Despite @AdamWuerl's sage advice, I'll give it a go.
I'll do a single-stage, massless rocket body where mass is only propellant and payload as a zeroth-order spherical-cow attempt. If you needed better agreement, you'll have to model the staging and the boosters more faithfully.
$$v = v_{ex} \log(m_0/m)$$
inverts to
$$m = m_0 * \exp(-v/v_{ex}).$$
Let's say that $C_3$ is the excess $v^2$ above escape velocity from Earth:
$$C_3 = v^2 - v_{esc}^2,$$
$$v^2 = C_3 + v_{esc}^2,$$
so:
$$m = m_0 * \exp\left(-\sqrt{\frac{C_e + v_{esc}^2}{v_{ex}^2}}\right).$$
Your plot show the Delta IV(4450-14) has a payload mass of 4,500 kg at $C_3$=0. If you plug that convenient $C_3=0$ point into:
$$m_0 = m * \exp\left( \frac{v_{esc}}{v_{ex}} \right)$$
and use an exhaust velocity of 2.4 km/s (from Wikipedia), you get:
$$m_0 = 4,600 kg * \exp\left( \frac{11.2 km/s}{2.4 km/s} \right)$$
or about 489,000 kg, which is roughly right. The article puts various versions between 250,000 and 733,000 kg.
Putting $C_3$ = 25 km/s back into:
$$m = m_0 * \exp\left(-\sqrt{\frac{C_3 + v_{esc}^2}{v_{ex}^2}}\right).$$
gives $m$ = 2951 kg, which is roughly the 2500 kg shown in your plot.
So within the constraints of a single-stage, massless rocket body where mass is only propellant and payload, this is how you might do it. If you needed better agreement, you'll have to model the staging and the boosters more faithfully.

