Change of momentum for a rocket A rocket with empty fuel has mass $m_i$ and with full fuel has mass $m_f$. The speed with which the fuel leaves is $v_{fuel}$. 
If the initial velocity of rocket is $0$ and has full fuel, what's the final velocity? 
The rocket is not influenced by gravitational forces and is accelerated by change of momentum gained by the dissipated fuel.
I am just confused with how to form the initial equation that relates change in $v$ and change in $m$ and $v_{fuel}$

Thrust force $R$ on rocket is given by $M\dfrac{\mathrm  dv}{dt}$ = $- v_{rel}\dfrac{\mathrm dm}{dt}$ where $\dfrac{\mathrm dm}{\mathrm dt}$ is the rate of fuel consumption and $ v_{rel}$ is the velocity of rocket relative to the ejected mass. 
Assuming absence of gravitational field,
Multiplying both sides by $dt$ we get
$$\mathrm dv= -v_{rel}\dfrac{\mathrm  dm}{M} \\\ \implies \int_{v_i}^{v_f} dv= -v_{rel} \int_{M_i}^{M_f} \dfrac{\mathrm dm}{M} \\ \implies v_f - v_i= v_{rel}\ln\dfrac{M_i}{M_f} $$
where $M_i$ is inital mass, $M_f$ is final mass. You are given relative velocity of rocket with respect to fuel as $-v_{fuel}$ . You can easily compute $M_f$.
If gravity was there too, you would get an additional term $gt$ in the final equation. 
