stoichiometry review spring final exam. pbo 2 hno 3 pb: 1 (207.2 g) = 207.2 g o: 2 (16.0 g) = 32.0 g...
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![Page 1: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/1.jpg)
Stoichiometry Review
Spring Final Exam
![Page 2: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/2.jpg)
PbO2
HNO3
Pb: 1 (207.2 g) = 207.2 g
O: 2 (16.0 g) = 32.0 g 239.3 g
H: 1 (1.0 g) = 1.0 g
N: 1 (14.0 g) = 14.0 g 63.0 g
O: 3 (16.0 g) = 48.0 g
Molar Massthe mass of one mole of a substance
![Page 3: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/3.jpg)
percentage composition: the mass % of each element in a compound
Find % composition.
PbO2
(NH4)3PO4
% of element =g element
molar mass of compoundx 100
207.2 g Pb 239.2 g : = 86.6% Pb
32.0 g O 239.2 g = 13.4% O
31.2 g P 149.0 g = 20.8% P
64.0 g O 149.0 g = 43.0% O
42.0 g N 149.0 g = 28.2% N
12.0 g H 149.0 g = 8.1% H
:
::
::
(see calcs above)
![Page 4: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/4.jpg)
Finding an Empirical Formula from Experimental Data
1. Find # of g of each element.
2. Convert each g to mol.
3. Divide each “# of mol” by the smallest “# of mol.”4. Use ratio to find formula.
A compound is 45.5% yttrium and 54.5% chlorine.Find its empirical formula.
YCl3
Yg 45.5
Yg 88.9 Ymol 1
Ymol 0.512 0.512 1
Cl g 54.5
Cl g 35.5Cl mol 1
Cl mol 1.535 0.512 3
![Page 5: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/5.jpg)
(How many empiricals “fit into” the molecular?)
To find molecular formula… A. Find empirical formula.
B. Find molar mass of empirical formula.C. Find x = MM molecular MM empirical D. Multiply all parts of empirical formula by x.
![Page 6: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/6.jpg)
A carbon/hydrogen compound is 7.7% H and has amolar mass of 78 g. Find its molecular formula.
emp. form. CH
mmemp = 13 g 78 g 13 g
= 6 C6H6
H g 7.7
H g 1.0H mol 1
H mol 7.7 7.69 1
C g 92.3
C g 12.0C mol 1
C mol 7.69 7.69 1
![Page 7: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/7.jpg)
A compound has 26.33 g nitrogen, 60.20 g oxygen,and molar mass 92 g. Find molecular formula.
mmemp = 46 g = 2 N2O4 92 g 46 g
NO2
N g 26.33
N g 14.0N mol 1
N mol 1.881 1.881 1
O g 60.20
O g 16.0O mol 1
O mol 3.763 1.881 2
![Page 8: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/8.jpg)
Island Diagram:
a. Diagram has four islands.
b. “Mass Island” for elements or compounds
c. “Particle Island” for atoms or molecules
d. “Volume Island”: for gases only
1 mol @ STP = 22.4 L = 22.4 dm3
1 mol = 6.02 x 1023 particles
MOLE(mol)
Mass(g)
Particle(at. or m’c)
1 mol = molar mass (in g)
Volume(L or dm3)
1 mol = 22.4 L
1 mol = 22.4 dm3
![Page 9: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/9.jpg)
1.29 mol
What mass is 1.29 mol ?
How many molecules is 415 L at STP?sulfur dioxide
Fe2+ NO31– Fe(NO3)2
( )1 mol179.8 g
= 232 g
sulfur dioxideSO2
415 L ( )1 mol22.4 L ( )1 mol
6.02 x 1023 m’c
= 1.12 x 1025 m’c
Iron (II) nitrate
![Page 10: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/10.jpg)
22.4 L
( )1 mol
87.3 L
What mass is 6.29 x 1024 m’cules ? Al3+ SO4
2– Al2(SO4)3
aluminum sulfate aluminum sulfate
= 3580 g
( )1 mol
342.3 g( )1 mol6.02 x 1023 m’c
6.29 x 1024 m’c
At STP, how many g is 87.3 L of nitrogen gas?N2
( )1 mol28.0 g = 109 g
![Page 11: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/11.jpg)
•During a chem. rxn.; atoms are rearranged (NOT created or destroyed!)•Chemical equations must be balanced to show the relative amounts of all substances.•Balanced means: each side of the equations has the same # of atoms of each element.CH4 + O2 —> H2O + CO2
CH4 + 2O2 —> 2H2O + CO2
Chemical Reactions--Chemical Equations--
Unbalanced
Balanced
![Page 12: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/12.jpg)
Cellulose reacts with oxygen gas to form carbon dioxide gas & liquid water.
C6H10O5 + 6O2 6CO2 + 5H2O
6 C 6 10 H 10 17 O 17 Balanced!!!
Balanced?
![Page 13: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/13.jpg)
Nitroglycerin decomposes to form nitrogen gas, oxygen gas, carbon dioxide gas & water vapor
2 C3H5(NO3)3 3N2 + O2 + 6CO2 + 5H2O
6 C 6 10 H 10 6 N 6 18 O 19
Not Balanced!
Balanced?
![Page 14: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/14.jpg)
• When balancing equations, you may change coefficients as much as you need
to, but you may never change subscripts because you can’t change what substances are involved.
Balancing Chemical Equat.
![Page 15: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/15.jpg)
H2 (g) + O2 (g) H2O (l)
1. Atom Inventory:
Reactants Products
2 H 2
2 O 12 Add coefficients to balance
3. Double check to make sure it is all BALANCED!
2 2
244
Example:
![Page 16: Stoichiometry Review Spring Final Exam. PbO 2 HNO 3 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g 239.3 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0](https://reader035.vdocuments.us/reader035/viewer/2022071710/56649dd25503460f94ac81d0/html5/thumbnails/16.jpg)
• Balancing Chemical Equations practice
Balancing Chemical Equat.