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…
Chapter 7: Chemical Formula Relationships
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+4C 3H2 C4H6
+4 moles of
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1 mole ofC4H6
For a mole of 1,3-Butadiene …
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
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)
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
• 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
• 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?
• 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
• 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
Chemical Formulas as Conversion Factors
• 1 spider 8 legs• 1 chair 4 legs• 1 H2O molecule 2 H atoms 1 O atom
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
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.
• 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
• 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).
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
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
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
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
Empirical FormulasHydrogen PeroxideMolecular Formula = H2O2
Empirical Formula = HOBenzeneMolecular Formula = C6H6
Empirical Formula = CHGlucoseMolecular Formula = C6H12O6
Empirical Formula = CH2O
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
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
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
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