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The molecular formula of aspartame, the artificial sweetener marketed as NutraSweet \(^{0}\), 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 \(3.40 \times 10^{-6} \: moles\) of aspartame in 1.00 mg. (c) There are \(2.05 \times 10^{18} \: molecules\) of aspartame in 1.00 mg. (d) There are \(3.68 \times 10^{19} \: hydrogen \: atoms\) in 1.00 mg of aspartame.

Step by step solution

01

Calculate the Molar Mass of Aspartame

Using the molecular formula, C14H18N2O5, count the number of each type of atom in aspartame and multiply this by their atomic masses. Add the resulting masses to find the molar mass of aspartame. A molar mass of aspartame is equal to the sum of the molar masses of all its elements multiplied by their respective quantities in the formula: Molar mass = \(14 \times (Mass\_C) + 18 \times (Mass\_H) + 2 \times (Mass\_N) +5 \times (Mass\_O)\) where, Mass_C = 12.01 u (atomic mass of carbon) Mass_H = 1.01 u (atomic mass of hydrogen) Mass_N = 14.01 u (atomic mass of nitrogen) Mass_O = 16.00 u (atomic mass of oxygen)
02

Find the Moles of Aspartame in 1.00 Mg

To convert the given mass of aspartame (1.00 mg) to moles, divide the mass by the molar mass of aspartame. Remember to convert milligrams to grams first. The formula for finding moles: moles = (given mass) / (molar mass)
03

Calculate the Number of Molecules in 1.00 Mg of Aspartame

To find the number of molecules present in 1.00 mg of aspartame, multiply the number of moles found in the previous step by Avogadro's number, \(6.022 \times 10^{23}\). The formula for finding the number of molecules: molecules = (moles) x (Avogadro's number)
04

Determine the Number of Hydrogen Atoms in 1.00 mg of Aspartame

Since we know the number of aspartame molecules in 1.00 mg, we can multiply this value by the number of hydrogen atoms per molecule to determine the total number of hydrogen atoms. The formula for finding hydrogen atoms: hydrogen atoms = (number of molecules) x (number of hydrogen atoms per molecule)

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Key Concepts

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

Molar Mass
The molar mass of a compound gives us insight into the weight of one mole of its molecules. It is an essential property in chemistry. To calculate the molar mass, you need the molecular formula of the compound and the atomic masses of each element.
In the case of aspartame, with the formula \(\text{C}_{14}\text{H}_{18}\text{N}_{2}\text{O}_{5}\), you perform the following steps:
  • Identify the number of each type of atom present.
  • Multiply the atomic mass of each element by its respective number of atoms.
  • Add up the products to get the total molar mass.
For example, for carbon in aspartame, you multiply 14 by its atomic mass of 12.01 u. Repeat this for hydrogen, nitrogen, and oxygen, and sum all these values to calculate the molar mass. Understanding this process helps in computation of other quantities like moles and molecules.
Moles Calculation
When we refer to calculating the number of moles, we're describing how to determine the amount of a substance present in a given mass. This is a bridge connecting macroscopic quantities that we measure directly and microscopic scales used in chemistry.
To calculate moles, you use the formula: \[\text{moles} = \frac{\text{given mass (g)}}{\text{molar mass (g/mol)}}\] Given the mass of aspartame is specified in milligrams, conversion to grams is necessary (1 mg = 0.001 g). This calculation tells us how many moles are in this small mass of aspartame, providing a key quantity for further chemical analysis and computations.
Avogadro's Number
Avogadro's number is the key to connecting moles to molecules. This constant, \(6.022 \times 10^{23}\), represents the number of atoms or molecules in one mole of a substance.
By using Avogadro’s number, you can determine the number of molecules in a given amount of a substance. For aspartame, once the moles have been calculated, multiply the moles by Avogadro's number to convert moles into number of molecules: \[\text{number of molecules} = \text{moles} \times 6.022 \times 10^{23}\]
This is a crucial step to bridge the gap between the quantity of substance and the number of particles, extending our understanding of molecular interactions and reactions.
Chemical Formula Analysis
Understanding the chemical formula of a compound is the foundation for calculating its properties. A formula like \(\text{C}_{14}\text{H}_{18}\text{N}_{2}\text{O}_{5}\) reveals the types and numbers of atoms present in a molecule of aspartame.
Analysis of this formula involves:
  • Identifying the kinds of atoms in the compound.
  • Counting the number of each type of atom in one molecule.
  • Using these counts to calculate molar mass and related chemical properties.
This formula serves as a map, allowing us to perform calculations determining molar mass, number of particles, and even specific atom counts within the compound. Such analysis is critical for any chemical computation or experimental procedure involving the compound.

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Most popular questions from this chapter

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\), where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\). When \(5.061 \mathrm{~g}\) of this hydrate is heated to \(250{ }^{\circ} \mathrm{C}\), all the water of hydration is lost, leaving \(2.472 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}\). What is the value of \(x\) ?

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(10.4 \% \mathrm{C}, 27.8 \% \mathrm{~S}\), and \(61.7 \% \mathrm{Cl}\) (b) \(21.7 \% \mathrm{C}, 9.6 \% \mathrm{O}\), and \(68.7 \% \mathrm{~F}\) (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}\), and \(54.19 \% \mathrm{~F}\)

Aluminum hydroxide reacts with sulfuric acid as follows: \(2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l)\) Which reagent is the limiting reactant when \(0.500 \mathrm{~mol}\) \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(0.500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(3.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 32 \mathrm{~g} \mathrm{O}_{2}\).

An element \(X\) forms an iodide \(\left(\mathrm{XI}_{3}\right)\) 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 \(\mathrm{XI}_{3}\) is treated, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element \(\mathrm{X}\) (b) Identify the element \(\bar{X}\).

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