Do semisimple algebraic groups always have faithful irreducible representations? For simplicity, I will be talking only about connected groups over an algebraically closed field of characteristic zero.
The basic theorem of affine algebraic groups is that they all admit faithful, finite-dimensional representations.  The fundamental theorem for semisimple groups is that these representations are all completely reducible, but unfortunately there is no reason that any irreducible summand of a faithful representation should be faithful, only that the kernels of all these representations intersect trivially.

My question is whether such a representation does, in fact, exist.
(Answered: iff the center is cyclic.)

This does not hold of general reductive groups for the following reason: if $T$ is any torus of rank $r > 1$, then its irreducible representations are all characters $\chi \colon T \cong \mathbb{G}_m^r \to \mathbb{G}_m$, which therefore have nontrivial kernels.  More generally, any reductive group $G$ has connected center a torus of some rank $r$, so by Schur's lemma this center acts by a character $\chi$ in any irreducible representation of $G$ and if $r > 1$, therefore does not act faithfully.
The exceptional case $r = 1$ does have an example, namely $\operatorname{GL}_n$, whose standard representation is faithful and irreducible and whose center has rank 1.  A more general version of this question might be, then:

Does any reductive group whose center has rank at most 1 have a faithful irreducible representation?
(Answered: when not semisimple, iff the center is connected.)

Another special case is that if $G$ is simple and of adjoint type, then its adjoint representation is irreducible and faithful by definition (or, depending on your definition, because the center is trivial).  A constructive version of this question for any $G$ (semisimple or reductive of central rank 1) is then:

Can we give a construction of a faithful, irreducible representation of $G$ from its adjoint representation?
(Not yet answered!)

This is deliberately a little vague since I don't want to restrict the possible form of such a construction, only that it not start out with "Throw away the adjoint representation and take another one such that..."
Finally, suppose the answer is "no".

What is the obstruction to such a representation existing?
(Answered: for $Z$ the center, it is the existence of a generator for $X^*(Z)$.)

 A: Spin groups in even dimensions have center a non-cyclic group (of order 4) and so have no faithful irreducible representations. 
A: This is an elaboration of George's answer (and my own deleted semi-answer), along with a comment on the further questions raised.  George has already addressed the basic question about existence of faithful irreducible representations for semisimple groups.   So this is just a supplement, not a separate answer.
For a connected semisimple algebraic group $G$ (over any algebraically closed field), the Chevalley classification determines $G$ up to isomorphism in terms of the way the weight lattice of a maximal torus of $G$ lies between the full weight lattice (of a simply connected group) and the root lattice.   The related classification of irreducible representations of $G$ by highest weights is also given in terms of dominant integral weights lying in this weight lattice.  In particular, if such a weight can be chosen to lie in no smaller intermediate lattice, it will produce a faithful irreducible representation of $G$: otherwise it gives a faithful irreducible representation of a proper quotient group and thus has highest weight in a proper sublattice.   
Such a weight exists unless the "co-fundamental group" (quotient of weight lattice by root lattice) is noncyclic.   For an irreducible root system this happens just for even type D, where you get a Klein 4-group (for Spin groups); in other cases the co-fundamental group (= center of $G$ in characteristic 0) is cyclic.   In general $G$ is an almost-direct product of groups with irreducible root systems; so one has to be careful about
factors of even type D.
ADDED: For any connected semisimple $G$ (and any characteristic), the argument just outlined leads routinely to the conclusion that a faithful irreducible representation exists precisely when no simple factor of $G$ is simply connected of even type D.
That exception was pointed out in earlier comments/answer and uses only Schur's Lemma.   To treat all simple types, the (Killing-Cartan/Chevalley) classification is used to avoid cases when the co-fundamental group is noncyclic.   Then the proof for a general $G$ reduces to this one: $G$ is an almost-direct product of simple groups, so you just have to find suitable highest weights for the individual
factors to get a suitable highest weight for $G$.  (In effect, you find an irreducible representation of the simply connected covering group, a direct product of simple groups, which induces a faithful irreducible representation of $G$ just in case $G$ has no Spin factors.) 
As Ryan points out there is a problem for reductive groups if the center is a torus of dimension $>1$.   Otherwise the situation for general linear groups can probably be imitated: start with a faithful irreducible representation of the derived group (if available), then throw in the scalars (which may of course already overlap the image of the derived group in a finite subgroup).
The further question about "construction" of $G$ (in a suitable representation) from knowledge of its adjoint representation depends on what
"construction" means in this context.   For semisimple groups, there are indirect ways to construct each irreducible representation of a given highest weight (in characteristic 0) by working with tensor products of fundamental representations.   This does not give direct information of the sort contained in Weyl's character and dimension formulas, however.   Anyway, if there exists a suitable highest weight (as discussed above) giving a faithful representation of $G$, this rather abstract procedure does "construct" the corresponding faithful irreducible representation.   But I think you are asking for a more "natural" construction based on the adjoint representation, which I can't visualize.
A: Edit: I now give the argument for general reductive $G$.
Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max
torus $T$ and write $X = X^*(T)$ for its group of characters.  Write $R$ for the
subgroup of $X$ generated by the roots of $G$.  Then the center $Z$
of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$. 

Claim: $G$ has a faithful irreducible representation if and only if the character 
  group $X/R$ of $Z$ is cyclic.

Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is  (non-canonically) isomorphic
to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.
In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.
As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.
For $(\Leftarrow)$ let me first treat the case where $G$ is almost simple; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since
the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant
and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...]
Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$"
will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups
are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the
$\lambda$ weight space of $L$ is faithful.
The general case is more-or-less the same, but with a bit more book-keeping.
Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi_i$
of its irreducible components. There is an isogeny
$$\pi:\prod_i G_{i,sc} \times T \to G$$
where $T$ is a torus and $G_{i,sc}$ is the simply connected almost simple group
with root system $\Phi_i$. Write $G_i$ for the image $\pi(G_{i,sc}) \subset G$.
The key fact is this: 
a representation $\rho:G \to \operatorname{GL}(V)$ has
$\ker \rho \subset Z$ if and only if the restriction $\rho_{\mid G_i}$ is non-trivial
for each $i$.
Now, as before pick $\lambda \in  X$ for which the coset of $\lambda$ generates
the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group
conjugate, we may suppose $\lambda$ to be dominant.  After possibly repeatedly replacing
$\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose
that $\lambda$ has the following property:
$$(*) \quad \text{for each $i$, there is $\beta_i \in \Phi_i$ with $\langle \lambda,\beta_i^\vee \rangle \ne 0$}$$
Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition
$(*)$ implies that $G_i$ acts nontrivially on $L$ for each $i$, so by the "key fact",
the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$
generates the group of characters of $Z$, the center $Z$ acts faithfully on the
$\lambda$ weight space of $L$.
