i have a question about finding the concentration of a substance then changing it to a percent. i check over my notes and i can't find it.
the concentration of an unkown substance is 0.831 M
and the two given substances are 4% and 5.5%
so i do not know howto calculate which one it would be
6 answers
I don't think you haven't provided enough information. The molar mass would help and the density would help.
alright! thank you, you just helped me figure it out! i didn't use all the info provided properly
I'm glad it helped but after reading my response I'm surprised it would help anyone. I really messed up that sentence. Double negatives and all.
(avg. ammount of substance)
( 28.0 mL + 27.3 mL + 27.5 mL ) / 3 = 27.7 mL
HC2H3O2 + NaOH => NaC2H3O2 + HOH
n= M x L
= 0.30 M x 0.0277 L
= 0.00831 mols
unknown:NaC2H3O2:x:1mol:NaC2H3O2:/
known:NaOH:0.00831 mols:1 mol:NaOH:
x=0.00831 mol
molar mass of HC2H302:
H x 4 = 4.04
C x 2= 24.02
O x 2= 32.00
60.06g/mol
m= n x M
= (0.00831 mol) ( 60.06g/mol)
= 0.4990986 g
= 0.4990 g
c= n / v
= (0.00831 mol) / (0.010 L)
= 0.831 M
i was just wondering if i did correct calculations and if im on the right track to finding the percent of the acetic acid
( 28.0 mL + 27.3 mL + 27.5 mL ) / 3 = 27.7 mL
HC2H3O2 + NaOH => NaC2H3O2 + HOH
n= M x L
= 0.30 M x 0.0277 L
= 0.00831 mols
unknown:NaC2H3O2:x:1mol:NaC2H3O2:/
known:NaOH:0.00831 mols:1 mol:NaOH:
x=0.00831 mol
molar mass of HC2H302:
H x 4 = 4.04
C x 2= 24.02
O x 2= 32.00
60.06g/mol
m= n x M
= (0.00831 mol) ( 60.06g/mol)
= 0.4990986 g
= 0.4990 g
c= n / v
= (0.00831 mol) / (0.010 L)
= 0.831 M
i was just wondering if i did correct calculations and if im on the right track to finding the percent of the acetic acid
I averaged the three readings and obtained 0.0276 L instead of 0.0277 L.
So that will make a little difference. Then you are ok until you get to the 0.00831/0.010 step.
If I wanted to calculate the percent of the acetic acid in the sample, I would use the grams (the previous step) and I found 0.4973. That is 0.4973 grams acetic acid in the 10 mL sample, so in 100 mL, that is 4.97%(mass/volume). I don't know if that is the percent you want or not but that is the easiest to calculate from the data you provided.
So that will make a little difference. Then you are ok until you get to the 0.00831/0.010 step.
If I wanted to calculate the percent of the acetic acid in the sample, I would use the grams (the previous step) and I found 0.4973. That is 0.4973 grams acetic acid in the 10 mL sample, so in 100 mL, that is 4.97%(mass/volume). I don't know if that is the percent you want or not but that is the easiest to calculate from the data you provided.
alright thank you:)