Avogadro Number Calculations II
How Many Atoms or Molecules?
- Avogadro's Number Practice And Answers
- Mole And Avogadro's Number Explained
- The Mole And Avogadro Number Khan Academy
- Mole And Avogadro's Number Calculator
- Relation Between A Mole And Avogadro's Number
- Mole And Avogadro's Number Practice
Avogadro proposed his hypothesis in 1811. At that time there was no data at all on the number of particles in a mole, or an agreement on any atomic weights or the standard. The first measurements which could give an approximate value for Avogadro's number were observations of brownian motion by Robert Brown in 1827. 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 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)
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/mole| 1 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
Avogadro's Number Practice And Answers
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):
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 1024Be 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:
Mole And Avogadro's Number Explained
0.614414 mol times 44.0962 g/mol = 27.1 g (to three sig figs)
In 1811 Amedeo Avogadro proposed the volume of a gas is dependent on the number of atoms of gas at a fixed temperature and pressure. If more gas is added at this fixed temperature and pressure, the volume will increase due to more atoms being added. He proposed this was the same for all gases.
Avogadro’s proposal lay dormant for approximately 50 years when other scientists began to see the value of his work. Namely the relation between the mass (grams) of an element in a container and the number of atoms of the element in that mass.
The number of atoms in a volume of gas was first determined by Johann Josef Loschmidt in 1865. Other attempts to determine the number followed.
Atomic weights were known at this time and it was decided to determine the number of atoms in the mass of an element equal to it’s atomic weight. For example, how many helium (He) atoms would be in 4.0026 g grams of He? 4.0026 is the gram atomic weight of He. The experimentally determined number has changed over the years as measurement methods improved. The currently accepted value is 6.022 X 1023 atoms. This is a very large number . Written out it would be 602200000000000000000000. The exponent representation (6.022 X 1023) is more convenient to use. (See Exponents and Exponents for help with exponents)
Other examples. There are 6.022 X 1023 atoms of C in 12.0107 grams of C. 12.0107 is the gram atomic weight of C. There are 6.022 X 1023 atoms of Al in 26.9812 grams of Al. 26.9816 is the gram atomic weight of Al.
The number 6.022 X 1023 was given the name Avogadro’s Number by Jean Perrin in 1901.
In 1902 Ostwald proposed the term “mole” as another way to express Avogadro’s Number. So a mole of carbon ( C ) would have a mass of 12.0107 g and contain 6.022 X 1023 atoms of carbon. This definition adds units to atomic weight values and a name change to gram atomic weights.
The relative atomic weight of carbon 12.0107 would become the gram atomic weight of carbon 12.0107 g/mole.
For any element, the mass of that element equal to its gram atomic weight would be 1 mole of that element and that mass would have 6.022 X 1023 atoms of that element.
The gram atomic weight of an element is a conversion factor between any mass of an element weighted on a balance and the number of moles of that element. The number of moles of an element is a representation of the number of atoms of that element.
The unit mole is comparable to unit dozen. It represents a number. A dozen represents 12 of something. A mole represents 6.022 X 1023 of something.
A dozen eggs would be 12 eggs, likewise a dozen doughnuts would be 12 doughnuts. A mole (Fe) of iron would be 6.022 X 1023 atoms of iron, likewise a mole of electrons would be 6.022 X 1023 electrons.
Just as one could have 0.5 (1/2) dozen eggs, one could have 0.5 (1/2) mole of copper (Cu). It is possible to have any fraction or any multiple of a mole of a chemical.
Calculations involving moles (Review conversion factors in the Math Skills section under the Basics tab.)
Example 1
A piece of copper (Cu) weighs 14.886 g. How many moles of Cu would be in this mass? How many atoms of Cu would be in this mass?
From the Periodic Table, the gram atomic weight of Cu is 63.546 g/mole. The conversion factor for copper (Cu) is 63.546 g Cu = 1 mole Cu. This can be expressed as:
63.546 g Cu or 1 mole Cu
1 mole Cu 63.546 g Cu
The Mole And Avogadro Number Khan Academy
In this problem we want moles to be the final unit, the conversion factor would be:
1 mole Cu
63.546 g Cu
The calculation would be:
Mole And Avogadro's Number Calculator
14.886 g x 1 mole Cu = 0.234 mole of Cu
63.546 g Cu
The number of atoms of Cu would be:
0.234 moles x 6.022 X 1023 atoms/mole = 1.409 x 1023 atoms
Example 2
If a beaker contained 0.75 mole of Fe (iron) and a second beaker contained 0.75 mole of Al (aluminum), would the number of atoms of Fe be less than, same as or greater than the number of atoms of Al in the second beaker? The mole unit represents the number of entities, in this case atoms of Fe and Al. Since the mole values are the same, the number of atoms of both metals would be the same.
The second part of the example. Would the mass of Fe in the first beaker be less than, same as or greater than the mass of Al in the second beaker?
To get the answer to this question, one needs to make a calculation similar to the calculation in Example 1 because this requires a conversion from moles to grams. To do this, the student needs to use a conversion factor for Fe and one for Al.
The conversion factor for Fe is: 55.845 g of Fe = 1 mole of Fe.
In this problem, the conversion factor would be:
55.845 g Fe
1 mole Fe
the calculation would be:
0.75 moles Fe x 55.845 g Fe = 41.88 g
1 mole Fe
The conversion factor for Al is 26.982 g of Al = 1 mole of Al. The calculation would be:
Relation Between A Mole And Avogadro's Number
0.75 moles Al x 26.982 g Al = 20.24 g
1 mole Al
The mass of Fe in the first beaker is greater than the mass of Al in the second beaker because the mass of Fe atoms is greater than the mass of Al atoms. The GAW (gram atomic weight) of Fe is greater than the GAW of Al.
Mole And Avogadro's Number Practice
The conversion between grams and moles is the most frequent calculation that will be done in this class. Rarely will the number of atoms be calculated. It is important however to establish that a mole represents a number of entities. One can talk about a mole of protons, a mole of electrons, a mole of atoms, a mole of ions (to be defined later) and a mole of molecules (to be defined later).