“Mole” Concept Review Answers
Name ____________________
1. Any measurable sample contains particles that are too SMALL and contain too MANY to count
directly.
2. _PT g = 1 mole =6.02 x1023 particles
3. Grams to # of particles g ( 6.02x1023)
( PTg
)
4.
Grams to # of moles of particles
g ( 1 mole )
PT g
5.
# of particles to grams
6.
# of particles to # of moles
7.
# of moles of particles to grams
8.
# of moles of particles to # of particles
9.
Empirical Formula
# ( PT g
)
(6.02 x1023)
# (1 mole )
(6.02 x1023)
mole ( PT g )
( 1mole )
mole ( 6.02 x1023 )
( 1mole
)
a) convert grams or % to moles
b) divide by SMALLEST number of
MOLES to determine ratios
c) If ratios are not WHOLE multiply by a common factor to make them WHOLE
d) Identify Polyatomic ions
Mg N2O6 = Mg (NO3)2
10. Molecular Formula
a) Determine Empirical formula
b) Determine Empirical formula mass
c) Divide the given molecular mass by Empirical formula mass to determine
factor that relates the molecular formula to the ___ Empirical formula.
d) Multiply Empirical formula by factor that relates the molecular formula to the
Empirical _ formula.
Empirical = simplest whole number ratio
(Conver to mole if necessary ) ( Mole Ration from coeffiecients) ( Convert to grams if necessary)
12. Sodium azide decomposes in an airbag to produce sodium metal and nitrogen gas that fills the
airbag.
A driver side airbag uses 50.00 g of sodium azide and produces 17.69 g of sodium and 32.31 g of
nitrogen gas.
a) The 17.69 g of sodium represents
a) more than 6.02 x 1023 Na atoms
b) less than 6.02 x 1023 Na atoms
c) equal to 6.02 x 1023 Na atoms
17.69g Na ( 6.02x1023) =
( 23.0 g Na )
b) Explain how you arrived at your answer.
Since 17.69g of sodium is less than the standard atomic mass of sodium the sample will
contain less than the standard number of sodium atoms ( 23.0g Na = 6.02 x 1023 Na
atoms)
13. How many moles nitrogen gas molecules (N2) are found in the 32.31 g of nitrogen gas released by
a driver side airbag?
a) Circle the correct factor label proportion that can be used to solve the problem.
i)
ii)
iii)
iv)
v)
32.31 g N2 ( 1 mole of N2)
(6.02 x 1023 N2)
(28.0g N2
)
( 1 mole of N2)
(6.02 x 1023 N2 )
( 1 mole of N2 )
( 1 mole of_N2)
(28.O g N2 )
(6.02 x 1023 N2 )
( 44.0g N2
)
vi)
( 28.0 g N2
)
(6.02 x 1023 N2)
b) What question does the following factor label solution answer?
__How many sodium azide particles are in a 3.08 mole sample of sodium azide
3.08 mole sodium azide ( 6.02 x 10 23 sodium azide ) = _1.85x1024 sodium azide____________
( 1 mole sodium azide
)
c) Calculate the answer to the question, label your answer
14. A 200 g sample of sodium azide is used in a passenger side airbag is found to contains 70.76 g of
sodium and 129.23 g of nitrogen. What is the empirical formula of sodium azide? Show all your
work, including all the appropriate units and labels.
70.76g ( 1 mole Na) = 3.08 mole Na1.0
( 23.0g Na ) 3.08
129.23g ( 1 mole N ) = 9.23 mole N3.0
( 14.0g N ) 3.08
70.76g Na x 100 = 35.4 % Na
200g
129.23g N x100 = 64.6 % N
200g
1. Count by changing mass to moles
2. NaN
Determine
ratio or subscripts by
3
dividing by the smallest number
of moles
3. If needed multiple by a common
factor to make the ratios whole
15. The monomer that may be used to make the plastic bag of an airbag has an empirical formula of is
composed of CH2 and a molecular mass of 42.0 amu. What is the molecular formula of the
monomer that may be used to make the plastic bag of the airbag?
Molecular Mass =
= 42.0 amu
Empirical Mass = (C=12.0+2(1.0) = 14.0 amu
= 3.0 ( CH2) = C3H6
16. We will make a simulated airbag in class. We may use the following reaction to produce carbon
dioxide gas
3 Na2CO3 (s) + 2 C6H8O7 yields 2Na3C6H5O7 + 3H2O + 3 CO2
The following data can be used to determine the percent yield of carbon dioxide produced
from the reaction between Na2CO3 and C6H8O7
Mass of cup and 100 ml of water =
Mass of cup, 100 ml water and Na2CO3 =
Mass of C6H8O7 added
=
Mass of cup, 100 ml water and Na3C6H5O7
121.00 g
130.98 g
12.00 g
138.97 g
a) How many grams of Na2CO3 reacted?
130.98g – 121.00g = 9.98 g
b) How many grams of CO2 were produced according to the data.
(130.98g + 12.00g ) – 138.97 g = 4.01 g
c) If 4.14 g of carbon dioxide should be produced what is the percent yield of CO2(g) in
this experiment?
Percent yield = 4.01 g CO2 (experimental yield) * 100 = 96.9 % yield
4.14 g CO2 (theoretical yield )
18. Given:
6Na(s) + Fe2O3(s) -> 3Na2O(s) + 2Fe(s)
a) What question does the following factor label method solution below answer?
How many grams of Na are needed to produce 100 g Fe?
100 g Fe ( 1 mole Fe ) ( 6 Na moles ) ( 23.0 g Na ) = __124.0 g Na__b) calculate the answer.
( 55.8 g Fe ) ( 2 Fe moles ) ( 1mole Na )
a) What does the middle step of this factor indicate?
a.
6 Na forms for every 2 Fe that react
b.
6 Na react for every 2 Fe that react
c.
6 Na forms for every 2 Fe that form
d.
6 Na react for every 2 Fe that form
The Na is a reactant and the Fe is a product. The Na reacts while the Fe forms.
17. The sodium produced by the airbag is highly reactive and needs to be combined with other
elements in order to lower the hazards associated with sodium. It is made to react with iron (III)
oxide according to the following reaction.
6Na(s) + Fe2O3(s) -> 3Na2O(s) + 2Fe(s)
b) If a passenger side airbag produces 3.08 moles of sodium how many grams of iron (III) oxide are
needed to react with the sodium to reduce the hazards associated with the sodium.
3.08 moles Na (
(
) ( 1 Fe2O3 ) ( 159.6 g Fe2O3 ) = 81.9 g Fe2O3
) ( 6 Na
) ( 1mol Fe2O3 )
c) How many moles of Fe are produced if 1.54 moles of Na2O are produced?
1.54 moles Na2O (
(
c)
) ( 2 Fe ) (
) (3Na2O) (
) = 1.03 mole Fe
)
Is 75.00 g of Fe2O3(s) enough to react with 50.00 g of Na? Which substance is in excess if
75.00 g of Fe2O3(s) reacts with 50.00 g of Na How many grams are in excess? Show all
your work including units and labels.
Have
Need
75.0 g of Fe2O3 ( 1 mole Fe2O3 ) ( 6 Na
) ( 23.0 g Na ) = 64.8 g Na
( 159.6 g Fe2O3) ( 1 Fe2O3) ( 1 mol Na )
50.0 g Na
( 1mole Na
( 23.0 g Na
) ( 1 Fe2O3 ) ( 159.6 g Fe2O3) = 57.8 g Fe2O3
) ( 6 Na
) ( 1 mole Fe2O3 )
(Have ) 75.00g g –(Need) 57.8g = 17.2 g Fe2O3 in excess
d)
If 50.00 g of Na(s) react with 75.00 g of Fe2O3(s) how many grams of Fe are produced?
The amount of Na you have is less than the Na needed therefore it limits the amount of product
formed.
50.0 g Na
( 1mole Na
( 23.0 g Na
) ( 2 Fe
) ( 6 Na
) (55.8 g Fe ) = 40.43 g Fe
) ( 1 mole Fe2O3 )
75.0 g of Fe2O3 ( 1 mole Fe2O3 ) ( 2 Fe
) ( 58.5 g Fe ) = 54.98 g Fe
( 159.6 g Fe2O3) ( 1 Fe2O3) ( 1 mol Fe )