How many bits are required to store MARIE's instruction set?

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I know that MARIE's instruction set has 16-bit per instruction, given that there are 9 instructions, doesn't it make sense that 16*9 = 144bits are needed to store all of the instruction? But apparently its wrong, whats wrong with my reasoning and can you guide me to the answer?

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I think you are looking for the following data structure

-------------------------------------
| Opcode  | Address                 |
-------------------------------------
| _ _ _ _ | _ _ _ _ _ _ _ _ _ _ _ _ |
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  Bit Bit   Bit                 Bit
   15  12    11                   0

as Paul R and Peter Cordes already mentioned in his comments.