Question

I am to conduct an experiment based on methyl ester methionine. There's an experimental procedure that I am to follow. The procedure states that they used a 1 to 1 equivalent, 5.9 grams of 33mM methyl ester of Methionine. I am to follow this except I have l-methyl ester of methionine hcl that is 98% (assuming 98 g of l-methyl ester methionine hcl over 100 g of solution). How do I figure out the amount of l-methyl esther methionine hcl I need in order to match the 8.7 grams of 33mM for the methyl ester methionine used in the procedure? I need the answer in grams.

My answer was 6.45 grams of l-methyl ester methionine hcl. Did I approach this correctly? Please let me know if more information is needed or if clarification is needed.

33 mmol methyl ester of methionine	10-3	1 mol l-methyl ester Methionine	199.70 g l-methyl ester methionine	98 g l-methyl ester methionine
L of solution	m	1 mol methyl ester of methionine	1 mol of -methyl ester methionine	100 g solution

More information- Molar mass for methyl ester of methionine and l-methyl ester of methionine seem to be 199.70 g/mol.

The whole procedure- " acid chloride of palmitic acid, C15H31COOH ( 1 equir, 8.7 grams, 30 mM) was dissolved in dry DCM. To this solution methyl ester of Methionine (1.1 equiv, 5.9 g, 33 mM) was added followed by pryridine (1.1 equiv.)"

Answer 1

Molar mass of L- methionine methyl ester hydrochloride is 199.70 grams/ mole

Molar mass of HCl is = 36.5 grams/ mole

So, in 199.70 grams of L- methionine methyl ester hydrochloride, there is (199.70-36.50) grams = 163.2 grams of L- methionine methyl este

So, you may say that: 163.2 grams of L- methionine methyl este is present in 199.70 grams of L- methionine methyl ester hydrochloride

You have to use 5.9 grams of L- methionine methyl ester which is present in (199.7* 5.9)/ 163.2 grams = 7.22 grams of L- methionine methyl ester hydrochloride

The answer is : 7.22 grams of L- methionine methyl ester hydrochloride