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Ch.2 - Atoms & Elements
Tro - Chemistry: A Molecular Approach 6th Edition
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Capitolo 2, Problema 101

How many carbon atoms are there in a diamond (pure carbon) with a mass of 83 mg?

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1
Convert the mass of the diamond from milligrams to grams by dividing by 1000, since there are 1000 milligrams in a gram.
Use the molar mass of carbon, which is approximately 12.01 g/mol, to convert the mass of the diamond in grams to moles of carbon. This is done by dividing the mass in grams by the molar mass of carbon.
Apply Avogadro's number, which is approximately \(6.022 \times 10^{23}\) atoms/mol, to convert the moles of carbon to the number of carbon atoms. Multiply the number of moles by Avogadro's number.
Ensure that all units cancel appropriately during the calculations to confirm that the final result is in atoms.
Review the steps to ensure that each conversion is correctly applied and that the logical flow from mass to moles to atoms is clear.

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Molar Mass of Carbon

The molar mass of carbon is approximately 12.01 g/mol. This value represents the mass of one mole of carbon atoms, which is essential for converting between mass and the number of atoms. In this question, knowing the molar mass allows us to calculate how many moles of carbon are present in the given mass of diamond.
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Molar Mass Concept

Avogadro's Number

Avogadro's number, approximately 6.022 x 10²³, is the number of atoms, ions, or molecules in one mole of a substance. This constant is crucial for determining the number of individual carbon atoms in the diamond once the number of moles is calculated from the mass. It provides a bridge between the macroscopic scale of grams and the microscopic scale of atoms.
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Mass to Moles Conversion

To find the number of carbon atoms in a diamond sample, we first convert the mass of the sample (in grams) to moles using the formula: moles = mass (g) / molar mass (g/mol). This conversion is fundamental in stoichiometry, allowing us to relate the mass of a substance to the number of particles it contains, which is necessary for answering the question.
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Mass and Moles Conversion