The following graph has vertices with total 183. There are 183 ways to remove a set of one of more connected vertices so that the remaining vertices are connected. These 183 sets have distinct totals 1 to 183. If stacks of coins were on the vertices, any amount of change is in a connected set that leaves a connected set.
Call a graph with this property a full change graph. Complete graph $K_n$ is a full change graph by distributing $2^0 .. 2^{n-1}$ among the vertices.
What other graphs are full change graphs?
Here's a partial result: If you allow for negative labels, every (labeled) tree on $n$ vertices can be labeled to be a full change graph in exactly $2^{n-1}(n-1)!$ ways.
Call a connected subset of a graph's vertices "biconnected" if its removal leaves the graph connected.
First note that the partitions of a tree into two connected subgraphs are exactly the "slices" by an edge, and in addition we have the removal of every vertex. So, there are exactly $2n-1$ ways to pick a nonempty biconnected set with $n$ vertices.
Label the vertices $V_1,\dots,V_n$, and for each "cut" into $\{S,\ T\backslash S\}$ assign sums at will among $\{1,2,\dots,2n-2\}$ so that the sum assigned to $S$ and the sum assigned to $T\backslash S$ add to $2n-1$. I believe there are $2^{n-1}(n-1)!$ ways to do this.
We will prove by induction on $n$ that each assignment of the sum $s$ over all the vertices (to any integer; here $2n-1$) and sums over the biconnected subsets to any integers so that any complementary biconnected subsets add to $s$ results in exactly one labeling of all the vertices.
The base case of $n=1$ is trivial -- there is exactly one vertex and we are given its label $s$.
For the inductive step, take a leaf $L$. We know its label (as it is a biconnected subset), so it is labeled uniquely. Now, consider $T'=T\backslash L$. The biconnected subsets of $T'$ are exactly that of $T$, except that those that contain $L$ no longer do -- in particular, we know all their sums. So we may apply the inductive hypothesis and relabel.
Edit: It was not originally clear why the sum of $2n-1$ must be assigned to the sum of all the vertices. To prove this, sum up the labels across all the biconnected subsets. On one hand, each vertex is counted exactly $n$ times, so you get $n$ times the sum of the labels of the vertices. On the other,
$$1+2+\dots+(2n-1)=n(2n-1).$$
Remark. This is probably not very well written up. Please don't hesitate to comment if something is not clear.
Now, if our graph is not a tree, there are no longer $n$ equations in $n$ variables; there are more. So, we expect that in most cases such labelings do not exist. However, they seem to a lot of the time, in the examples of the wheel graph and $K_n$ given in the above post. I'll update this answer if I come up with anything more.
It does, however, seem that the cycle graph $C_n$ has such a labeling. Examples:
$$(1,2,4),(1,2,6,4),(1,3,2,7),(1,3,10,2,5),(1,2,5,4,6,13),(1,2,7,4,12,5)$$