Download Unit 3 2 Basic Mole Conversions and Mole Maps

NAME _____________________________ NOTES: UNIT 3 (2): THE MOLE & STOICHIMETRY
APPLYING DIMENSIONAL ANALYSIS TO MOLE THEORY
I-II) Mole and Mole Mass
III) Stoichiometry (Greek) * measure of the element (measuring the balance of the element)
stoicheion: element
This connects to the Law of the Conservation of Matter (MR #1) … since
the process is concerned solely with the conversion of one measurement to a
different frame of reference, then the process is really concerned with ensuring
that whatever you “put into the reaction”, you somehow, “get out” with products.
A) RECAP: The basic Mole Map and Simple Conversions
1) Every beginning to a stoichiometry problem should look like:
Desiredunit = Given#unit' |
|
unit'
22.4 Liters
1 mole
6.02 x1023species
? grams
22.4 L of gas only at STP
1 mole of a substance
use formula &P.T. : changes w/ substance
grams
(constant)
(constant)
6.02 x 1023 species
(constant)
2) Calculate the number of grams equal to 0.050 moles of NI3(s).
1 mole of NI3(s)
use the PT
_____grams
3) Calculate the number of molecules equal to 1,245 grams of O2 gas at STP.
1 mole of O2(g)
use the PT
_____grams
4) When do I use just the single mole map? Well, draw a single mole map when you are
*
dealing with only 1 substance
*
doing any conversion problem when there is no balanced equation
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DIRECTIONS: Below are a number of simple mole conversion problems. I have provided you with the
conversion factors you require. Focus upon becoming comfortable with the way questions are constructed
and the application of your knowledge of unit cancellation. All answers should have 3 sig figs.
What we know…
1) How many grams of I2(s) equal 0.0290 moles of iodine?
1 mole
6.02 x1023
molecules
254 g
all are absolute numbers
What we know…
22.4 Liters
2) Calculate the mass of 9.73 Liters of H2(g) at STP.
1 mole
2 g
6.02 x1023
molecules
all are absolute numbers
What we know…
3) How many moles of Ca(NO3)2(s) equal 17.0 grams ?
1 mole
164 g
6.02 x1023
formula units
all are absolute numbers
4) How many molecules of O2(g) are equivalent to 40.6 liters of oxygen gas at STP ?
22.4 Liters
1 mole
32g
6.02 x1023
molecules
all are absolute numbers
Answers: 1) 7.37 grams
2) 0.869 grams
3) 0.104 mol
4) 1.09 x 1024 molecules of O2
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5) How many grams of Fe(s) equal 0.0750 moles of iron?
1 mole
6.02 x1023
atoms
56g
all are absolute numbers
6) Calculate the mass, in grams of 1.40 liters of He(g) at STP.
What is the mass, in kilograms?
7) How many moles of Al2(SO4)3(s) equal 24.0 grams ?
8) How many molecules of F2(g) are equivalent to 7.89 Liters of fluorine gas at STP ?
9) Challenge: Calculate the gram equivalent of 0.827 mol NH4Cl(s)
(what do you need to know???)
*
1 mol NH4Cl(s)
* grams NH4Cl = 0.827 mol NH4Cl | 53 grams NH4Cl | =
1 mol NH4Cl
6.02 x 1023
53 g
formula units
10) Challenge: Calculate the mole equivalent of 75.3 grams of Fe2O3(s)
(what do you need to know???)
*
* mol Fe2O3(s) = 75.3 grams | 1 mol Fe2O3(s)
| =
160 grams Fe2O3(s)
Answers:
5) 4.20 gram
6) 0.250 g & 2.50 x10 -4 kg 7) 0.0702 mol
8) 2.12 x 1023 molecules
1 mol Fe2O3(s)
160 g
9) 43.8 grams
6.02 x 1023
formula units
10) 0.471 mol
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NAME _____________________________
STOICHIOMETRY: EXTRA 1 TO 2 STEP MOLE PRACTICE
These are just a few extra practice problems. I urge you to keep to the technique …. of
Desired unit = Given unit’|
| and then designing a mole map, to keep you organized.
unit’
1) Calculate the volume, in liters equivalent to 71.92 grams of H2(g) at STP
2) Calculate the number of molecules equivalent to 38.9 grams of NH3(g).
Did you draw a mole map?
3) Calculate the mass, in grams of 1.20 x 1024 molecules of O2(g)
Did you draw a mole map?
4) Calculate the mole equivalent of 454 grams of CO2(g)
5) How many moles are equivalent to in 13.00 grams of NaHCO3(s)?
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6) Calculate the number of formula units equivalent to 327.89 grams of MgSO4
Did you draw a mole map?
7) Calculate the number of moles equivalent to 0.9201 Liters of He(g) at STP.
8) Which has a greater number of molecules A) 0.507 L of CO2(g) at STP OR B) 0.987 grams of C6H12O6?
Did you draw 2 mole maps?
Answers:
22.4 L at STP
1) LitersH = 71.92 grams H | 22.4 L | = 805.5 L
2 grams
2(g)
1 mol
2(g)
2 grams
1 mole
2)moleculesNH = 38.9grams NH | 6.02 x 1023 molecules| = 1.38x 1024 molecules
17 grams NH3
3(g)
3(g)
3) gramsO2 = 1.20 x 1024 molecules O2 |
32 grams
| = 63.8 grams
6.02 x 1023 molecules
6.02 x1023molecules
17 grams
1 mol
32 grams
6.02 x 1023molecules
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1 mol
4) mol = 454 grams CO2 | 1 mol
| = 10.3 mol
44 grams
44 grams
5) mol = 13.00 grams NaHCO | 1mol | = 0.1548
84 grams
1 mol
3
84 grams
6) formula units = 327.89 grams MgSO4 | 6.02 x 1023 formula units | 1.6449 x 1024 formula units
120 grams
1 mol
7) moles = 0.9201 L He | 1 mol | = 0.04108 mol
22.4 L
22.4 L at STP
120 grams
6.02 x 1023formula units
1 mol
2 grams
22.4 L
8) to solve, run both conversions…whichever gives the larger value is the answer
44 g
1 mol
6.02 x 1023molecules
A) molecules = 0.507L CO2 | 6.02 x 1023 molecules| =1.36 x 1022
22.4 L
1 mol
B) molecules = 0.987 grams C6H12O6 |6.02 x 1023 molecules| = 3.30 x 1021
180 grams
180 g
6.02 x 1023molecules
ANS: Thus, the 0.507L of CO2 has the greater number of molecules.
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IV) More Complex Stoichiometry and the Double Mole Map
A) When should you get the “hint” that you need a double mole map?


When given a balanced chemical reaction equation
When asked to deal with two different substances, in a single problem
B) You'll notice that the "double mole map" requires the use of a balanced equation.
We've introduced the issues of balancing in light of the Law of the Conservation of Matter.
Now, very briefly, I would like to introduce the role of the balanced equation in terms of
stoichiometry.
I am quite aware that you may not yet know how to balance an equation ... but I wish to discuss
what a balanced equation is.
1) First, the coefficients of the balanced equation represent the mole ratios between each of the
reactants, each of the products and each reactant to each product.
2) The coefficients tend to be in the simplest whole number ratio
3) Many chemistry teachers approach the balanced equation using a recipe...
For instance ... take the basic recipe for brownies:
For 1 batch of brownies:
2 sticks of butter (B)
4 eggs (E)
2 cups sugar (S)
2 cups flour (F)
5 squares unsweetened chocolate (C)
... and assorted flavorings...
So the recipe goes like this:
2 B + 4 E + 2 S + 2 F + 5 C +....  1 batch of brownies
When I have only 2 eggs ...how much butter is required?
*1
how many cups of sugar ?
*1
how many cups of flour?
*1
how many squares of chocolate? *2.5
How many batches of brownies can I make? *1/2 batch
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Given a chemical reaction:
2 C2H6(g) + 7 O2(g) 
4 CO2(g) + 6 H2O(l) + 3170 kJ
Meaning: For every 2 mol of ethane, 7 mol of dioxygen molecules are consumed in the
combustion (a ratio of 2 to 7) This produces 4 moles of carbon dioxide, 6 moles of
water and releases 3,170 kJ of energy from the bonding chemicals to the environment.
This set of relationships can indicate, at a glance that:
Twice as many moles of CO2 are produced as moles of ethane consumed
Three times as many moles of H2O are produced as moles of ethane used
3.5 times more moles of dioxygen are required than moles of ethane used
For every 2 mol of ethane combusted, 3,170 kJ of energy are produced
So ... Here's that balanced equation again.... Try to answer the following questions:
2 C2H6(g) + 7 O2(g) 
4 CO2(g) + 6 H2O(l) + 3,170 kJ
example 1) When 4 mol of ethane are combusted completely,
example 2)
* 14
mol of dioxygen are required.
* 8
mol of carbon dioxide are produced.
* 12
mol of water are produced.
* 6,340
kJ energy are released to the environment.
Imagine a container, that has ONLY 1 mole of ethane molecules but there are still
7 moles of dioxygen present. Assume that all of the ethane molecules are combusted.
* 3.5
mol of dioxygen are required.
and * 3.5
mole of dioxygen molecules are left in the container, unreacted
* 2
mol of carbon dioxide are produced.
* 3
mol of water are produced.
* 1,585
kJ energy are released to the environment.
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