Post a New Question

Chemistry

posted by .

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.

  • Chemistry -

    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



  • Chemistry -

    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.

  • Chemistry -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. math

    A bond has an average return of 6.8 percent and a standard deviation of 4.6 percent. What range of returns would you expect to see 68 percent of the time?
  2. chemistry 106

    Dry air near sea level has the following composition by volume: N2, 78.08 percent; O2, 20.94 percent; Ar, 0.93 percent; CO2, 0.05 percent. The atmospheric pressure is 1.00 atm. (Hint: Since volume is proportional to the number of moles …
  3. law of definite composition

    when performing an experiment using potassium chlorate and heating it to form potassium chloride what reasons can you offer for any differences between your results and another lab group in your class?
  4. chemistry

    The decomposition of potassium chlorate, KClO3, gives oxygen gas and solid potassium chloride. How many grams of KClO3 are needed to produce 5.00 L of Oxygen gas at STP?
  5. Chem

    A sample of an unknown biochemical compound is found to have a percent composition of 55.03 percent carbon, 5.54 percent hydrogen, 32.10 percent nitrogen and the balance oxygen. What is the simplest formula for this compound ?
  6. Chemistry

    Consider the thermal decomposition of potassium chlorate. 2 KClO3 (s)-----> 2 KCl (s) + 3 O2 (g) A mixture containing KCl and KClO3 weighing 1.80 g was heated producing 1.40 x 102 mL of O2 gas at STP. What percent of the original …
  7. chemistry

    can anyone help with this problem please..... Potassium chlorate decomposes according to the following equation: 2KClO3 (s) → 2KCl (s) + 3O2 (g). If a 3.00 g sample of KClO3 is decomposed and the oxygen is collected at 24.0 °C …
  8. Chemistry

    Consider the thermal decomposition of potassium chlorate. 2 KClO3(s) ---> 2 KCl(s) + 3 O2(g) A mixture containing KCl and KClO3 weighing 1.80 g was heated producing 1.40 x 102 mL of O2 gas at STP. What percent of the original mixture …
  9. Chemistry

    At STP potassium chlorate (KClO3) decomposes to produce solid potassium chloride (KCL) and oxygen gas (O2) according to the balanced chemical equation: 2KClO3(s)->2KCl(s)+3O2(g) What volume of oxygen gas, measured at 40 degrees …
  10. Chemistry

    A 26.3 gram sample of solid potassium chlorate decomposed and produced 9.45 grams of oxygen gas. What is the percent yield of oxygen?

More Similar Questions

Post a New Question