# How many miles are in a single gram of hydrogen cyanide?

Nov 13, 2016

I got $4.59 \times {10}^{9}$ $\text{mi}$.

First, we assume that all molecules of $\text{HCN}$ are placed in one straight line. Now, we must figure out how to calculate the length of $1$ molecule of $\text{HCN}$. We'll go from there.

$\text{H"-"C"-="N}$ contains one $\text{H}$ atom, one $\text{C"-"H}$ single bond, one $\text{C}$ atom, one $\text{C"-="N}$ triple bond, and one $\text{N}$ atom. From the NIST online database for the bond lengths, and wikipedia for the radii:

• R_("H","atomic") = "52.9 pm" = 5.29xx10^(-11) $\text{m}$
• R_("N","atomic") = "56.0 pm" = 5.60xx10^(-11) $\text{m}$
• r_("C"-"H","HCN") = "106.4 pm" = 1.064xx10^(-10) $\text{m}$
• r_("C"-="N","HCN") = "115.6 pm" = 1.156xx10^(-10) $\text{m}$

Note that bond length is defined as the internuclear distance. That's why we don't need the atomic radius of carbon---that is already accounted for by the two bond lengths.

So, the length of one molecule of $\text{HCN}$ is approximately:

l_"HCN" ~~ R_("H","atomic") + r_("C"-"H","HCN") + r_("C"-="N","HCN") + R_("N","atomic")

$= 5.29 \times {10}^{- 11}$ $\text{m}$ $+$ $1.064 \times {10}^{- 10}$ $\text{m}$ $+$ $1.156 \times {10}^{- 10}$ $\text{m}$ $+$ $5.60 \times {10}^{- 11}$ $\text{m}$

$= 3.31 \times {10}^{- 10}$ $\text{m}$ (or about $\text{331 pm}$)

So now, we take a look at how many $\text{HCN}$ molecules are in $\text{1 g}$:

1 cancel"g HCN" xx cancel"1 mol HCN"/("1.0079 + 12.011 + 14.007" cancel"g HCN") xx (6.022xx10^(23) "molecules")/(cancel"1 mol")

$= 2.23 \times {10}^{22}$ $\text{molecules HCN}$

So, the full length of this many molecules of $\text{HCN}$ in $\text{m}$ would be:

${l}_{\text{1 g HCN" = (3.31xx10^(-10) "m")/cancel"molecule HCN" xx 2.23xx10^(22) cancel"molecules HCN}}$

$= 7.38 \times {10}^{12}$ $\text{m}$

and this length in miles of $\text{1 g HCN}$ would be:

color(blue)(l_"1 g HCN") = 7.38xx10^(12) cancel("m") xx (100 cancel("cm"))/cancel"1 m" xx cancel"1 in"/(2.54 cancel("cm")) xx cancel"1 ft"/(12 cancel("in")) xx "1 mi"/(5280 cancel("ft"))

$= \textcolor{b l u e}{4.59 \times {10}^{9}}$ $\textcolor{b l u e}{\text{mi}}$