Ch. 6: Chemical CompositionCh. 6: Chemical Composition

Dr. Namphol Sinkaset

Chem 152: Introduction to General Chemistry

I. Chapter OutlineI. Chapter Outline

I. Introduction

II. Counting by Weighing

III. Chemical Formulas as Conversion Factors

IV. Mass Percent Composition

V. Finding Formulas from Mass Data

I. IntroductionI. Introduction

• The FDA RDA for Na = 2.4 g.

• Na usually consumed as NaCl.

• But 2.4 g NaCl ≠ 2.4 g Na.

• How do we figure out a problem like this?

II. Selling NailsII. Selling Nails

• Nails can be sold two ways: as individual units or by the pound.

• By the pound much more convenient if want to buy hundreds of nails.

• Analogy extends to atoms/molecules.

II. Sample ProblemII. Sample Problem

• If a dozen large nails weighs 0.275 lb., how many nails are contained in 5.5 lb?

II. Counting AtomsII. Counting Atoms

• Finding the number of atoms in a sample of matter is similar to selling nails by the pound.

• We used a dozen as a unit of count, but 12 is way too small for atom.

• In chemistry, the mole is used instead – it’s the chemist’s “dozen.”

II. The MoleII. The Mole• The mole is the SI unit for amount and

is defined as the amount of material containing 6.0221421 x 1023

particles.

• This # is known as Avogadro’s number.

• This # is defined by the number of atoms in exactly 12 g of carbon-12.

• One mole of atoms, ions, or molecules generally makes up a macroscopically meaningful size.

II. How Much is a Mole?II. How Much is a Mole?

II. Avogadro’s NumberII. Avogadro’s Number

• Every time you hear “mole,” you should think 6.022 x 1023.

• We can use Avogadro’s number as a conversion factor to calculate # of atoms.

6.022 x 1023 atoms

1 mole atoms 6.022 x 1023 atoms

1 mole atomsOR

II. Sample ProblemII. Sample Problem

• How many atoms of gold are in a pure gold ring containing 8.83 x 10-2 moles of Au?

II. Grams and MolesII. Grams and Moles• In the nail example, we had a

relationship between a dozen nails and a mass.

• We need the same relationship for the mole.

• The mass of 1 mole of atoms of an element is its molar mass. The value of an element’s molar mass in g/mole is numerically equal to the element’s atomic mass in amu (from periodic table).

II. Mole of Atoms & CompoundsII. Mole of Atoms & Compounds

• 1 Fe atom weighs 55.85 amu, so 1 mole of Fe atoms weighs 55.85 g.

• 1 O atom weighs 16.00 amu, so 1 mole of O atoms weighs 16.00 g.

• Same applies for compounds. 1 molecule H2O weighs 18.02 amu, so 1

mole H2O weighs 18.02 g.

II. Sample ProblemII. Sample Problem

• Calculate the molar mass of calcium sulfate.

II. The Mass of a Mole II. The Mass of a Mole Depends on Unit SizeDepends on Unit Size

II. Mole EquivalenciesII. Mole Equivalencies

• Since mass and moles are related, we can set up conversion factors.

• e.g. 1 mole Fe weighs 55.85 g.

55.85 g Fe

1 mole Fe 55.85 g Fe

1 mole FeOR

II. Sample ProblemII. Sample Problem

• Graphite, a crystalline form of carbon, is used in pencils. How many moles of carbon are in 0.315 g of graphite?

II. Sample ProblemII. Sample Problem

• How many grams of sulfur are in 2.78 moles of sulfur?

II. Sample ProblemII. Sample Problem

• How many moles of NO2 are in 1.18 g of NO2?

II. Sample ProblemII. Sample Problem

• How many molecules of H2O are in a sample of water with a mass of 3.64 g?

II. Sample ProblemII. Sample Problem

• If a sample of molecular bromine weighs 2100 g, how many molecules of molecular bromine are in the sample?

III. Inherent Conversion FactorsIII. Inherent Conversion Factors

• Any object that can be broken down into parts or pieces has an inherent conversion factor.

• For example, a clover.

III. 3 Leaves : 1 CloverIII. 3 Leaves : 1 Clover

• Since there are 3 leaves in one clover, we can write a conversion factor between them.

III. Other ExamplesIII. Other Examples

III. Scaling UpIII. Scaling Up

• The ratios hold as long as we keep the unit the same for both the parts and the whole.

• 3 leaves : 1 clover 3 dozen leaves : 1 dozen clovers.

• In chemistry, we scale up to the mole.

• 2 H atoms : 1 H2O molecule 2 moles H atoms : 1 mole H2O molecules.

III. Breaking Down CClIII. Breaking Down CCl44

III. Sample ProblemIII. Sample Problem

• List all the possible atom : formula unit mole relationships in barium nitrate.

III. Mole RelationshipsIII. Mole Relationships

• A chemical formula gives the relationships between moles of substances, NOT grams.

• Always convert to moles then use a mole relationship to get to the substance of interest.

III. Sample ProblemIII. Sample Problem

• Determine the number of moles of oxygen in 2.45 moles of nickel(II) phosphate.

III. Sample ProblemIII. Sample Problem

• Calculate the mass of carbon in 25.0 g of C6H12O6.

III. Sample ProblemIII. Sample Problem

• How many grams of sodium chloride contains 2.00 grams of sodium?

IV. Mass PercentIV. Mass Percent

• The mass percent expresses an element’s percentage of the total mass of the compound.

IV. Sample ProblemIV. Sample Problem

• Calculate the mass percent of nitrogen in ammonium nitrate.

IV. Mass Percent as a IV. Mass Percent as a Conversion FactorConversion Factor

• The key to using mass percents is realizing that they can be expressed as conversion factors.

• NaCl is 39% Na can be expressed as 39 g Na : 100 g NaCl.

39 g Na

100 g NaCl

100 g NaCl

39 g Na

IV. Sample ProblemIV. Sample Problem

• If CCl2F2 is 58.64% Cl by mass, how many grams of CCl2F2 contains 22 g of Cl?

V. Finding Formulas from V. Finding Formulas from Mass DataMass Data

• Given elemental mass % data of a compound, it’s possible to find the formula of the compound.

• Elemental analysis is a common test performed on newly synthesized compounds.

V. Empirical vs. MolecularV. Empirical vs. Molecular

• Often experiments will yield the relative masses of each element in a compound.

• With just the relative masses, the best we can do is find the empirical formula.

• Additional data is needed to go from an empirical formula to a molecular formula.

V. The Key to Finding V. The Key to Finding Empirical FormulasEmpirical Formulas

• In any formula, subscripts represent the number of atoms of a given element in that compound.

• However, these ratios scale up!

• Therefore, the subscripts can also be interpreted as mole ratios between atoms of a given element in the compound.

V. Finding Empirical FormulasV. Finding Empirical Formulas

• To find an empirical formula, you must calculate the number of moles of each atom present in a certain sample of the compound.

• These moles become the temporary subscripts in the formula.

• You then use math to convert to whole numbers.

V. Sample ProblemV. Sample Problem

• Calculate the empirical formula of ethyl butyrate (pineapple oil) if its mass composition is 62.04% C, 10.41% H, and 27.55% O.

V. Sample ProblemV. Sample Problem

• A certain compound is 50.66% C, 4.25% H, and 45.09% S by mass. Determine the empirical formula.

V. Determining Molecular V. Determining Molecular FormulasFormulas

• To find a molecular formula from an empirical formula, you must know the molecular molar mass.

• The molecular molar mass is always a multiple of the empirical molar mass.

• To find the multiple, we divide the molecular molar mass by the empirical molar mass.

V. Sample ProblemV. Sample Problem

• Butane has an empirical formula of C2H5. If its molar mass is 58.12 g/mole, determine the molecular formula of butane.

V. Sample ProblemV. Sample Problem

• The carcinogen benzo[a]pyrene (MW = 252.30 g/mole) is found to be 95.21% C and 4.79% H by mass. What is its molecular formula?