# Avogadro's Number Particles

Posted : admin On 15.08.2021- Avogadro's Full Number
- Avogadro Number Represents How Many Particles
- Avogadro's Number Symbol
- Formula Avogadro's Number

### Avogadro Number Calculations II

How Many Atoms or Molecules?

The value I will use for Avogadro's Number is 6.022 x 10^{23} mol¯^{1}.

How to find number of particles. Number of given moles x 6.0x10²³ particles/mol. The mass of one mole of any pure substance (compound or element) Mole is. The amount of a pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12. The number of units in one mole of any substance is called Avogadro's number or Avogadro's constant. Avogadro’s number is approximately 6.022140857(74)×1023 mol−1. It tells us the number of particles in 1 mole (or mol) of a substance. Chemists also need a unit for counting very small particles like molecules, formula units and atoms. This unit is called the mole. A mole is the SI base unit for measuring the amount of a substance. One mole is 6.02. 10^23 particles. This number is called Avogadro's number, after Amedeo Avogadro. Problem Solving Using Moles, Number of Particles, and Avogadro Number The Problem: Bo the Biologist has been studying the effect of chloride ions, Cl -, on plant cells. Bo has asked Chris the Chemist to make 1 litre of a solution containing 0.50 moles of chloride ions, Cl -, dissolved in water. Jul 03, 2019 Avogadro's Number Problem Key Takeaways. Avogadro's number is 6.02 x 10 23. It is the number of particles in a mole. You can use Avogadro's number to convert between mass and the number of molecules of any pure substance. If you are given the mass of a sample (such as a snowflake), convert the mass to moles, and then use Avogadro's number to.

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 H

_{2}O contains how many molecules? (Example #2)

0.450 gram of Fe contains how many atoms? (Example #3)

0.200 gram of H_{2}O 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 H_{2}O contains how many atoms?

0.200 gram of H_{2}O 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:**

0.450 mol x 6.022 x 10^{23} mol¯^{1} = see below for answer

**Example #2:** 0.200 mole of H_{2}O contains how many molecules?

**Solution:**

0.200 mol x 6.022 x 10^{23} 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 P_{4} and S_{8}.)

So, doing the calculation and rounding off to three sig figs, we get 2.71 x 10^{23} 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 10^{23} 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 10^{23} 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 10^{23}mol¯^{1}= 2.71 x 10^{23}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 H_{2}O 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 Two (moles ---> how many): (0.0080573 mol) (6.022 x 10^{23} atoms/mol) = 4.85 x 10^{21} atoms

**Solution to Example #4:**

Step Two: (0.01110186 mol) (6.022 x 10^{23} molecules/mol) = 6.68 x 10^{21} molecules

**Example #5:** Calculate the number of molecules in 1.058 mole of H_{2}O

**Solution:**

(1.058 mol) (6.022 x 10^{23}mol¯^{1}) = 6.371 x 10^{23}molecules

**Example #6:** Calculate the number of atoms in 0.750 mole of Fe

**Solution:**

(0.750 mol) (6.022 x 10^{23}mol¯^{1}) = 4.52 x 10^{23}atoms (to three sig figs)

**Example #7:** Calculate the number of molecules in 1.058 gram of H_{2}O

**Solution:**

^{23}molecules/mole)

Here is the solution set up in dimensional analysis style:

1 mol | 6.022 x 10^{23} | |||

1.058 g x | ––––––––– | x | –––––––––– | = 3.537 x 10^{22} 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

^{23}atoms/mole

1 mol | 6.022 x 10^{23} | |||

0.750 g x | ––––––––– | x | –––––––––– | = 8.09 x 10^{21} atoms (to three sig figs) |

55.85 g | 1 mol |

**Example #9:** Which contains more molecules: 10.0 grams of O_{2} or 50.0 grams of iodine, I_{2}?

**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 10^{23} | |||

10.0 g x | ––––––––– | x | –––––––––– | = number of O_{2} molecules |

31.998 g | 1 mol |

1 mol | 6.022 x 10^{23} | |||

50.0 g x | ––––––––– | x | –––––––––– | = number of I_{2} molecules |

253.8 g | 1 mol |

**Example #10:** 18.0 g of H_{2}O is present. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present?

**Solution:**

### Avogadro's Full Number

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 10^{23}mol¯^{1}) = 6.02 x 10^{23}molecules

3) Determine number of atoms of oxygen present:

(6.02 x 10^{23}molecules) (1 O atom / 1 H_{2}O molecule) = 6.02 x 10^{23}O atoms

4) Determine number of atoms of hydrogen present:

(6.02 x 10^{23}molecules) (2 H atoms / 1 H_{2}O molecule) = 1.20 x 10^{24}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 H_{2}O, 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 10^{23}molecules) (3 atoms/molecule) = 1.81 x 10^{24}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 C_{6}H_{12}O_{6}

(b) 2 mol of (NH_{4})_{2}CO_{3}

(c) 4 mol of H_{2}O

(d) 5 mol of CH_{3}COOH

**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:

### Avogadro Number Represents How Many Particles

(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 N_{2}O_{5} 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 N_{2}O_{5}:

(1.21353 mol N_{2}O_{5}) (5 mol O / 1 mol N_{2}O_{5}) = 6.06765 mol O

3) Use Avogadro's Number:

(6.06765 mol O) (6.022 x 10^{23}atoms O / mole O) = 3.65 x 10^{24}atoms O (to three sig figs)

**Example #13:** How many carbon atoms are in 0.850 mol of acetaminophen, C_{8}H_{9}NO_{2}?

**Solution:**

1) There are 8 moles of C in every mole of acetaminophen:

(0.850 mol C_{8}H_{9}NO_{2}) (8 mol C / mol C_{8}H_{9}NO_{2}) = 6.80 mol C

2) Use Avogadro's Number:

(6.80 mol C) (6.022 x 10^{23}atoms C / mole C) = 4.09 x 10^{24}atoms C (to three sig figs)

**Example #14:** How many atoms are in a 0.460 g sample of elemental phosphorus?

**Solution:**

_{4}. (Not P!!)

0.460 g / 123.896 g/mol = 0.00371279 mol

(6.022 x 10^{23} molecules/mol) (0.00371279 mol) = 2.23584 x 10^{21} molecules of P_{4}

(2.23584 x 10^{21} molecules) (4 atoms/molecule) = 8.94 x 10^{21} atoms (to three sig figs)

Set up using dimensional analysis style:

1 mol | 6.022 x 10^{23} molecules | 4 atoms | ||||

0.460 g x | –––––––– | x | –––––––––––––––––– | x | ––––––––– | = 8.94 x 10^{21} atoms |

123.896 g | 1 mol | 1 molecule |

**Example #15:** Which contains the most atoms?

(a) 3.5 molecules of H_{2}O

(b) 3.5 x 10^{22}molecules of N_{2}

(c) 3.5 moles of CO

(d) 3.5 g of water

**Solution:**

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 10^{22} atoms. We continue to analyze the answer choices.

Choice (c): Use Avogadro's number (3.5 x 10^{23} 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 N_{2} 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 C_{3}H_{8} has 2.96 x 10^{24} H atoms.

(a) How many carbon atoms does the sample contain?

(b) What is the total mass of the sample?

**Solution to (a):**

1) The ratio between C and H is 3 to 8, so this:

3 | y | |

––––––– | = | –––––––––––––––– |

8 | 2.96 x 10^{24} H atoms |

2) will tell us the number of carbon atoms present:

y = 1.11 x 10^{24}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 10^{24}

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 C_{3}H_{8} present.

2.96 x 10^{24}/ 8 = 3.70 x 10^{23}molecules of C_{3}H_{8}

2) Divide by Avogadro's Number:

3.70 x 10^{23}/ 6.022 x 10^{23}mol¯^{1}= 0.614414 mol <--- I'll keep some guard digits

3) Use the molar mass of C_{3}H_{8}:

0.614414 mol times 44.0962 g/mol = 27.1 g (to three sig figs)

In chemistry the mole is a fundamental unit in the Système International d'Unités, the SI system, and it is used to measure the amount of substance. This quantity is sometimes referred to as the * chemical amount. * In Latin * mole * means a 'massive heap' of material. It is convenient to think of a chemical mole as such.

Visualizing a mole as a pile of particles, however, is just one way to understand this concept. A sample of a substance has a mass, volume (generally used with gases), and number of particles that is proportional to the chemical amount (measured in moles) of the sample. For example, one mole of oxygen gas (O _{ 2 } ) occupies a volume of 22.4 L at standard temperature and pressure (STP; 0°C and 1 atm), has a mass of 31.998 grams, and contains about 6.022 × 10 ^{ 23 } molecules of oxygen. Measuring one of these quantities allows the calculation of the others and this is frequently done in stoichiometry.

The * mole * is to the * amount of substance * (or chemical amount) as the * gram * is to * mass. * Like other units of the SI system, prefixes can be used with the mole, so it is permissible to refer to 0.001 mol as 1 mmol just as 0.001 g is equivalent to 1 mg.

## Formal Definition

According to the National Institute of Standards and Technology (NIST), the Fourteenth Conférence Générale des Poids et Mesures established the definition of the mole in 1971.

The mole is the amount of a substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12; its symbol is 'mol.' When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

### Avogadro's Number Symbol

## One Interpretation: A Specific Number of Particles

### Formula Avogadro's Number

When a quantity of particles is to be described, mole is a grouping unit analogous to groupings such as pair, dozen, or gross, in that all of these words represent specific numbers of objects. The main differences between the mole and the other grouping units are the magnitude of the number represented and how that number is obtained. One mole is an amount of substance containing Avogadro's number of particles. Avogadro's number is equal to 602,214,199,000,000,000,000,000 or more simply, 6.02214199 × 10 ^{ 23 } .

Unlike pair, dozen, and gross, the exact number of particles in a mole cannot be counted. There are several reasons for this. First, the particles are too small and cannot be seen even with a microscope. Second, as naturally occurring carbon contains approximately 98.90% carbon-12, the sample would need to be purified to remove every atom of carbon-13 and carbon-14. Third, as the number of particles in a mole is tied to the mass of exactly 12 grams of carbon-12, a balance would need to be constructed that could determine if the sample was one atom over or under exactly 12 grams. If the first two requirements were met, it would take one million machines counting one million atoms each second more than 19,000 years to complete the task.

Obviously, if the number of particles in a mole cannot be counted, the value must be measured indirectly and with every measurement there is some degree of uncertainty. Therefore, the number of particles in a mole, Avogadro's constant ( * N *_{ A } ), can only be approximated through experimentation, and thus its reported values will vary slightly (at the tenth decimal place) based on the measurement method used. Most methods agree to four significant figures, so * N *_{ A } is generally said to equal 6.022 × 10 ^{ 23 } particles per mole, and this value is usually sufficient for solving textbook problems. Another key point is that the formal definition of a mole does not include a value for Avogadro's constant and this is probably due to the inherent uncertainty in its measurement. As for the difference between Avogadro's constant and Avogadro's number, they are numerically equivalent, but the former has the unit of mol ^{ −1 } whereas the latter is a pure number with no unit.

## A Second Interpretation: A Specific Mass

Atoms and molecules are incredibly small and even a tiny chemical sample contains an unimaginable number of them. Therefore, counting the number of atoms or molecules in a sample is impossible. The multiple interpretations of the mole allow us to bridge the gap between the submicroscopic world of atoms and molecules and the macroscopic world that we can observe.

To determine the chemical amount of a sample, we use the substance's * molar mass, * the mass per mole of particles. We will use carbon-12 as an example because it is the standard for the formal definition of the mole. According to the definition, one mole of carbon-12 has a mass of exactly 12 grams. Consequently, the molar mass of carbon-12 is 12 g/mol. However, the molar mass for the element carbon is 12.011 g/mol. Why are they different? To answer that question, a few terms need to be clarified.

On the Periodic Table, you will notice that most of the atomic weights listed are not round numbers. The atomic weight is a weighted average of the atomic masses of an element's natural isotopes. For example, bromine has two natural isotopes with atomic masses of 79 u and 81 u. The unit * u * represents the atomic mass unit and is used in place of grams because the value would be inconveniently small. These two isotopes of bromine are present in nature in almost equal amounts, so the atomic weight of the element bromine is 79.904. (i.e., nearly 80, the arithmetic mean of 79 and 81). A similar situation exists for chlorine, but chlorine-35 is almost three times as abundant as chlorine-37, so the atomic weight of chlorine is 35.4527. Technically, atomic weights are ratios of the average atomic mass to the unit * u * and that is why they do not have units. Sometimes atomic weights are given the unit * u * , but this is not quite correct according to the International Union of Pure and Applied Chemistry (IUPAC).

To find the molar mass of an element or compound, determine the atomic, molecular, or formula weight and express that value as g/mol. For bromine and chlorine, the molar masses are 79.904 g/mol and 35.4527 g/mol, respectively. Sodium chloride (NaCl) has a formula weight of 58.443 (atomic weight of Na + atomic weight of Cl) and a molar mass of 58.443 g/mol. Formaldehyde (CH _{ 2 } O) has a molecular weight of 30.03 (atomic weight of C + 2 [atomic weight of H]) + atomic weight of O] and a molar mass of 30.03 g/mol.

The concept of molar mass enables chemists to measure the number of submicroscopic particles in a sample without counting them directly simply by determining the chemical amount of a sample. To find the chemical amount of a sample, chemists measure its mass and divide by its molar mass. Multiplying the chemical amount (in moles) by Avogadro's constant ( * N *_{ A } ) yields the number of particles present in the sample.

Occasionally, one encounters gram-atomic mass (GAM), gram-formula mass (GFM), and gram-molecular mass (GMM). These terms are functionally the same as molar mass. For example, the GAM of an element is the mass in grams of a sample containing * N *_{ A } atoms and is equal to the element's atomic weight expressed in grams. GFM and GMM are defined similarly. Other terms you may encounter are formula mass and molecular mass. Interpret these as formula weight and molecular weight, respectively, but with the units of * u. *

## Avogadro's Hypothesis

Some people think that Amedeo Avogadro (1776–1856) determined the number of particles in a mole and that is why the quantity is known as Avogadro's number. In reality Avogadro built a theoretical foundation for determining accurate atomic and molecular masses. The concept of a mole did not even exist in Avogadro's time.

Much of Avogadro's work was based on that of Joseph-Louis Gay-Lussac (1778–1850). Gay-Lussac developed the law of combining volumes that states: 'In any chemical reaction involving gaseous substances the volumes of the various gases reacting or produced are in the ratios of small whole numbers.' (Masterton and Slowinski, 1977, p. 105) Avogadro reinterpreted Gay-Lussac's findings and proposed in 1811 that (1) some molecules were diatomic and (2) 'equal volumes of all gases at the same temperature and pressure contain the same number of molecules' (p. 40). The second proposal is what we refer to as Avogadro's hypothesis.

The hypothesis provided a simple method of determining relative molecular weights because equal volumes of two different gases at the same temperature and pressure contained the same number of particles, so the ratio of the masses of the gas samples must also be that of their particle masses. Unfortunately, Avogadro's hypothesis was largely ignored until Stanislao Cannizzaro (1826–1910) advocated using it to calculate relative atomic masses or atomic weights. Soon after the 1 ^{ st } International Chemical Congress at Karlsrule in 1860, Cannizzaro's proposal was accepted and a scale of atomic weights was established.

To understand how Avogadro's hypothesis can be used to determine relative atomic and molecular masses, visualize two identical boxes with oranges in one and grapes in the other. The exact number of fruit in each box is not known, but you believe that there are equal numbers of fruit in each box (Avogadro's hypothesis). After subtracting the masses of the boxes, you have the masses of each fruit sample and can determine the mass ratio between the oranges and the grapes. By assuming that there are equal numbers of fruit in each box, you then know the average mass ratio between a grape and an orange, so in effect you have calculated their relative masses (atomic masses). If you chose either the grape or the orange as a standard, you could eventually determine a scale of relative masses for all fruit.

## A Third Interpretation: A Specific Volume

By extending Avogadro's hypothesis, there is a specific volume of gas that contains * N *_{ A } gas particles for a given temperature and pressure and that volume should be the same for all gases. For an ideal gas, the volume of one mole at STP (0°C and 1.000 atm) is 22.41 L, and several real gases (hydrogen, oxygen, and nitrogen) come very close to this value.

## The Size of Avogadro's Number

To provide some idea of the enormity of Avogadro's number, consider some examples. Avogadro's number of water drops (twenty drops per mL) would fill a rectangular column of water 9.2 km (5.7 miles) by 9.2 km (5.7 miles) at the base and reaching to the moon at perigee (closest distance to Earth). Avogadro's number of water drops would cover the all of the land in the United States to a depth of roughly 3.3 km (about 2 miles). Avogadro's number of pennies placed in a rectangular stack roughly 6 meters by 6 meters at the base would stretch for about 9.4 × 10 ^{ 12 } km and extend outside our solar system. It would take light nearly a year to travel from one end of the stack to the other.

## History

Long before the mole concept was developed, there existed the idea of chemical equivalency in that specific amounts of various substances could react in a similar manner and to the same extent with another substance. Note that the historical equivalent is not the same as its modern counterpart, which involves electric charge. Also, the historical equivalent is not the same as a mole, but the two concepts are related in that they both indicate that different masses of two substances can react with the same amount of another substance.

The idea of chemical equivalents was stated by Henry Cavendish in 1767, clarified by Jeremias Richter in 1795, and popularized by William Wollaston in 1814. Wollaston applied the concept to elements and defined it in such a way that one equivalent of an element corresponded to its atomic mass. Thus, when Wollaston's equivalent is expressed in grams, it is identical to a mole. It is not surprising then that the word 'mole' is derived from 'molekulargewicht' (German, meaning 'molecular weight') and was coined in 1901 or 1902.

** SEE ALSO ** Avogadro, Amedeo ; Cannizzaro, Stanislao ; Cavendish, Henry ; Gay-Lussac, Joseph-Louis .

* Nathan J. Barrows *

## Bibliography

Atkins, Peter, and Jones, Loretta (2002). * Chemical Principles * , 2nd edition. New York: W. H. Freeman and Company.

Lide, David R., ed. (2000). * The CRC Handbook of Chemistry & Physics * , 81st edition. New York: CRC Press.

Masterton, William L., and Slowinski, Emil J. (1977). * Chemical Principles * , 4th edition. Philadelphia: W. B. Saunders Company.

### Internet Resources

National Institute of Standards and Technology. 'Unit of Amount of Substance (Mole).' Available from http://www.nist.gov .