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Diamond is a natural form of pure carbon. What number of atoms of carbon are in a 1.00-carat diamond \((1.00 \text { carat }=\) \(0.200 \mathrm{g}) ?\)

Short Answer

Expert verified
There are \( 1.00 \times 10^{22} \) carbon atoms in a 1.00-carat diamond.

Step by step solution

01

Calculate moles of carbon

To find the moles of carbon present in the diamond, we'll use the formula: Moles = Mass / Molar Mass The molar mass of carbon is 12.01 g/mol. So, we can find the moles of carbon using the mass of the diamond (0.200 g) and the molar mass (12.01 g/mol). Moles of Carbon = \( \frac{0.200\: g}{12.01\: g/mol} \)
02

Calculate the number of carbon atoms

Now that we have the moles of carbon, we can find the number of carbon atoms using Avogadro's number. Avogadro's number is approximately \( 6.022 x 10^{23} \) atoms/mol. Number of Carbon Atoms = Moles of Carbon × Avogadro's Number Number of Carbon Atoms = \( \frac{0.200\: g}{12.01\: g/mol} \times 6.022\times10^{23}\: atoms/mol \) Now, calculate the number of carbon atoms: Number of Carbon Atoms = \( \frac{0.200}{12.01} \times 6.022\times10^{23} \) = \( 1.00 \times 10^{22} \: atoms \) So, there are \( 1.00 \times 10^{22} \) carbon atoms in a 1.00-carat diamond.

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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 is an essential concept in chemistry, as it relates to the mass of one mole of a substance, which is equivalent to its average atomic or molecular mass in grams. For any element, you can find the molar mass on the periodic table, which is expressed in grams per mole (g/mol).

For example, the molar mass of carbon is crucial when calculating the composition of a diamond. All carbon atoms have a molar mass of 12.01 g/mol, a value based on the weighted average of all naturally occurring isotopes of carbon. The precision of this mass allows chemists to calculate how many moles of carbon are in a given mass of a sample, establishing a bridge between the microscopic scale of atoms and the macroscopic world we can measure.
Avogadro's Number
Avogadro's number is a constant named after the Italian scientist Amedeo Avogadro. It represents the number of units (atoms, molecules, ions, or other particles) in one mole of any substance. The value of Avogadro's number is approximately \(6.022 \times 10^{23}\) particles per mole.

This constant is integral in converting moles to discrete particles, which is necessary when working with individual atoms, as in the case of atoms in diamond. Avogadro's number bridges the gap between the macroscopic amounts we work with in the lab and the atomic scale, helping students and chemists visualize and compute the vast numbers of atoms or molecules present in tangible quantities of matter.
Moles of Carbon
In chemistry, a mole is a unit that represents a specific number of particles, much like a 'dozen' represents twelve items. One mole of any substance contains Avogadro's number of particles, corresponding to its molecular or atomic mass expressed in grams.

To determine the moles of carbon in a diamond, one divides the mass of the diamond by the molar mass of carbon. This relationship allows us to move from mass, an easily measurable property, to moles, which connects us to the number of atoms or molecules via Avogadro's number. Understanding how to calculate moles is fundamental to stoichiometry, the study of the amounts of substances involved in chemical reactions.
Atoms Calculation
The final aim of calculations in exercises like determining the number of carbon atoms in a diamond is to bridge theoretical concepts with practical application. By calculating the moles of carbon first and then multiplying by Avogadro's number, students can transform a minuscule mass of diamond into an understanding of the quantity of carbon atoms within.

This step is the culmination of applying the concepts of molar mass and Avogadro's number, turning the abstract into the concrete. Estimating the number of atoms in a given substance requires an understanding of both the macroscopic (mass and moles) and microscopic (atoms and molecules) worlds, emphasizing the relevance and practicality of stoichiometric calculations in understanding the material world.

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Most popular questions from this chapter

In the production of printed circuit boards for the electronics industry, a 0.60-mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ A plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{cm}\) in area. An average of \(80 . \%\) of the copper is removed from each board (density of copper \(=8.96 \mathrm{g} / \mathrm{cm}^{3}\)). What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

Consider a mixture of potassium chloride and potassium nitrate that is \(43.2 \%\) potassium by mass. What is the percent KCl by mass of the original mixture?

Para-cresol, a substance used as a disinfectant and in the manufacture of several herbicides, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a 0.345-g sample of \(p\)-cresol produced 0.983 g carbon dioxide and \(0.230 \mathrm{g}\) water. Determine the empirical formula for \(p\)-cresol.

Consider the following unbalanced reaction: $$\mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g)$$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a \(78.1 \%\) yield?

In using a mass spectrometer, a chemist sees a peak at a mass of \(30.0106 .\) Of the choices \(^{12} \mathrm{C}_{2}\) \(^{1} \mathrm{H}_{6},\) \(^{12} \mathrm{C}\) \(^{1} \mathrm{H}_{2}\) \(^{16} \mathrm{O},\) and \(^{14} \mathrm{N}^{16} \mathrm{O},\) which is responsible for this peak? Pertinent masses are \(^{1} \mathrm{H}\), \(1.007825 ;^{16} \mathrm{O}, 15.994915 ;\) and \(^{14} \mathrm{N}, 14.003074.\)

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