I have 99.9% of ethanol
MW is 46.07
Density is 0.789
Based on above values, how can I convert to mg's?
I need 1 mg/ml for this. How much of ethanol do I have to add to the water to get 1 mg/ 1 ml or 0.5 mg/ ml concentration?
99.9% means 100ml contains 99.9ml Ethanol
But since density is 0.789, so presumably specific gravity would also be the same, that would mean that 1ml of ethanol contains 0.789g, so 99.9 ml would contain 0.789 x 99.9 = 78.82 gm of ethanol.
In other words, your 100 ml Ethanol contains 78.82 gm of ethanol or 0.7882 g/ml or 788.2mg/ml. So to take 1 mg you have to take 1/788.2 ml or 0.0012ml or 1.2 microL and mix it in 1 ml of water.
99.9% means 100ml contains 99.9ml Ethanol
But since density is 0.789, so presumably specific gravity would also be the same, that would mean that 1ml of ethanol contains 0.789g, so 99.9 ml would contain 0.789 x 99.9 = 78.82 gm of ethanol.
In other words, your 100 ml Ethanol contains 78.82 gm of ethanol or 0.7882 g/ml or 788.2mg/ml. So to take 1 mg you have to take 1/788.2 ml or 0.0012ml or 1.2 microL and mix it in 1 ml of water.
Dear Golla, if 99.9 % is m/m percentage, than molar concentration of your stock ethanol solution is approximately 17.11 M. 1mg/mL = 1g/L and it means you will have 0,0217 M dilute solution. By using C1.V1 = C2.V2 eguation you can calculate the volume of stock solution that you need to use for your dilute solution which is 1.27 mL for a total volume of 1 L. I hope this recipe will be helpful for you.
I think you have to do this:
First= you have 99.9% (v/v) ethanol so you have 99.9 ml ethanol in 100 ml volume
the density of ethanol is 0.789 g/ml so
0.789g/1ml * 99.9ml= 78.82gr ethanol is in 100 ml of your volume
so you have 788.2 mg/ml ethanol concentration
finally we use this equation : C1V1=C2V2
we have 100 ml ethanol with 788.2 mg/ml concentration(C1) and we want 1000 ml (V2)ethanol solution with 1mg/ml concentration(C2):
788.2*V1=1000*1
V1=1.26ml
so we have to take 1.26 ml from ethanol 99.9(v/v)% and adding water to make 1000ml solution
you can use of C1*V1 = C2*V2 equation to calculate the volume of stock solution that you need to use for your dilute solution. it make 1.26 milliliter ethanol 99%
Density=Mass/volume
Volume=Mass/Density
Your Ethanol is practically absolute, i.e., you may avoid this percentage.
Volume=0,001g/0.789=0.0012674271229404 L
If you will put 1.27 micro liters of your ethanol to 998.73 micro liters of H2O you will have ethanol 1 mg/1 ml
It depends on how many mL you want (V2) and can accurately and conveniently measure (V1). The basic equation is V2 = V1 x (C1/C2). Your original concentration is 789 mg/mL alcohol ("99.9 %"), or (C1). If you want 1 mg/mL, you have to dilute 10 mL (V1 of C1) to 7,890 mL (V2), or you could dilute 1 mL(V1) to 789 mL (V2). If you want 5 mg/mL(C2), you could dilute 10 mL (V1) of C1 to 1580 mL(V2), or 1 mL (V1) to 158 mL(V2).
All above are alrigth, but let me tell something more. Since ethanol water mixtures present volume contraction, the well known formula CiVi= constant is valid for diluted concentration but, can not be applied for all concentrations. That is why ethanol concentration in this unit (g/l) is very hard to work. If you consult Perry's Chemical Engineering Handbook, you will find density tables of ethanol water mixtures with concentrations expressed as % volume and % by weigth. If you suppose that density variation is negligible between 15°C and 15,56°C (60F), tou can convert volume concentration in weigth concentration looking for the same density value in the other table.
You know more about the solution already with these additional details. The weight percents make much more sense if the volumes are not additive, as they most famously are not for alcohol-water mixtures. You are obviously on the right track here, but you are looking for beginning with nearly 100 % alcohol to quite diluted alcohol-water solutions. Why even bring volumes into it (yes, more convenient) if weights will give you what you want? HM
can anyone help with how to create a 5mM concentrations of this ethanol. I need to get 5,10,10,30 and 40 mM concentrations in stock. say about 20ml of each.
How should I calculate the butanol 2 wt percentage ? 99.99 % pure
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