Chapter 6: Chemical Composition

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+4 Wood Boards 6 Nails 1 Fence Panel

+4 DozenWood Boards

6 DozenNails

1 DozenFence Panels

For a 12 panels fence…

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+4C 3H2 C4H6

+4 moles ofC

3 moles ofH2

1 mole ofC4H6

For a mole of 1,3-Butadiene …

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

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)

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)2Atoms Atomic Weights

1 Ca 1 X 40.078 40.078

2 N 2 X 14.007 28.014

6 O 6 X 15.999 95.994

Total: 164.086 g/mol

• 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 Ag moles Ag

atoms Ag10022.6Ag mole 1

23

• 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 H2OFind: ? g H2OC F: 1 mole H2O = 18.02 g H2O

mol H2O g H2O

OH mol 1OH 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 O1 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 1C g 0112.

OHC g 50.21OHC mol 1

1410

1410

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

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).

molC10H14O

molC

gC

OHC mol 1C 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%wholepartPercentage

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 100Na g 39 NaCl g NaCl g

Na g 39NaCl 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