Mole Worksheet (Key)
1. 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
2. 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
3. What is the mass of 6.656 moles of C2H6?
6.656 moles C 2 H6 x
30.068g C 2 H6
= 200.1g C 2 H 6
moles C 2 H6
4. 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
= 2.75x1012 pennies
km
m
1.40cm
5. 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
6. 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
7. 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
8. What is the mass of the average calcium atom? Answer in grams.
40.08g Ca
mole Ca
x
= 6.656x10 -23 g Ca/atom Ca
mole Ca 6.022x10 23 Ca atoms
9. 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
= 6.414174186x10 22 Al atoms = 6.414x10 22 Al atoms
101.96g Al 2O3 mole Al 2O3
mole Al
10. 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
11. 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
1000mL 66 kernals
x
=7620 kernals
L
13.0mL
2
c. 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
12. 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 Mgx
Mole Mg
6.022x10 23Mg atoms
x
= 8.84x10 20 Mg atoms
24.31g Mg
Mole Mg
13. Using the answer from the previous problem determine the mass of oxygen needed to react
with all the magnesium to for magnesium oxide. Hint: what is the ratio of magnesium atoms
to oxygen atoms in magnesium oxide?
8.84x10 20 Mg atomsx
O atom
Mole O
16.00g O
x
x
= 0.0235g O
Mg atom 6.022x10 23O atoms Mole O
14. What mass of chlorine contains 1.396 x 10
1.396x10 21Cl atomsx
21
chlorine atoms?
Mole Cl
35.45g Cl
x
= 0.08218g Cl
6.022x10 23Cl atoms Mole Cl
15. 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
16. 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
17. A vial contains (NH4)2Fe(SO4)2⋅6H2O. The empty vial has a mass of 11.3045 grams.
Measure the mass of the vial containing the compound and determine the number of
each atom present in the 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
Number of Compounds = 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
392.21g(NH 4 )2 Fe(SO4 )2 ⋅6H 2O
Mole (NH 4 )2 Fe(SO4 )2 ⋅6H 2O
Number of Compounds = 4.871062696x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds
Number of Compounds = 4.871x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds
Number of N atoms = 4.871062696x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x
2 N atoms
= 9.742x10 21 N atoms
(NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds
Number of H atoms = 4.871062696x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x
20 H atoms
= 9.742x10 22 H atoms
(NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds
Number of Fe atoms = 4.871062696x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x
Fe atom
= 4.871x10 21 Fe atoms
(NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds
Number of S atoms = 4.871062696x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x
2 S atoms
= 9.742x10 21 S atoms
(NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds
Number of O atoms = 4.871062696x10 21 (NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds x
14 O atoms
= 6.819x10 22 O atoms
(NH 4 )2 Fe(SO4 )2 ⋅6H 2O compounds
18. 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
19. 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