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Chapter 3: Problem 38

The molecular formula of aspartame, the artificial sweetener marketed as NutraSweet \(^{\oplus}\), is \(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{5} .\) (a) What is the molar mass of aspartame? (b) How many moles of aspartame are present in \(1.00 \mathrm{mg}\) of aspartame? (c) How many molecules of aspartame are present in \(1.00 \mathrm{mg}\) of aspartame? (d) How many hydrogen atoms are present in \(1.00 \mathrm{mg}\) of aspartame?

Short Answer

Expert verified
(a) The molar mass of aspartame is 294.30 g/mol. (b) There are \(3.40\times10^{-6}\) moles of aspartame in 1.00 mg. (c) There are \(2.04\times10^{18}\) molecules of aspartame in 1.00 mg. (d) There are \(3.67\times10^{19}\) hydrogen atoms in 1.00 mg of aspartame.

Step by step solution

01

Calculate the molar mass of aspartame

To calculate the molar mass, sum up the individual atomic masses of the elements in aspartame: Molar mass = (14 × atomic mass of C) + (18 × atomic mass of H) + (2 × atomic mass of N) + (5 × atomic mass of O) We are given the atomic masses of the elements: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.01 g/mol, Nitrogen (N) = 14.01 g/mol, Oxygen (O) = 16.00 g/mol Molar mass = (14 × 12.01) + (18 × 1.01) + (2 × 14.01) + (5 × 16.00)
02

Find the number of moles in 1.00 mg of aspartame

To find the number of moles in 1.00 mg of aspartame, we need to convert the mass from mg to g and use the molar mass from Step 1. Number of moles = (mass in grams) / (molar mass) To convert from mg to g, divide by 1000: 1.00 mg = 0.00100 g Now, we divide the mass in grams by the molar mass of aspartame we found in the previous step: Number of moles = 0.00100 g / molar mass
03

Calculate the number of molecules in 1.00 mg of aspartame

Now we use Avogadro's number to find the number of molecules in 1.00 mg of aspartame. Number of molecules = (number of moles) × (Avogadro's number) We are given that Avogadro's number = \(6.022\times10^{23}\) molecules/mol Now, multiply the number of moles from Step 2 by Avogadro's number: Number of molecules = number of moles × \(6.022\times10^{23}\)
04

Calculate the number of hydrogen atoms in 1.00 mg of aspartame

Since there are 18 hydrogen atoms per aspartame molecule, we can find the total number of hydrogen atoms by: Number of hydrogen atoms = (number of molecules) × (hydrogen atoms per molecule) We have the number of molecules from Step 3. Multiply the number of molecules by 18 to find the total number of hydrogen atoms in 1.00 mg of aspartame.

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Most popular questions from this chapter

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol \((\mathrm{THC}),\) the active ingredient in marijuana, is required to produce intoxication. The molecular formula of \(\mathrm{THC}\) is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\). How many moles of THC does this 25 \mug represent? How many molecules? (b) Caffeine, a stimulant found in coffee, contains \(49.5 \% \mathrm{C}\), \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N},\) and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\). (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \% \mathrm{C}, 4.77 \% \mathrm{H}, 37.85 \% \mathrm{O},\) \(8.29 \% \mathrm{~N},\) and \(13.60 \% \mathrm{Na},\) and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\)

If \(1.5 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, 1.5 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8},\) and \(1.5 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2}\) \(\mathrm{COCH}_{3}\) are completely combusted in oxygen, which produces the largest number of moles of \(\mathrm{H}_{2} \mathrm{O} ?\) Which produces the least? Explain.

(a) What is Avogadro's number, and how is it related to the mole? (b) What is the relationship between the formula weight of a substance and its molar mass?

When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right)\), bromobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)\) is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) When \(30.0 \mathrm{~g}\) of benzene reacts with \(65.0 \mathrm{~g}\) of bromine, what is the theoretical yield of bromobenzene? (b) If the actual yield of bromobenzene is \(42.3 \mathrm{~g}\), what is the percentage yield?

An element \(X\) forms an iodide \(\left(\mathrm{Xl}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right)\). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{Xl}_{3}\) is treated, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.
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