# 6.1: Counting Nails by the Pound

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## Counting by Weighing and Avogadro's number

The size of molecule is so small that it is physically difficult if not impossible to directly count out molecules (Figure \(\PageIndex{1}\)). However, we can count them indirectly by using a common trick of "counting by weighing".

Consider the example of counting nails in a big box at a hardware store. You need to estimate the number of nails in a box. The weight of an empty box is \(213 \,g\) and the weight of the box plus a bunch of big nails is \(1340\, g\). Let's assume we know that the weight of one big nail is 0.450 g. I hope you aren't going to tear open the package and count the nails. We agree that

\[\text{mass of big nails} = 1340\, g - 213\, g = 1227 \,g\]

Therefore

\[\text{Number of big nails in box} = \dfrac{1227\, g}{0.450\, g/ \text{big nail}} = 2,726.6\, \text{big nails} = 2,730 \,\text{big nails}. \label{eq2}\]

You have just counted the number of big nails in the box by weighing them (rather than counting them individually).

Now consider if the box of nails weighed the same, but box were filled with small nails with an individual mass of \(0.23\, g/\text{small nail}\) instead? You would do the same math, but use a different denominator in Equation \ref{eq2}:

\[\text{Number of small nails in box} = \dfrac{1227\, g}{0.230\, g/\text{small nail}} = 5,334.7\, \text{small nails} = 5,335 \, \text{small nails}. \label{eq3}\]

The individual mass is the conversion factor used the calculation and changes based on the nature of the nail (big or small). Let's ask ask a different question? How many *dozens *of nails are their in that same box of small nails described above.

If we knew the information from in Equation \ref{eq3}, we can just use the conversion of how many nails is in a dozen

\[\dfrac{5,335 \,\text{small nails} }{12 \text{small nails/dozen}} = 444.6 \,\text{dozen small nails} \label{eq4}\]

If we wanted to get this value from weighing we use the "dozen mass" instead of individual mass of

\[12 \times 0.23 g = 2.76\, g/\text{dozen small nails}. \label{eq5}\]

So following Equation \ref{eq3} we get

\[\text{Number of dozens of small nails} = \dfrac{1227\, g}{2.76\, g/\text{dozen small nails}} = 444.6 \,\text{dozen small nails} \label{eq6}\]

and this is the same result as from Equation \ref{eq4}. These calculations demonstrate the difference between individual mass (i.e., per individual) and collective mass (e.g., per dozen or per gross). The collective mass of most importance to chemistry is *molar mass* (i.e., mass per mole or mass per \(6.022 \times 10^{23}\)).

## Molar Mass for Elements

You are able to read the periodic table and determine the average atomic mass for an element like carbon. The average mass is 12.01 amu. This mass is a ridiculously tiny number of grams. It is too small to handle normally. The molar mass of carbon is defined as the mass in grams that is numerically equal to the average atomic weight. This means

\[1 g/ mole carbon = 12.01 \,g \,carbon\]

this is commonly written

\[1\, mol\, carbon = 12.01\, grams\, carbon.\]

This is the mass of carbon that contains \(6.022 \times 10^{23}\) carbon atoms.

Avogadro's number is \(6.022 \times 10^{23}\) particles.

This same process gives us the molar mass of any element. For example

- \(1\, mol\, neon = 20.18\, g\, neon\, Ne\)
- \(1 \,mol\, sodium = 22.99\, g \,sodium\, Na\)