Avogadro's number, the number of particles in one mole of substance, was accurately determined in the early 1900s following Millikan's oil-drop experiment.

Image showing Millikan's oil-drop experiment.

In a sophisticated series of experiments started in 1908, Robert Millikan working with graduate student Harvey Fletcher determined the charge of the electron to be −1.592 × 10−19 C. To review Millikan’s oil-drop experiment, see the ChemTour in Chapter 2 titled Millikan Oil-Drop Experiment.

In 1834, Michael Faraday determined that the mass of a substance liberated at an electrode was directly proportional to the total electric charge passed through the substance. At a later date, the results of Faraday’s experiments allowed the determination of the magnitude of electric charge per one mole of electrons—the Faraday constant. In the early 1900s, the accepted value of the Faraday constant was 96,536.9 C/mol. Combining the Faraday constant with the electron charge determined by Millikan allowed the first accurate determination of the number of particles, in this case electrons, in one mole of substance.

96536.9 C/mol1.592 × 10−19 C/electron = 6.064 × 1023 electrons/mol

As the accuracies of the electron charge and the Faraday constant improved, the accuracy of Avogadro’s number improved. The currently accepted value for Avogadro’s number using electron charge and the Faraday constant is 6.022141067 × 1023 particles/mole. Avogadro’s number is designated with the symbol, NA.

Two, more accurate methods are currently used to determine Avogadro’s number. The most accurate method determines Avogadro’s number from the ratio of the mass of one mole of electrons to its rest mass. Another method determines Avogadro’s number from the ratio of the density of an ultrapure sample of material to its density on an atomic scale measured using single-crystal, X-ray diffraction. The number of atoms in a unit cell and the dimensions of the unit cell are determined from X-ray crystallography to calculate the density of the sample at an atomic scale.

an image showing an atomic-level view of a silicon wafer

To appreciate the incredibly small size of an atom and the magnitude of Avogadro’s number, let’s look at an atomic-level image of silicon (Si). Shown is a photo of a silicon wafer. Silicon is widely used to make computer chips and photovoltaic cells for solar panels. A zoom-in feature shows a scanning tunneling microscopy (STM) image of the silicon wafer surface at the atomic level.

In the image, the blurry spheres are individual silicon atoms, each having a radius of only 117 pm or 117 trillionths of a meter (1 pm = 1 × 10−12 m). Given the exceedingly tiny size of a silicon atom, a silicon wafer must contain an incredibly large number of silicon atoms. A typical silicon wafer, used to make hundreds of computer chips, is 300 mm (11.8 in) in diameter and 775 µm thick. The wafer has a mass of approximately 127 g (4.52 mol) and therefore contains approximately 2.72 × 1024 Si atoms.

Avogadro’s number allows us to move from the submicroscopic world of atoms and molecules into the macroscopic (visible) world where chemists work. Chemists work with measurable quantities of substances that contain an enormous number of particles.

an image of a scale with 12 grams of carbon

One mole of any substance is defined as the quantity of that substance containing the same number of particles as the number of carbon (C) atoms in exactly 12 g of 12C.

12 g 12C = 1 mole 12C = 6.022 × 1023 atoms 12C

One mole of any substance contains 6.022 × 1023 particles. If you tried to count from 1 to 6.022 × 1023 at a rate of one number each second, it would take you over 19,000 trillion years. This amount of time is more than a million times longer than the age of the universe (13.85 billion years).

In the remainder of this ChemTour, we will use Avogadro’s number and other conversion factors to link the submicroscopic world to the macroscopic world.

We can use Avogadro’s number to convert from atoms or molecules (particles) of a substance to moles of substance or vice versa. As an example, let’s say that we want to calculate the number of atoms in 5.00 moles of argon (Ar) gas. Note that Avogadro’s number, the number of particles in 1 mole of a substance, contains the two units of interest, atoms and moles.

1 mole Ar = 6.022 × 1023 atoms Ar

From this equality, we can write two conversion factors.

1 mol Ar6.022 × 1023 atoms Ar    or     6.022 × 1023 atoms Ar1 mol Ar

We set up our unit conversion as follows, starting with 5.00 moles of Ar with a goal of determining atoms of Ar.

Which of the two conversion factors is needed to complete the conversion? Drag the appropriate conversion factor into the box.

We can use molar mass, the mass of one mole of a substance, to convert from grams of substance to moles of substance or vice versa. As an example, let’s say that we want to determine the number of moles of carbon dioxide in 22.5 grams of CO2 gas. Note that the molar mass, the mass (in grams) of 1 mole of a substance, contains the two units of interest, moles and grams.

1 mole CO2 = 44.01 g CO2

From this equality, we can write two conversion factors.

1 mole CO244.01 g CO2    or     44.01 g CO21 mole CO2

We set up our unit conversion as follows, starting with grams of CO2 with a goal of determining moles of CO2.

Which of the two conversion factors is needed to complete the conversion? Drag the appropriate conversion factor into the box.

We can use subscripts in a chemical formula to convert from molecules of substance to the number of atoms of a particular element in the substance or vice versa. As an example, let’s say that we want to determine the number of chlorine (Cl) atoms in 1.75 × 1023 molecules of carbon tetrachloride, CCl4. Note that the subscript "4" for Cl tells us there are 4 atoms of Cl in 1 molecule of CCl4.

From this information, we can write two conversion factors.

4 atoms Cl1 molecule CCl4     or     1 molecule CCl44 atoms Cl

We set up our unit conversion as follows, starting with molecules of CCl4 with a goal of determining atoms of chlorine.

Which of the two conversion factors is needed to complete the conversion? Drag the appropriate conversion factor into the box.

Let’s put all three conversion factors, Avogadro’s number, molar mass, and the atoms of an element in one molecule of a substance, together into one calculation. As an example, let’s calculate the number of hydrogen (H) atoms in 144.2 grams of water (H2O).

The appropriate conversion factors, Avogadro’s number, molar mass, and the number of H atoms in 1 molecule of H2O are shown below.

We set up our unit conversion as follows, starting with 144.2 g of H2O with a goal of determining atoms of hydrogen.

We cannot convert directly from grams of H2O to atoms of hydrogen, but if we can convert to moles of H2O, we can use Avogadro’s number to convert to molecules of water. Drag the appropriate conversion factor into the box.

We can now convert from moles of H2O to molecules of H2O. Drag the appropriate conversion factor into the box.

We’re almost there. We now have the number of molecules of H2O in 144.2 g of water, and we want to determine the number of hydrogen atoms in that same 144.2 g of H2O. Drag the final conversion factor needed to complete the conversion into the box.

The number of atoms of hydrogen contained in 144.2 g of H2O is 9.638 × 1024 atoms H.

Question 1:

How many CO2 molecules are in 0.5 moles of carbon dioxide gas?

× 10 molecules

Question 2:

What is the mass of one atom of Al?

× 10 g

Question 3:

What is the mass of 4.2 × 1015 molecules of CCl4?

× 10 g

Question 4:

How many hydrogen atoms are in 5.0 g of NH3

× 10 atoms

Question 5:

A sample of pure ethylene, C2H4, contains 1.8 × 1024 carbon atoms. How many moles of ethylene are in this sample?

moles

Question 6:

Part A:

2.26 g of compound PX3 contains 0.800 g P. What is the molar mass of element X?

g/mol

Part B:

X represents which element?

element symbol =

Avogadro’s number allows chemists to convert from the submicroscopic units of atoms and molecules to the macroscopic unit of moles.

Moles are very practical units because chemists most often deal with macroscopic quantities of substances. The mole is as useful to chemists as the dozen is to the baker, the bushel is to the farmer, and the light-year is to the astrophysicist.

We use Avogadro’s number to convert from moles CO2 to molecules of CO2.

0.5 mol CO2 × 6.022 × 1023 molecules CO2 1 mol CO2 = 3 × 1023 molecules CO2

First, we convert from atoms Al to moles Al using Avogadro’s number, then we use molar mass as a conversion factor to convert from moles Al to grams Al.

1 atom Al × 1 mol Al 6.022 × 1023 atoms Al × 26.98 g Al 1 mol Al = 4.480 × 10-23 g/atom

First, we convert from molecules CCl4 to moles CCl4 using Avogadro’s number, then we use molar mass as a conversion factor to convert from moles CCl4 to grams CCl4.

4.2 × 1015 molecules CCl4 × 1 mol CCl4 6.022 × 1023 molecules CCl4 × 153.8 g CCl4 1 mol CCl4
= 1.1 × 10-6 g CCl4

First, we use molar mass as a conversion factor to convert from grams NH3 to moles NH3, then we use Avogadro’s number to convert from moles NH3 to molecules NH3. Finally, we can use the subscript of hydrogen (H) in the formula of NH3 to convert from molecules of NH3 to atoms of hydrogen (H).

5.0 g NH3 × 1 mol NH3 17.03 g NH3 × 6.022 × 1023 molecules NH3 1 mol NH3 × 3 atoms H 1 molecule NH3
= 5.3 × 1023 atoms H

First, we use the subscript of carbon (C) in the formula of C2H4 to convert from atoms of C to molecules of C2H4, and then we use Avogadro’s number to convert from molecules C2H4 to moles C2H4.

1.8 × 1024 atoms C × 1 molecule C2H4 2 atoms C × 1 mol C2H4 6.022 × 1023 molecules C2H4 = 1.5 mol C2H4

First, we use the molar mass of phosphorus and the subscript ratio to solve for moles of X.

0.800 g P × 1 mol P 30.97 g P × 3 mol X 1 mol P = 0.0775 mol X

Now we can calculate the molar mass of X.

(2.26 − 0.800) g X 0.0775 mol X = 18.8 g/mol

Element symbol = F