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(a) The molecular formula of acetylsalicylic acid (aspirin), one of the most common pain relievers, is \(\mathrm{CgH}_{8} \mathrm{O}_{4} .\) How many moles of \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) are in a \(0.500-\mathrm{g}\) tablet of aspirin? Assume the tablet is composed entirely of aspirin. (b) How many molecules of \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) are in this tablet? (c) How many carbon atoms are in the tablet?

Short Answer

Expert verified
In the \(0.500\,\mathrm{g}\) tablet of aspirin, there are \(2.775 \times 10^{-3}\, \mathrm{mol}\) of \(\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4\), \(1.67 \times 10^{21}\, \mathrm{molecules}\), and \(1.50 \times 10^{22}\, \mathrm{carbon\, atoms}\).

Step by step solution

01

Calculate the molar mass of aspirin

To calculate the molar mass of aspirin, we need to sum up the molar masses of all the atoms that form the molecule: \(9\) carbon atoms, \(8\) hydrogen atoms, and \(4\) oxygen atoms. Using the molar masses from the periodic table, we have: \[ M (\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4) = 9M (\mathrm{C}) + 8M (\mathrm{H}) + 4M (\mathrm{O}) \] \[ M (\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4) = 9 \times 12.01\,\mathrm{g/mol} + 8 \times 1.01\,\mathrm{g/mol} + 4 \times 16.00\,\mathrm{g/mol} \] \[ M (\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4) = 180.18\,\mathrm{g/mol} \]
02

Calculate the number of moles of aspirin

Now that we have the molar mass of aspirin, we can find the number of moles in \(0.500\,\mathrm{g}\) of aspirin by dividing the mass of the tablet by the molar mass of aspirin: \[ \text{moles of}\,\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4 = \frac{\text{mass of aspirin}}{M (\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4)} = \frac{0.500\,\mathrm{g}}{180.18\,\mathrm{g/mol}} = 2.775 \times 10^{-3} \,\mathrm{mol} \]
03

Calculate the number of molecules in the tablet

To find the number of molecules of \(\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4\) in the tablet, we can multiply the number of moles by Avogadro's number: \[ \text{molecules of}\,\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4 = (\text{moles of}\,\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4) \times N_A \] \[ \text{molecules of}\,\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4 = (2.775 \times 10^{-3} \,\mathrm{mol}) \times (6.022 \times 10^{23} \, \mathrm{mol^{-1}}) \] \[ \text{molecules of}\,\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4 = 1.67 \times 10^{21} \,\mathrm{molecules} \]
04

Calculate the number of carbon atoms in the tablet

Each molecule of aspirin contains \(9\) carbon atoms, so we need to multiply the number of molecules we found in step 3 by \(9\) to find the total number of carbon atoms in the tablet: \[ \text{carbon atoms in tablet} = (\text{molecules of}\,\mathrm{C}_9\mathrm{H}_8\mathrm{O}_4) \times 9 \] \[ \text{carbon atoms in tablet} = (1.67 \times 10^{21} \,\mathrm{molecules}) \times 9 \] \[ \text{carbon atoms in tablet} = 1.50 \times 10^{22} \,\mathrm{carbon\, atoms} \] In summary, there are \(2.775 \times 10^{-3}\, \mathrm{mol}\) of aspirin, \(1.67 \times 10^{21}\, \mathrm{molecules}\), and \(1.50 \times 10^{22}\, \mathrm{carbon\, atoms}\) in the \(0.500\,\mathrm{g}\) tablet.

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Key Concepts

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

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It is based on the principle that matter is conserved during a reaction. To perform stoichiometric calculations, a balanced chemical equation is necessary as it indicates the proportions of reactants and products. In real-world applications, such as calculating the amount of a substance in a medication like aspirin, stoichiometry helps us determine the number of moles, which is a counting unit similar to a dozen but for atoms and molecules.

When we calculate the number of moles in a sample, we're essentially converting mass, which is easily measurable, into number of particles, which relates directly to chemical reactions at the particle level. This relates to the problem given, where we first find the molar mass of aspirin to understand the mass-to-mole relationship and then use this to find out the moles in a 0.500-g tablet.
Avogadro's Number
Avogadro's number, denoted as \(N_A\), is a fundamental constant in chemistry and represents the number of constituent particles, usually atoms or molecules, that are contained in one mole, which is equal to \(6.022 \times 10^{23}\). It enables chemists to count atoms or molecules by weighing them. With Avogadro's number, we bridge the macroscopic world that we can measure to the microscopic world of atoms and molecules.

In practical terms, if you have a mole of any substance, you have \(6.022 \times 10^{23}\) of its particles. This is crucial when we want to calculate the number of molecules in a known mass of substance, as was required in the exercise for aspirin. By knowing the number of moles from stoichiometry, we multiply by Avogadro's number to find the exact number of molecules, which helps chemists understand the quantity of a substance on a molecular level.
Molecular Formula
The molecular formula provides the number and type of atoms in a single molecule of a compound. For example, the molecular formula for aspirin is \(C_9H_8O_4\), indicating that each molecule contains 9 carbon (C), 8 hydrogen (H), and 4 oxygen (O) atoms. The molecular formula is vital because it offers a clear picture of the molecule's composition and is directly related to its molar mass, which is the sum of the atomic masses of all atoms in the molecule.

This information is foundational when performing stoichiometric calculations. Since the molecular formula tells us the number of each kind of atom in a molecule, we use this data in conjunction with Avogadro's number to calculate the total number of individual atoms in a given sample, as was done to find the number of carbon atoms in the aspirin tablet in the exercise. The molecular formula directly links the type and number of atoms to macroscopic quantities measured in laboratory or real-world applications.

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Most popular questions from this chapter

Determine the formula weights of each of the following compounds: (a) nittrous oxide, \(\mathrm{N}_{2} \mathrm{O}\) , known as laughing gas and used as an anesthetic in dentistry; (b) benzoic acid; \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) a substance used as a food preservative; \((c) \mathrm{Mg}(\mathrm{OH})_{2},\) the active ingredient in milk of magnesia; (d) urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO},\) a compound used as a nitrogen fertilizer; (e) isopentyl acetate, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{5} \mathrm{H}_{11},\) responsible for the odor of bananas.

Without doing any detailed calculations (but using a periodic table to give atomic weights, rank the following samples in order of increasing numbers of atoms: 42 gof NaHCO \(_{3}, 1.5 \mathrm{mol} \mathrm{CO}_{2}, 6.0 \times 10^{24} \mathrm{Ne}\) atoms.

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (b) } \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)} \\\ {\text { (e) } \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(a q)}\end{array} $$

Serotonin is a compound that conducts nerve impulses in the brain. It contains \(68.2 \% \mathrm{C}, 6.86 \% \mathrm{H}, 15.9 \% \mathrm{N},\) and 9.08\(\% \mathrm{O}\) . Its molar mass is 176 \(\mathrm{g} / \mathrm{mol} .\) Determine its molecular formula.

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with 0.350 \(\mathrm{kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) . (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when 0.350 \(\mathrm{kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

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