# LB_MnCl2

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In previous post, I had quoted Daniel Zeigler with "It's been 35 years since I did the experiments, but if memory serves me correctly, if I added trace amounts of MnCl2 to LB, I got an efficiency of about 30%, as opposed to closer to 80"% with 2xSG." I need some spores for transformation experiments, so this is precisely what I'm going to do.

There is 1mL of 0.1M MnCl2 added to 1L of 2xSG media, so I asked ChatGPT to give me how much that is in grams:

### ChatGPT answers

To calculate the number of grams of Manganese (II) Chloride Tetrahydrate in 1mL of a 0.1M solution, we need to first calculate the number of moles of the compound in the solution, and then use its molar mass to convert the moles to grams.

The molar mass of Manganese (II) Chloride Tetrahydrate is:

- Manganese (II) Chloride Tetrahydrate = MnCl2 * 4H2O
- Molecular weight of MnCl2 = 125.84 g/mol
- Molecular weight of 4H2O = 72.09 g/mol
- Molecular weight of MnCl2 * 4H2O = (125.84 + 4 * 18.02) g/mol = 197.88 g/mol

To calculate the number of moles in 1 mL of a 0.1M solution, we use the formula:

- moles = concentration (in M) x volume (in liters)
- 1 mL = 0.001 L (since 1 L = 1000 mL)
- moles of MnCl2 * 4H2O = 0.1 M x 0.001 L = 0.0001 moles

Finally, to calculate the mass of MnCl2 * 4H2O in grams, we use the formula:

- mass (in grams) = moles x molar mass
- mass of MnCl2 * 4H2O = 0.0001 moles x 197.88 g/mol = 0.0198 g

Therefore, there are 0.0198 grams of Manganese (II) Chloride Tetrahydrate in 1mL of a 0.1M solution.

### Plan

To make ~0.1M MnCl2 in 200mL I should add 3.96 grams. Then, I add 1uL of MnCl2 per mL of LB agar that I want to create. In particular, I want to use solid rather than liquid media because it is more efficient than liquid media and much easier to handle.

The Edgewood Chemical Biological Center PDF on sporulation showing agar is more effective for spore creation

### Results

TODO