- Avogadro's Number Of Particles
- Avogadro's Number Atoms To Moles
- Avogadro's Number Particles Are Made
- Avogadro's Number Particles
- Avogadro's Number Is Equal To One __ Of Particles
Converting Between Moles and Particles. To convert between moles and the number of particles we will use the factor label system from the first unit. Remember that a mole of anything is defined as 6.022x1023 of that item. In other words a mole of cats is 6.022x1023 cats and 6.022x1023 water molecules is a mole of water. Let's try a few problems. Okay so continuing the history of the ideal gas equation here we get to the 19th century with an Italian chemist named Amedeo Avogadro and actually his name was Lorenzo Romano Amedeo Carlo Avogadro de quoi regne add that sheddeth though but we're going to call him Amedeo and Amedeo spent a bit of his time experimenting with tiny particles and in honor of his experiments the number of particles.
Avogadro Number Calculations II
How Many Atoms or Molecules?
The value I will use for Avogadro's Number is 6.022 x 1023 mol¯1.
Types of problems you might be asked look something like these:
0.450 mole of Fe contains how many atoms? (Example #1)0.200 mole of H2O contains how many molecules? (Example #2)
0.450 gram of Fe contains how many atoms? (Example #3)
0.200 gram of H2O contains how many molecules? (Example #4)
0.200 gram of H2O contains how many molecules? (Example #4)
When the word gram replaces mole, you have a related set of problems which requires one more step.
And, two more:
0.200 mole of H2O contains how many atoms?
0.200 gram of H2O contains how many atoms?
When the word gram replaces mole, you have a related set of problems which requires one more step. In addition, the two just above will have even another step, one to determine the number of atoms once you know the number of molecules.
Here is a graphic of the procedure steps:
Pick the box of the data you are given in the problem and follow the steps toward the box containing what you are asked for in the problem.
Example #1: 0.450 mole of Fe contains how many atoms?
Solution:
Start from the box labeled 'Moles of Substance' and move (to the right) to the box labeled 'Number of Atoms or Molecules.' What do you have to do to get there? That's right - multiply by Avogadro's Number.0.450 mol x 6.022 x 1023 mol¯1 = see below for answer
Example #2: 0.200 mole of H2O contains how many molecules?
Solution:
Start at the same box as Example #1.0.200 mol x 6.022 x 1023 mol¯1 = see below for answer
The answers (including units) to Examples #1 and #2
The unit on Avogadro's Number might look a bit weird. It is mol¯1 and you would say 'per mole' out loud. The question then is WHAT per mole?
The answer is that it depends on the problem. In the first example, I used iron, an element. Almost all elements come in the form of individual atoms, so the correct numerator with most elements is 'atoms.' (The exceptions would be the diatomic elements plus P4 and S8.)
So, doing the calculation and rounding off to three sig figs, we get 2.71 x 1023 atoms. Notice 'atoms' never gets written until the end. It is assumed to be there in the case of elements. If you wrote Avogadro's Number with the unit atoms/mol in the problem, you would be correct.
The same type of discussion applies to substances which are molecular in nature, such as water. So the numerator I would use in example #2 is 'molecule' and the answer is 1.20 x 1023 molecules.
Once again, the numerator part of Avogadro's Number depends on what is in the problem. Other possible numerators include 'formula units,' ions, or electrons. These, of course, are all specific to a given problem. When a general word is used, the most common one is 'entities,' as in 6.022 x 1023 entities/mol.
Keep this in mind: the 'atoms' or 'molecules' part of the unit is often omitted and simply understood to be present. However, it will often show up in the answer. Like this:
0.450 mol x 6.022 x 1023 mol¯1 = 2.71 x 1023 atoms
It's not that a mistake was made, it's that the 'atoms' part of atoms per mole was simply assumed to be there.
Example #3: 0.450 gram of Fe contains how many atoms?
Example #4: 0.200 gram of H2O contains how many molecules?
Look at the solution steps in the image above and you'll see we have to go from grams (on the left of the image above) across to the right through moles and then to how many atoms or molecules.
Solution to Example #3:
Step One (grams ---> moles): 0.450 g / 55.85 g/mol = 0.0080573 molStep Two (moles ---> how many): (0.0080573 mol) (6.022 x 1023 atoms/mol) = 4.85 x 1021 atoms
Solution to Example #4:
Step One: 0.200 g / 18.015 g/mol = 0.01110186 molStep Two: (0.01110186 mol) (6.022 x 1023 molecules/mol) = 6.68 x 1021 molecules
Example #5: Calculate the number of molecules in 1.058 mole of H2O
Solution:
(1.058 mol) (6.022 x 1023 mol¯1) = 6.371 x 1023 molecules
Example #6: Calculate the number of atoms in 0.750 mole of Fe
Solution:
(0.750 mol) (6.022 x 1023 mol¯1) = 4.52 x 1023 atoms (to three sig figs)
Example #7: Calculate the number of molecules in 1.058 gram of H2O
Solution:
(1.058 g / 18.015 g/mol) (6.022 x 1023 molecules/mole)Here is the solution set up in dimensional analysis style:
1 mol | 6.022 x 1023 | |||
1.058 g x | ––––––––– | x | –––––––––– | = 3.537 x 1022 molecules (to four sig figs) |
18.015 g | 1 mol | |||
↑ grams to moles ↑ | ↑ moles to ↑ molecules |
Example #8: Calculate the number of atoms in 0.750 gram of Fe
(0.750 gram divided by 55.85 g/mole) x 6.022 x 1023atoms/mole1 mol | 6.022 x 1023 | |||
0.750 g x | ––––––––– | x | –––––––––– | = 8.09 x 1021 atoms (to three sig figs) |
55.85 g | 1 mol |
Example #9: Which contains more molecules: 10.0 grams of O2 or 50.0 grams of iodine, I2?
Solution:
Basically, this is just two two-step problems in one sentence. Convert each gram value to its mole equivalent. Then, multiply the mole value by Avogadro's Number. Finally, compare these last two values and pick the larger value. That is the one with more molecules.
1 mol | 6.022 x 1023 | |||
10.0 g x | ––––––––– | x | –––––––––– | = number of O2 molecules |
31.998 g | 1 mol |
1 mol | 6.022 x 1023 | |||
50.0 g x | ––––––––– | x | –––––––––– | = number of I2 molecules |
253.8 g | 1 mol |
Example #10: 18.0 g of H2O is present. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present?
Solution:
1) Convert grams to moles:
18.0 g / 18.0 g/mol = 1.00 mol
2) Convert moles to molecules:
(1.00 mol) (6.02 x 1023 mol¯1) = 6.02 x 1023 molecules
3) Determine number of atoms of oxygen present:
(6.02 x 1023 molecules) (1 O atom / 1 H2O molecule) = 6.02 x 1023 O atoms
4) Determine number of atoms of hydrogen present:
(6.02 x 1023 molecules) (2 H atoms / 1 H2O molecule) = 1.20 x 1024 H atoms (to three sig figs)
Notice that there is an additional step (as seen in step 3 for O and step 4 for H). You multiply the number of molecules times how many of that atom are present in the molecule. In one molecule of H2O, there are 2 atoms of H and 1 atom of O.
Sometimes, you will be asked for the total atoms present in the sample. Do it this way:
(6.02 x 1023 molecules) (3 atoms/molecule) = 1.81 x 1024 atoms (to three sig figs)
The 3 represents the total atoms in one molecule of water: one O atom and two H atoms.
Example #11: Which of the following contains the greatest number of hydrogen atoms?
(a) 1 mol of C6H12O6
(b) 2 mol of (NH4)2CO3
(c) 4 mol of H2O
(d) 5 mol of CH3COOH
Solution:
1) Each mole of molecules contains N number of molecules, where N equals Avogadro's Number. How many molecules are in each answer:
(a) 1 x N = N
(b) 2 x N = 2N
(c) 4 x N = 4N
(d) N x 5 = 5N
2) Each N times the number of hydrogen atoms in a formula equals the total number of hydrogen atoms in the sample:
(a) N x 12 = 12N(b) 2N x 8 = 16N
(c) 4N x 2 = 8N
(d) 5N x 4 = 20N
(d) is the answer.
Example #12: How many oxygen atoms are in 27.2 L of N2O5 at STP?
Solution:
1) Given STP, we can use molar volume:
27.2 L / 22.414 L/mol = 1.21353 mol
2) There are five moles of O atoms in one mole of N2O5:
(1.21353 mol N2O5) (5 mol O / 1 mol N2O5) = 6.06765 mol O
3) Use Avogadro's Number:
(6.06765 mol O) (6.022 x 1023 atoms O / mole O) = 3.65 x 1024 atoms O (to three sig figs)
Example #13: How many carbon atoms are in 0.850 mol of acetaminophen, C8H9NO2?
Solution:
1) There are 8 moles of C in every mole of acetaminophen:
(0.850 mol C8H9NO2) (8 mol C / mol C8H9NO2) = 6.80 mol C
2) Use Avogadro's Number:
(6.80 mol C) (6.022 x 1023 atoms C / mole C) = 4.09 x 1024 atoms C (to three sig figs)
Example #14: How many atoms are in a 0.460 g sample of elemental phosphorus?
Solution:
Phosphorus has the formula P4. (Not P!!)0.460 g / 123.896 g/mol = 0.00371279 mol
(6.022 x 1023 molecules/mol) (0.00371279 mol) = 2.23584 x 1021 molecules of P4
(2.23584 x 1021 molecules) (4 atoms/molecule) = 8.94 x 1021 atoms (to three sig figs)
Set up using dimensional analysis style:
1 mol | 6.022 x 1023 molecules | 4 atoms | ||||
0.460 g x | –––––––– | x | –––––––––––––––––– | x | ––––––––– | = 8.94 x 1021 atoms |
123.896 g | 1 mol | 1 molecule |
Example #15: Which contains the most atoms?
(a) 3.5 molecules of H2O
(b) 3.5 x 1022 molecules of N2
(c) 3.5 moles of CO
(d) 3.5 g of water
Solution:
The correct answer is (c). Now, some discussion about each answer choice.Choice (a): You can't have half of a molecule, so this answer should not be considered. Also, compare it to (b). Since (a) is much less than (b), (a) cannot ever be the answer to the most number of atoms.
Choice (b): this is a viable contender for the correct answer. Since there are two atoms per molecule, we have 7.0 x 1022 atoms. We continue to analyze the answer choices.
Choice (c): Use Avogadro's number (3.5 x 1023 mol¯1) and compare it to choice (b). You should be able to see, even without the 3.5 moles, choice (c) is already larger than choice (b). Especially when you consider that N2 and CO both have 2 atoms per molecule.
Choice (d): 3.5 g of water is significantly less that the 3.5 moles of choice (c). 3.5 / 18.0 equals a bit less that 0.2 moles of water.
Bonus Example: A sample of C3H8 has 2.96 x 1024 H atoms.
(a) How many carbon atoms does the sample contain?
(b) What is the total mass of the sample?
Solution to (a):
Avogadro's Number Of Particles
1) The ratio between C and H is 3 to 8, so this:
3 | y | |
––––––– | = | –––––––––––––––– |
8 | 2.96 x 1024 H atoms |
2) will tell us the number of carbon atoms present:
y = 1.11 x 1024 carbon atoms
3) By the way, the above ratio and proportion can also be written like this:
3 is to 8 as y is to 2.96 x 1024Avogadro's Number Atoms To Moles
Be sure you understand that the two different ways to present the ratio and proportion mean the same thing.
Solution to (b) using hydrogen:
1) Determine the moles of C3H8 present.
2.96 x 1024 / 8 = 3.70 x 1023 molecules of C3H8
2) Divide by Avogadro's Number:
3.70 x 1023 / 6.022 x 1023 mol¯1 = 0.614414 mol <--- I'll keep some guard digits
3) Use the molar mass of C3H8:
0.614414 mol times 44.0962 g/mol = 27.1 g (to three sig figs)
(Redirected from Avogadro's Law)
Avogadro's law (sometimes referred to as Avogadro's hypothesis or Avogadro's principle) or Avogadro-Ampère's hypothesis is an experimental gas law relating the volume of a gas to the amount of substance of gas present.[1] The law is a specific case of the ideal gas law. A modern statement is:
Avogadro's law states that 'equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.'[1]
For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.
The law is named after Amedeo Avogadro who, in 1812,[2][3] hypothesized that two given samples of an ideal gas, of the same volume and at the same temperature and pressure, contain the same number of molecules. As an example, equal volumes of gaseous hydrogen and nitrogen contain the same number of atoms when they are at the same temperature and pressure, and observe ideal gas behavior. In practice, real gases show small deviations from the ideal behavior and the law holds only approximately, but is still a useful approximation for scientists.
Mathematical definition[edit]
The law can be written as:
or
where
- V is the volume of the gas;
- n is the amount of substance of the gas (measured in moles);
- k is a constant for a given temperature and pressure.
This law describes how, under the same condition of temperature and pressure, equal volumes of all gases contain the same number of molecules. For comparing the same substance under two different sets of conditions, the law can be usefully expressed as follows:
The equation shows that, as the number of moles of gas increases, the volume of the gas also increases in proportion. Similarly, if the number of moles of gas is decreased, then the volume also decreases. Thus, the number of molecules or atoms in a specific volume of ideal gas is independent of their size or the molar mass of the gas.
Relationships between Boyle's, Charles's, Gay-Lussac's, Avogadro's, combined and ideal gas laws, with the Boltzmann constantkB = R/NA = n R/N (in each law, properties circled are variable and properties not circled are held constant)
Derivation from the ideal gas law[edit]
The derivation of Avogadro's law follows directly from the ideal gas law, i.e.
- ,
where R is the gas constant, T is the Kelvin temperature, and P is the pressure (in pascals).
Solving for V/n, we thus obtain
- .
Compare that to
which is a constant for a fixed pressure and a fixed temperature.
An equivalent formulation of the ideal gas law can be written using Boltzmann constantkB, as
- ,
where N is the number of particles in the gas, and the ratio of R over kB is equal to the Avogadro constant.
In this form, for V/N is a constant, we have
- .
If T and P are taken at standard conditions for temperature and pressure (STP), then k′ = 1/n0, where n0 is the Loschmidt constant.
Historical account and influence[edit]
Avogadro's hypothesis (as it was known originally) was formulated in the same spirit of earlier empirical gas laws like Boyle's law (1662), Charles's law (1787) and Gay-Lussac's law (1808). The hypothesis was first published by Amadeo Avogadro in 1811,[4] and it reconciled Dalton atomic theory with the 'incompatible' idea of Joseph Louis Gay-Lussac that some gases were composite of different fundamental substances (molecules) in integer proportions.[5] In 1814, independently from Avogadro, André-Marie Ampère published the same law with similar conclusions.[6] As Ampère was more well known in France, the hypothesis was usually referred there as Ampère's hypothesis,[note 1] and later also as Avogadro–Ampère hypothesis[note 2] or even Ampère–Avogadro hypothesis.[7]
Experimental studies carried out by Charles Frédéric Gerhardt and Auguste Laurent on organic chemistry demonstrated that Avogadro's law explained why the same quantities of molecules in a gas have the same volume. Nevertheless, related experiments with some inorganic substances showed seeming exceptions to the law. This apparent contradiction was finally resolved by Stanislao Cannizzaro, as announced at Karlsruhe Congress in 1860, four years after Avogadro's death. He explained that these exceptions were due to molecular dissociations at certain temperatures, and that Avogadro's law determined not only molecular masses, but atomic masses as well.
Ideal gas law[edit]
Avogadro's Number Particles Are Made
Boyle, Charles and Gay-Lussac laws, together with Avogadro's law, were combined by Émile Clapeyron in 1834,[8] giving rise to the ideal gas law. At the end of the 19th century, later developments from scientists like August Krönig, Rudolf Clausius, James Clerk Maxwell and Ludwig Boltzmann, gave rise to the kinetic theory of gases, a microscopic theory from which the ideal gas law can be derived as an statistical result from the movement of atoms/molecules in a gas.
Avogadro constant[edit]
Avogadro's law provides a way to calculate the quantity of gas in a receptacle. Thanks to this discovery, Johann Josef Loschmidt, in 1865, was able for the first time to estimate the size of a molecule.[9] His calculation gave rise to the concept of the Loschmidt constant, a ratio between macroscopic and atomic quantities. In 1910, Millikan'soil drop experiment determined the charge of the electron; using it with the Faraday constant (derived by Michael Faraday in 1834), one is able to determine the number of particles in a mole of substance. At the same time, precision experiments by Jean Baptiste Perrin led to the definition of Avogadro's number as the number of molecules in one gram-molecule of oxygen. Perrin named the number to honor Avogadro for his discovery of the namesake law. Later standardization of the International System of Units led to the modern definition of the Avogadro constant.
Molar volume[edit]
Taking STP to be 101.325 kPa and 273.15 K, we can find the volume of one mole of gas:
Avogadro's Number Particles
For 101.325 kPa and 273.15 K, the molar volume of an ideal gas is 22.4127 dm3⋅mol−1.
See also[edit]
- Boyle's law – Relationship between pressure and volume in a gas at constant temperature
- Charles's law – Relationship between volume and temperature of a gas at constant pressure
- Gay-Lussac's law – Relationship between pressure and temperature of a gas at constant volume.
- Ideal gas – Mathematical model which approximates the behavior of real gases
Notes[edit]
Avogadro's Number Is Equal To One __ Of Particles
- ^First used by Jean-Baptiste Dumas in 1826.
- ^First used by Stanislao Cannizzaro in 1858.
References[edit]
- ^ abEditors of the Encyclopædia Britannica. 'Avogadro's law'. Encyclopædia Britannica. Retrieved 3 February 2016.CS1 maint: extra text: authors list (link)
- ^Avogadro, Amedeo (1810). 'Essai d'une manière de déterminer les masses relatives des molécules élémentaires des corps, et les proportions selon lesquelles elles entrent dans ces combinaisons'. Journal de Physique. 73: 58–76.English translation
- ^'Avogadro's law'. Merriam-Webster Medical Dictionary. Retrieved 3 February 2016.
- ^Avogadro, Amadeo (July 1811). 'Essai d'une maniere de determiner les masses relatives des molecules elementaires des corps, et les proportions selon lesquelles elles entrent dans ces combinaisons'. Journal de Physique, de Chimie, et d'Histoire Naturelle (in French). 73: 58–76.
- ^Rovnyak, David. 'Avogadro's Hypothesis'. Science World Wolfram. Retrieved 3 February 2016.
- ^Ampère, André-Marie (1814). 'Lettre de M. Ampère à M. le comte Berthollet sur la détermination des proportions dans lesquelles les corps se combinent d'après le nombre et la disposition respective des molécules dont les parties intégrantes sont composées'. Annales de Chimie (in French). 90 (1): 43–86.
- ^Scheidecker-Chevallier, Myriam (1997). 'L'hypothèse d'Avogadro (1811) et d'Ampère (1814): la distinction atome/molécule et la théorie de la combinaison chimique'. Revue d'Histoire des Sciences (in French). 50 (1/2): 159–194. doi:10.3406/rhs.1997.1277. JSTOR23633274.
- ^Clapeyron, Émile (1834). 'Mémoire sur la puissance motrice de la chaleur'. Journal de l'École Polytechnique (in French). XIV: 153–190.
- ^Loschmidt, J. (1865). 'Zur Grösse der Luftmoleküle'. Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften Wien. 52 (2): 395–413.English translation.
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