Is this a pure sample of KClO3 or a sample given to you that is KClO3 + some inert material?
-Mass of glassware: 120.3340g
-Mass of glassware+sample (Before heat)= 122.3240g
mass KClO3 sample = 122.3240-120.3340 = ?g
-Mass of Glassware+sample (After heat)= 121.6070g
mass O2 = 122.3240-121.6070 = ?
-Mass of beaker=299.60g
-Mass of Beaker+H2O=874.20g
-Density of water=0.99733g
-Volume of H2O Displaced=576.14mL
This 576.14 comes from
mass H2O = 874.20-299.60 = 574.60g, then
volume = mass/density = 574.60/0.99733 = 576.138 which rounds to 576.14 mL and this is V1.
-Atmospheric Pressure=760.5mmHg
-Barometer Temperature=22.5*C
-Vapor Pressure of H2O=22.4 torr
-Temperature of O2 gas=24.0*C
Pressure wet O2 gas is 760.5 mm. Pressure dry O2 gas (at 24.0C) is 760.5-22.4 = ? mm and this = P1
I don't know if you are to make a correction for the different T of barometer and O2 but I will assume no. We will call T of O2 gas (24.0) T1. So you have P1, V1, and T1 above along with mass sample etc.
You want to use P1,V1 and T1 and correct to V2 at P2 (760 mm) and T2 (273).
(P1V1/T1) = (P2V2/T2). All of that will give you V2 at STP.
Use gram sample from above, convert to volume oxygen at STP at 100% yield (just a stoichiometry problem)and
(V2 @ STP/V sample @ 100% yield)*100 = %KClO3
Hi, I'm doing a chem lab of the Percent Composition of Potassium Chlorate-A Gas Law Experiment and we need to calculate the percent of KClO3 in the sample volumetrically and gravimetrically. I need help calculating the Volumetric percent of KClO3 in sample given the information we know:
-Mass of glassware: 120.3340g
-Mass of glassware+sample (Before heat)= 122.3240g
-Mass of Glassware+sample (After heat)= 121.6070g
-Mass of beaker=299.60g
-Mass of Beaker+H2O=874.20g
-Density of water=0.99733g
-Volume of H2O Displaced=576.14mL
-Atmospheric Pressure=760.5mmHg
-Barometer Temperature=22.5*C
-Vapor Pressure of H2O=22.4 torr
-Temperature of O2 gas=24.0*C
I know how to calculate the percent of KClO3 gravimetrically but Im lost as to how to calculate it volumetrically using gas laws. We also need to calculate the theoretical percent of KClO3 in the sample which we need to calculate the percent error.
3 answers
Thanks for replying.
No, this isn't a pure sample of KClO3. The mixture is 90% KClO3 and 10% MnO4 (catalyst) for the decomposition of KClO3. I understood most of what you said but you lost me at the part where you talked about using the gram sample and converting it to volume oxygen @ STP at 100% yield. The way that I thought of doing it was once we correct the volume of O2 to STP, we plug it into PV=nRT and solve for n (Number of O2 moles evolved in the reaction) Then we use that value and perform stoichiometry using the balanced chemical equation of 2KClO3---> 2KCl + 3O2. Once we have the moles of KClO3 we can find the mass of KClO3 by multiplying it by the molar mass. Then I got stuck. I also need to calculate the theoretical percent of KClO3 in the sample in order to calculate the percent error.
No, this isn't a pure sample of KClO3. The mixture is 90% KClO3 and 10% MnO4 (catalyst) for the decomposition of KClO3. I understood most of what you said but you lost me at the part where you talked about using the gram sample and converting it to volume oxygen @ STP at 100% yield. The way that I thought of doing it was once we correct the volume of O2 to STP, we plug it into PV=nRT and solve for n (Number of O2 moles evolved in the reaction) Then we use that value and perform stoichiometry using the balanced chemical equation of 2KClO3---> 2KCl + 3O2. Once we have the moles of KClO3 we can find the mass of KClO3 by multiplying it by the molar mass. Then I got stuck. I also need to calculate the theoretical percent of KClO3 in the sample in order to calculate the percent error.
What you outlined is ok and that will work for the theoretical percent KClO3 but as soon as you find n and convert to mass O2 I assume that is not a volume percent anymore but a mass percent. So I avoided that route and stuck with volume.
1 answer