[Lynda92]

Apologies if I've put this question is the wrong folder.

I would like to use 5.18 mmol of sodium ethoxide for a reaction

In my lab I have a 21 wt % solution of sodium ethoxide in ethanol. Thus I know that doing a direct calculation of mol= Mass/molecular weight and density= mass/ volume will not give me the correct amount of sodium ethoxide I actually need. I would really like to put this reaction on tomorrow so any help would be much appreciated

MW of NaOEt = 68.05. Density of NaOEt solution =0.868g/ml   

[sjb]

Apologies if I've put this question is the wrong folder.

I would like to use 5.18 mmol of sodium ethoxide for a reaction

In my lab I have a 21 wt % solution of sodium ethoxide in ethanol. Thus I know that doing a direct calculation of mol= Mass/molecular weight and density= mass/ volume will not give me the correct amount of sodium ethoxide I actually need. I would really like to put this reaction on tomorrow so any help would be much appreciated

MW of NaOEt = 68.05. Density of NaOEt solution =0.868g/ml

Consider one litre of solution. What mass of ethoxide is there in there?

[Lynda92]

Apologies if I've put this question is the wrong folder.

I would like to use 5.18 mmol of sodium ethoxide for a reaction

In my lab I have a 21 wt % solution of sodium ethoxide in ethanol. Thus I know that doing a direct calculation of mol= Mass/molecular weight and density= mass/ volume will not give me the correct amount of sodium ethoxide I actually need. I would really like to put this reaction on tomorrow so any help would be much appreciated

MW of NaOEt = 68.05. Density of NaOEt solution =0.868g/ml

Consider one litre of solution. What mass of ethoxide is there in there?

This is the calculation I did. I hope it's correct

I need 5.18 mmol of sodium ethoxide

21wt% solution

21g of sodium ethoxide present in every 100g of solution

MW= 68.05 ( MW of sodium ethoxide) d=0.868 g/ml ( density of solution)

5.18mmol = 0.35g of sodium ethoxide

I need 0.35g of sodium ethoxide

If there Is 21g of sodium ethoxide in every 100g solution.

0.35g will be present in 1.6g of solution

D=m/v

V=1.6g/0.868g/ml

V=1.84ml of sodium ethoxide solution I need

[Dan]

Your method is fine, but you are rounding off the numbers too early and have rounded the numbers in the wrong direction at least once; specifically, check this:

0.35g will be present in 1.6g of solution

Overall, you are out by about 5% in your calculation.

You can verify this by going the other way without rounding until the end: 1.84 mL of this solution comes out at 4.93 mmol.