Assuming you know the exact shape, you do know the mass, as it is equal to the density of water multiplied by the volume displaced. I will assume you mean that you don't know the distribution of the mass.
You know (or can calculate) the horizontal location of the center of mass--it is vertically aligned with the center of buoyancy, which you get from the centroid of the submerged portion (if it was not so aligned, there would be a torque which would tilt the house until alignment was achieved). So imagine a vertical line passing through this position, and know that the CG of the house lies on that line.
Add a known amount of weight (let's make it equal to the weight of the house--this is not necessary but makes the arithmetic a little simpler) in a known spot and let the house come to equilibrium. Then calculate the new center of buoyancy (it's different now because the submerged portion is different). The center of gravity of the house+weight is now vertically aligned with this center of buoyancy.
The center of gravity of the house alone still lies along the line you determined before, now no longer vertical, as it tilted over with the house when you added the weight. The CG of the house+weight lies at the midpoint of the line connecting the weight with the CG of the house.
There is only one spot along the tilted line containing the CG of the house that satisfies this requirement: the midpoint of the line connecting it to the weight lies directly above the new center of buoyancy. This spot is the CG of the house.