Chemistry help needed
posted by tricia .
How do I prepare 90% formic acid from conc. formic acid (98%-100% of density 1.22g/ml)?
Please help me check if my calculations are correct.
Assuming conc. formic acid is 99%,
0.99 X 1.22 = 1.2078
Add 1.2078 / 9 = 0.1342ml H2O to 1ml of conc. acid to have a solution that is 90% formic acid by mass.
Let V be the volume of conc. formic acid needed.
0.1342V = 1000 – V
i.e. V = 881.7ml
Hence I will need to add 881.7ml of conc formic acid & make up to 1L with water?
(0.99)(1.22g.soln/ml) = 1.2078 g pure acid / mL original soln.
To get 100.0 g of pure acid:
(100.0g) / 1.2078 g acid / mL = 82.8 mls original solution. The total weight of that solution is:
(82.8 mls)(1.22g/ml) = 101g)
(Let the diluted solution contain 100.0g pure acid = m)
100.0g /m = 0.900
m = 111 g total new solution
Total mass - 99% cid mass = mass of H2O
111g - 101g = 10.0g water added
The recipe for making the new solution is
10.0mls of water added to 82.8 mls 99% solution
to make 92.8 mls 90% solution.
Now I will let you scale up the recipe to 1000 mls of 90% solution.
Thank you for your explanation.Is there any formula that we can apply universally for preparations of this nature(w/w)?