iv stoichiometry. stoichiometry the relationship (mole ratio) between elements in a compound and...
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IV Stoichiometry
Stoichiometry
• The relationship (mole ratio) between elements in a compound and between elements and compounds in a reaction
e.g. H2O
means: 1 molecule H2O contains 2 atom H and 1 atom
O
and 1 mole H2O contains 2 moles H atoms and 1 mole O atoms
Chemical Compounds
• Combination of elements– A compound is made up of specific
elements in a specific ratio - Law of Constant Composition
• Chemical Formula– Written representation of a chemical
compound. So, a specific compound has a specific formula
Terms
• Formula Unit– Involves the lowest subscript which
describes the ratio.
– e.g. H2O; CH; CH2; CH4
– May or may not actually exist
• Formula Weight– The mass of the formula unit
• Empirical Formula– Written representation of the formula unit
Terms continued
• Molecule– Integral multiple of the formula unit (integer
may be 1) that actually exists
– e.g. C2H2; C2H4; C2H6
• Molecular Weight (Molar Mass)– Mass of the molecule
• Molecular Formula– Written representation of the molecule
• Mass to mole conversions
• Stoichiometry is in mole ratios
• Most measurements are made in grams
• So, you need to be able to get from grams to moles and moles to grams
• The atomic weight listed on the periodic table is listed without units. Why?
• Units depend on what you want. • If you are looking on the atomic scale,
atoms or molecules, units are amu.• If you are working on the macroscopic
scale, moles of material, units are in grams.
Convert 34.0 grams NH3 to moles.
1) Determine M of NH3.M = AW N + 3(AW H)
= 14.0 + 3(1.01)= 17.0
2) Determine the moles of NH3
34.0gNH3
1molNH3
17.0gNH3
2.00 molNH3
How many molecules in 32.0 g of oxygen gas.
A) 2.0
B) 1.0
C) 0.5
D) 6.02 x 1023
E) 1.20 x 1024
How many molecules in 32.0 g of oxygen gas.
Oxygen occurs as O2 gas.
AW of O = 16.00, so O2 = 32.00
32.0gO2
1molO2
32.00g
6.022x1023 molecules
1mol
6.02 x1023 moleculesO2
• % Composition (% weight/ weight; %w/w)
%composition grams of element
grams of compoundX100
1. Calculate the % composition form the Molecular Formula (or the Empirical Formula)
What is the % composition by weight of C2H4O2? C = 12.01; H = 1.01; O = 16.00
Step 1: Find the molecular weight of the compound.
MW = 2(AW C) + 4(AW H) + 2(AW O)
= 2(12.01) + 4(1.01) + 2(16.00)
= 60.06 g/ mol
Step 2: Find the % of each element.
%CgramsC
MWcpdx100
2(AW C)
60.06x100
2(12.01)60.06
x100= 39.99%
%HgramsH
MWcpdx100
4(AW H)
60.06x100
4(1.01)
60.06x100=6.73%
%OgramsO
MWcpdx100
2(AW O)
60.06x100
2(16.00)60.06
x100=53.28%
or: %O100 %C %H
%O100 39.99 6.73=53.28%
• Find the Chemical Formula from the % composition
• Which Chemical formula will you get?– Empirical Formula, to get the molecular
formula you would need more information than just the % composition
A compound containing only carbon, hydrogen, and oxygen was found to contain 62.02 % C and 10.42% H. What is the formula of the compound?
C = 12.01 H = 1.01 O = 16.00
1) Find the amount of O
%O = 100 - %C - %H
= 100 - 62.02 - 10.42
= 27.57 % O
2) Determine the moles of each element
(assume a 100 g sample)
molCgramsC
MW C
62.02gC
12.01g/mol5.163molC
molHgramsH
MWH
10.42gH
1.01g/mol10.3168molH
molOgramsO
MWO
27.57gO
16.00g/mol1.723molO
3) Divide by the smallest number to get whole number ratio
Mol C = 5.163 mol
Mol H = 10.3168 mol
Mol O = 1.723 mol
/ 1.723 = 2.996 = 3
/ 1.723 = 5.9877 = 6
/ 1.723 = 1
So, the compound has the ratio 3C:6H:1 O
And the Empirical formula: C3H6O
A further analysis of the compound found that the molar mass was 115.99 g/mol. What is the molecular formula of the compound?
Molecular formula =
(empirical formula)(# of formula units)
You can find the # of formula units from
MW/ FW
Formula weight = weight of empirical formula: C3H6O
FW = 3(12.01) + 6(1.01) + 16.00 = 58.048
# formula units =
So the molecular formula contains 2 formula units. Multiply the subscripts by 2
2(C3H6O) = C6H12O2
115.99g
58.048g1.9982
A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound?
U = 238.0 O = 15.9994
A) U2O5
B) U3O8
C) UO3
D) UO2
E) U3O
A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound?
U = 238.0 O = 15.9994
1. Determine the amount of O
100 - 84.80 = 15.20
2. Determine moles and mole ratio
molU84.80gU
238.0g/mol0.3563molU
molO15.20gO
15.9994 g/mol0.9500 molO
/ 0.3563 = 1
/ 0.3563 = 2.666
0.666 = 2/3 so O is 2 2/3 = 8/3 U1O8/3
Clear denominator, multiply by 3 U3O8
3) Determine an unknown element (X) from the % composition
A compound XO2 is 78.8% X. What element is X? O = 16.0
A) Ni
B) Co
C) P
D) Sn
3) Determine and unknown element (X) from the % composition
A compound XO2 is 78.8% X. What element is X? O = 16.0
% O = 100 - 78.8 = 21.2
21.2gO
1molO
16.0g
1.325 molO
1.325 molO
1molX
2 molO
0.6625 molX
Since Compound is 78.8% X, 0.6625 mol X = 78.8 g X
78.8g X
0.6625molX
? g X
1molX
so :?118.9 Go to Periodic Table
118.9 = Sn
Reactions
• Chemical Reaction is represented by a Chemical Equation
• General Form:aA + bB cC + dD
A/B are ?
Reactants
C/D are ?
Products
a,b,c,d are ? Stoichiometric coefficients
• Law of Conservation of Mass says?– No mass lost or gained
• Total mass of reactants = total mass of products
• Elements are re - arranged not changed
• This allows us to “Balance” Equations
• The number and kinds of atoms in the reactants have to show up as the same number and kind of atoms in the product
Ca + H2O Ca(OH)2 + H2
1 Ca + H2O 1 Ca(OH)2 + H2
1Ca + 2H2O 1 Ca(OH)2 + H2
1 Ca + 2 H2O 1Ca(OH)2 + 1 H2
Ca + 2 H2O Ca(OH)2 + H2
C3H7OH + O2 CO2 + H2O
C3H7OH + O2 3 CO2 + H2O
C3H7OH + O2 3 CO2 + 4 H2O
C3H7OH + 9/2 O2 3 CO2 + 4 H2O
2 C3H7OH + 9 O2 6 CO2 + 8 H2O
SiF4 + H2O HF + SiO2
SiF4 + H2O 4 HF + SiO2
SiF4 + 2 H2O 4 HF + SiO2
When the reaction:
C2H8N2 + N2O4 N2 + H2O + CO2
Is balanced using the smallest whole numbers, what is the coefficient of N2?
• 1
• 2
• 3
• 4
• 5
When the reaction:
C2H8N2 + N2O4 N2 + H2O + CO2
Is balanced using the smallest whole numbers, what is the coefficient of N2?
C2H8N2 + N2O4 N2 + H2O + CO2242 3
• Hydrates
• A compound (solid) that contains intact water molecules as part of the compound.
• The water can be removed by heating to leave an anhydrous residue (solid).
• Water can then be re - added to the anhydrate to yield the original hydrate
CaSO4 • 2H2OCalcium sulfate dihydrateEach mole of compound contains:1 mole calcium sulfate and 2 mole water
or1 mole Ca 1 mole S6 mol O4 mol H
CaSO4 • 2H2O(s) CaSO4(s) + 2 H2O(g)
172 136 + 2(18)
CaSO4(s) CaSO4 • 2H2O(s)
heat
Water vapor
17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left?
A) 13.4
B) 13.44
C) 3.56
D) 15.0
E) 15.03
17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left?
17.0gCaSO4 2H2O
1molCaSO4 2H2O
172.0g
1molCaSO4
1molCaSO4 2H2O
136.0gCaSO4
1molCaSO4
13.44g
13.4 g
What happened to the difference (17.0 - 13.4) = 3.6 grams of material?
Lost as water vapor
2. 15.00 grams of the hydrate
Na2SO4 • XH2O(s) was heated to remove the water. After heating, 7.95 grams of material remained. What is the formula of the hydrate: (find the value of X)
A) Na2SO4 • H2O
B) Na2SO4 • 2H2O
C) Na2SO4 • 3H2O
D) Na2SO4 • 5H2O
E) Na2SO4 • 7H2O
Na2SO4 • XH2O(s) Na2SO4(s) + XH2O 15.00 7.95 15.00-7.95
7.05Find moles of both products and compare
molNa 2SO4 (7.95g) 1mol
142g
0.0559 mol
molH2O(7.05g) 1mol
18.0g
0.391mol
/ 0.0559 = 1
/ 0.0559 = 6.99 = 7
So, X = 7, formula = Na2SO4 • 7H2O
Limiting Reactant
• Limiting Reactant is that element or compound that determines the amount of product that you get.
• It is the reactant that is used up
You are the owner of a bike shop. A shipment came in with 183 frames, 150 seats, 252 pedals, 131 brake assemblies. How many bikes can you sell? (enter the number)
Each bike needs, 1 frame, 1 seat, 2 pedals, 1 brake.
Given: 183 frames
150 seats
252 pedals
131 brake assemblies
Pedals will allow you to make only 126 bikes so that is the limiting reactant
NH3 + O2 N2 + H2O
Balance
NH3 + O2 N2 + H2O
NH3 + O2 N2 + H2O
2 33/2
4 63 2
4 63 2 NH3 + O2 N2 + H2O
Remember: Stoichiometry is mole ratios
How many moles of N2 can be formed from 4 mol NH3 and 4 mol O2?
1. Determine the LR.
4 molNH3
3molO2
4 molNH3
3molO2 required
Mol O2 given > mol O2 required. So, NH3 is LR
2. Determine amount of product
4 molNH3
2 molN2
4 molNH3
2 molN2 produced
4 63 2 NH3 + O2 N2 + H2O
How many moles of N2 can be formed from 6 mol NH3 and 4 mol O2?
1. Determine the LR.
6 molNH3
3molO2
4 molNH3
4.5 molO2 required
Mol O2 given < mol O2 required. So, O2 is LR
2. Determine amount of product
4 molO2
2 molN2
3molO2
2.67 molN2 produced
4 63 2 NH3 + O2 N2 + H2O
How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2?
A) 5B) 4C) 3.75D) 2.67E) 2.5
4 63 2 NH3 + O2 N2 + H2O
How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2?
1. Determine the LR.
5 molNH3
3molO2
4 molNH3
3.75 molO2 required
Mol O2 given > mol O2 required. So, NH3 is LR
2. Determine amount of product
5 molNH3
2 molN2
4 molNH3
2.5 molN2 produced
If 10.0 grams each of NH3 and O2 are reacted, how many grams of water and N2 are formed?
1. Find moles of each reactant.
2. Determine the Limiting Reactant Determine mole of O2 needed
Compare to what was given: 0.442 mol required > 0.312 mol given
So, O2 = LR
4 63 2 NH3 + O2 N2 + H2O
10.0gNH3
1molNH3
17.0g
0.589 molNH3
10.0gO2
1molO2
32.0g
0.312 molO2
0.589molNH3
3molO2
4 mol NH3
0.442 molO2 required
3. Determine the amount of product based on the LR
total mass of products = 17.0 g
What happened to conservation of mass?
3.0 g un-reacted NH3
0.312molO2
6 molH2O
3 mol O2
18.0 gH2O
1 mol H2O
11.2 gH2O
0.312 molO2
2 molN2
3 mol O2
28.0 gN2
1 mol N2
5.8gN2
10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed?
Cr = 52.0 S = 32.0
A) 20.9 g
B) 20.0 g
C) 19.2 g
D) 3.12 g
E) 0.80 g
10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed?
Cr = 52.0 S = 32.0
1. Write and balance the equation:
2Cr + 3S --> Cr2S3
2. Determine moles of each
10.0gCr
1molCr
52.0g
0.192 molCr
10.0gS
1molS
32.0g
0.313molS
3. Determine LR
4. Determine the amount of product based on the LR
0.192 molCr
3molS
2 mol Cr
0.288 molS required
0.288 molS required < 0.313 molS given
so : Cr LR
0.192 mol Cr
1molCr2S3
2 mol Cr
200.0gCr2S3
1molCr2S3
19.2gCr2S3
• Determination of an unknown element
• A metal oxide has the formula XO3 and reacts with H2 to form free metal X and H2O. If 15.99 grams XO3 yields 6.00 grams H2O, what element is X?
A) Nd
B) Ti
C) S
D) Mo
E) H
• Determination of an unknown element
• A metal oxide has the formula XO3 and reacts with H2 to form free metal X and H2O. If 15.99 grams XO3 yields 6.00 grams H2O, what element is X?
XO3 + H2 --> X + H2O3 3
Find moles XO3 from stoichiometry and moles of H2O
6.00gH2O
1molH2O
18.0g
1molXO3
3molH2O
0.111 molXO3
Determine MW XO3
15.99g XO3
0.111mol
? g
1mol
?143.9
AW X = 143.9 - 3(16.0) = 95.9 = MO
Theoretical Yield
• Theoretical Yield– Maximum amount of product that can
obtained if all the reactant is converted to product
• Actual Yield– Actual amount of product obtained
• % Yield
%Yield ActualYield
TheoreticalYieldx100
%Yield AY
TYx100
One hundred grams of potassium chlorate was heated. What is the final state of affairs?
K = 39.1 Cl = 35.45 O = 16.00
One hundred grams of potassium chlorate was heated. What is the final state of affairs?
K = 39.1 Cl = 35.45 O = 16.00
1. Determine the formula for potassium chlorate
A. KClO B. KClO2 C. KClO3 D. KClO4
One hundred grams of potassium chlorate was heated. What is the final state of affairs?
K = 39.1 Cl = 35.45 O = 16.00
1. Determine the formula for potassium chlorate
A. KClO B. KClO2 C. KClO3 D. KClO4
2. Write and Balance the reaction
3. Determine Theoretical Yields of Products
KClO3 --> KCl + O2 2 2 3
KClO3 --> KCl + O2 2 2 3
100. grams Convert to moles-mole ratio-convert to g
100.g KClO3 1molKClO3
122.5g
3molO2
2 molKClO3
32.0gO2
1molO2
39.2gO2
100.g KClO3 1molKClO3
122.5g
2 molKCl
2 molKClO3
74.5 g KCl
1molKCl
60.8g KCl
Or: Conservation of mass: 100.0 g - 39.2 g = 60.8 g KCl
Part B. If the reaction only produced 50.37 g KCl, 1) What is the %Yield and 2) how many grams of O2 2) were produced.
1) %Yield AY
TYx100
%Yield 50.37
60.8x100 =82.8%
2) gO2 %Yield
100x TY
gO2 82.8%
100x 39.2= 32.4 gO2
Consecutive Reactions
• Consecutive Reactions– Sequence of reactions (steps) that are
required to reach desired products
FeS2 + O2 --> Fe2O3 + SO2
SO2 + O2 --> SO3
SO3 + H2O --> H2SO4
FeS2 + O2 --> Fe2O3 + SO2
SO2 + O2 --> SO3
SO3 + H2O --> H2SO4
4 11 2 8
2 2
4 11 2 8
8 8
4(
4
8(
8 8 8
4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4
Now you can have any kind of problem
4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4
How many moles of sulfuric acid can be made from 5 mol FeS2 and 17 mol O2?
Have LR problem.
1) Determine the LR
5 molFeS2 15 molO2
4 molFeS2
18.75 molO2 required
Mol O2 given < mol O2 required so O2 LR
2) Determine the amount of product
17 molO2 8 molH2SO4
15 molO2
9.1molH2SO4
Combustion Analysis
• Combustion analysis– Reaction of a compound that contains
carbon, hydrogen and sometimes oxygen burned in air (O2) to produce CO2 and H2O
• Can be used to determine empirical formula and % composition
A 0.3000 gram sample containing only carbon, hydrogen and oxygen was burned in air to produce 0.440 grams CO2 and 0.180 grams H2O. Determine the empirical formula of the sample.
CxHyOz + O2 --> CO2 + H2Ox
y
2?
Determine moles of C and H from moles of CO2 and H2O
molC0.440gCO2 1molCO2
44.0g
1molC
1molCO2
0.01molC
molH0.180gH2O 1molH2O
18.0g
2 molH
1molH2O
0.02molH
Determine moles of O from g of O in sampleg O = g sample - g C - g Hg O = 0.3000 -(0.01 mol x 12.0 g/mol) - (0.02 mol x 1.01g/mol)g O = 0.16 g mol O = 0.16g/ 16.0 g/mol = 0.01 mol
Mol C = 0.01
Mol H = 0.02
Mol O = 0.01
/0.01 = 1
/0.01 = 2
/0.01 = 1
Empirical formula = CH2O
• A) impossible
• B) Very hard
• C) able to be done
• D) un understandable
• E) hard but workable
How will the exam on Thursday be?