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Chapter 3: Problem 39

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? (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 162.3 g/mol. (b) There are \(3.08 \times 10^{-5}\) moles of allicin in 5.00 mg of the substance. (c) There are \(1.86 \times 10^{19}\) molecules of allicin in 5.00 mg of the substance. (d) There are \(3.72 \times 10^{19}\) sulfur atoms in 5.00 mg of allicin.

Step by step solution

01

(a) Calculate the molar mass of allicin

To find the molar mass of allicin, add the molar masses of all the atoms in the molecular formula. The molecular formula of allicin is given as \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{OS}_{2}\). The molar mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.01 g/mol, oxygen (O) is 16.00 g/mol, and sulfur (S) is 32.07 g/mol. Molar mass of allicin = \((6 \times 12.01) + (10 \times 1.01) + (1 \times 16.00) + (2 \times 32.07)\) g/mol = \(72.06 + 10.10 + 16.00 + 64.14\) g/mol = \(162.3\) g/mol The molar mass of allicin is 162.3 g/mol.
02

(b) Calculate the number of moles of allicin

To find the number of moles of allicin in 5.00 mg, convert the mass to grams and use the molar mass we calculated in part (a). 5.00 mg = \(5.00 \times 10^{-3}\) g Number of moles = \(\frac{mass}{molar \, mass}\) Number of moles = \(\frac{5.00 \times 10^{-3}\,g}{162.3\,g/mol}\) = \(3.08 \times 10^{-5}\) mol There are \(3.08 \times 10^{-5}\) moles of allicin in 5.00 mg of the substance.
03

(c) Calculate the number of molecules of allicin

To find the number of molecules of allicin, we multiply the number of moles by Avogadro's number (6.022 x 10^23). Number of molecules = (number of moles) x (Avogadro's number) Number of molecules = \((3.08 \times 10^{-5}\,mol) \times (6.022 \times 10^{23}\,mol^{-1})\) = \(1.86 \times 10^{19}\) molecules There are \(1.86 \times 10^{19}\) molecules of allicin in 5.00 mg of the substance.
04

(d) Calculate the number of sulfur atoms in 5.00 mg of allicin

To determine the number of sulfur atoms present in 5.00 mg of allicin, we must first determine the number of molecules in that sample (which we already calculated in part (c)) and then account for the fact that there are two sulfur atoms in each allicin molecule. Number of S atoms = (number of allicin molecules) x (number of S atoms per molecule) Number of S atoms = \((1.86 \times 10^{19}\,molecules) \times (2\,S\,atoms/molecule)\) = \(3.72 \times 10^{19}\) S atoms There are \(3.72 \times 10^{19}\) sulfur atoms in 5.00 mg of allicin.

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Most popular questions from this chapter

When hydrocarbons are burned in a limited amount of air, both CO and \(\mathrm{CO}_{2}\) form. When 0.450 g of a particular hydrocarbon was burned in air, 0.467 \(\mathrm{g}\) of \(\mathrm{CO}, 0.733 \mathrm{g}\) of \(\mathrm{CO}_{2},\) and 0.450 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of O \(_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

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