# Difference between revisions of "Talk:M9 salts"

(<math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math>) |
(→<math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math>) |
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== <math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math> == | == <math>Na_2HPO_4 \cdot 7H_2O</math> Versus <math>Na_2HPO_4</math> == | ||

− | I think that this may be explained by the additional mass of 7 waters associated with each <math>Na_2HPO_4</math>. The molecular weight of <math>Na_2HPO_4</math> is about 142g/mol, while that of <math>Na_2HPO_4 \cdot 7H_2O</math> is about 268g/mol. So 5x M9 salts could be made with either 64g/L <math>Na_2HPO_4 \cdot 7H_2O</math> (according to Sambrook) or 33.9g/L <math>Na_2HPO_4</math>--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 <math>Na_2HPO_4</math>. The molecular weight of <math>Na_2HPO_4</math> is about 142g/mol, while that of <math>Na_2HPO_4 \cdot 7H_2O</math> is about 268g/mol. So 5x M9 salts could be made with either 64g/L <math>Na_2HPO_4 \cdot 7H_2O</math> (according to Sambrook) or 33.9g/L <math>Na_2HPO_4</math>--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) |

## Revision as of 08:48, 9 June 2007

The 5x M9 salts recipe here seems to be almost identical to the 10x M9 salts protocol we use in our lab and which I also found elsewhere (see e.g. http://www.changbioscience.com/protocols/recipe/M9salts10X.htm and http://www.atcc.org/mediapdfs/2057.pdf). It could be that bacteria don't mind if they get the double of salts, but it might nevertheless be interesting to know what conc. of salts would be best to use... (MDolinar)

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Na_2HPO_4 \cdot 7H_2O}**
Versus **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Na_2HPO_4}**

I think that this may be explained by the additional mass of 7 waters associated with each **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Na_2HPO_4}**
. The molecular weight of **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Na_2HPO_4}**
is about 142g/mol, while that of **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Na_2HPO_4 \cdot 7H_2O}**
is about 268g/mol. So 5x M9 salts could be made with either 64g/L **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle Na_2HPO_4 \cdot 7H_2O}**
(according to Sambrook) or 33.9g/L