StoichiometryRelationships between reactants and products in a chemical reaction

General definitionsGeneral definitions

Element- a pure substance which cannot be Element- a pure substance which cannot be broken down by any ordinary chemical means. broken down by any ordinary chemical means.

Molecule- two or more atoms chemically bonded Molecule- two or more atoms chemically bonded together. Example Otogether. Example O22 , H , H22OO

Compound- two or more elements chemically Compound- two or more elements chemically bonded together. Physical and chemical bonded together. Physical and chemical properties change.properties change.

Solution: homogeneous,Solution: homogeneous,containing 2 or more containing 2 or more substances. No reaction to makesubstances. No reaction to make

Mixtures: Heterogeneous, The substances Mixtures: Heterogeneous, The substances are not evenly distributed. Ex. Flour in are not evenly distributed. Ex. Flour in water, water,

Physical Change: There is no change in Physical Change: There is no change in the chemical formula of the substance. Ex. the chemical formula of the substance. Ex. HH22O O (s) (s) H H22O (l) have same formulas, O (l) have same formulas, chemical and some physical properties.chemical and some physical properties. Color, texture, density, taste, rate of Color, texture, density, taste, rate of

evaporation, boiling and melting points evaporation, boiling and melting points

Chemical Change: In a chemical change there is Chemical Change: In a chemical change there is a change in the arrangement of the atoms, that a change in the arrangement of the atoms, that is, the chemical formula changes. is, the chemical formula changes.

Chemical properties: flammability, temp. at Chemical properties: flammability, temp. at which something ignites, oxidation (rusting), which something ignites, oxidation (rusting), acidity, pH, reducing ability, etc. acidity, pH, reducing ability, etc.

Some indications of chemical changes: heat Some indications of chemical changes: heat produced, change in color or smell, gas or produced, change in color or smell, gas or flames produced, precipitation.flames produced, precipitation.

Classification of MatterClassification of Matter

Matter

Pure Substances Mixtures

Elements CompoundsHomogeneous

MixturesHeterogeneous

Mixtures

Solutions

The MoleThe Mole

What is it?What is it? A unit of measurement which equals 6.0 x A unit of measurement which equals 6.0 x

10102323 of something of something Similar to 1 dozen = 12 of somethingSimilar to 1 dozen = 12 of something

Why do we use it?Why do we use it? Other conventional quantities such as “one Other conventional quantities such as “one

dozen” are too small a numberdozen” are too small a number to represent to represent atomsatoms

Gives us a way to “count” atomsGives us a way to “count” atoms

ex. 1g H = 1mole of H = 6.0 x 10ex. 1g H = 1mole of H = 6.0 x 1023 23 atomsatoms

How the mole was developed.How the mole was developed. The mass of each type of atom was determined using The mass of each type of atom was determined using

a mass spectrometer.a mass spectrometer.

ex. H = 1.66 x 10ex. H = 1.66 x 10-24-24g/atomg/atom The masses of the atoms were compared and ratios The masses of the atoms were compared and ratios

were developed.were developed.

ex. ex. HeHe: : 6.64 x 106.64 x 10-24-24 = 4.0 (amu) = 4.0 (amu) relative mass ratio relative mass ratio

H 1.66 x 10H 1.66 x 10-24-24

Mass SpectrometerMass Spectrometer

Mass spectrometerMass spectrometer

The Mole cont’d.The Mole cont’d.

The ratios were used to make standard The ratios were used to make standard masses.masses.• Made “H” have a mass of 1g. Other elements Made “H” have a mass of 1g. Other elements

masses are determined by multiplying 1g by the masses are determined by multiplying 1g by the ratio.ratio.ex. 1.0 g Hex. 1.0 g H x 4.0 for He (since He is 4 times heavier)x 4.0 for He (since He is 4 times heavier) 4.0g for He4.0g for He

These masses represent These masses represent one moleone mole of the of the element and are A.M.U. or atomic mass units.element and are A.M.U. or atomic mass units.

6.02 x 106.02 x 102323 was determined by: was determined by:

Atomic massAtomic mass = = # of atoms in mass# of atoms in mass

Mass of one atomMass of one atom

ex. ex. 1.00g H1.00g H = 6.02 x 10 = 6.02 x 102323atomsatoms

1.66 x 101.66 x 10-24-24g/atomg/atom

Amedeo AvogadroAmedeo Avogadro

Mole conversionsMole conversions

Chemical FormulasChemical Formulas

Molecular Formula Empirical Formula

C6H12O6 CH2O

-Elements are written in their # of atoms found in nature.-This is often not the lowest ratio.-Hard to determine in the lab.

-Elements are written in the lowest ratio.-Easy to experimentally determine.

Empirical FormulaEmpirical Formula Ex. You are heating a 0.2636g sample of Ni and reacting Ex. You are heating a 0.2636g sample of Ni and reacting

with oxygen. After reacting the nickel with oxygen. After reacting the nickel oxideoxide weighed weighed 0.3354g. Find the empirical formula.0.3354g. Find the empirical formula.

Find the mass of each elementFind the mass of each element .3354g Ni ox..3354g Ni ox.

- .2636g Ni.2636g Ni masses masses0.0718g O 0.0718g O mass mass

Convert to moles.Convert to moles..2636g Ni x .2636g Ni x 1mol Ni1mol Ni = .004491mol Ni = .004491mol Ni

58.69g58.69g.0718g O x .0718g O x 1mol O1mol O = .004490mol O = .004490mol O

16.00g16.00g

Empirical Formula cont’d.Empirical Formula cont’d.

Make ratio – use smallest mole value on the Make ratio – use smallest mole value on the bottom.bottom.

.004491mol Ni.004491mol Ni = = 1mol Ni1mol Ni

.004490mol O 1mol O.004490mol O 1mol O Use the ratio to determine the formula.Use the ratio to determine the formula.

NiONiO

Ex. An Aluminum oxide compound was found to Ex. An Aluminum oxide compound was found to contain 4.151g Al and 3.692g O. What is contain 4.151g Al and 3.692g O. What is Aluminum oxides empirical formula?Aluminum oxides empirical formula?• Determine masses.Determine masses.

4.151g Al, 3.692g O4.151g Al, 3.692g O• 4.151g Al4.151g Al x x 1mol Al1mol Al = 0.1539mol Al = 0.1539mol Al

26.98g26.98g

3.692g O3.692g O x x 1mol O1mol O = 0.2308mol O = 0.2308mol O

11 16.0016.00

RatioRatio0.2308mol O0.2308mol O = = 1.5mol O1.5mol O0.1539mol Al 1mol Al0.1539mol Al 1mol Al

Multiply the ratio by the smallest integer in Multiply the ratio by the smallest integer in order to get whole #s.order to get whole #s.2 x (2 x (1.51.5) = ) = 3mol O3mol O

1 2mol Al1 2mol Al Formula.Formula.

AlAl22OO33

Empirical Formula cont’d.Empirical Formula cont’d.

Ex. A Platinum compound was found to contain by mass: Ex. A Platinum compound was found to contain by mass: 65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl. 65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl. Determine its empirical formula.Determine its empirical formula.

Find the masses.Find the masses.65.02g Pt, 9.34g N, …65.02g Pt, 9.34g N, …

65.02g Pt65.02g Pt x x 1mol Pt1mol Pt = 0.3333m Pt = 0.3333m Pt 195.1g195.1g

9.34g N x 9.34g N x 1mol N1mol N = 0.667m N = 0.667m N 14.014.0

2.02g H x 2.02g H x 1mol H1mol H = 2.00m H = 2.00m H 1.001.00

continuedcontinued

RatiosRatios

NN : : 0.667m0.667m = = 2 N2 N HH = = 2.00m H2.00m H = = 6H 6H

Pt 0.333m 1 PtPt 0.333m 1 Pt Pt 0.3333m Pt Pt 0.3333m Pt 1 1 ClCl = = 0.6667mol C0.6667mol C = = 2.0 Cl2.0 Cl

Pt 0.3333mol Pt 1 PtPt 0.3333mol Pt 1 Pt Formula?Formula?

PtNPtN22HH66ClCl22

Percent MassPercent Mass Calculating the % by mass of an element in a molecular Calculating the % by mass of an element in a molecular

formula/empirical formula.formula/empirical formula. Equation: Equation: Mass of element (in 1mole)Mass of element (in 1mole) x 100% = x 100% =

Molecular massMolecular mass Find the % mass of Chromium in the molecular formula KFind the % mass of Chromium in the molecular formula K22CrCr22OO7.7.

Find molecular mass:Find molecular mass:KK22CrCr22OO77 = 294.2g = 294.2g

Find mass of element in compound.Find mass of element in compound.CrCr22 = 104g = 104g

Substitute.Substitute.Mass of elementMass of element x 100% x 100%Molecular massMolecular mass104g Cr104g Cr x 100 = 35.4% Cr x 100 = 35.4% Cr294.2g294.2g

% mass can be useful for determining the empirical formula of a % mass can be useful for determining the empirical formula of a compound.compound.

DefinitionsDefinitionsex. 2Hex. 2H2(g)2(g) + O + O2(g)2(g) 2H 2H22OO(g)(g) everything on the left of the arrow is a reactant, everything everything on the left of the arrow is a reactant, everything on the right of the arrow is a product.on the right of the arrow is a product.

states of matterstates of matter(g) = gas(g) = gas(l) = liquid(l) = liquid(s) = solid(s) = solid(aq) = aqueous (dissolved in H(aq) = aqueous (dissolved in H22O)O)

Chemical EquationsChemical Equations

symbol can be interpreted as makes, symbol can be interpreted as makes, produces, yields, decomposesproduces, yields, decomposes

materials above and below the materials above and below the arrow:arrow:

MnOMnO22

ex. Hex. H22OO22 H H22O + OO + O22

chemicals = catalystschemicals = catalystse- = electricitye- = electricityΔΔ = heat = heat

Types of Chemical ReactionsTypes of Chemical Reactions Single DisplacementSingle Displacement

CuCu(s)(s) + 2AgNO + 2AgNO3(aq)3(aq) 2Ag 2Ag(s)(s) + Cu(NO + Cu(NO33))2(aq)2(aq)

Double DisplacementDouble Displacement

CoClCoCl2(aq)2(aq) + 2NaOH + 2NaOH(aq)(aq) Co(OH) Co(OH)2(s)2(s) + 2NaCl + 2NaCl(aq)(aq)

DecompositionDecomposition

2H2H22OO22 2H 2H22O + OO + O22

(NH(NH44))22CrCr22OO7(s)7(s) N N2(g)2(g) + Cr + Cr22OO3(s)3(s) + 4H + 4H22OO(g)(g)

Redox (Oxidation-reduction)Redox (Oxidation-reduction)FeFe22OO33 + 3CO + 3CO 2Fe + 3CO 2Fe + 3CO22

SynthesisSynthesis2H2H2(g)2(g) + O + O2(g)2(g) 2H 2H22OO(g)(g)

Law of Conservation of MassLaw of Conservation of Mass

This law states that matter is neither This law states that matter is neither created or destroyed in a chemical created or destroyed in a chemical reaction. The mass used in the beginning reaction. The mass used in the beginning of a reaction equals the mass made at the of a reaction equals the mass made at the end of a reaction.end of a reaction.

Balancing: This is a process by which the Balancing: This is a process by which the written reaction is modified so the number written reaction is modified so the number of atoms on the reactant side equals the of atoms on the reactant side equals the number of atoms on the product side.number of atoms on the product side.

Antoine LavoisierAntoine Lavoisier

Percent YieldPercent Yield

Theoretical and actual yieldTheoretical and actual yield Theoretical yield – maximum amount of product that Theoretical yield – maximum amount of product that

can be produced. Determined by math.can be produced. Determined by math. Actual yield – the “actual” amount of product Actual yield – the “actual” amount of product

produced.produced.

Percent yieldPercent yield Actual yieldActual yield x 100 = % yield x 100 = % yield

Theoretical yieldTheoretical yield

Problems with % yield. (Either the % yield or the Problems with % yield. (Either the % yield or the actual yield must be given in the problem.)actual yield must be given in the problem.)