Mole Worksheet (Key)

Mole Worksheet (Key). 1. The average distance between the Earth and Moon is 384,403 km. Using material provided in lab, calculate how many pennies wou...

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Mole Worksheet (Key) 1. The average distance between the Earth and Moon is 384,403 km. Using material provided in lab, calculate how many pennies would be needed to stack head to tails to span this distance. *I measured 10 pennies to have a height of 1.40cm.

384403km x

 

1000m 100cm 10 pennies x x = 2.75x1011 pennies km m 1.40cm

2. Determine the mass of a 5.00 cm line drawn with a number 2 pencil. -10 a. Assuming the pencil “lead” is made of carbon atoms (diameter = 1.54 x 10 m), calculate the length of these atoms if they were lined end to end. *Determine the mass of blank piece of paper to be 0.3040 grams. Then drew a line on the paper and determined the mass to be 0.3045 grams. Mass of line = (Paper + line) - (Paper with no line) Mass of line = 0.3045 g - 0.3040g = 0.0005g

0.0005g C x

mole C 6.022x10 23C atoms 1.54x10 -10 m x x = 3.860899251x10 9 m = 4x10 9 m 12.01gC moleC C atom

b. If the atoms are lined end to end will they reach the moon? If so how many times? 3.860899251x10 9 m x

km trips to moon x =10 trips to moon 1000m 384403km

3. Using material provided in lab, determine the mass (to five significant figures) and volume (to three significant figures) of the average popcorn kernel. *I measured the mass of 18 kernals to be 2.351 grams, and the volume of 66 kernals to be 13.0 mL.

a. Calculate the number of popcorn kernels are present in a 1.00 lb bag of popcorn? (2.2046 lb = 1 kg) 1.00 lb x

kg 1000 g 18 kernals x x = 3470 kernals 2.2046 lb kg 2.351g

b. I buy popcorn in bulk. I take a 2.00 liter bag with me to the store and fill it 75.0% full. How many popcorn kernels am I purchasing?   (0.750)(2.00L) x

c.

1000mL 66 kernals x =7620 kernals L 13.0mL

2

The surface area of Texas is 696241 km . How many miles high would 1.00 moles of popcorn kernels stack on top of Texas? Assume the stack goes straight up and keeps the shape of Texas. (1 mi = 1.60943 km)

Volume = (height)(surface area) 3

3

3

$ 13.0mL 6.022x10 23 kernals cm 3 ! m $ ! km $ ! mi x x x# & # & # & Volume 66 keranls mole mL "100cm % "1000m % "1.60943km % height = = 2 surface area ! $ mi 696241km 2 x # & "1.60943km %

height =

28452726.6mi 3 =106 mi 268791.4249mi 2

4. At the front of the classroom is a vial containing (NH4)2Fe(SO4)2⋅6H2O. The empty vial prior to containing the solid sample had a mass of 11.3045 grams. Measure the mass of the vial containing the compound and answer the following questions. a. How many compounds are present in this sample? *I measured the mass of the vial with contents to be 14.4770 grams. Mass of compound = (Mass of vial and contents) - (Mass of empty vial) Mass of compound = (14.4770g) -(11.3045g) Mass of compound = 3.1725g

3.1725g (NH 4 )2 Fe(SO4 )2 ⋅6H 2O x

mole(NH 4 )2 Fe(SO4 )2 ⋅6H 2O 6.022x10 23 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x 392.17 0g(NH 4 )2 Fe(SO4 )2 ⋅6H 2O mole (NH 4 )2 Fe(SO4 )2 ⋅6H 2O

4.871559528x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds 4.872x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds

  b. How many nitrogen atoms are in the sample? 4.871559528x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x

2 N atoms = 9.743x10 21 N atoms (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds

  c.

How many hydrogen atoms are in the sample?

4.871559528x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x

20 H atoms = 9.743x10 22 H atoms (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds

  d. What is the mass of water in the sample? 3.1725g (NH 4 )2 Fe(SO4 )2 ⋅6H 2O x

mole (NH 4 )2 Fe(SO4 )2 ⋅6H 2O 6 moles H 2O 18.016g H 2O x x 392.17 0g(NH 4 )2 Fe(SO4 )2 ⋅6H 2O mole (NH 4 )2 Fe(SO4 )2 ⋅6H 2O moles H 2O

0.874453833g H 2O 0.8745g H 2O

  e. How many water molecules are in the sample? 0.874453833g H 2O x

moles H 2O 6.022x10 23 H 2O molecules   x = 2.923x10 22 H 2O molecules 18.016 gH 2O moles H 2O

  f.

How many hydrogen atoms in this compound are part of water molecules?

0.874453833g H 2O x

moles H 2O 6.022x10 23 H 2O molecules 2H atoms x x = 5.846x10 22 H atoms 18.016 gH 2O moles H 2O H 2O molecules

5. Pour approximately 50 mL of water into a 100 mL beaker. Place the beaker on the analytical balance and record the initial mass. Allow the beaker to sit on the balance for two minutes and then record the final mass. Calculate the number of water molecules which evaporated in those two minutes. *I measured the intial mass to be 186.7327 grams. After two minutes the mass was 186.7259 grams. Mass of water that evaporated = 186.7327 grams - 186.7259 grams = 0.0068 grams

0.0068g H 2Ox

mole H 2O 6.022x10 23 H 2O molecules x = 2.3x10 20 H 2O molecules 18.016g H 2O mole H 2O

  6. A sample of sodium metal is determined to have a mass of 5.43 grams. How many moles of sodium are in this sample? 5.43g NaCl x

mole NaCl = 0.236 mole NaCl 22.99g NaCl

7. What is the molar mass of C2H6? Molar Mass = 2(12.01g) + 6(1.008g)   Molar Mass = 24.02g + 6.048g Molar Mass = 30.068g Molar Mass = 30.07g

  8. What is the mass of 0.1346 moles of C2H6? 0.1346 moles C 2 H6 x

30.068g C 2 H6   = 4.047 g C 2 H 6 moles C 2 H6

9. A balloon contains 4.29 grams of oxygen. How many oxygen molecules are present in the balloon? 4.29g O 2 x

mole O 2 6.022x10 23 molecules O 2 x = 8.07x10 22 molecules O 2 32.00g O 2 mole O 2

10. How many oxygen atoms are present in the balloon described in the previous problem? 4.29g O 2 x

mole O 2 6.022x10 23 molecules O 2 2 atoms O   x x = 1.61x10 23 atoms O 32.00g O 2 mole O 2 molecule O 2

    11. What is the mass of the average helium atom? Answer in grams. 4.003 g He mole He x = 6.647x10 -24 g He/atom He   mole He 6.022x10 23 He atoms

12. How many aluminum atoms are there in a 5.430g sample of aluminum oxide? 5.430g Al 2O3 x

mole Al 2O3 2 moleAl 6.022x10 23 Al atoms x x 101.96g Al 2O3 mole Al 2O3 mole Al

6.414174186x10 22 Al atoms 6.414x10 22 Al atoms

13. What mass of aluminum bromide has the same number of aluminum atoms as 5.430g of aluminum oxide? 6.414174186x10 22 Al atoms x

1AlBr3 compounds mole AlBr3 266.68g AlBr3 x x = 28.40g AlBr3 1Al atom 6.022x10 23 AlBr3 compounds mole AlBr3

14. In lab you will be reacting a strip of magnesium with oxygen from the air to form magnesium oxide. If the mass of magnesium used is 0.0357 grams, how many magnesium atoms are being used in this reaction? 0.0357g Mg x

mole Mg 6.022x10 23Mg atoms x = 8.84x10 20Mg atoms 24.31g Mg mole Mg

15. What mass of oxygen is needed to react completely with all the magnesium from the previous problem? Hint: what is the ratio of magnesium atoms to oxygen atoms in magnesium oxide?   mole Mg mole O 16.00g O   0.0357g Mg x x x = 0.0235g O 24.31g Mg

mole Mg

mole O

16. What mass of chlorine contains 1.396 x 10 1.396x10 21Cl atoms x

21

chlorine atoms?

mole Cl mole Cl 2 70.90g Cl 2 x x = 0.08218g Cl 2 6.022x10 23Cl atoms 2 mole Cl mole Cl 2

17. How many electrons are there in 2.45 grams of calcium?   2.45g Ca x

mole Ca 6.022x10 23Ca atoms 20 electrons   x x = 7.36x10 23electrons 40.08g Ca mole Ca Ca atom

    18. The distance (based on a direct path) between San Francisco and Los Angeles is 558.68 km. -10 Calculate how many iodine atoms (diameter = 2.66 x 10 m) would be needed to stack end to end to span this distance. 558.68km x

1000m I atom x = 2.10x1015I atoms   km 2.66x10 -10 m

19. A compound is composed of lead and oxygen. The mass of a sample of this compound is 1.0245g. The sample contains 0.8874 grams of lead. a. How much oxygen is in this sample? Mass of oxygen = Sample Mass - Mass of lead Mass of oxygen = 1.0245g - 0.8874g Mass of oxygen = 0.1371g

b. How many moles of oxygen atoms are in this sample? 0.1371g O x

mole O   = 8.569x10 -3 moles O 16.00g O

    c.

How many moles of lead atoms are in this sample? 0.8874g Pb x

mole Pb   = 4.283x10 -3 moles Pb 207.2g Pb

  d. What is the chemical formula of this sample? O/Pb Molar Ratio =

8.569x10 -3 = 2.000, there are 2 oxygen atoms for every 1 lead atom, (PbO 2 ) 4.283x10 -3

e. What is the chemical name of this sample? Lead (IV) oxide