6.1: Counting Nails by the Pound
- Page ID
- 369347
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Counting by Weighing
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".
![](https://chem.libretexts.org/@api/deki/files/126200/Biological_and_technological_scales_compared-en.svg.png?revision=1)
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 \(1340\, g\). Assume that we know that the weight of one big nail is 0.450 g. Hopefully it's not necessary to tear open the package and count the nails. We agree that
\[\text{mass of big nails} = 1340\, g - 213\, g = 1227 \,g\]
Therefore
\[\text{Number of big nails in box} = 1227\, g \dfrac{\text{1 big nail}}{0.450\, g} = 2,726.6\, \text{big nails} = 2,730 \,\text{big nails}. \label{eq2}\]
You have just counted the number of big nails in the box by weighing them (rather than by counting them individually).
![Metal nails.](https://chem.libretexts.org/@api/deki/files/126194/800px-NailsCloseup.jpg?revision=1&size=bestfit&width=428&height=284)
Now consider if the box of nails weighed the same, but the 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}:
\[\text{Number of small nails in box} = 1227\, g \: \dfrac{\text{1 small nail}}{0.230\, g} = 5,334.7\, \text{small nails} = 5,335 \, \text{small nails}. \label{eq3}\]
The individual mass is the conversion factor used in the calculation and changes, based on the nature of the nail (big or small). Let's ask a different question: how many dozens of nails are there in the same box of small nails described above?
If we know the information from Equation \ref{eq3}, we can just use the conversion of how many nails are in a dozen:
\[\text{5,335 small nails} \: \dfrac{\text{1 dozen}}{12 \text{small nails}} = 444.6 \,\text{dozen small nails}. \label{eq4}\]
If we want to get this value from weighing, we use the "dozen mass" instead of individual mass:
\[12 \times 0.23 g = 2.76\, g/\text{dozen small nails}. \label{eq5}\]
So following Equation \ref{eq3}, we get:
\[\text{Number of dozens of small nails} = 1227\, g \: \dfrac{\text{1 dozen small nails}}{2.76\, g} = 444.6 \,\text{dozen small nails} \label{eq6}\]
and this is the same result as 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 \(6.022 \times 10^{23}\)), which will covered in more detail in the next section.