Let $G_{sc}$ be as in the answer by Victor Protsak and let $\varpi_1$, $\ldots$, $\varpi_l$ be the fundamental dominant weights.
Let $\lambda$ be a dominant weight and write $\lambda = \sum_{i = 1}^l a_i \varpi_i$ for $a_i \in \mathbb{Z}_{\geq 0}$.
Then the irreducible representation of $G_{sc}$ with highest weight $\lambda$ is faithful precisely in the following cases:
- Type $A_{l}$ ($l \geq 1$): $\gcd(l+1, a_1+2a_2+\cdots+la_l) = 1$.
- Type $B_l$ ($l \geq 2$): $a_l$ is odd.
- Type $C_l$ ($l \geq 2$): $a_1 + a_3 + a_5 + \cdots$ is odd.
- Type $D_l$ ($l \geq 4$): $l$ is odd and $a_{l-1} + a_l$ is odd
- Type $G_2$: always
- Type $F_4$: always
- Type $E_6$: $m_1 - m_3 + m_5 - m_6$ is not divisible by $3$.
- Type $E_7$: $m_2 + m_5 + m_7$ is odd
- Type $E_8$: always
This can be determined by a direct computation. The kernel of any irreducible representation of $G_{sc}$ lies in $Z(G_{sc})$, which is finite. Furthermore, you can describe $Z(G_{sc})$ explicitly and compute the action of any $z \in Z(G_{sc})$ in an irreducible representation (it is of course always multiplication by some scalar). See for example Chapter 3, Lemma 28 in "Lectures on Chevalley Groups" by Steinberg.
Note above that in types $G_2$, $F_4$ and $E_8$ the center of $Z(G_{sc})$ is trivial so every irreducible representation is faithful.
Also, for type $D_{2l}$ the center is not cyclic so no irreducible representation is faithful.