Loadingcome357

• Chemistry Avogadro's Number Worksheet
• Mole Chemistry Avogadro

The Chem 13 News and Avogadro Exams are challenging online exams written by more than 6000 grade 11 and 12 Chemistry high school students every May, and consist of 40 multiple choice questions written in a 75 minute exam period.These exams are only available for purchase by a teacher. The contributions of the Italian chemist Amedeo Avogadro (1776–1856) relate to the work of two of his contemporaries, Joseph Louis Gay-Lussac and John Dalton. Gay-Lussac’s law of combining volumes (1808) stated that when two gases react, the volumes of the reactants and products—if gases—are in whole number ratios. Amedeo Avogadro: A Scientific Biography (Chemists and Chemistry, 1) Hardcover – February 29, 1984 by M. Morselli (Author) See all formats and editions Hide other formats and editions. Amedeo Avogadro, in full Lorenzo Romano Amedeo Carlo Avogadro, conte di Quaregna e Cerreto, (born August 9, 1776, Turin, in the Kingdom of Sardinia and Piedmont Italy—died July 9, 1856, Turin), Italian mathematical physicist who showed in what became known as Avogadro’s law that, under controlled conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. The specific number of molecules in one gram-mole of a substance, defined as the molecular weight in grams, is 6.02214076 × 10 23, a quantity called Avogadro’s number, or the Avogadro constant.For example, the molecular weight of oxygen is 32.00, so that one gram-mole of oxygen has a mass of 32.00 grams and contains 6.02214076 × 10 23 molecules. The volume occupied by one gram-mole of gas.

The number of moles in a system can be determined using the atomic mass of an element, which can be found on the periodic table. This mass is usually an average of the abundant forms of that element found on earth. An element's mass is listed as the average of all its isotopes on earth.

Avogadro's Constant

One mole of oxygen atoms contains (6.02214179 times 10^{23}) oxygen atoms. Also, one mole of nitrogen atoms contains (6.02214179 times 10^{23}) nitrogen atoms. The number (6.02214179 times 10^{23}) is called Avogadro's number ((N_A)) orAvogadro's constant, after the 19^th century scientist Amedeo Avogadro.

Each carbon-12 atom weighs about (1.99265 times 10^{-23}; g); therefore,

[(1.99265 times 10^{-23}; g) times (6.02214179 times 10^{23}; atoms) = 12; g; text{ of carbon-12} nonumber ]

Applications of the Mole

The mass of a mole of substance is called the molar mass of that substance. The molar mass is used to convert grams of a substance to moles and is used often in chemistry. The molar mass of an element is found on the periodic table, and it is the element's atomic weight in grams/mole (g/mol). If the mass of a substance is known, the number of moles in the substance can be calculated. Converting the mass, in grams, of a substance to moles requires a conversion factor of (one mole of substance/molar mass of substance).

The mole concept is also applicable to the composition of chemical compounds. For instance, consider methane, CH₄. This molecule and its molecular formula indicate that per mole of methane there is 1 mole of carbon and 4 moles of hydrogen. In this case, the mole is used as a common unit that can be applied to a ratio as shown below:

[2 text{ mol H } + 1 text{ mol O }= 1 text{ mol } ce{H2O} nonumber]

In this this chemical reactions, the moles of H and O describe the number of atoms of each element that react to form 1 mol of (ce{H_2O}).

To think about what a mole means, one should relate it to quantities such as dozen or pair. Just as a pair can mean two shoes, two books, two pencils, two people, or two of anything else, a mole means 6.02214179×10²³ of anything. Using the following relation:

[text{1 mole} = 6.02214179 times 10^{23}]

is analogous to saying: Hp probook 6450b drivers for windows 10.

[text{1 Dozen} = text{12 eggs}]

It is quite difficult to visualize a mole of something because Avogadro's constant is extremely large. For instance, consider the size of one single grain of wheat. If all the people who have existed in Earth's history did nothing but count individual wheat grains for their entire lives, the total number of wheat grains counted would still be much less than Avogadro's constant; the number of wheat grains produced throughout history does not even approach Avogadro's Number.

Example (PageIndex{1}): Converting Mass to Moles

How many moles of potassium ((ce{K})) atoms are in 3.04 grams of pure potassium metal?

Solution

In this example, multiply the mass of (ce{K}) by the conversion factor (inverse molar mass of potassium):

[dfrac{1; mol; K}{39.10; grams ;K} nonumber ]

39.10 grams is the molar mass of one mole of (ce{K}); cancel out grams, leaving the moles of (ce{K}):

[3.04; cancel{g; K} left(dfrac{1; mol; K}{39.10; cancel{g; K}}right) = 0.0778; mol; K nonumber ]

Similarly, if the moles of a substance are known, the number grams in the substance can be determined. Converting moles of a substance to grams requires a conversion factor of molar mass of substance/one mole of substance. One simply needs to follow the same method but in the opposite direction.

Example (PageIndex{2}): Converting Moles to mass

How many grams are 10.78 moles of Calcium ((ce{Ca}))?

Wireless keyboard for mac not working. Solution

Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.

[10.78 cancel{;mol; Ca} left(dfrac{40.08; g; Ca}{1; cancel{mol; Ca}}right) = 432.1; g; Ca nonumber ]

The total number of atoms in a substance can also be determined by using the relationship between grams, moles, and atoms. If given the mass of a substance and asked to find the number of atoms in the substance, one must first convert the mass of the substance, in grams, to moles, as in Example (PageIndex{1}). Then the number of moles of the substance must be converted to atoms. Converting moles of a substance to atoms requires a conversion factor of Avogadro's constant (6.02214179×10²³) / one mole of substance. Verifying that the units cancel properly is a good way to make sure the correct method is used.

Example (PageIndex{3}): Atoms to Mass

How many atoms are in a 3.5 g sample of sodium (Na)?

Solution

[3.5; cancel{g; Na} left(dfrac{1; mol; Na}{22.98; cancel{g; Na}}right) = 0.152; mol; Na nonumber ]

[0.152; cancel{mol; Na} left(dfrac{6.02214179times 10^{23}; atoms; Na}{1;cancel{ mol; Na}}right) = 9.15 times 10^{22}; atoms; of; Na nonumber ]

In this example, multiply the grams of Na by the conversion factor 1 mol Na/ 22.98 g Na, with 22.98g being the molar mass of one mole of Na, which then allows cancelation of grams, leaving moles of Na. Then, multiply the number of moles of Na by the conversion factor 6.02214179×10²³ atoms Na/ 1 mol Na, with 6.02214179×10²³ atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.

Using Avogadro's constant, it is also easy to calculate the number of atoms or molecules present in a substance (Table (PageIndex{1})). By multiplying the number of moles by Avogadro's constant, the mol units cancel out, leaving the number of atoms. The following table provides a reference for the ways in which these various quantities can be manipulated:

Example (PageIndex{4}): Mass to Moles

How many moles are in 3.00 grams of potassium (K)?

Solution

[3.00 ; cancel{g; K} left(dfrac{1; mol; K}{39.10; cancel{g; K}}right) = 0.0767; mol; K nonumber ]

In this example, multiply the mass of K by the conversion factor:

[dfrac{1; mol; K}{39.10; grams; K} nonumber ]

39.10 grams is the molar mass of one mole of K. Grams can be canceled, leaving the moles of K.

Example (PageIndex{5}): Moles to Mass

How many grams is in 10.00 moles of calcium (Ca)?

Solution

This is the calculation in Example (PageIndex{2}) performed in reverse. Multiply moles of Ca by the conversion factor 40.08 g Ca/ 1 mol Ca, with 40.08 g being the molar mass of one mole of Ca. The moles cancel, leaving grams of Ca:

[10.00; cancel{mol; Ca} left(dfrac{40.08; g; Ca}{1;cancel{ mol; Ca}}right) = 400.8; grams ;of ;Ca nonumber ]

The number of atoms can also be calculated using Avogadro's Constant (6.02214179×10²³) / one mole of substance.

Example (PageIndex{6}): Mass to Atoms

How many atoms are in a 3.0 g sample of sodium (Na)?

Solution

Convert grams to moles

[3.0; cancel{g; Na} left(dfrac{1; mol; Na}{22.98; cancel{g; Na}}right) = 0.130; mol; Na nonumber ]

Convert moles to atoms

[0.130548; cancel{ mol; Na} left(dfrac{6.02214179 times 10^{23}; atoms ;Na}{1; cancel{ mol; Na}}right) = 7.8 times 10^{22} ; atoms; of; ; Na nonumber ]

Chemistry Avogadro's Number Worksheet

Summary

The mole, abbreviated mol, is an SI unit which measures the number of particles in a specific substance. One mole is equal to (6.02214179 times 10^{23}) atoms, or other elementary units such as molecules.

Problems

1. Using a periodic table, give the molar mass of the following:
   1. H
   2. Se
   3. Ne
   4. Cs
   5. Fe
2. Convert to moles and find the total number of atoms.
   1. 5.06 grams of oxygen
   2. 2.14 grams of K
   3. 0.134 kg of Li
3. Convert the following to grams
   1. 4.5 mols of C
   2. 7.1 mols of Al
   3. 2.2 mols of Mg
4. How many moles are in the product of the reaction
   1. 6 mol H + 3 mol O → ? mol H₂O
   2. 1 mol Cl + 1 mol Cl → ? mol Cl₂
   3. 5 mol Na + 4 mol Cl → ? mol NaCl

Answers

1. Question 2
   1. 1.008 g/mol
   2. 78.96 g/mol
   3. 20.18 g/mol
   4. 132.91g/mol
   5. 55.85 g/mol
2. Question 2

2. 5.06g O (1mol/16.00g)= 0.316 mol of O

0.316 mols (6.022x10²³ atoms/ 1mol) = 1.904x10²³ atoms of O

Mole Chemistry Avogadro

3. 2.14g K (1mol/39.10g)= 0.055 mol of K

0.055 mols (6.022x10²³ atoms/ 1mol) = 3.312x10²² atoms of K

4. 0.134kg Li (1000g/1kg)= 134g Li (1mol/6.941g)= 19.3 mols Li

19.3 (6.022x10²³ atoms/ 1mol) = 1.16x10²⁵ atoms of Li

3. Question 3
   1. 4.5 mols of C (12.011g/1mol) = 54.05 g of C
   2. 7.1 mols of Al (26.98g/1mol) = 191.56 g of Al
   3. 2.2 mols of Mg (24.31g/1mol) = 53.48 g of MG
4. Question 4
   1. 8. 6 mol H + 3 mol O → 3 mol H₂O
   2. 9. 1 mol Cl + 1 mol Cl → 1 mol Cl₂
   3. _{10. 5 mol Na + 4 mol Cl → 4 mol NaCl + 1 mol Na (excess)}

References

1. Keenan, Charles W. and Wood, Jesse H. . General College Chemistry. 4th ed. New York: Haper and Row, 1971.
2. Mortimer, Charles E. Chemistry a Conceptual Approach. 2nd ed. New York: Van Nostrand Reinhold, 1971.
3. Jones, Loretta and Atkins, Peter. Chemistry: Molecules, Matter, and Change. 4th ed. New York: W.H. Freeman, 2000.
4. Petrucci, Ralph H., Herring, Goeffrey F., Madura, Jeffrey D., and Bissonnette, Carey. General Chemistry: Principles and Modern Applications. 10th ed. New Jersey: Pearson Canada, 2011.

Contributors and Attributions

• Ryan Benoit (UCD), Michael Thai (UCD), Charlie Wang (UCD), Jacob Gomez (UCD)

Contrary to the beliefs of generations of chemistry students, Avogadro’s number—the number of particles in a unit known as a mole—was not discovered by Amadeo Avogadro (1776-1856). Avogadro was a lawyer who became interested in mathematics and physics, and in 1820 he became the first professor of physics in Italy. Avogadro is most famous for his hypothesis that equal volumes of different gases at the same temperature and pressure contain the same number of particles.

The first person to estimate the actual number of particles in a given amount of a substance was Josef Loschmidt, an Austrian high school teacher who later became a professor at the University of Vienna. In 1865 Loschmidt used kinetic molecular theory to estimate the number of particles in one cubic centimeter of gas at standard conditions. This quantity is now known as the Loschmidt constant, and the accepted value of this constant is 2.6867773 x 10²⁵ m^-3.

The term “Avogadro’s number” was first used by French physicist Jean Baptiste Perrin. In 1909 Perrin reported an estimate of Avogadro’s number based on his work on Brownian motion—the random movement of microscopic particles suspended in a liquid or gas. In the years since then, a variety of techniques have been used to estimate the magnitude of this fundamental constant.

Accurate determinations of Avogadro’s number require the measurement of a single quantity on both the atomic and macroscopic scales using the same unit of measurement. This became possible for the first time when American physicist Robert Millikan measured the charge on an electron. The charge on a mole of electrons had been known for some time and is the constant called the Faraday. The best estimate of the value of a Faraday, according to the National Institute of Standards and Technology (NIST), is 96,485.3383 coulombs per mole of electrons. The best estimate of the charge on an electron based on modern experiments is 1.60217653 x 10^-19 coulombs per electron. If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value of Avogadro’s number of 6.02214154 x 10²³ particles per mole.

Another approach to determining Avogadro’s number starts with careful measurements of the density of an ultrapure sample of a material on the macroscopic scale. The density of this material on the atomic scale is then measured by using x-ray diffraction techniques to determine the number of atoms per unit cell in the crystal and the distance between the equivalent points that define the unit cell (see Physical Review Letters, 1974, 33, 464).