Chapter 3 Stoichiometry Part II
Percent Composition from Formulas
First you calculate the molar mass of the compound/
molecule, which represent the whole of your percentage.
Then you take the mass of each individual atom that it contributes to the whole molar mass, which would represent the part, and finally divide that by the mass of the compound/molecule You multiply by 100 to create the %age then you are done!
Part/Whole x 100 = percent composition Same Problem using Dimensional Analysis (Your book uses) Find the % composition of lead(II) chloride.
First, write the formula: PbCl2
Mass of Pb = 1 mol Pb x 207.19 g= 207.19 g
1 mol Pb
Mass of Cl = 2 mol Cl x 35.45g = 70.9 g
1 mol Cl
1 mole of PbCl2 = (207.19 + 70.9) = 278.09 g/mol
Pb = 207.19 g/mol x 100 = 74.50%
278.09 g/mol
Cl = 70.9 g/mol
x 100 = 25.5%
278.09 g/mol
Example:
Find the % composition of lead(II) chloride.
First, write the formula: PbCl2
Molar mass = 207.19 g/mol + (2 x 35.45 g/mol) = 278.09
g/mol
Pb = 207.19 g/mol x 100 = 74.50%
278.09 g/mol
Cl = 70.9 g/mol
x 100 = 25.5%
278.09 g/mol
3.6 Determing the Formula of A Compound From % Composition
When a new compound is prepared, one of the first items of interest is the formula of the compond.
The percent composition of a compound leads directly to its empirical formula.
An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts.
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Chapter 3 Stoichiometry Part II
Benzoic Acid is a white, crystalline powder used as a food preservative which is 68.8% C, 5.00 % H, and 26.2% O by mass. What is its empirical formula?
For 100.0 g of benzoic acid:
68.8 g C x 1 mol C = 5.73 mol C
12.0 g C
5.00 g H x 1mol H = 4.95 mole H
1.01 g H
26.2 g O x 1 mol O = 1.638 mol O
16.0 g O
divide by 1.638 (its the lowest value)
Empirical formula is C3.5H3O (have to use chart) to come up with C7H6O2
Molecular Formulas
To obtain a molecular formula for a substance we compare the calculated empirical formula mass to the molar mass and use the following equation.
Caffeine, a stimulant found in coffee, tea, and chocalate, contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula for caffeine.
49.48g C x 1 mol C
Molar Mass
= 4.120 mol C
12.01 g C
Empirical formula Mass
The value that you come up with is multiplied by each atom in the empirical formula. The result is your molecular formula
5.15g H
x 1 mol H
= 5.10 mol H
1.01 g H
28.87g N x 1 mol N
= 2.061 mol N
14.01 g
16.49g O x 1 mol O
= 1.031 mol H
16.00 g
3.7 Chemical Equations
C = 4.120 mol/1.031 mol = 4
H = 5.10 mol/1.031 mol = 5
N = 2.061 mol/1.031 mol = 2
O = 1.031 mol/1.031 mol = 1
C 4H 5N 2O
Molecular Formula = Molar mass (194.2 )
= 2
Empirical formula mass (97.10)
Molecular formula = 2(C4H5N2O) = C8H10N4O2
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Chapter 3 Stoichiometry Part II
Information Conveyed by the Balanced Equation for the Combusion of Methane
The Meaning of a Chemical Equation
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Chapter 3 Stoichiometry Part II
More Examples Balancing Formula Equations ___KOH + ___HBr => ___KBr + ___ H 2O
___Al + ___S8 => ___Al 2S3
__ Al(OH)3 + __ H2SO4 __ Al2(SO4)3 + __ H2O
Writing and Balancing Word Equations
The elements H, Br, O, N, Cl, I, and F all appear as diatomic molecules in chemical equations/reactions!
H2 Br2 O2 N2 Cl2 I2 and F2
Way To Help You Remember:
H. BrONCLIF
Harry Bronclif
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Chapter 3 Stoichiometry Part II
Balancing Word Equations
Example:
Silver reacts with hydrogen nitrate to yield silver (I) nitrate, nitrogen (II) oxide and water. Balance the equation for this reaction.
Step 1: In chemical equation form, write down the atoms and compounds involved.
Ag+ + H+ (NO3 )­ Ag+ (NO3 )­ + N2+ O2­ + H2 O
Step 2: Make sure the chemical formulas are the correct ones by using your ion sheet, book or your periodic table.
Another Example:
Solid magnesium hydroxide reacts with chlorine gas to form solid magnesium chloride, Solid magnesium chlorate and liquid water. Balance the equation for this reaction.
Step 1:
Ag + H(NO3 )
Ag(NO3 ) + NO + H2 O
It is determined that the equation’s compounds are correct, so no criss­crossing is required.
Step 3: Balance the equation
3Ag + 4H(NO3 ) 3Ag(NO3 ) + NO + 2H2 O
NOTE: If the states of the reactants and products are given in the question you must indicate the state of matter of each reactant and product in the final balanced equation.
3.9 Stoichiometric Calculations: Amounts of Reactants and Pdts
Mg2+ (OH)­(s) + Cl2(g) Mg2+ (ClO3 )­(s) + Mg2+ Cl­(s) + H2 O(l)
(Remember H. BrONClIF)
Step 2: Mg(OH)2 + Cl2 Mg(ClO3 )2 + MgCl2 + H2 O
Step 3:
6Mg(OH)2 (s) + 6Cl2 (g) Mg(ClO3)2 (s) + 5MgCl2(s) + 6H2O (l)
Problem? What mass of oxygen will react with 96.1 grams of propane (C3H8)? (note: any combusion reaction will produce water and carbon dioxide as it products)
Write the equation and balance it
C3H8 + O2 (bronclif element)
C3H8 + 5O2 96.1g C3H8
3CO2
x 1 mol C3H8
44.10 g C3H8
CO2
96.1g C3H8
1 mol C3H8
x 1 mol C3H8
44.10 g C3H8
+ H2O
In steps,
+ 4H2O
x 5 mol O2
What if, you were asked what the mass of carbon dioxide produced when 96.1 g of propane is combused with oxygen?
x 3 mol CO2 x 44.01g CO2
1 mol C3H8
1 mol CO2
= 288 g CO2
x 32.00g O2
1 mol O2
= 349 g O2
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Chapter 3 Stoichiometry Part II
What is a limiting reactant or reagent?
3.10 Calculations Involving Limiting Reactants
• The reactant used up first is called the limiting reactant, once used up no more product can be formed!!!!
• The substance that is not used up completely in a reaction is called the excess reactant.
• Note: the limited reactant may be referred to as the limiting reagent.
• Reactant A is a test tube. I have 20 of them.
• Reactant B is a stopper. I have 30 of them.
• Product C is a stoppered test tube.
• The reaction is: A + B ­­­> C
• Or: test tube plus stopper gives stoppered test tube.
• So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and we are making stoppered test tubes (the amount goes up).
• Suddenly, we run out of one of the "reactants." We run out of test tubes first. • We had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted.
• So, which "reactant" is limiting and which is in excess?
After balancing the equation, we need to compare the amount of both reactants to the amount of product they would produce. (Pick any product)
18.1g NH3 x 1 mol NH3 x 1 mol N2
90.4 g CuO x
17.03g NH3
2 mol NH3 1 mol CuO
x 1 mol N2
79.55g CuO
= 0.531 mol N2 produced
3 mol CuO
Problem?
Balanced equation
2NH3(g) + 3CuO(s)
90.4 g CuO x
1 mol CuO
79.55g CuO
N2(g)
+ 3Cu(s) + 3H2O(g)
x 1 mol N2
3 mol CuO
x 28.0g N2
1 mol N2
= 10.6 g N2
= 0.379 mol N2 produced
Since CuO produced the least amount of product it is the limiting reagent or reactant!
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Chapter 3 Stoichiometry Part II
If you are asked to find the amount or mass of the excess reactant, you take the mass of the limiter and use the mole ratio between the other reactant (the one in excess) to find moles of NH3 produced and then convert to grams.
90.4 g CuO x
1 mol CuO
x 2 mol NH3
x 17.03g NH3
3 mol CuO
1 mol NH3
79.55g CuO
Percent Yield
= 12.9g NH3 Amount of Excess = (18.1g (starting) ­ 12.9g (consumed) = 5.19g excess NH3
First, figure the the reactant that is limiting product production.
68.6 kg CO x 1000g CO x 1 mol CO x 1 mol CH3OH
1 kg CO
8.60 kg H2 x 1000g H2
1 kg H2
2H2(g)
+ CO(g)
28.02 g CO
x 1 mol H2
2.02 g H2
1 mol CO
= 2.45 x 103 mol CH3OH
x 1 mol CH3OH
2 mol H2
= 1.95 x 103 mol CH3OH
CH3OH(l)
H2 produces the least amount of pdt, so is the limter!
Now figure out the thoerhetical yield by using the limiter.
8.60 kg H2 x 1000g H2 x 1 mol H2 x 1 mol CH3OH x 32.04g CH3OH 1 kg H2
2.02 g H2
2 mol H2
1 mol CH3OH
= 6.26 x 104 g CH3OH
% yield = actual/theo x 100
3.57 x 104g
x 100 = 52.3 %
6.26 x 10 g
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Attachments
AVagadro's Number History
Faraday Constant