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(a) What is the mass, in grams, of a mole of \({ }^{12} \mathrm{C}\) ? (b) How many carbon atoms are present in a mole of \({ }^{12} \mathrm{C}\) ?

Short Answer

Expert verified
(a) The mass of a mole of \({ }^{12} \mathrm{C}\) is 12 grams. (b) There are \(6.022 \times 10^{23}\) carbon atoms in a mole of \({ }^{12} \mathrm{C}\).

Step by step solution

01

Part (a) - Finding the mass of a mole of \({ }^{12} \mathrm{C}\) in grams

To find the mass of a mole of \({ }^{12} \mathrm{C}\) in grams, we simply need to recall that the atomic mass of carbon is approximately 12 atomic mass units (amu) or 12 grams/mole. Since we are asked to find the mass of a mole of \({ }^{12} \mathrm{C}\), the mass is simply: Mass = Atomic mass of Carbon * Number of moles In this case, we are asked about 1 mole of \({ }^{12} \mathrm{C}\): Mass = 12 g/mole * 1 mole Mass = 12 grams
02

Part (b) - Finding the number of carbon atoms in a mole of \({ }^{12} \mathrm{C}\)

To find the number of carbon atoms in a mole of \({ }^{12} \mathrm{C}\), we need to use Avogadro's number. Avogadro's number is the number of atoms or particles in one mole of any substance and given by: Avogadro's number = \(6.022 \times 10^{23}\) atoms/mole Since we are asked to find the number of carbon atoms in a mole of \({ }^{12} \mathrm{C}\), we multiply the number of moles by Avogadro's number: Number of carbon atoms = Number of moles * Avogadro's number In this case, the number of moles is 1: Number of carbon atoms = 1 mole * \(6.022 \times 10^{23}\) atoms/mole Number of carbon atoms = \(6.022 \times 10^{23}\) atoms

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Key Concepts

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

Atomic Mass
When diving into the world of chemistry, one foundational concept we come across is that of atomic mass. This term refers to the mass of an individual atom, expressed in atomic mass units (amu), which is essentially a scale for weighing atoms.

The atomic mass provides us with a handy way to compare the weights of different atoms. It is calculated based on the weighted average of all naturally occurring isotopes of an element. To simplify calculations in chemistry, however, we often use the relative atomic mass, where elements are compared to a standard: the carbon-12 isotope

One atomic mass unit is defined as one twelfth of the mass of a carbon-12 atom. In the given exercise, carbon has an atomic mass of approximately 12 amu. This means that one mole of carbon atoms, just like a dozen eggs in a carton, has a mass of 12 grams.
Avogadro's Number
Another cornerstone of the mole concept in chemistry is Avogadro's number. It's essentially a bridge connecting the macroscopic world we observe to the microscopic realm of atoms and molecules. Avogadro's number is defined as exactly \(6.022 \times 10^{23}\) entities per mole, whether those entities are atoms, molecules, electrons, or any other particles.

This constant allows chemists to count particles in a given sample by simply weighing it. When we say there's a mole of something, we're saying there are \(6.022 \times 10^{23}\) of it, which is exactly the number of carbon atoms present in one mole of the carbon-12 isotope as noted in the exercise.

Understanding Avogadro's number is crucial because it underpins the concept of stoichiometry, allowing precise calculations of reactants and products in chemical reactions.
Carbon-12 Isotope
The carbon-12 isotope holds a special place in chemistry, particularly in the context of atomic masses and mole calculations. An isotope is a variant of an element with a differing number of neutrons in the nucleus. For carbon, the carbon-12 isotope has six protons and six neutrons, adding up to a mass number of 12.

Carbon-12 is the basis upon which the atomic mass unit is founded. While discussing atomic masses, we often say that the atomic mass of carbon-12 is exactly 12 amu by definition. This makes calculations much smoother because we can directly equate the mass of a mole of the carbon-12 atoms to 12 grams, providing a tangible way to measure atoms in terms of grams as done in the solution to the exercise provided.
Stoichiometry
Stoichiometry is the part of chemistry that is concerned with the measurement and quantitative relationships of reactants and products in chemical reactions. It's like a recipe: if you know the proportions of each ingredient, you can predict the amount of the final product.

By using the atomic masses of elements and Avogadro's number, stoichiometry provides a systematic approach to solving problems like the one in our exercise. It requires a clear understanding of the mole concept, which connects the countable nature of atoms and molecules to measurable quantities. In the exercise, stoichiometry was indirectly applied: we used the atomic mass of carbon and Avogadro's number to find the mass and number of atoms in a mole of carbon-12, demonstrating a basic stoichiometric conversion.

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Most popular questions from this chapter

Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily HCl: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Balance this equation. (b) Calculate the number of grams of \(\mathrm{HCl}\) that can react with \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\). (c) Calculate the number of grams of \(\mathrm{AlCl}_{3}\) and the number of grams of \(\mathrm{H}_{2} \mathrm{O}\) formed when \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\). A \(0.1005-\mathrm{g}\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

A sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), contains \(1.250 \times 10^{21}\) carbon atoms. (a) How many atoms of hydrogen does it contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it contain? (d) What is the mass of this sample in grams?

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(1.85 \mathrm{~mol}\) \(\mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

The fizz produced when an Alka-Seltzer \({ }^{B}\) tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) $$ \begin{gathered} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \longrightarrow \\ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{gathered} $$ In a certain experiment \(1.00 \mathrm{~g}\) of sodium bicarbonate and \(1.00 \mathrm{~g}\) of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

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