# Talk:M9 salts

## $\displaystyle Na_2HPO_4 \cdot 7H_2O$ Versus $\displaystyle Na_2HPO_4$
I think that this may be explained by the additional mass of 7 waters associated with each $\displaystyle Na_2HPO_4$ . The molecular weight of $\displaystyle Na_2HPO_4$ is about 142g/mol, while that of $\displaystyle Na_2HPO_4 \cdot 7H_2O$ is about 268g/mol. So 5x M9 salts could be made with either 64g/L $\displaystyle Na_2HPO_4 \cdot 7H_2O$ (according to Sambrook) or 33.9g/L $\displaystyle Na_2HPO_4$ --the concentration will be the same. Then if you double this for 10x salts, you would end up needing 67.8g/L, which is close to the 60g/L given in the standard 10x recipes. I can only wave my hands at the 7.8g discrepancy and say that someone goofed or altered the recipe slightly.