# Talk:M9 salts

(Difference between revisions)
 Revision as of 11:45, 9 June 2007 (view source) ($Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$)← Previous diff Revision as of 11:48, 9 June 2007 (view source) (→$Na_2HPO_4 \cdot 7H_2O$ Versus $Na_2HPO_4$)Next diff → Line 4: Line 4: == $Na_2HPO_4 \cdot 7H_2O$  Versus  $Na_2HPO_4$ == == $Na_2HPO_4 \cdot 7H_2O$  Versus  $Na_2HPO_4$ == - I think that this may be explained by the additional mass of 7 waters associated with each $Na_2HPO_4$.  The molecular weight of $Na_2HPO_4$ is about 142g/mol, while that of $Na_2HPO_4 \cdot 7H_2O$ is about 268g/mol.  So 5x M9 salts could be made with either 64g/L $Na_2HPO_4 \cdot 7H_2O$ (according to Sambrook) or 33.9g/L $Na_2HPO_4$--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. + I think that this may be explained by the additional mass of 7 waters associated with each $Na_2HPO_4$.  The molecular weight of $Na_2HPO_4$ is about 142g/mol, while that of $Na_2HPO_4 \cdot 7H_2O$ is about 268g/mol.  So 5x M9 salts could be made with either 64g/L $Na_2HPO_4 \cdot 7H_2O$ (according to Sambrook) or 33.9g/L $Na_2HPO_4$--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. (dcekiert)
## $Na_2HPO_4 \cdot 7H_2O$ Versus Na2HPO4
I think that this may be explained by the additional mass of 7 waters associated with each Na2HPO4. The molecular weight of Na2HPO4 is about 142g/mol, while that of $Na_2HPO_4 \cdot 7H_2O$ is about 268g/mol. So 5x M9 salts could be made with either 64g/L $Na_2HPO_4 \cdot 7H_2O$ (according to Sambrook) or 33.9g/L Na2HPO4--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. (dcekiert)