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Gold has a molar (atomic) mass of \(197 \mathrm{~g} / \mathrm{mol}\). Consider a 2.56\(g\) sample of pure gold vapor. (a) Calculate the number of moles of gold present. (b) How many atoms of gold are present?

Short Answer

Expert verified
The number of moles is \(0.013\) and the number of atoms is \(7.81 \times 10^{21}\)

Step by step solution

01

Calculation of number of moles

The number of moles of a substance is given by the formula: \[ number of moles = \frac{mass of substance}{molar mass of substance}\] For gold, this becomes \[number of moles = \frac {2.56 g}{197 g/mol} \]
02

Evaluation of the number of moles

Calculate the value on the right hand side of the equation to get the number of moles.
03

Calculation of number of atoms

The number of atoms in a substance is given by the formula: \[ number of atoms = number of moles \times Avogadro's number\] Substituting the values, we get \[ number of atoms = number of moles \times 6.022 \times 10^{23} \]
04

Evaluation of the number of atoms

Calculate the value on the right hand side of the equation to get the number of atoms.

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Key Concepts

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

Calculation of Moles
One of the fundamental concepts in chemistry is the calculation of moles. The mole represents a standard quantity in chemistry used to measure the amount of any substance. Think of it as the chemist's dozen. To calculate the number of moles, you need to know two things:
  • The mass of the substance you have.
  • The molar mass of the substance.
The formula is simple: \[\text{number of moles} = \frac{\text{mass of substance}}{\text{molar mass of substance}}\]In the given exercise, the mass of gold is 2.56 grams, and its molar mass is 197 grams per mole. By substituting these values into the formula, you can determine the number of moles of gold in the sample. This step is crucial because it sets the stage for determining the number of atoms using Avogadro's Number.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry and physics. It is used to convert moles into numbers of atoms, molecules, or other particles, depending on the substance involved. Named after the Italian scientist Amedeo Avogadro, this number is:\[6.022 \times 10^{23}\]This amazingly large number represents the number of atoms or molecules in one mole of a substance. When calculating the number of atoms in a given sample, Avogadro's number becomes indispensable. Once you calculate the number of moles in a gold sample, multiplying it by Avogadro's number will give the total number of atoms in that sample. This concept is key for translating moles into real-world quantities of atoms.
Atomic Mass
Atomic mass, sometimes referred to as atomic weight, is a measure of the mass of an atom. It is expressed in atomic mass units (amu) or grams per mole (g/mol) when dealing with the mole concept. The atomic mass of gold, as given in the problem, is 197 g/mol. Understanding atomic mass is crucial because it serves as the basis for calculating the number of moles. Once you know the atomic mass of a substance, you can easily determine how much one mole of it weighs. Knowing this allows chemists to measure and calculate chemical quantities precisely, using the mole as a bridge between the mass of a substance and the number of entities of which it is composed.
Number of Atoms
The number of atoms in a sample can be determined by using the relationship between moles and Avogadro's number. Once you have calculated the number of moles from the mass and molar mass, the next step is to find out how many atoms are in that mole count.The formula to determine the number of atoms is:\[\text{number of atoms} = \text{number of moles} \times \text{Avogadro's number}\]This calculation reveals the actual number of individual atoms in a sample, giving you a sense of scale that goes beyond just mass or moles. For a 2.56 gram sample of gold, this process allows you to see just how many atoms make up that seemingly small mass, providing a tangible understanding of the enormity of molecular world.

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Most popular questions from this chapter

Estimate the van der Waals constant \(b\) for \(\mathrm{H}_{2} \mathrm{O}\) knowing that one kilogram of water has a volume of \(0.001 \mathrm{~m}^{3}\). The molar mass of water is \(18 \mathrm{~g} / \mathrm{mol}\).

You are given the following group of particles \(\left(N_{n}\right.\) represents the number of particles that have a speed \(v_{n}\) ): $$\begin{array}{lc}N_{n} & v_{n}(\mathrm{~km} / \mathrm{s}) \\\\\hline 2 & 1.0 \\\4 & 2.0 \\\6 & 3.0 \\\8 & 4.0 \\\2 & 5.0\end{array}$$ (a) Compute the average speed \(v_{\mathrm{av}} .(b)\) Compute the rootmean- square speed \(v_{\text {rms. }}\) (c) Among the five speeds shown, which is the most probable speed \(v_{\mathrm{p}}\) for the entire group?

A cylindrical container of length \(56.0 \mathrm{~cm}\) and diameter \(12.5 \mathrm{~cm}\) holds \(0.350\) moles of nitrogen gas at a pressure of \(2.05\) atm. Find the rms speed of the nitrogen molecules.

(a) Ten particles are moving with the following speeds: four at \(200 \mathrm{~m} / \mathrm{s}\), two at \(500 \mathrm{~m} / \mathrm{s}\), and four at \(600 \mathrm{~m} / \mathrm{s}\). Calculate the average and root-mean-square speeds. Is \(v_{\mathrm{rms}}>v_{\mathrm{av}} ?(b)\) Make up your own speed distribution for the ten particles and show that \(v_{\mathrm{rms}} \geq v_{\mathrm{av}}\) for your distribution. ( \(c\) ) Under what condition (if any) does \(v_{\mathrm{rms}}=v_{\mathrm{av}} ?\)

Show that the constant \(a\) in the van der Waals equation can be written in units of \(\frac{\text { energy per particle }}{\text { particle density }}\)

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