How much gas does a car use to carry its own gas? I have always been curious about this one.
Since the gas has some weight, the car will have to burn some extra gas to carry it's own fuel around.
How can I calculate how much that extra gas is?
Assumptions:


*

*car lifetime of 300,000km

*50lt tank, topped up on each refuel

*steady gas price along the years

*steady gas density despite climate conditions

*linear relationship between car weight and fuel consumption


What would be the formula to calculate the extra fuel needed?

In my specific case, the car weighs 1200kg and burns 7.2lt/100km.
 A: With the information given, we can't do better than a good guestimate. 
Fuel consumption comes mostly from three components: Rolling resistance (linear with speed), the energy used to run the motor (square of rpm), and wind resistance (fourth power of speed). Each of these has a constant which depends on the weight, on the construction of the engine, and the shape of the car. Note that the distance also grows linear with speed, so going faster doesn't increase the energy needed to overcome the rolling resistance over a mile, just over a second or minute. 
The fuel consumption was given as 7.2 ltr / 100km. This is shared between the three components. 7.2 is a lifetime average. If you drive to save fuel (hyper-miling), you can't reduce the rolling resistance, but wind resistance and engine resistance by driving slow and in a high gear. The fuel used for the rolling resistance could be approximated by taking the fuel consumption when the driver tries their best to save fuel (minimising engine and wind resistance), and then subtracting some more because there is still engine and wind resistance left. With 7.2 ltr average, 5.5 ltr / 100 km is probably achievable, and I would estimate 4 ltr / 100 km for rolling resistance. 
The car weighs 1200 kg. A full tank weighs 37.5 kg. On average, the fuel in the car will weigh about 20 kg (because nobody empties their fuel tank completely). Additional weight for driver and passengers may be 100kg on average (people mostly drive alone, but not always). So the total weight on average is 1320 kg, fuel weight on average is 20 kg, and with 300,000 km I'd estimate the fuel spent to carry the fuel is 
(4.0 / 100) * 300,000 * (20 / 1320) = 12,000 / 66 = 182 liter. 
So the guesswork involved was: How much of the fuel consumption is weight dependent? (Less than the total fuel consumption). What percentage of the total weight is fuel (need to take into account weight of car + passengers + stuff in the car). 
A: This is NOT an answer. I put my thought in here just because the description has exceeded the comment limit.
I have seen the two valid answers but the question was treated in an arithmetical way. I, somehow, think that the question should be treated in a calculus manner as follows.
We assume that (1) the car is travelling at constant speed at all times; (2) in every $d$ seconds, the car has traveled a fixed distance of s km.
Initially, when $t = 0$, the total weight of the car is $1200 + 37.5$ kg. Suppose that it takes $x$ liters of gas to move the car (weighing $1237.5$ kg) forward for that interval.
At the start of the next interval, i.e. $t = \Delta t$, the car now weighs $1237.5 – x$ kg. For the next fixed interval, because the car is lighter now, it requires less fuel, (say $x – \Delta x$) liters to push it forward for the same distance.
The scenario goes on ….
A calculus problem?
A: Assuming the fuel tank is half full on average, the density of fuel is $0.75 \space kg/l$ (ref.), so the mass of fuel is $25 \times 0.75 = 18.75 \space kg$
The car and fuel weighs $1218.75 \space kg$.  The amount of energy used to transport the fuel alone is $18.75 \div 1218.75 = 1.5385\%$
The fuel used to carry fuel over the car's lifetime is $7.2 \times 3000 \times 0.015385 = 332.3 \space l$
A: Gas is about $3/4$ the density of water, so a $50$ liter tank (on average half full) has a mass of about $18$ kg.  If your dry car has a mass of $1200$ kg (easy to calculate with) and (as you ask) we assume fuel consumption is proportional to mass, you are using about $1.5\%$ of your fuel to move fuel.  In fact, much of the fuel goes to overcome friction and air resistance, so the real loss will be lower.
