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Chapter 9: Problem 169

How many molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\), would be in a tablet that contained \(250 \mathrm{mg}\) of aspirin? How many atoms of carbon would be in the aspirin in that tablet?

Short Answer

Expert verified
In a 250 mg tablet of aspirin, there are approximately \(8.37 \times 10^{20}\) molecules of aspirin, and there are approximately \(7.53 \times 10^{21}\) carbon atoms in the aspirin in that tablet.

Step by step solution

01

Calculate the molar mass of aspirin

First, find the molar mass of aspirin (C9H8O4): Molar mass of C = 12.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of aspirin = (9 × 12.01) + (8 × 1.01) + (4 × 16.00) = 180.16 g/mol
02

Convert the mass of aspirin in the tablet to moles

Next, convert the mass of aspirin in the tablet from mg to g and then to moles. To do this, divide the mass in grams by the molar mass of aspirin: 250 mg = 0.250 g Moles of aspirin = Mass in grams / Molar mass Moles of aspirin = 0.250 g / 180.16 g/mol ≈ 0.00139 mol
03

Find the number of molecules of aspirin using Avogadro's Number

To find the number of molecules of aspirin, multiply the moles of aspirin by Avogadro's Number (6.022 × 10^23 molecules/mol): Number of molecules = Moles × Avogadro's Number Number of molecules ≈ 0.00139 mol × 6.022 × 10^23 molecules/mol ≈ 8.37 × 10^20 molecules
04

Calculate the number of carbon atoms in these molecules

Finally, calculate the number of carbon atoms in the aspirin molecules. Since each aspirin molecule contains 9 carbon atoms, multiply the number of aspirin molecules by 9: Carbon atoms = Number of molecules × Carbon atoms per molecule Carbon atoms ≈ 8.37 × 10^20 molecules × 9 ≈ 7.53 × 10^21 carbon atoms In conclusion, there are approximately \(8.37 \times 10^{20}\) molecules of aspirin in a 250 mg tablet, and there are approximately \(7.53 \times 10^{21}\) carbon atoms in the aspirin in that tablet.

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Key Concepts

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

Molar Mass Calculation
Understanding the molar mass of a compound is fundamental in chemistry. It can be thought of as the mass of one mole of a substance. To calculate it, we look at the chemical formula and sum up the masses of all the atoms that compose the molecule.

For example, aspirin has a formula \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\). Here's how we calculate its molar mass: we take the number of carbon (\(C\)) atoms and multiply that by the molar mass of carbon, do the same for hydrogen (\(H\)) and oxygen (\(O\)), and then add up all those products. This gives us a combined molar mass that represents the mass of one mole of aspirin molecules.
Avogadro's Number
Avogadro's number, denoted as \(6.022 \times 10^{23}\), is the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. It is a bridge between the microscopic world of atoms and the macroscopic world of grams and moles.

When we calculate the number of molecules in a sample, we use this constant to convert moles to the actual number of molecules. It is much like converting dozens to individual eggs; one dozen equals twelve eggs, one mole equals Avogadro's number of molecules.
Stoichiometry
Stoichiometry is the aspect of chemistry that pertains to the calculation of the quantities of reactants and products in chemical reactions. It is based on the law of conservation of mass and the concept of the mole.

In our exercise, stoichiometry allows us to convert the mass of aspirin to moles and then use Avogadro's number to determine the number of molecules or atoms present in the sample. It is a central concept that bridges the gap between the molecular scale and measurable quantities.
Chemical Formula
A chemical formula gives us the proportions of elements in a compound using symbols and numbers. It serves as a shorthand to express the type and number of atoms present in a molecule. For instance, aspirin's chemical formula, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\), tells us that each molecule contains 9 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms.

This formula allows chemists to understand the composition of the molecule and to calculate properties such as molar mass and the number of molecules or atoms, which can be deduced using stoichiometry.
Molecular Mass
While molar mass is the mass of a mole of a given substance, molecular mass refers to the mass of a single molecule, measured in atomic mass units (amu). For calculations like the one in our exercise, molecular mass and molar mass can be used interchangeably because they numerically equal each other when the molar mass is expressed in grams per mole (g/mol).

Knowing the molecular mass of a compound along with Avogadro's number enables us to move from the scale of single molecules to an understandable macroscopic scale, which is necessary for practical applications in chemistry.

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Most popular questions from this chapter

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is \(25.6 \%\) methane by mass. When \(1.50 \mathrm{~g}\) of this mixture is burned, what is the mass of carbon dioxide produced?

For the reaction \(\mathrm{BF}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{BO}_{3}+\mathrm{HBF}_{4}\) (a) Balance the equation. (b) If you react \(24.2 \mathrm{~g}\) of \(\mathrm{BF}_{3}\) with an excess of water and generate \(14.8 \mathrm{~g}\) of \(\mathrm{HBF}_{4}\), what is your percent yield?

Consider the following unbalanced chemical equation: \(\mathrm{H}_{2}+\mathrm{N}_{2} \rightarrow \mathrm{NH}_{3}\) (a) To run this reaction in a balanced fashion, how much nitrogen is required if you start with \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{2} ?\) (b) How many grams of ammonia \(\left(\mathrm{NH}_{3}\right)\) will you produce if you run the reaction with the masses calculated in part (a)? (c) How many molecules of ammonia will you produce?

What is the empirical formula of a compound that is \(17.552 \% \mathrm{Na}, 39.696 \% \mathrm{Cr}\), and \(42.752 \% \mathrm{O} ?\)

A \(2.230-\mathrm{g}\) sample of a solid is subjected to combustion analysis, yielding \(76.59 \% \mathrm{C}\) and \(6.39 \% \mathrm{H}\). It may also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(94 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?
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