Yes there is.
The question is ambiguous because you don't say if you need to use ALL the letters. If you only want 9 letter words, then the formula is the permutation formula, nPr, divided by repeated elements.
For formula for the number of permutations of r objects picked from n is n!/(nr)!. I.e. 9!/0! in this case, or just 9! in the case when you use all 9 letters (since 0! = 1).
Where there are repeated elements, we must divided by the number of each repeated element, i.e. 2! and 2!. 2! is the same as 2, so it's just 9!/2/2 (which is the same as what Okanagan says)  there are nine letters, and two groups of two, so you take 9! (factorial) and divide by 2! for each group
If you want all the answers, i.e. oneletter words up to nineletter words, it is much more complicated:
There are 7 different letters, and two of the letters are repeated twice, so you've got 5 single letters and 2 double letters
Therefore with 8 letters you can have:
2 Cs and an O, or 2 Os and a C, plus all the other letters, which is: = 8!/2! * 2 (times two for the Cs and O vs Os and C) or 2 Cs, 2 Os, and 4 of the 5 other letters = 8!/2!/2! * 5 (times five because there are obviously five identical patterns, each missing one of the undoubled letters)
Which is
(8!/2!)* (2 + 5/2!) = 20160 * 4.5 = 90,720
Which is of course equivalent to 9!/1!/2!/2!, which is identical to 9P8 / 2!/2!
Going down to 7 letters, you have the following patterns:
5 single letters, a C and an O = 7! * 1 (only one such pattern) 2 Cs or 2 Os, 5 single letters = 7!/2! * 2 (either 2 Cs or 2 Os) 2 Cs and 2 Os, 3 single letters = 7!/2!/2! * (5!/3!/2!) = 7!/2!/2! * 10 [10 possible ways of picking 3 single letters from 5] 2 Cs, 1 O, or vice versa, 4 single letters =7!/2! * 2 * 5 [as before, 5 possible missing letters]
Which is
7! * (1 + 2/2! + 10/2!/2! + 5*2/2!) = 7! * (1 + 1 + 2.5 + 5) = 7! * 9.5 = 47,880
This however is larger than 9P7 / 2!/2!, which only comes to 45,360. Obviously the fewer letters you use the less significant the repetition comes, so when you go down to 6, 5, 4 letters, the difference between nPr/2!/2! and the actual number of permutations.
Note that the formula 5!/3!/2! which is from the formula for combinations (nCr), for picking 3 letters from 5.
You could write them out by hand, assuming there are 5 letters, ABCDE, and we must pick 3 of them, we can easily see that there are 10 possible options:
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
which is identical to 5!/3!/2!. If you consider combinations/permutations as being based around factorials, then the 5! represents the permutations of 5 letters, and the 3! is the permutations of the 3 letters we are choosing, and the 2! is the permutations of the letters that we aren't.
To get the correct answer to the number of possible words we can make from chocolate, we would have to continue all the way down to 1 letter 'words' (of which there are obviously 7), and then sum the results  9letter 'words', plus 8letter words, plus 7letter words, etc.
