In this lab we will observe the products of decomposition of potassium perchlorate (KClO4). We will then predict from our results the correct chemical reaction equation.

Can We Help with Your Assignment?

Let us do your homework! Professional writers in all subject areas are available and will meet your assignment deadline. Free proofreading and copy-editing included.


1.  Weigh out about 4.0g of KClO4 in a test tube.  Record the accurate weight below.

Product                                      Weight Before     Weight After

Mass of Test tube + KClO4     41.5g                            39.8g

"Be Bold" No-Essay $10,000 Scholarship

The $10,000 “Be Bold” Scholarship is a no-essay scholarship that will be awarded to the applicant with the boldest profile. To us, boldest does not mean “best”, or “most accomplished”. Being bold means being: Earnest, Determined, Moving. The scholarship will be awarded to the student whose profile is most bold, according to these characteristics.

Mass of Test tube                       37.5g                           37.5g

Mass of KClO4                             4.0g                               2.3g

2.  Set up the apparatus shown below.

3.  Gently heat the test tube containing the potassium perchlorate.  Gas should begin to collect in the collection bottle.  Record all observation.

4.  Once the reaction is complete, no more gas give off, allow the test tube to cool.  While the test tube is cooling test the gas in the collection bottle with glowing splint.

Caution: Do not leave the rubber tubing down in the water trough during cooling or you will experience back-up.

5.  After the test tube has cooled weigh it on a balance.  What is the change in mass?

Othello Act 3 Analysis and Summary


Oxygen flowed from the test tube into the bottle of water, forcing the water out.

Burning ember re-ignited when placed into the bottle of O2.


1.  The number of moles of KClO4 that we began with is .03 moles. 4.0g / 138.6g/mol = .03 moles

2.  The number of moles of O2 that were present in our sample of KClO4 was .06 moles. 1.9g / 32g/mole = .06 moles

3.  The number of moles of O2 lost is .02 moles. 1.7g / 32g/mol = .05 moles

4.  KClO4 ->KCl + 202

4.0g / 138.6g/mol = .03 moles……………202 / KClO4 = .06 moles x 32g = 1.9g

5   Percent Yield: 89% O2 lost 1.7g ¸ O2 Expected 1.9g

Inline Feedbacks
View all comments