Challenge
Given the high resolution molecular mass of an organic molecule, output the molecule's molecular formula.
Explanation
The input will be a single number to three decimal places of precision, the relative molecular mass of the molecule.
Here, the molecular mass is defined as the sum of the masses of the atoms in the compound. Since you only are finding the molecular formulae of organic compounds, the atomic masses you need to know are:
- C, Carbon: 12.011
- H, Hydrogen: 1.008
- O, Oxygen: 15.999
- N, Nitrogen: 14.007
Your formula should only ever contain carbon, hydrogen, oxygen or nitrogen.
When writing the formula, it should take the form:
CaHbOcNd
Where the elements must be in that order (C -> H -> O -> N
, so C2O8N4H6
should be C2H6O8N4
) and a
, b
, c
and d
are numbers of the preceding element in the molecule (i.e. C2
means that there are two carbon atoms in the molecule).
If a
, b
, c
or d
are zero, that element should not be included in the formula (e.g. C2H6O2N0
should be C2H6O2
). Finally, if a
, b
, c
or d
are one, you should not include the number in the formula (e.g. C1H4
should be CH4
).
The input will always be valid (i.e. there will be a molecule with that mass). If the input is ambiguous (multiple molecules have the same mass), you must only output one of the molecules. How you choose this molecule is up to you.
Worked Example
Suppose the input is 180.156
, there is only one combination of the elements which can have this molecular mass:
12.011*6 + 1.008*12 + 15.999*6 + 14.007*0 = 180.156
So there are:
- 6 Carbons
- 12 Hydrogens
- 6 Oxygens
- 0 Nitrogens
Therefore, your output should be:
C6H12O6
More Examples
Input -> Output
28.054 -> C2H4
74.079 -> C3H6O2
75.067 -> C2H5O2N
18.015 -> H2O
Winning
Shortest code in bytes wins.
28054
) \$\endgroup\$12.011
is the relative atomic mass of carbon, which is a weighted average of the relative isotopic masses of the isotopes. In a mass spectrometer, where different isotopes are distinguished, you should see exactly12
. Similar for other atoms. \$\endgroup\$672.336
has 24 possible solutions, including a pure-nitrogen and a pure-hydrogen solution. \$\endgroup\$