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How many atoms are in \(6.00 \mathrm{~g}\) of carbon-12?

Short Answer

Expert verified
There are approximately 3.011 x 10^23 atoms in 6.00 g of carbon-12.

Step by step solution

01

Calculate moles of carbon-12

First, determine the amount of carbon-12 in moles using the molar mass of carbon-12, which is exactly 12.00 grams per mole by definition. Use the formula: moles of carbon = mass of carbon / molar mass of carbon. In this case, moles of carbon = 6.00 g / 12.00 g/mol.
02

Use Avogadro's Number to find the number of atoms

Avogadro's number, which is approximately 6.022 x 10^23 atoms/mol, is used to convert moles to atoms. To find the number of atoms, multiply the moles of carbon-12 by Avogadro's number: Number of atoms = moles of carbon x Avogadro's number.
03

Calculate the total number of atoms

Multiply the moles of carbon calculated in Step 1 by Avogadro's number to get the total number of atoms in 6.00 g of carbon-12. If there are 0.500 moles of carbon-12, the calculation will be: Number of atoms = 0.500 mol x 6.022 x 10^23 atoms/mol.

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Key Concepts

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

Understanding Molar Mass
Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance. It is usually expressed in grams per mole (g/mol). Every element has a unique molar mass, which can be found on the periodic table. For carbon-12, the molar mass is precisely 12.00 g/mol. This value is crucial for converting between the mass of a substance and the number of moles.

For instance, if you have a 6.00 g sample of carbon-12, you can determine the number of moles by dividing the mass by the molar mass. The molar mass serves as a conversion factor between grams and moles, enabling chemists to count particles (like atoms or molecules) by weighing them. Understanding how to use molar mass helps in calculating the amount of reactants and products in chemical reactions.
Moles Calculation Made Simple
The calculation of moles is an essential step in stoichiometry, which is the study of the quantities of substances involved in chemical reactions. Moles provide a way to express amounts of a chemical substance that can be related back to the number of atoms or molecules. To calculate moles, simply divide the given mass of a substance by its molar mass.

The formula to calculate moles is: \( \text{moles} = \frac{\text{mass of substance}}{\text{molar mass of substance}} \). In our case with carbon-12, the calculation would be \( 0.500 \) moles = \( \frac{6.00 \text{ g}}{12.00 \text{ g/mol}} \). This step is the bridge between the laboratory scale and the microscopic world of atoms and molecules.
How to Count Atoms with Avogadro's Number
Avogadro's number, approximately \(6.022 \times 10^{23}\) atoms/mol, is a constant that represents the number of particles in one mole of substance. This number is named after Amedeo Avogadro and is a cornerstone of chemistry because it allows chemists to count atoms by weighing them. When you have the number of moles, multiply it by Avogadro's number to get the actual number of atoms or molecules.

For example, using the calculated moles of carbon-12 from our exercise, we can find the number of atoms as follows: \( \text{Number of atoms} = \text{moles of carbon-12} \times \text{Avogadro's number} \) which equals \( 0.500 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \). This calculation tells us exactly how many individual carbon-12 atoms are present in a 6.00 g sample.
Chemical Quantity and Its Significance
Chemical quantity is a measure of the amount of a substance. In chemistry, the quantitative relationships between reactants and products in a chemical reaction are analyzed in terms of moles. This approach is used to balance chemical equations and to calculate reactant and product quantities.

Quantitative analysis in chemistry often requires the conversion of mass to moles, and subsequently to number of particles using Avogadro's number. Used together, these concepts allow scientists and students alike to perform calculations that predict the outcomes of chemical reactions. By understanding the relationship between mass, moles, Avogadro's number, and chemical quantity, one can accurately scale reactions, design experiments, and interpret results in the proper context.

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Most popular questions from this chapter

Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed by its reaction with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), as follows. $$ \begin{aligned} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)+4 \mathrm{Cl}_{2}(g)+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \\ 2 \mathrm{NaHSO}_{4}(a q)+8 \mathrm{HCl}(a q) \end{aligned} $$ (a) How many moles of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) are needed to react with \(0.12 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) ? (b) How many moles of \(\mathrm{HCl}\) can form from \(0.12 \mathrm{~mol}\) of \(\mathrm{Cl}_{2} ?\) (c) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) are required for the reaction $$ \text { of } 0.12 \mathrm{~mol} \text { of } \mathrm{Cl}_{2} ? $$ (d) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) react if \(0.24 \mathrm{~mol} \mathrm{HCl}\) is formed?

The following are empirical formulas and the masses per mole for three compounds. What are their molecular formulas? (a) \(\mathrm{NaS}_{2} \mathrm{O}_{3} ; 270.4 \mathrm{~g} / \mathrm{mol}\) (c) \(\mathrm{C}_{2} \mathrm{HCl} ; 181.4 \mathrm{~g} / \mathrm{mol}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{Cl} ; 147.0 \mathrm{~g} / \mathrm{mol}\)

A compound of \(\mathrm{Ca}, \mathrm{C}, \mathrm{N},\) and \(\mathrm{S}\) was subjected to quantitative analysis and formula mass determination, and the following data were obtained. A \(0.250 \mathrm{~g}\) sample was mixed with \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to convert all of the Ca to \(0.160 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\). A \(0.115 \mathrm{~g}\) sample of the compound was carried through a series of reactions until all of its \(S\) was changed to \(0.344 \mathrm{~g}\) of \(\mathrm{BaSO}_{4}\). A \(0.712 \mathrm{~g}\) sample was processed to liberate all of its \(\mathrm{N}\) as \(\mathrm{NH}_{3}\), and \(0.155 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) was obtained. The formula mass was found to be 156 . Determine the empirical and molecular formulas of this compound.

The incandescent white of a fireworks display is caused by the reaction of phosphorus with \(\mathrm{O}_{2}\) to give \(\mathrm{P}_{4} \mathrm{O}_{10}\). (a) Write the balanced chemical equation for the reaction. (b) How many grams of \(\mathrm{O}_{2}\) are needed to combine with \(6.85 \mathrm{~g}\) of \(\mathrm{P} ?\) (c) How many grams of \(\mathrm{P}_{4} \mathrm{O}_{10}\) can be made from \(8.00 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\) (d) How many grams of \(\mathrm{P}\) are needed to make \(7.46 \mathrm{~g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10} ?\)

Lead compounds are often highly colored and are toxic to mold, mildew, and bacteria, properties that in the past were useful for paints used before 1960 . Today we know lead is very hazardous and it is not used in paint; however, old paint is still a problem. If a certain lead-based paint contains \(14.5 \% \mathrm{PbCr}_{2} \mathrm{O}_{7}\) and \(73 \%\) of the paint evaporates as it dries, what mass of lead will be in a paint chip that weighs \(0.15 \mathrm{~g}\) ?

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