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

I got `4.59xx10^9` `"mi"`.

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

`"H"-"C"-="N"` contains one `"H"` atom, one `"C"-"H"` single bond, one `"C"` atom, one `"C"-="N"` triple bond, and one `"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)` `"m"`
• `R_("N","atomic") = "56.0 pm" = 5.60xx10^(-11)` `"m"`
• `r_("C"-"H","HCN") = "106.4 pm" = 1.064xx10^(-10)` `"m"`
• `r_("C"-="N","HCN") = "115.6 pm" = 1.156xx10^(-10)` `"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 `"HCN"` is approximately:

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

`= 5.29xx10^(-11)` `"m"` `+` `1.064xx10^(-10)` `"m"` `+` `1.156xx10^(-10)` `"m"` `+` `5.60xx10^(-11)` `"m"`

`= 3.31xx10^(-10)` `"m"` (or about `"331 pm"`)

So now, we take a look at how many `"HCN"` molecules are in `"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")`

`= 2.23xx10^(22)` `"molecules HCN"`

So, the full length of this many molecules of `"HCN"` in `"m"` would be:

`l_"1 g HCN" = (3.31xx10^(-10) "m")/cancel"molecule HCN" xx 2.23xx10^(22) cancel"molecules HCN"`

`= 7.38xx10^(12)` `"m"`

and this length in miles of `"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"))`

`= color(blue)(4.59xx10^9)` `color(blue)("mi")`