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6.02 x 10 23 tpks x (^) 750 tpks1 box = 8.03 x 10 20 boxes
CHE
A Mole of Toothpicks Experiment
Introduction
The mole is a very useful counting unit when dealing with atoms or molecules. The value of the mole is 6.02 x 10 23 units. A very large number that is required when we deal with atoms or molecules. A mole of aluminum atoms contains 6.02 x 10 23 Al atoms, weighs 26.98 g and has a volume of 10 cm^2. This is a very manageable amount of aluminum. To get an idea of the size of a mole of particles it is often useful to use objects that are very much larger than atoms. In this laboratory we will use toothpicks to help us gain an understanding of the mole.
Procedure
- Measure the length width and height of a box of toothpicks in cm and calculate the volume of 1 box of flat toothpicks in cm^3. length ____7.3___________ cm width _____6.5___________ cm height ____2.8___________ cm 6.5 cm x 2.8 cm x 7.3 cm = 1.3 3 x 10^2 cm^3
answer = 125 to 1331 box^ cm 3
2) Calculate the number of boxes 750 picks1 box of toothpicks required to provide one mole of
toothpicks (6.02 x 10 23 toothpicks).
number of boxes = 8.03 x 10mole toothpicks^ boxes 20
2.75 x 10 cm 1 room x
1 box tpks 133 cm = 2.07 x 10^ boxes
8 3 3 6
8.03 x 10 20 boxes x (^) 2.07 x 101 room 6 boxes = 3.88 x 10 14 rooms
- Measure the length, width and height of this room in cm and calculate the volume of this room in cm^3. length ___1337____________ cm width ____832____________ cm height ____247____________ cm 1337 cm x 832 cm x 247 cm = 2.75 x 10^8 cm^3
answer = 2.75 x 101 room^ cm 8 3
- Calculate the number of boxes of toothpicks required to fill this room.
answer = 2.07 x 10room^6 boxes
- Calculate the number of rooms full of toothpick boxes required to store one mole toothpicks.
answer = (^) mole toothpicks3.88 x 10^14 rooms
6) Form a team of four persons and carefully count the number of toothpicks in one box. Count
rapidly but accurately to avoid error. Have one person time the process accurately to the nearest second while the others count toothpicks. Record the time here.
__450_____sec.
43,527 flat 51,350 total x 100% = 84.77% flat; 0.8477 x 0.055 g = 0.04662 g 7,823 round 51,350 total x 100% = 15.23% round; 0.1523 x 0.135 g = 0.02056 g 0.04662 g + 0.02056 g = 0.06718 g
6.02 x 10 23 tpks x 0.0672 gtpk = 4.05 x 10 22 g
4.05 x 10 22 g x (^) 454 g1 lb = 8.92 x 10 19 lbs
8.92 x 10 19 lbs x (^) 2000 lbs1 ton = 4.46 x 10^16 tons
A sample of flat and round toothpicks contains 43,527 flat toothpicks and 7823 round toothpicks. Using the mass of flat and round toothpicks from numbers 9 & 10, calculate the “atomic mass” (weighted average) of the toothpicks in this sample of 51,350 toothpicks. % of flat toothpicks in the sample _____84.77_____________ % % of round toothpicks in the sample _____15.23_____________ % weighted mass of flat toothpicks ______0.0466____________ g weighted mass of round toothpicks ______0.0206____________ g weighted average of toothpicks ______0.0672____________ g
Using the weighted mass of toothpicks as calculated in #11, determine the weight of 1 mole (6.02 x 10 23 ) of toothpicks in (a) grams, (b) pounds and (c) tons. a)
_______4.05 x 10^22 __________ g b)
_____8.92 x 10 19 __________ lbs c)
_____4.46 x 10 16 __________ tons
27 g Al x (^) 0.030 g Al foil1 in^ Al foil = 900 in Al foil
(^22)
A = L x W; 900 in Al foil = L X 12 in Al foil
L = 900 in12 in Al foil^ Al foil in
75 in x (^) 12 in1 ft ft
2 2 =
=
Aluminum atoms have a mass such that it takes 27 g of aluminum to equal 6.02 x 10 23 or 1 mole, of Al atoms. Obtain a piece of Al foil which is 6 inches square. Determine the mass of the Al foil and calculate the mass of 1 in 2 of Al foil. mass of Al foil (6" x 6") _______1.080___________ g square inches of foil ________36_____________ in^2 6 in x 6 in = 36 in 2 answer 0.030in 2 g
How many square inches of Al foil are required to provide 27 grams of aluminum or 1 mole (6.02 x 10 23 ) of Al atoms?
______9.0_x 10^2 __________ in^2
- If Al foil is available in 12 inch wide rolls, how long a piece of foil, in feet, would be required to provide 6.02 x 10 23 Al atoms (1 mole Al atoms)?
_________6.25____________ ft
