Unit 1: Stoichiometry and Reactions

Unit 1: Stoichiometry Unit 1: Stoichiometry and Reactionsand Reactions

Sarah & SaraSarah & Sara

NomenclatureNomenclature

Cations—electron-deficient, (+) charge, Cations—electron-deficient, (+) charge, charge is group #charge is group # transition metals-charge shown with Roman transition metals-charge shown with Roman

Numeral in nameNumeral in name exceptions: Znexceptions: Zn2+2+ Ni Ni2+2+ Ag Ag++

the less charged cation of each atom, such the less charged cation of each atom, such as Copper uses the –ous ending, while the as Copper uses the –ous ending, while the one with greater charge uses the –ic endingone with greater charge uses the –ic ending

Copper (II) ion can also be called cuprousCopper (II) ion can also be called cuprous


Anions—excess electrons, (-) charge, Anions—excess electrons, (-) charge, charge is # of columns from noble gases charge is # of columns from noble gases (**exceptions), (**exceptions), monatomic –ide suffixmonatomic –ide suffix polyatomic oxyanions –ate suffix most polyatomic oxyanions –ate suffix most

common (-ite with fewer O)common (-ite with fewer O) oxyanions modified by Hoxyanions modified by H++

**cyanide = CN**cyanide = CN-- hydroxide = OH hydroxide = OH--


Neutral ionic compound = a salt = metal Neutral ionic compound = a salt = metal cation + nonmetal anioncation + nonmetal anion [empirical formula with net charge of zero][empirical formula with net charge of zero] to name: modify cation and/or anion with to name: modify cation and/or anion with

subscripts [name the cation then the anion, subscripts [name the cation then the anion, the subscript is not stated, write transition the subscript is not stated, write transition metals with Roman Numerals for charge]metals with Roman Numerals for charge]


Acids—have HAcids—have H++ cation, name is based on cation, name is based on anionanion

-ide = hydro_______ic acid-ide = hydro_______ic acid-ate = ____________ic acid-ate = ____________ic acid-ite = ____________ous acid-ite = ____________ous acid

EX: Chloride= hydrocholric acidEX: Chloride= hydrocholric acid Chlorate= chloric acidChlorate= chloric acid Chlorite= chlorous acidChlorite= chlorous acid


Covalent (or binary) compounds = 2 nonmetals (name Covalent (or binary) compounds = 2 nonmetals (name farthest left first with subscripts stated in name, farthest left first with subscripts stated in name, exceptions—Fluorine is always last, Oxygen is last exceptions—Fluorine is always last, Oxygen is last unless with F)unless with F)

Organic compounds = hydrocarbons, only C and H; Organic compounds = hydrocarbons, only C and H; others can include O, N, and/or Sothers can include O, N, and/or S alkanes – C “backbone,” all single bonds, as many H as alkanes – C “backbone,” all single bonds, as many H as

necessary to fill bonds, end with –anenecessary to fill bonds, end with –ane Name with # of C: 1=meth 2=eth 3=prop 4=but 5+ = binary prefixesName with # of C: 1=meth 2=eth 3=prop 4=but 5+ = binary prefixes

alcohols – O-H “functional group” in place of 1 or more H alcohols – O-H “functional group” in place of 1 or more H [-ane becomes -anol] [-ane becomes -anol]

Number position in functional group in front (or to which C is Number position in functional group in front (or to which C is it bonded)it bonded)


The –ates: The –ates: Sulfate (SOSulfate (SO44

--))

Chlorate (ClOChlorate (ClO33--))

Phosphate (POPhosphate (PO443-3-))

Carbonate (COCarbonate (CO332-2-) )

Nitrate (NONitrate (NO33--))

Acetate (CAcetate (C22HH33OO22-- = =

CHCH33COOCOO--))

Chromate (CrOChromate (CrO442-2-))

Dichromate (CrDichromate (Cr22OO772-2-))

Permanganate (MnOPermanganate (MnO44--))

Hydride = HHydride = H-- Hydrogen ion = HHydrogen ion = H++ ammonium ion = ammonium ion = NHNH44

++

Atomic StructureAtomic Structure

angstrom = 10angstrom = 10-10-10 m = Å = size of atom in m = Å = size of atom in metersmeters

Weight of 1 electron = (1/1800) protonWeight of 1 electron = (1/1800) proton size of the electron cloud is the full atomic size and size of the electron cloud is the full atomic size and

is about (1/4000) of an atom’s mass is about (1/4000) of an atom’s mass 1 x 101 x 10-10-10 to 5 x 10 to 5 x 10-10-10 m = 1 to 5 Å = .1 to .5 nm = 100 m = 1 to 5 Å = .1 to .5 nm = 100

to 500 pmto 500 pm

nucleus contains 99.9% mass of an atom; nucleus contains 99.9% mass of an atom; 1 x 10-15 to 1 x 10-14 m = .00001 to .0001 Å1 x 10-15 to 1 x 10-14 m = .00001 to .0001 Å

Isotope NotationIsotope Notation

A = mass # = protons + neutronsA = mass # = protons + neutrons Z = atomic # = protonsZ = atomic # = protons q = protons – electronsq = protons – electrons

XA

Z

q

Reactions and Other StuffReactions and Other Stuff

Decomposition reaction: Y Decomposition reaction: Y C + D C + D Synthesis/composition: A + B Synthesis/composition: A + B X X Combustion: CCombustion: CxxHHyyOOzz CO CO22 + H + H22OO

Avogadro’s Number = NA = 6.02 x 1023 Avogadro’s Number = NA = 6.02 x 1023 Spectrometer Spectrometer average mass – average mass –

weighted averageweighted average Diatomic elements = H, O, N, Cl, Br, I, FDiatomic elements = H, O, N, Cl, Br, I, F

StoichiometryStoichiometry

Stoichiometric Relationships:Stoichiometric Relationships: grams of A grams of A moles of A moles of A moles of B moles of B grams of B grams of B

Limiting reactants: practical solutions (amounts of Limiting reactants: practical solutions (amounts of reactants are mismatched)reactants are mismatched) all calculations are based on limiting reactantall calculations are based on limiting reactant limiting reactant is completely consumedlimiting reactant is completely consumed Method A: use one reactant to find how much of other reactant Method A: use one reactant to find how much of other reactant

is neededis needed fewer calculations, most efficient if only finding limiting reactantfewer calculations, most efficient if only finding limiting reactant

Method B: use both to find out how much product is madeMethod B: use both to find out how much product is made more calculations, fool proof strategic planningmore calculations, fool proof strategic planning

Yield!Yield! Theoretical Yield = grams of limiting reactant Theoretical Yield = grams of limiting reactant

moles of limiting reactant moles of limiting reactant moles of moles of product product grams of product grams of product

% Yield=(actual or experimental) / theoretical% Yield=(actual or experimental) / theoretical

Example Problem 1Example Problem 1

CC66HH66 + Br + Br22 C C66HH55Br +HBr +H22

You have 30.0g of CYou have 30.0g of C66HH66 and 65.0g of Br and 65.0g of Br22

present for a reaction. present for a reaction. A) Find the limiting reactant, and state A) Find the limiting reactant, and state

how much Chow much C66HH55Br will be produced?Br will be produced?

B) Calculate the percent yield if 56.7g of B) Calculate the percent yield if 56.7g of CC66HH55Br are produced by the reaction.Br are produced by the reaction.

Solution to Example 1Solution to Example 1 A) First calculate the moles of each reactant you have A) First calculate the moles of each reactant you have

present at the beginning of the reaction.present at the beginning of the reaction. 30.0g C30.0g C66HH66 x (1 mol C x (1 mol C66HH66 / 78.11g C / 78.11g C66HH66) = .384 mol ) = .384 mol

CC66HH66

65.0g Br65.0g Br22 x (1 mol Br x (1 mol Br22 / 159.8g Br / 159.8g Br22) = .407 mol Br) = .407 mol Br22

Because CBecause C66HH66 and Br and Br22 are in a 1:1 ratio, C are in a 1:1 ratio, C66HH66 is the is the

limiting reactant and determines the theoretical yield.limiting reactant and determines the theoretical yield. .384 mol C.384 mol C66HH66 x (1 mol C x (1 mol C66HH55Br / 1 mol CBr / 1 mol C66HH66) x (157.0g ) x (157.0g

CC66HH55Br / 1 mol CBr / 1 mol C66HH55Br) = 60.3g CBr) = 60.3g C66HH55BrBr

Solution fo Example 1Solution fo Example 1

B) % yield = actual or experimental/ B) % yield = actual or experimental/ theoreticaltheoretical

% yield = (56.7g C% yield = (56.7g C66HH55Br actual / 60.3g Br actual / 60.3g

CC66HH55Br theoretical) x 100 = 94.0%Br theoretical) x 100 = 94.0%


Calculate the percent composition of Calculate the percent composition of Carbon in each of the following Carbon in each of the following molecules.molecules.

A) CA) C77HH66OO

B )CB )C88HH88OO33

C) CC) C77HH1414OO22

Solution to Example 2Solution to Example 2

A) CA) C77HH66OO

FW: 7(12.0 amu) + 6(1.0 amu) + FW: 7(12.0 amu) + 6(1.0 amu) + 1(16.0 amu) = 106.0 1(16.0 amu) = 106.0

amuamu

%C = ( 7(12.0 amu) / 106.0 amu ) %C = ( 7(12.0 amu) / 106.0 amu ) x x 100 = 79.2%100 = 79.2%


B )CB )C88HH88OO33

FW: 8(12.0 amu) + 8(1.0 amu) + FW: 8(12.0 amu) + 8(1.0 amu) + 3(16.0 amu) = 152.0 amu3(16.0 amu) = 152.0 amu

%C = ( 8(12.0 amu) / 152.0 amu ) x %C = ( 8(12.0 amu) / 152.0 amu ) x 100 = 63.2%100 = 63.2%


C) CC) C77HH1414OO22

FW: 7(12.0 amu) + 14(1.0 amu) + FW: 7(12.0 amu) + 14(1.0 amu) + 2(16.0 amu) = 130.0 amu2(16.0 amu) = 130.0 amu

%C = ( 7(12.0 amu) / 130.0 amu ) x %C = ( 7(12.0 amu) / 130.0 amu ) x 100 = 64.6%100 = 64.6%


Caffeine is readily available in common foods Caffeine is readily available in common foods and drinks. By mass, caffeine contains 49.48% and drinks. By mass, caffeine contains 49.48% C, 5.15% H, 16.49% O, and 28.87% N. Its C, 5.15% H, 16.49% O, and 28.87% N. Its molar mass is 194.2g/mol. molar mass is 194.2g/mol.

A) Determine the empirical formula for caffeine.A) Determine the empirical formula for caffeine. B) Determine the molecular formula for B) Determine the molecular formula for

caffeine.caffeine.

Solution to Example 3Solution to Example 3 A) Assume the sample is 100gA) Assume the sample is 100g 49.48g C x ( 1 mol C/ 12.011 g C) = 4.1195 mol C49.48g C x ( 1 mol C/ 12.011 g C) = 4.1195 mol C 5.15g H x (1 mol H/ 1.00794g H) = 5.1096 mol H5.15g H x (1 mol H/ 1.00794g H) = 5.1096 mol H 16.49g O x (1 mol O/ 15.9994g O) = 1.03066 mol O16.49g O x (1 mol O/ 15.9994g O) = 1.03066 mol O 28.87g N x (1 mol N/14.00674g N) = 2.06115 mol N28.87g N x (1 mol N/14.00674g N) = 2.06115 mol N DIVIDE ALL BY THE LOWEST MOLE AMOUNT DIVIDE ALL BY THE LOWEST MOLE AMOUNT

(1.03066 mole)(1.03066 mole) Gives us: ~4 mol C, 5 mol H, 1 mol O, and 2 mol NGives us: ~4 mol C, 5 mol H, 1 mol O, and 2 mol N Therefore, our empirical formula is CTherefore, our empirical formula is C44HH55ONON22

Solution fo Example 3Solution fo Example 3

B) molecular formulaB) molecular formula 4(12.011g) + 5(1.00794g) + 15.9994g + 4(12.011g) + 5(1.00794g) + 15.9994g +

2(14.00674g)= 97.09658 g/ mol empirical2(14.00674g)= 97.09658 g/ mol empirical 194.2 g/mol / 97.09658 g/mol = 2194.2 g/mol / 97.09658 g/mol = 2 Therefore, the empirical equation needs Therefore, the empirical equation needs

to be double to get the proper molecular to be double to get the proper molecular formula of caffeine, which is formula of caffeine, which is CC88HH1010OO22NN44..

ZE END!ZE END!

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Unit 1: Stoichiometry and Reactions