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.

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

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.

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.

To calculate the volumetric percent of KClO3 in the sample, you can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's calculate the number of moles of oxygen gas (O2) evolved from the KClO3 decomposition reaction. From the balanced chemical equation, it is known that 2 moles of KClO3 produce 3 moles of O2 gas.

1. Calculate the mass of oxygen gas evolved:
Mass of oxygen gas = Mass of glassware + sample (Before heat) - Mass of glassware + sample (After heat)

2. Convert the mass of oxygen gas to moles:
Moles of oxygen gas = mass of oxygen gas / molar mass of oxygen gas

3. Calculate the moles of KClO3:
Moles of KClO3 = (3/2) * moles of oxygen gas

Next, let's calculate the volume of oxygen gas evolved using the ideal gas law equation PV = nRT.

1. Convert the atmospheric pressure from mmHg to atm:
Atmospheric pressure (in atm) = Atmospheric pressure (in mmHg) / 760

2. Convert the volume of water displaced from mL to L:
Volume of water displaced (in L) = Volume of water displaced (in mL) / 1000

3. Convert the temperature to Kelvin:
Temperature (in K) = Temperature (in °C) + 273.15

4. Calculate the moles of oxygen gas:
Moles of oxygen gas = atmospheric pressure * volume of water displaced / (R * temperature)

5. Calculate the volume of oxygen gas:
Volume of oxygen gas (in L) = moles of oxygen gas * RT / atmospheric pressure

Finally, calculate the volumetric percent of KClO3 using the mass of KClO3 and the mass of the sample.

1. Calculate the mass of KClO3:
Mass of KClO3 = Mass of glassware + sample (Before heat) - Mass of glassware - Mass of beaker - Mass of water

2. Calculate the volumetric percent of KClO3:
Volumetric percent of KClO3 = (mass of KClO3 / mass of the sample) * 100

To calculate the theoretical percent of KClO3 in the sample, you need to know the molar mass of KClO3. Divide the molar mass of KClO3 by the molar mass of the sample and multiply by 100 to get the theoretical percent. Then, calculate the percent error by subtracting the experimental percent from the theoretical percent and dividing by the theoretical percent, multiplied by 100.

I hope this explanation helps you calculate the volumetric percent of KClO3 in the sample.