Why does a calculation to count objects covering a certain area seem to give nonsensical units? Suppose you want to estimate the number of atoms in a rectangular sheet of graphene. You might estimate the sheet to have $10^{7}$ atoms along one edge and $2*10^{7}$ atoms along the other edge. Multiplying while keeping track of units, we get
$$10^{7}\text{atoms} * 2*10^{7} \text{atoms} = 2*10^{14} \text{atoms}^2$$
But obviously, there are $2*10^{14}$ atoms, not $2*10^{14} \text{atoms}^2$. What is wrong with this calculation?
 A: Your "dimensions" are not quite "correct".  The calculation should be something like $10^{7} \frac{\hbox{atoms}}{\hbox{row}}\times 2\times 10^7 \hbox{row}=2\times 10^{14}$ atoms.  In fact, atoms are objects to be counted and added, like cars or pears. 

I believe I remember from a math course that the Greeks could not (apparently) abstract numbers and so would always think of "$5$" as associated to objects: $5$ apples, $5$ pebbles, etc.  Thus you could add apples: $5$ apples + $5$ apples + $5$ apples  is $15$ apples.
Multiplication was different and considered as a geometrical operation.  A rectangle of sides $3$ m  and $4$ m  had an area of $3\times 4 =12\hbox{m}^2$.  
As a result, they (apparently) never "discovered" the general abstract result that $a\times b=b+b+b\ldots$ ($a$ times) since the two operations were in some sense "incompatible".  Moreover, since we live in $\mathbb{R}^3$, it didn't make sense to them to multiply more than $3$ numbers together.
The OP likewise would like to equate two "incompatible" operations (in the sense of the Greeks), the outcome of which numerically agree because one needs to sum all atoms of all rows rather than "multiply" atoms together.
Unfortunately, I cannot find a source to confirm this.
A: Your units aren't atoms, but atom-bounding-widths; the square of atom-bounding-widths naturally comes in units of atom-bounding-areas.
A: There is no such unit as an atom$^2$.  
You are counting the number of objects (atoms) and so adding 
$10^{7} \,\rm atoms\, + 10^{7} \,\rm atoms\,+10^{7} \,\rm atoms\,+10^{7} \,\rm atoms\,+ . . . . . + 10^{7} \,\rm atoms\,$  
with $2*10^{7}$ terms in the summation and you write this in "shorthand" as 
$(10^{7} * 2*10^{7})\,  \rm atoms = 2*10^{14} \,atoms$.
A: Atoms are not really a unit. They don't combine in the right way when you multiply. If you are careful, you can make them work in some situations, as both other answers show (+1 to both). 
But as you saw, they don't work everywhere. Not like a length would. If the distance from atom to atom is 1 Angstrom, there is no problem with an area of $10^{14}$ Angstrom$^2$.
So what is the unit? A count is dimensionless. 
Even so, people often will use it as a unit where it works. You just have to be careful not to use it where it doesn't. 
Physicists are sometimes sloppy in ways like this where mathematicians are much more careful. For example, a function can have a value of $0$ most places, but have a tall thin spike near $0$, so the area under the spike is 1. This is useful in some situations. Physicists find they need the spike to be infinitely narrow. So they created the Dirac delta function, which is $0$ everywhere except at $0$. The value at $0$ is infinite. The area under the spike is $1$. 
A mathematician would find problems with such a "function" and say it doesn't exist. A physicist is careful to use it where it works. 
A: What is wrong with your calculation, is that you substituted a "count" for a dimension.
The correct way to do it, would be to determine the length of $10^7$ atoms. Assuming they take 1cm, then $1cm^2$ would have $1x10^{14}$ atoms, and the sheet (1 cm by 2 cm, or 2 $cm^2$) would contain ($1x10^{14}$ atoms/$cm^2$ x 2 $cm^2$ =) $2x10^{14}$ atoms.     
