It is known that, when $\mathcal{A}$ is a $C^{*}$-algebra with an identity element, the space $\mathcal{S}$ of states of $\mathcal{A}$ is a convex subset of the topological dual $\mathcal{A}^*$ of $\mathcal{A}$ which is compact in the weak* topology of $\mathcal{A}^*$.
Furthermore, $\mathcal{S}$ turns out to be the closed convex hull $\overline{co}(\mathcal{P})$ of the space $\mathcal{P}\subset\mathcal{S}$ of pure states on $\mathcal{A}$ in the weak* topology of $\mathcal{A}^*$.
On the other hand, when $\mathcal{A}$ is a non-unital $C^{*}$-algebra, the space $\mathcal{S}$ of states of $\mathcal{A}$ is a convex subset of the topological dual $\mathcal{A}^*$ of $\mathcal{A}$ which is not compact in the weak* topology of $\mathcal{A}^*$.
However, we may still compute the closed convex hull $\overline{co}(\mathcal{P})$ of the space of pure states on $\mathcal{A}$, and I would like to have some kind of intuition about it.
Since $\overline{co}(\mathcal{P}\cup\{\mathbf{0}\})$ (in the weak* topology), with $\mathbf{0}$ the continuous linear functional on $\mathcal{A}$ which is constantly $0$, is the weak*-compact set of positive linear functionals on $\mathcal{A}$ with norm $\leq 1$ (see theorem 2.3.15 at page 53 in this book), I would say that $\overline{co}(\mathcal{P})$ is a weak*-closed subset of $\overline{co}(\mathcal{P}\cup\{\mathbf{0}\})$, and thus it is weak*-compact.
By the Krein-Milman theorem, this would imply that $\mathcal{P}$ is the set of extremal points of $\overline{co}(\mathcal{P})$.
Apart from that, I am not able to say anything else, and I would like to know if every element of $\overline{co}(\mathcal{P})$ is a state, and if there is an explicit representation of $\overline{co}(\mathcal{P})$ in the commutative case where $\mathcal{A}$ is the $C^*$-algebra $C_{\infty}(\mathcal{X})$ of complex-valued, continuous functions vanishing at infinity on some locally compact, non-compact, Hausdorff space $\mathcal{X}$.
Best Answer
You don't say how you define state when $\mathcal A$ is not unital. I'll go with "positive and norm equal to 1". This implies that a state $f$ satisfies $\lim_jf(e_j)=1$ for all approximate units; and the converse also holds (see for instance Lemma I.9.9 in Davidson's C$^*$-Algebras By Example).
So if $f=\sum_{k=1}^n t_kf_k$, where $f_k$ are states, $t_k\geq0$ and $\sum_kt_k=1$, , we will have $f(e_j)\to1$ for all approximate units, and so $\|f\|=1$. And $f$ is also positive, so it is a state.
When you consider $C_0(\mathcal X)$ with $\mathcal X$ locally compact Hausdorff, the states are precisely the regular probability measures (this is the Riesz-Markov Theorem). The pure states are precisely the Dirac measures (i.e., point evaluations).
But now the following happens. Let $\mathcal F$ be the family of compact subsets of $\mathcal X$, ordered by inclusion. For each $K\in \mathcal F$, fix $t_K\in\mathcal X\setminus K$. Now consider the net $\{t_K\}_{k\in\mathcal F}$. For any $f\in C_0(\mathcal X)$, fix $\varepsilon>0$; then there exists $K\in\mathcal F$ with $|f|<\varepsilon$ on $\mathcal X\setminus K$. In particular, $|\delta_{t_K}f|=|f(t_K)|<\varepsilon$. Thus $\delta_{t_K}f\to0$, which shows that $\delta_{t_K}\to0$ in the weak$^*$-topology.
Now we have that $0\in\overline{\operatorname{conv}}^{w^*}\mathcal P$, which immediately gives that $r\phi\in\overline{\operatorname{conv}}^{w^*}\mathcal P$ for all $r\in[0,1]$ and state $\phi$. With a little more work one shows that $$\overline{\operatorname{conv}}^{w^*}\mathcal P=\{\phi\in\mathcal A^*:\ \phi\geq0,\ \|\phi\|\leq1\}.$$