If I understand your question correctly, you have 4 elements (e.g. C, N, O, and H) and you want to find all possible simple products (e.g. CH4 etc.) and not longer, more complicated molecules like ethanol or octane.
If this is correct, you should be able to use a for loop to check each combination without much concern for performance. The following code snippet will loop through and check the charge of each combination of elements. If the total charge sum is equal to zero AND if the total number of elements is more than one and less than or equal to five, then this code will record the number of each elements that can be used to make each product. The variable "out" contains the number of each element that corresponds to C, N, O, and H respectively.
elements = {'C','N','O','H'};
charge = [-4 -3 -2 1];
sizeprob = 5^4;
out = [];
for ii = 0:sizeprob-1
letternum = [mod(ii,5), mod(floor(ii/5),5), mod(floor(ii/25),5), ...
mod(floor(ii/125),5)];
chargesum = sum(letternum.*charge);
if chargesum == 0
if sum(letternum) <= 5 && sum(letternum) > 0
out(end+1,:) = letternum;
end
end
end
If you need to examine more complicated molecules with an unknown number of side chains, then this approach may not be suitable given the number of iterations that would be required. You would need to apply constraints to your problem to limit the number possibilities your script would need to check. In any case, this should hopefully give you a starting point for creating a script to solve your problem.
One additional note I have is that this code will not identify CO2, given that summing the ionic charge does not equal zero. For products such as these, you may need to incorporate additional checks in a similar manner to summing the charges. Hope this helps.
