(l10)stoichiometry part1
TRANSCRIPT
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Stoichiometry
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The relationship between the number of reactant and product molecules in a chemical equation.
Quantitative
Stoichiometry
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Atomic Weights
Weighted average of the masses of the constituent isotopes if an element. Tells us the atomic masses
of every known element. Lower number on periodic
chart.
How do we know what the values of these numbers are?
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The Mole
A number of atoms, ions, or molecules that is large enough to see and handle.A mole = number of things Just like a dozen = 12 things One mole = 6.022 x 1023 things
Avogadro’s number = 6.022 x 1023 Symbol for Avogadro’s number is NA.
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The Mole
How do we know when we have a mole? count it out weigh it out
Molar mass - mass in grams numerically equal to the atomic weight of the element in grams.H has an atomic weight of 1.00794 g 1.00794 g of H atoms = 6.022 x 1023 H
atoms
Mg has an atomic weight of 24.3050 g 24.3050 g of Mg atoms = 6.022 x 1023 Mg
atoms
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The Mole
Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.
Mgg ?
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The Mole
Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.
atom Mg1 Mgg ?
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The Mole
Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.
atoms Mg 106.022
atoms Mg mol 1atom Mg 1Mg g ?
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The Mole
Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures.
Mg g 104.04atoms Mg mol 1
Mgg24.30
atoms Mg 106.022
atoms Mg mol 1atom Mg 1Mg g ?
23
23
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The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
atoms Mg ?
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The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
Mg mol 1.67atoms Mg ?
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The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
Mg mol 1
atoms Mg 106.022 Mg mol 1.67atoms Mg ?
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The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
atoms Mg 101.00
Mg mol 1
atoms Mg 106.022 Mg mol 1.67atoms Mg ?
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23
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The Mole
Example 2-3. How many atoms are contained in 1.67 moles of Mg?
atoms Mg101.00
Mgmol 1
atoms Mg106.022Mg mol 1.67atoms Mg?
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The Mole
Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?
You do it!
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The Mole
Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?
Mg g 4.73Mg mol ?
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The Mole
Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?
Mg g 24.30
atoms Mg mol 1 Mg g 4.73Mg mol ?
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The Mole
Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?
Mg mol 02.3
Mg g 24.30
atoms Mg mol 1 Mg g 4.73Mg mol ?
IT IS IMPERATIVE THAT YOU KNOWHOW TO DO THESE PROBLEMS
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Formula Weights, Molecular Weights, and Moles
How do we calculate the molar mass of a compound? add atomic weights of each atom
The molar mass of propane, C3H8, is:
amu 44.11 mass Molar
amu 8.08 amu 1.01 8H 8
amu 36.03amu 12.01 3C 3
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Formula Weights, Molecular Weights, and Moles
The molar mass of calcium nitrate, Ca(NO3)2 , is:
You do it!
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Formula Weights, Molecular Weights, and Moles
amu 164.10 massMolar
amu 96.00 amu 16.006 O6
amu 28.02 amu 14.012 N2
amu 40.08 amu 40.081 Ca1
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Formula Weights, Molecular Weights, and Moles
One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2
molecules
2(6.022 x 1023 ) Cl atoms
C3H8 You do it!
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Formula Weights, Molecular Weights, and Moles
One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2
molecules 2(6.022 x 1023 ) Cl
atoms C3H8 or 44.11 g 6.022 x 1023 C3H8
molecules 3 (6.022 x 1023 ) C
atoms 8 (6.022 x 1023 ) H
atoms
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Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 8383 HC g 74.6molecules HC ?
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Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.
83
83
8383
HC g 44.11
HC mole 1
HC g 74.6molecules HC ?
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Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.
83
8323
83
83
8383
HC g 44.11
molecules HC 106.022
HC g 44.11
HC mole 1
HC g 74.6molecules HC ?
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Formula Weights, Molecular Weights, and Moles
Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.
molecules 10 02.1
HC g 44.11
molecules HC 106.022
HC g 44.11
HC mole 1
HC g 74.6molecules HC ?
24
83
8323
83
83
8383
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Formula Weights, Molecular Weights, and Moles
Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.
You do it!
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Formula Weights, Molecular Weights, and Moles
Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.
atoms O 106.49
COLi unit formula 1
atoms O 3
COLi mol 1
COLi unitsform.106.022
COLi g 73.8
COLi mol 1COLi g 26.5atoms O ?
23
3232
3223
32
3232
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Formula Weights, Molecular Weights, and Moles
Occasionally, we will use millimoles. Symbol - mmol 1000 mmol = 1 mol
For example: oxalic acid (COOH)2 1 mol = 90.04 g 1 mmol = 0.09004 g or 90.04 mg
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Formula Weights, Molecular Weights, and Moles
Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2.
You do it!
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Formula Weights, Molecular Weights, and Moles
Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2.
22
2
22
(COOH) mmol 2.60(COOH) g 0.09004
(COOH) mmol 1
(COOH) g 234.0(COOH) mmol ?
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Percent Composition and Formulas of Compounds
% composition = mass of an individual element in a compound divided by the total mass of the compound x 100%Determine the percent composition of C in C3H8.
81.68%
100%g 44.11
g 12.013
100%HC mass
C mass C %
83
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Percent Composition and Formulas of Compounds
What is the percent composition of H in C3H8?
You do it!
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Percent Composition and Formulas of Compounds
What is the percent composition of H in C3H8?
81.68%100%18.32%
or
%18.32100%g 44.11
g 1.018
100%HC
H8
100%HC mass
H massH %
83
83
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Percent Composition and Formulas of Compounds
Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures.
You do it!
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Percent Composition and Formulas of Compounds
Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 sig. fig.
100% Total
O 48.0% 100%g 399.9
g 16.012 100%
)(SOFe
O12 O %
S 24.1% 100%g 399.9
g 32.13 100%
)(SOFe
S3 S %
Fe 27.9% 100%g 399.9
g 55.82 100%
)(SOFe
Fe2Fe %
342
342
342
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Derivation of Formulas from Elemental Composition
Empirical Formula - smallest whole-number ratio of atoms present in a compound
CH2 is the empirical formula for alkenes No alkene exists that has 1 C and 2 H’s
Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound
Ethene – C2H4
Pentene – C5H10
We determine the empirical and molecular formulas of a compound from the percent composition of the compound.
percent composition is determined experimentally
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Derivation of Formulas from Elemental Composition
Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula?Make the simplifying assumption that we have 100.0 g of compound. In 100.0 g of compound there are: 24.74 g of K 34.76 g of Mn 40.50 g of O
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Derivation of Formulas from Elemental Composition
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
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Derivation of Formulas from Elemental Composition
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
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Derivation of Formulas from Elemental Composition
rationumber wholesmallest obtain
O mol 2.531O g 16.00
O mol1O g 40.50 O mol ?
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
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Derivation of Formulas from Elemental Composition
Mn 10.6327
0.6327Mnfor K 1
0.6327
0.6327Kfor
rationumber wholesmallest obtain
O mol 2.531O g 16.00
O mol1O g 40.50 O mol ?
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
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Derivation of Formulas from Elemental Composition
4KMnO is formula chemical thethus
O 40.6327
2.531Ofor
Mn 10.6327
0.6327Mnfor K 1
0.6327
0.6327Kfor
rationumber wholesmallest obtain
O mol 2.531O g 16.00
O mol1O g 40.50 O mol ?
Mn mol 0.6327Mn g 54.94
Mn mol 1Mn g 34.76 Mn mol ?
K mol 0.6327K g 39.10
K mol 1K g 24.74 K mol ?
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Derivation of Formulas from Elemental Composition
Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?
You do it!
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Derivation of Formulas from Elemental Composition
Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?
ratio number wholesmallest find
O mol 0.1480O g 16.00
O mol1O g 2.368O mol ?
Co mol 0.1110Cog 58.93
Co mol 1Co g 6.541Co mol ?
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Derivation of Formulas from Elemental Composition
Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?
43OCo
:is formula scompound' theThus
O 43O 1.333 Co 33Co 1
number wholetofraction turn to 3 by both multipy
O1.3330.1110
0.1480O for Co 1
0.1110
0.1110Co for
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Determination of Molecular Formulas
Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? short cut method
H of g 8.100.1437g 56.1
C of g 48.00.8563g 56.1
H is 14.37% and C is 85.63%
g 56.1 contains mol 1
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Determination of Molecular Formulas
84HC
:is formula theThus
H mol 8H g 1.01
H mol 1H of g 8.10
C mol 4C g 12.0
C mol 1C of g 48.0
moles tomassesconvert
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Some Other Interpretations of Chemical Formulas
Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?
N mol 1.07N g 14.0
N mol 1N of g 15.0N mol ?
g/mol 149.0PO)(NH of massmolar 434
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Some Other Interpretations of Chemical Formulas
Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?
434434
434
PO)(NH mol 0.357N mol 3
PO)(NH mol 1N mol 1.07
N mol 1.07N g 14.0
N mol 1N of g 15.0N mol ?
g/mol 149.0PO)(NH of massmolar
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Some Other Interpretations of Chemical Formulas
Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?
434434
434434
434434
434
PO)(NH g 53.2PO)(NH mol 1
PO)(NH g 149.0PO)(NH mol 0.357
PO)(NH mol 0.357N mol 3
PO)(NH mol 1N mol 1.07
N mol 1.07N g 14.0
N mol 1N of g 15.0N mol ?
g/mol 149.0PO)(NH of massmolar
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Purity of Samples
The percent purity of a sample of a substance is always represented as
impurities includes sample of mass
%100sample of mass
substance pure of mass =purity %
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Purity of SamplesExample 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4?
sample g 100.0
PONa g 98.3factor unit 43
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Purity of SamplesExample 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4?
43
4343
43
PONa g 246=
sample g 100.0
PONa g 98.3sample g 0.250PONa g?
sample g 100.0
PONa g 98.3factor unit
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Purity of SamplesExample 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4?
impurities g 4.00 =
PONa g 246 - sample g 250.0 = impurities g?
PONa g 246=
sample g 100.0
PONa g 98.3sample g 0.250PONa g?
sample g 100.0
PONa g 98.3factor unit
43
43
4343
43