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In 78.20 g of potassium, there are ______ potassium atoms present.

Short Answer

Expert verified
In 78.20 g of potassium, there are \(1.999 × 10^{24}\) potassium atoms present.

Step by step solution

01

1. Find the molar mass of potassium

The molar mass of potassium can be found on the periodic table. Potassium has an atomic number of 19 and a molar mass of approximately 39.1 g/mol.
02

2. Convert the given mass to moles

To convert grams of potassium to moles, we'll divide the given mass by the molar mass of potassium. Let's denote the given mass of potassium as m. m = 78.20 g number of moles (n) = \(\frac{m}{molar~mass}\) n = \(\frac{78.20~g}{39.1~g/mol}\)
03

3. Calculate the number of potassium atoms

Now that we have the number of moles, we can multiply it by Avogadro's number (6.022 x 10^23/mol) to find the number of potassium atoms. Number of potassium atoms = number of moles × Avogadro's number Number of potassium atoms = \(n × 6.022 × 10^{23}~atoms/mol\) Now, perform the calculations to find the final answer.

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Key Concepts

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

Molar Mass of Potassium
Understanding the molar mass of an element is fundamental in chemistry. The molar mass refers to the mass of one mole of atoms of the element and it's measured in grams per mole (g/mol). For potassium, which is denoted with the symbol 'K' on the periodic table, its molar mass is approximately 39.1 g/mol.

This value is crucial for conversions between the mass of a substance and the amount of substance in moles. By definition, the molar mass of potassium is numerically equivalent to the atomic mass of a single potassium atom, but it's expressed in g/mol rather than atomic mass units (amu).

When solving problems related to the amount of substance, like the number of atoms in a given mass, the molar mass serves as a conversion factor. This allows us to move from the macroscopic scale of grams to the microscopic scale of moles, bridging the gap between mass and the amount of atoms or molecules.
Avogadro's Number
Avogadro's number, named after Amedeo Avogadro, is a fundamental constant in chemistry that provides the relationship between microscopic and macroscopic scales. It is defined as the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. Avogadro's number is precisely 6.02214076 × 1023 particles per mole.

This number allows chemists to count atoms or molecules in a tangible amount of material by simply knowing the amount in moles. When dealing with large quantities of tiny particles, like atoms, Avogadro's number becomes an indispensable tool for scientists. After calculating the number of moles, multiplying it by Avogadro's number provides the total number of atoms in that sample.

Without this constant, tasks such as determining the number of potassium atoms in a 78.20 g sample would be incredibly complex, since counting each atom individually is not feasible. Avogadro's number streamlines this process and connects the macroscopic world with the atomic scale.
Mole Concept
The mole concept is one of the cornerstones of chemistry. It is a unit of measurement used to express amounts of a chemical substance. One mole contains exactly 6.02214076 × 1023 (which is Avogadro's number) of particles, whether they are atoms, ions, or molecules.

The beauty of the mole concept lies in its ability to provide a bridge between the atomic scale and the real world. By measuring substances in moles, chemists can work with amounts that are both comprehensible and convenient for laboratory work. For any substance, the number of moles is the mass of the substance divided by its molar mass.

For instance, if we have 78.20 g of potassium, we can use its molar mass to find out how many moles that quantity represents. This step is the key to unlocking all further calculations related to the amount of substance, including finding the total number of atoms or the volume of a gas at standard temperature and pressure. The mole concept simplifies complex quantitative relationships in chemistry, making it an essential part of problem-solving.

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Most popular questions from this chapter

Calculate the mass of carbon in grams, the percent carbon by mass, and the number of individual carbon atoms present in each of the following samples. a. 7.819 g of carbon suboxide, \(\mathrm{C}_{3} \mathrm{O}_{2}\) b. \(1.53 \times 10^{21}\) molecules of carbon monoxide c. 0.200 mol of phenol, \(C_{6} \mathrm{H}_{6} \mathrm{O}\)

Using the average atomic masses given inside the front cover of this text, calculate the number of atoms present in each of the following samples. a. \(2.89 \mathrm{g}\) of gold b. 0.000259 mol of platinum c. 0.000259 g of platinum d. 2.0 lb of magnesium e. \(1.90 \mathrm{mL}\) of liquid mercury (density \(=13.6 \mathrm{g} / \mathrm{mL}\) ) f. 4.30 mol of tungsten g. \(4.30 \mathrm{g}\) of tungsten

A compound has the following percentages by mass: aluminum, 32.13\%; fluorine, 67.87\%. Calculate the empirical formula of the compound.

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds. a. methane, \(\mathrm{CH}_{4}\) b. sodium nitrate, \(\mathrm{NaNO}_{3}\) c. carbon monoxide, \(\mathrm{CO}\) d. nitrogen dioxide, \(\mathrm{NO}_{2}\) e. 1 -octanol, \(\mathrm{C}_{8} \mathrm{H}_{18} \mathrm{O}\) f. calcium phosphate, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) g. 3 -phenylphenol, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{O}\) h. aluminum acetate, \(\mathrm{Al}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{3}\)

When barium metal is heated in chlorine gas, a binary compound forms that consists of \(65.95 \%\) Ba and \(34.05 \%\) Cl by mass. Calculate the empirical formula of the compound.

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