According to http://en.wikipedia.org/wiki/De_Rham_cohomology,
one defines the $k$-th de Rham cohomology group $H^{k}_{\mathrm{dR}}(M)$ to be the set of equivalence classes, that is, the set of closed forms in $\Omega^k(M)$ modulo the exact forms.
On the other hand, the de Rham cohomology groups of a $n$-dimensional sphere $H_{dR}^q(S^n)$ is $\mathbb{R}$ if $q=0,n$ and 0 otherwise.
I am not sure to understand the link between $\mathbb{R}$ and the equivalence groups. Does that mean that to generate the de Rham cohomology groups of $S^n$, one can take any constant function $\omega$ (case $q=0$) or non-zero $n$-differential form $\omega$ (case $q=n$), and that each equivalence class can be generated from the product of $\omega$ by a particular member of $\mathbb{R}$ ?
For $q=0$ the isomorphism between $H^n_{dR}(M)$ and $\mathbb{R}$ is fairly canonical, but for $q=n$ this is no longer the case. Consider for example a nontrivial fibration with fiber $M$; typically there is no natural way of identifying top-dimensional cohomology of each fiber with $\mathbb{R}$, since the determinant bundle will generally not be trivial.