Let X be a topological space and $C=(C_n(X))$ be the singular complex. If there is a contracting chain homotopy, i.e. chain homotopy between $\text{id}_C$ and $0$, then $H_n(X)=0$. But I know that $H_0(X)=\tilde{H_0}(X)\oplus \Bbb Z$. So $H_0(X)=0$, which seems to be a contradiction.
What is the problem?
The formula $H_0(X) = \tilde{H}(X) \oplus \mathbb{Z}$ is only valid when $X$ is nonempty; when $X = \emptyset$, $H_0(X) = 0$. So you've correctly shown that there cannot be a contracting chain homotopy on $C_n(X)$ when $X \neq \emptyset$. When $X = \emptyset$ then $C_n(X) = 0$ from the beginning anyway.
What there can be, though, is a contracting chain homotopy in positive degrees. This would imply $H_n(X) = 0$ for $n > 0$. If in addition $X$ is path-connected, then $X$ is called an acyclic space. For example if $X$ is contractible then $X$ is acyclic, but there are more examples.