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Chapter 7: Chemical Formula Relationships
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+4 Wood Boards 6 Nails 1 Fence Panel
+4 Dozen
Wood Boards6 Dozen
Nails1 Dozen
Fence Panels
For a 12 panels fence…
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Chapter 7: Chemical Formula Relationships
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+4C 3H2 C4H6
+4 moles of
C3 moles of
H2
1 mole ofC4H6
For a mole of 1,3-Butadiene …
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Chapter 7: Chemical Formula Relationships
1 Dozen = 12 items
• The mole is defined (since 1960) as the amount of substance of a system that contains as many entities as there are atoms in 12 g of carbon-12.• Symbol: mol.• Coined by Wilhelm Ostwald in 1893
1 mol = 12 g of carbon-12
1 mole = 6.0221415×1023 items
# of Molecules = # of moles X 6.022×1023
• A mole of carbon contains 6.0221415×1023 atoms of carbon, but the same is true for anyother element or molecule; in general:
• 6.0221415×1023 is the Avogadro’s Number
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Chapter 7: Chemical Formula Relationships
m = number of nails X mass of 1 nailExample:500g = 100 nails X 5g
m = number of moles X mass of 1 mole of the substance
m = n X M.M.
Where:m = mass (g)n = number of moles (mol)M.M. = Molecular Mass (g/mol) = mass of 1 mole of the substance
Sometime the Molecular Mass is measured in Daltons (1Da = 1g/mol)
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Chapter 7: Chemical Formula Relationships
The molar mass of an element is numerically the same as the atomic weight of the element. They are not however the same; they have different units:• The atomic weight is defined as one twelfth of the mass of an isolatedatom of carbon-12 and is therefore dimensionless• The molar mass is measured in g/mol.
The molar mass of a compound is given by the sum of the atomic weights of the atoms which form the compound.
Example: molar mass of Ca(NO3)2
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• Write a Solution Map for converting the units :
InformationGiven: 1.1 x 1022 Ag atomsFind: ? molesConv. Fact.: 1 mole = 6.022 x 1023
Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
atoms Agatoms Ag moles Agmoles Ag
atoms Ag10022.6
Ag mole 123
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• Check the Solution:
1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag
The units of the answer, moles, are correct.The magnitude of the answer makes sense
since 1.1 x 1022 is less than 1 mole.
InformationGiven: 1.1 x 1022 Ag atomsFind: ? molesConv. Fact.: 1 mole = 6.022 x 1023
Sol’n Map: atoms mole
Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
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• Write a Solution Map for converting the units :
InformationGiven: 1.75 mol H2O
Find: ? g H2O
C F: 1 mole H2O = 18.02 g H2O
mol H2Omol H2O g H2Og H2O
OH mol 1
OH g 18.02
2
2
Example:Calculate the mass (in grams) of 1.75 mol of water
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• Check the Solution:
1.75 mol H2O = 31.5 g H2O
The units of the answer, g, are correct.The magnitude of the answer makes sense
since 31.5 g is more than 1 mole.
InformationGiven: 1.75 mol H2OFind: ? g H2OC F: 1 mole H2O = 18.02 g H2OSol’n Map: mol g
Example:Calculate the mass (in grams) of 1.75 mol of water
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Chemical Formulas as Conversion Factors
• 1 spider 8 legs• 1 chair 4 legs• 1 H2O molecule 2 H atoms 1 O atom
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Mole Relationships inChemical Formulas
• since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound
Moles of Compound Moles of Constituents1 mol NaCl 1 mole Na, 1 mole Cl1 mol H2O 2 mol H, 1 mole O
1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O
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Example:• Carvone, (C10H14O), is the main component in spearmint
oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquers, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.
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• Write a Solution Map for converting the units :
gC10H14O
C mol 1
C g 0112.
OHC g 50.21
OHC mol 1
1410
1410
InformationGiven: 55.4 g C10H14O
Find: g CCF: 1 mol C10H14O = 150.2 g
1 mol C10H14O 10 mol C
1 mol C = 12.01 g
Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
molC10H14O
molC
gC
OHC mol 1
C mol 10
1410
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• Check the Solution:55.4 g C10H14O = 44.3 g C
The units of the answer, g C, are correct.The magnitude of the answer makes sense since
the amount of C is less than the amount of C10H14O.
InformationGiven: 55.4 g C10H14OFind: g CCF: 1 mol C10H14O = 150.2 g
1 mol C10H14O 10 mol C 1 mol C = 12.01 g
SM: g C10H14O mol C10H14O mol C g C
Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).
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Percent Composition• Percentage of each element in a compound– By mass
• Can be determined from 1. the formula of the compound2. the experimental mass analysis of the
compound• The percentages may not always total to 100%
due to rounding
100%wholepart
Percentage
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Mass Percent as a Conversion Factor
• the mass percent tells you the mass of a constituent element in 100 g of the compound– the fact that NaCl is 39% Na by mass means that
100 g of NaCl contains 39 g Na
• this can be used as a conversion factor– 100 g NaCl 39 g Na
Na g NaCl g 100
Na g 39 NaCl g NaCl g
Na g 39
NaCl g 100 Na g
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Example - Percent Composition from the Formula C2H5OH
1. Determine the mass of each element in 1 mole of the compound
2 moles C = 2(12.01 g) = 24.02 g6 moles H = 6(1.008 g) = 6.048 g1 mol O = 1(16.00 g) = 16.00 g
2. Determine the molar mass of the compound by adding the masses of the elements
1 mole C2H5OH = 46.07 g
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Sample - Percent Composition from the Formula C2H5OH
3. Divide the mass of each element by the molar mass of the compound and multiply by 100%
52.14%C100%46.07g24.02g
13.13%H100%46.07g6.048g
34.73%O100%46.07g16.00g
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Empirical FormulasHydrogen PeroxideMolecular Formula = H2O2
Empirical Formula = HOBenzeneMolecular Formula = C6H6
Empirical Formula = CHGlucoseMolecular Formula = C6H12O6
Empirical Formula = CH2O
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Finding an Empirical Formula1) convert the percentages to grams
a) skip if already grams2) convert grams to moles
a) use molar mass of each element3) write a pseudoformula using moles as subscripts4) divide all by smallest number of moles5) multiply all mole ratios by number to make all whole
numbersa) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply
all by 3, etc. b) skip if already whole numbers
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All these molecules have the same Empirical Formula. How are the
molecules different?Name Molecular
FormulaEmpiricalFormula
glyceraldehyde C3H6O3 CH2O
erythrose C4H8O4 CH2O
arabinose C5H10O5 CH2O
glucose C6H12O6 CH2O
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All these molecules have the same Empirical Formula. How are the
molecules different?Name Molecular
FormulaEmpiricalFormula
MolarMass, g
glyceraldehyde C3H6O3 CH2O 90
erythrose C4H8O3 CH2O 120
arabinose C5H10O5 CH2O 150
glucose C6H12O6 CH2O 180
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Molecular Formulas
• The molecular formula is a multiple of the empirical formula
• To determine the molecular formula you need to know the empirical formula and the molar mass of the compound