Finding mass of unbalanced meter stick from torque

[sp3sp2sp]

Homework Statement

Unbalanced meter stick has hanging mass of 100g at 0m point. I then moved fulcrum to balance it and it was balanced at the 0.33m point. (The hanging mass and 0m point is on left side)
I want to get the mass of the portion of the meter stick on right side, but I am confused as to what the moment arm on right side would be.
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Homework Equations

ccw torque = cw torque (balanced stick)

The Attempt at a Solution

The torque on left side is 0.1kg(9.8m/s/s)(0.33m) = .3234NmI think the moment arm for right side is (1 - 0.33)/2 = 0.34m, because center of mass will be at 0.67m point on meter stick.
But when I calculate mass from this, it does not agree with my already weighed mass of stick
mass of meter stick on right side of fulcrum = 0.3234Nm / (9.8m/s/s * 0.34m) = 0.0971
mass of entire meter stick = 0.19kg (weighing it on scale)

I know that the mass of only part of the meter stick will be less than the mass of the whole meter stick, but my values are different by almost 50%. Thanks for any help

 

[haruspex]

What about the torque from the left side of the stick?

 

[Chandra Prayaga]

A diagram would help.