# How many atoms exist within a continuum body?

Materials, such as solids, liquids and gases, are composed of molecules separated by "empty" space. On a microscopic scale, materials have cracks and discontinuities. However, certain physical phenomena can be modelled assuming the materials exist as a continuum, meaning the matter in the body is continuously distributed and fills the entire region of space it occupies.

How many atoms exist within a continuum body?

• There's not a lot of "empty space" in solids and liquids. The atoms are bumping into each other. (or more exactly the outer electrons of the atoms.) – George Herold Oct 27 '14 at 14:11
• This question is a bit self-contradictory. Continuum mechanics is an approximation that explicitly ignores that matter is quantized. There are no atoms in continuum mechanics. – David Hammen Oct 27 '14 at 19:47

The number of atoms (or molecules) in a body is given by Avogadro's constant, or $6.022 \times 10^{23}$ per mole. A mole is the amount of material, in grams, equal to the atomic or molecular mass of the substance in question.

For example, for water ($H_2O$), 1 mole equals 18 grams. To get this number, remember that hydrogen ($H$) has an atomic mass of $1$. Oxygen ($O$) has an atomic mass of $16$, so the total mass of a water molecule is equal to $2\times1+16=18$ atomic mass units. It follows from Avogardo's constant that 18 grams of water have $6.022 \times 10^{23}$ molecules of water.

Similar calculations can be done for other elements or compounds, and for different masses. For instance, the atomic mass of iron is $55.845$, so there are $6.022 \times 10^{23}$ iron atoms in $55.845$ grams of iron, or $0.107 \times 10^{23}$ iron atoms per gram.

With those kind of numbers, normal bodies can easily be considered to be continuous, as the number of atoms in them is near enough to infinite for everyday practical purposes.

The simple answer to find the average number of atoms/molecules per unit volume is....

N/V (average atoms or molecules/$m^3$ ) = density ($kg/m^3$) * 1000 / atomic(or molecular) mass * $N_a$

where $N_a$ is Avogadro's number (~$6 \times 10^{23}$)

In general in solid or liquid the distance between the nuclei of atoms is approximately 1 Angstrom = $10^{-10}$ m

• Tom, here is the time to work to learn latex formatting. – peterh - Reinstate Monica Oct 27 '14 at 10:07
• I took the liberty to add some basic MathJax formatting - take a look a the source code to see how it's done. Great summary can be found [at this link](meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) – Floris Oct 27 '14 at 10:13
• thanks for the comments and edit - so I am quite new here - I have added some more latex formatting – tom Oct 27 '14 at 18:02