mass of C}_{13}\text{H}_{18}\text{O}_2 = 13\times 12.01 + 18\times 1.008 + 2\times 16.00 = 206.28 \text{ g/mol}\]
05

Calculate Moles of Ibuprofen in Advil Dose

Convert the mass of ibuprofen in the Advil dose (200 mg) to grams (200 mg = 0.200 g) and then find the number of moles using the molar mass.\[\text{Moles of ibuprofen} = \frac{0.200 \text{ g}}{206.28 \text{ g/mol}} = 9.70 \times 10^{-4} \text{ mol}\]
06

Find Number of Molecules in Advil Dose

Use Avogadro's number to find the number of molecules in the moles of ibuprofen.\[\text{Molecules} = 9.70 \times 10^{-4} \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 5.84 \times 10^{20} \text{ molecules}\]

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Key Concepts

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

Molar Mass
The molar mass is basically the weight of one mole of a compound, and it's measured in grams per mole (g/mol). Imagine it like a measuring cup for molecules.
To figure out the molar mass of a molecule like acetaminophen (C鈧圚鈧塏O鈧), you'll need the periodic table for the atomic weights: carbon (C) is about 12.01 g/mol, hydrogen (H) is 1.008 g/mol, nitrogen (N) is 14.01 g/mol, and oxygen (O) is 16.00 g/mol. Then, you simply add up the total weight using these atomic masses based on how many of each atom the molecule contains.
For example:
  • 8 carbons: 8 脳 12.01 = 96.08
  • 9 hydrogens: 9 脳 1.008 = 9.072
  • 1 nitrogen: 1 脳 14.01 = 14.01
  • 2 oxygens: 2 脳 16.00 = 32.00
So, the molar mass of acetaminophen is 151.17 g/mol!
The same process can be followed for ibuprofen (C鈧佲們H鈧佲倛O鈧) to find its molar mass, leading to 206.28 g/mol. Understanding molar mass is the key first step in any molecule calculation because it connects the amount of substance to its mass.
Avogadro's Number
Avogadro's number is like a giant, cosmic scoop that helps us count molecules. Scientists use it to make sense of the mind-bogglingly large number of molecules even in a tiny sample of a substance. Imagine if you had to count grains of sand!
The magic number is 6.022 脳 10虏鲁 molecules/mol. It tells us how many molecules are in one mole of any substance. This is super useful because, with moles, we can convert between the measurable mass of a substance and its molecular count.
For instance, once we know how many moles of acetaminophen are in a typical Tylenol dose, we multiply those moles by Avogadro's number to find the number of individual acetaminophen molecules. This conversion is crucial for comparing doses or understanding the potency of medications.
Chemical Formulas
Chemical formulas are like the recipe cards for molecules. They tell us what atoms are present and how many of each are tied together in one molecule of a compound. For instance, acetaminophen has the chemical formula C鈧圚鈧塏O鈧, which reads like:
  • 8 Carbon (C) atoms
  • 9 Hydrogen (H) atoms
  • 1 Nitrogen (N) atom
  • 2 Oxygen (O) atoms
Each chemical formula helps us determine not only the structure of the molecule but also allows calculation of the molar mass. It's as much about understanding the composition as it is about enabling calculations.
Grasping the formula is like having a decoder to unravel how many atoms are chitter-chattering in a dose of your chosen pain reliever. This foundational knowledge empowers you to calculate masses, moles, and ultimately, the number of molecules present in any given sample.
Pain Reliever Dosages
Pain reliever dosages may seem straightforward at first鈥攋ust measure the mass and administer, right? But the science within scrambles down to molecules and their specific interactions in your body. Here鈥檚 why knowing about molecules matters for dosages:
Take acetaminophen and ibuprofen, commonly debated over for their dose effectiveness. The actual weight of the compound gives initial insight, but counting molecules is what makes comparison precise.
When manufacturers boast about the potency of their headache remedy or others, they are often focusing on the number of active molecules you'll consume per dose. In our example, even if Tylenol has more molecules per dose than Advil, knowing this detail helps in understanding how each works in relieving pain and why specific dosing matters.
Ultimately, knowing the molecular breakdown of dosages assists in making informed choices about medicine鈥攅specially when comparing different brands or adjusting doses.

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Most popular questions from this chapter

When a gas is diffusing through air in a diffusion channel, the diffusion rate is the number of gas atoms per second diffusing from one end of the channel to the other end. The faster the atoms move, the greater is the diffusion rate, so the diffusion rate is proportional to the rms speed of the atoms. The atomic mass of ideal gas \(\mathrm{A}\) is \(1.0 \mathrm{u}\), and that of ideal gas \(\mathrm{B}\) is \(2.0 \mathrm{u}\). For diffusion through the same channel under the same conditions, find the ratio of the diffusion rate of gas \(A\) to the diffusion rate of gas B.

The dimensions of a room are \(2.5 \mathrm{m} \times 4.0 \mathrm{m} \times 5.0 \mathrm{m} .\) Assume that the air in the room is composed of \(79 \%\) nitrogen \(\left(\mathrm{N}_{2}\right)\) and \(21 \%\) oxygen \(\left(\mathrm{O}_{2}\right) .\) At a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of \(1.01 \times 10^{5} \mathrm{Pa}\) what is the mass (in grams) of the air?

A tube has a length of \(0.015 \mathrm{m}\) and a cross-sectional area of \(7.0 \times\) \(10^{-4} \mathrm{m}^{2} .\) The tube is filled with a solution of sucrose in water. The diffusion constant of sucrose in water is \(5.0 \times 10^{-10} \mathrm{m}^{2} / \mathrm{s} .\) A difference in concentration of \(3.0 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{3}\) is maintained between the ends of the tube. How much time is required for \(8.0 \times 10^{-13} \mathrm{kg}\) of sucrose to be transported through the tube?

Compressed air can be pumped underground into huge caverns as a form of energy storage. The volume of a cavern is \(5.6 \times 10^{5} \mathrm{m}^{3}\), and the pressure of the air in it is \(7.7 \times 10^{6} \mathrm{Pa}\). Assume that air is a diatomic ideal gas whose internal energy \(U\) is given by \(U=\frac{5}{2} n R T .\) If one home uses \(30.0 \mathrm{kW} \cdot \mathrm{h}\) of energy per day, how many homes could this internal energy serve for one day?

A young male adult takes in about \(5.0 \times 10^{-4} \mathrm{m}^{3}\) of fresh air during a normal breath. Fresh air contains approximately \(21 \%\) oxygen. Assuming that the pressure in the lungs is \(1.0 \times 10^{5} \mathrm{Pa}\) and that air is an ideal gas at a temperature of \(310 \mathrm{K}\), find the number of oxygen molecules in a normal breath.

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