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Caffiene has a molecular weight of 194 . If it contains \(28.9 \%\) by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is : (a) 4 (b) 6 (c) 2 (d) 3

Short Answer

Expert verified
The number of atoms of nitrogen in one molecule of caffeine is 4.

Step by step solution

01

Identify the Mass of Nitrogen in One Molecule

Calculate the mass of nitrogen in caffeine by using the percent by mass. Multiply the total molecular weight of caffeine by the percentage of nitrogen by mass (expressed as a decimal).
02

Convert Mass of Nitrogen to Moles

Use the molar mass of nitrogen to convert the mass of nitrogen to moles. The molar mass of nitrogen (N) is 14 g/mol, as nitrogen atoms are typically found in pairs (N2) in nature, but we are interested in individual nitrogen atoms.
03

Calculate Number of Nitrogen Atoms

One mole of any substance contains Avogadro's number of atoms or molecules. Since there's only one molecule of caffeine under consideration, use the mole ratio between nitrogen and caffeine to find the number of nitrogen atoms.

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Key Concepts

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

Molecular Weight Calculation
Understanding molecular weight is crucial for interpreting the composition and reactions of a substance in chemistry. Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. It is typically expressed in atomic mass units (amu) or grams per mole (g/mol).

To calculate the molecular weight of a compound like caffeine, one must add up the atomic weights of all the atoms present in its molecular formula. For example, if caffeine contains carbon (C), hydrogen (H), nitrogen (N), and oxygen (O) atoms, you would sum the atomic weights of these atoms, taking into account the number of each type present in a molecule, to arrive at the molecular weight of caffeine.
Percent Composition by Mass
Percent composition by mass is a way of expressing the relative quantities of elements in a compound. It indicates the mass percentage of each element compared to the total mass of the compound. To find the percent composition by mass, you would divide the mass of each element in one mole of the compound by the total molecular weight of the compound, and then multiply the result by 100 to get a percentage.

For instance, in our caffeine example, if nitrogen represents 28.9% of the molecule's mass, this tells us that nitrogen contributes significantly to the compound's mass. Knowing the percent composition is useful for predicting reactivity and for calculating how much of each element is present in a given sample of the compound.
Mole Concept
The mole is a foundational concept in chemistry that provides a bridge between the microscopic world of atoms and the macroscopic world we observe. One mole of any substance contains the same number of particles—usually atoms, molecules, or ions—as there are in 12 grams of carbon-12. This number is Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities per mole.

Using the mole concept allows chemists to count atoms by weighing. When working with a substance like caffeine, knowing the moles of nitrogen involved can help solve for the number of nitrogen atoms in the molecule, as seen in our original exercise. The moles of an element in a compound are calculated by dividing the mass of the element in the sample by its atomic weight.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\), is the cornerstone of stoichiometry in chemistry. It represents the quantity of atoms, ions, or molecules contained in one mole of a substance. This vast number makes it possible to use grams and moles for measuring atoms, which would otherwise be impracticable due to their minuscule size.

Applying this concept to our caffeine problem, once we have the moles of nitrogen, we can determine the number of nitrogen atoms by multiplying by Avogadro's number. For a single molecule of caffeine, the number of nitrogen atoms can be deduced without directly using Avogadro's number since we are looking at the molecular level, not the macroscopic level.

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Most popular questions from this chapter

The relation between molarity \((M)\) and molality \((m)\) is given by : \(\left(\rho=\right.\) density of solution \((\mathrm{mg} / \mathrm{mL}), M_{1}=\) molecular weight of solute) (a) \(m=\frac{1000 M}{1000 \rho-M_{1}}\) (b) \(m=\frac{1000 \rho M}{1000 \rho-M M_{1}}\) (c) \(m=\frac{1000 \mathrm{MM}}{1000 \rho-M M_{1}}\) (d) \(m=\frac{1000 M}{1000 \rho-M M_{1}}\)

A \(25.0 \mathrm{~mm} \times 40: 0 \mathrm{~mm}\) piece of gold foil is \(0.25 \mathrm{~mm}\) thick. The density of gold is \(19.32 \mathrm{~g} / \mathrm{cm}^{3}\). How many gold atoms are in the sheet? (Atomic weight : \(\mathrm{Au}=197.0\) ) (a) \(7.7 \times 10^{23}\) (b) \(1.5 \times 10^{23}\) (c) \(4.3 \times 10^{21}\) (d) \(1.47 \times 10^{22}\)

The formula of an acid is \(\mathrm{H} X \mathrm{O}_{2}\). The mass of \(0.0242\) moles of the acid is \(1.657 \mathrm{~g}\). What is the atomic weight of \(X ?\) (a) \(35.5\) (b) \(28.1\) (c) 128 (d) \(19.0\)

40 milligram diatomic volatile substance \(\left(X_{2}\right)\) is converted to vapour that displaced \(4.92 \mathrm{~mL}\) of air at 1 atm and \(300 \mathrm{~K}\). Atomic weight of element \(X\) is nearly: (a) 400 (b) 240 (c) 200 (d) 100

A sample of peanut oil weighing \(2 \mathrm{~g}\) is added to \(25 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{KOH}\). After saponification is complete, \(8.5 \mathrm{~mL}\) of \(0.28 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is needed to neutralize excess of \(\mathrm{KOH}\). The saponification number of peanut oil is : (saponification number is defined as the milligrams of KOH consumed by \(1 \mathrm{~g}\) of oil) (a) \(146.72\) (b) \(223.44\) (c) \(98.9\) (d) None of these

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