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Chapter 3: Problem 37

The molecular formula of allicin, the compound responsible for the characteristic smell of garlic, is \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{OS}_{2} .\) (a) What is the molar mass of allicin? (b) How many moles of allicin are present in \(5.00 \mathrm{mg}\) of this substance? \((\mathrm{c})\) How many molecules of allicin are in \(5.00 \mathrm{mg}\) of this substance? (d) How many S atoms are present in \(5.00 \mathrm{mg}\) of allicin?

Short Answer

Expert verified
(a) The molar mass of allicin is 162.27 g/mol. (b) There are 3.08 x 10^-5 moles of allicin in 5.00 mg of the substance. (c) There are 1.85 x 10^19 molecules of allicin in 5.00 mg of the substance. (d) There are 3.70 x 10^19 S atoms in 5.00 mg of allicin.

Step by step solution

01

Find the atomic masses of the elements in allicin

Let's start by locating the atomic masses of the constituent elements. Using the periodic table, we find: Carbon (C): 12.01 g/mol Hydrogen (H): 1.01 g/mol Oxygen (O): 16.00 g/mol Sulfur (S): 32.07 g/mol
02

Calculate the molar mass of allicin

Now, we can calculate the molar mass of allicin using the atomic masses and the given molecular formula C_6H_10OS_2: Molar mass of allicin = (6 * 12.01 g/mol) + (10 * 1.01 g/mol) + (1 * 16.00 g/mol) + (2 * 32.07 g/mol) = 162.27 g/mol. (a) The molar mass of allicin is 162.27 g/mol.
03

Convert mass to moles

For part (b), we need to convert 5.00 mg of allicin to moles using the molar mass: 1. Convert the mass to grams: 5.00 mg * (1 g / 1000 mg) = 0.00500 g 2. Divide the mass by the molar mass: 0.00500 g / 162.27 g/mol = 3.08 x 10^-5 mol (b) There are 3.08 x 10^-5 moles of allicin in 5.00 mg of the substance.
04

Calculate the number of molecules

For part (c), we need to convert moles to the number of molecules using Avogadro's number (6.022 x 10^23 molecules/mol): Molecules in 5.00 mg allicin = (3.08 x 10^-5 mol) * (6.022 x 10^23 molecules/mol) = 1.85 x 10^19 molecules (c) There are 1.85 x 10^19 molecules of allicin in 5.00 mg of the substance.
05

Find the number of S atoms in 5.00 mg of allicin

For part (d), we need to find the number of S atoms. Since there are two S atoms per molecule of allicin, we can multiply the number of allicin molecules by 2: S atoms in 5.00 mg allicin = (1.85 x 10^19 molecules) * (2 S atoms/molecule) = 3.70 x 10^19 S atoms (d) There are 3.70 x 10^19 S atoms in 5.00 mg of allicin.

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Most popular questions from this chapter

The compound \(\mathrm{XCl}_{4}\) contains \(75.0 \% \mathrm{Cl}\) by mass. What is the element \(\mathrm{X} ?\)

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Balance the following equations: (a) \(\mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\) (d) \(\mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)\) (e) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (f) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (g) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\)
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