What is the weight of the Philae lander on the Churyumov–Gerasimenko comet compared to earth? We know the payload mass of the Philae lander was 21kg. 
We know the mass of the Churyumov–Gerasimenko comet is roughly 1 x 10^13kg. 
We know the mass of Earth is roughly 5.9x10^24kg. 
I've heard one estimate of the weight of the lander as 100 earth grams. Looking at the ratios  (1/(1 x 10^11)) that doesn't seem right to me. 
My question is: What is the weight of the Philae lander on the Churyumov–Gerasimenko comet compared to earth? (And does it matter that the comet is Duck shaped? ie does the weight change depending on where on the comet it is?)
 A: As you rightly pointed out, the fact that 67P is oddly-shaped should alter its gravitational attraction on various parts of the comet.
That said, if we were to go by Wikipedia in a rather off-hand manner, we find that the lander is $100 kg$ (as @fibonatic rightly pointed out) and 67P has an acceleration due to gravity of $\textbf{g'} = 10^{-3} m/s^2$. Its weight $W$ would therefore be simply a calculation of $W = m\,\textbf{g'}$, giving us $W = \frac{10^2}{10^3} kg = 0.1 kg$ or $100 \,\,\verb+earth+\, g$.
[P.S. I will update this answer with better sources than Wikipedia as soon as I find time.]
Edit 1: This ESA webpage seems to confirm the figures.
Edit 2: Calculating $\textbf{g'}$
I made some calculations:
Using the formula $\textbf{F} = M\,\textbf{g'} = \frac{GmM}{r^2} \Rightarrow \textbf{g'} = \frac{GM}{r^2}$ we can calculate the acceleration due to gravity on 67P (m being the comet's mass and M that of our lander). The above ESA page gives us this figure:

Seeing how the dimensions vary wildly, I decided to consider a mean of, say, 3.5km as diameter and 1.75km as $r$. 67P's mass is, of course, $10^{13}\,kg$ which gives us,
$$
\textbf{g'} = \frac{6.67 \times 10^{-11} \times 10^{13}}{\left( 1.75 \times 10^3 \right)^2} \approx 10^{-3} ms^{-2}
$$
A more precise answer, is, of course, $0.217 \times 10^{-3} ms^{-2}$ but since we have been very liberal in our assumptions of mass and radius, I think we ought to simply consider the order of magnitude, $10^{-3} ms^{-2}$. This pdf file contains some simulation data that agrees with our result.
A: The appeared weight of an object does not only depend on the mass of the celestial body by which it is attracted. If you simplify to spherical symmetry, which is not definitely not that case for the comet 67P (in to a lesser extend also not for the Earth) you can approximate the ratios of weight by using Newton's law of universal gravitation:
$$
g=\frac{GM}{R^2},
$$
where $g$ is the acceleration due to gravity at the surface, $G$ is the gravitational constant, $M$ and $R$ are the mass and (mean) radius of the celestial body.
Depending on where you are on 67P you might also want to include the centripetal acceleration, which can be neglected for on Earth, since its much smaller relative to its gravity. If the radius of the landing site is roughly at the equator (the vector between the landing site and center of mass close to perpendicular to the direction of the rotation of the comet) then then following equation can be used for the surface acceleration (as seen from the rotating reference frame of the comet):
$$
a=\frac{GM}{R^2}-\omega^2R,
$$
where $\omega$ is the angular velocity with which the body rotates.
The intended landing site J of Philae (it eventually landed about one kilometer away) is about 2.8 km away from the center of 67P (for this I combined this scale image and this video showing its center) and the angular velocity can be derived from the rotational period of 12.4 hours. Combining this yields an approximated acceleration at the surface of the landing site of:
$$
a_J=3.0 \times 10^{-5}\ ms^{-2}.
$$
This is approximately a factor $3\times 10^{-6}$ times smaller than the surface gravity on Earth (which is approximately $9.8\ ms^{-2}$). So according to these estimations the 100 kilogram lander would at the landing site weigh about 0.3 grams, which is quite close the to mentioned 1 gram in this article. The error most likely for the most part comes from the assumption that the comet is spherical.
