Chapter 3Molecules, Compounds, and Chemical Equations

Elements and CompoundsElements combine together to make an almost unlimited number of compoundsThe properties of a compound are totally different from its constituent elements*

Formation of Water from Its Elements*

Chemical BondsCompounds are made of atoms held together by chemical bondsattractive forces between each atoms protons and the bonding electrons*

Two general Bond TypesIonic and CovalentIonic bonds result when electrons have been transferred between a metal and a nonmetal atom, resulting in oppositely charged ions that attract each otherCovalent bonds result when two atoms share one or more of their electronsgenerally found when nonmetal atoms bond together

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Formulas Describe CompoundsEach element is represented by its letter symbolThe number of atoms of each element is written to the right of the elements symbol as a subscript

A compound is a distinct substancecomposed of atoms from two or more elementsthe compound is defined by the number and type of each atomcompounds can be a neutral molecule or an ion*

An element is made up of*Multiple types of atomsCompoundsA single type of atom

Representing Compounds with Chemical FormulasCompounds are generally represented with a chemical formulaAll chemical formulas tell what elements are in the compound through use of the letter symbol of the elementAll formulas and models convey a limited amount of information none are perfect representations*

Types of Formulas: Empirical FormulaA compounds empirical formula gives the relative number of atoms of each element reduced to the lowest possible denominatorit doesnt describe the # atoms, the order of attachment, or the shapeformulas of ionic compounds are empiricalFluorspar (CaCl2) is an ionic compound. There is 1 Ca2+ ion for every 2 Cl ions in the compound.molecular compounds can also be written as an empirical formulae.g. the molecular formula for oxalic acid is C2H2O4 but the empirical formula is CHO2

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Types of Formula: Molecular FormulaA molecular formula gives the actual number of atoms of each element in a moleculeit does not describe the order of attachment, or the shapeThe molecular formula C2H2O4.tells you that there are 2 carbon atoms, 2 hydrogen atoms, and 4 oxygen atomsDoesnt tell you how the carbon, oxygen and hydrogen atoms are attached together*

Types of Formula: Structural FormulaA structural formula tells what atoms are present and how the atoms in a molecule are attached together. It uses lines to represent covalent bondsit does not describe the 3-dimensional shape, but an experienced chemist can make a good guess at iteach line describes the number of electrons shared by the bonded atomssingle line = 2 shared electrons, a single covalent bonddouble line = 4 shared electrons, a double covalent bondtriple line = 6 shared electrons, a triple covalent bond*

Representing Compounds Molecular ModelsMolecular models show the 3-d structure of the molecule plus all the information given by the structural formulaBall-and-stick models use balls to represent the atoms and sticks to represent the bonds between themSpace-filling models use interconnected spheres to show the electron clouds of atoms connecting together*

Structural Formula of Oxalic Acid*Ball and Stick ModelSpace fillingModel

The following formulas represent:*/ CO2H/ C2O4H2Empirical / Molecular / StructuralMolecular / Empirical / StructuralStructural / Molecular / EmpiricalEmpirical / Structural / MolecularMolecular / Structural / EmpiricalStructural / Empirical / Molecular

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Write the empirical formula for each of the following Al2O3CH2OC2H2NCH3O*

Classification of Elements and Compounds*

Classifying Elements*Atomic elementsSingle atomsMost elements are atomic elementsDiatomic elements2 atoms (e.g. H2)Polyatomic elements more than 2 atoms (e.g. P4, S8, Se8)

Molecular Elements*7 elements occur as diatomic molecules (2 atoms)shaped like a 7 on the periodic tablestart with element 7, N group V, go across the table to group VIIA, F, then go down group VIIA to IHydrogen stands aloneOther elements occur as polyatomic moleculesP4, S8, Se81A

Molecular Elements*

Classifying CompoundsIonic compounds a 3-D array of anions surrounded by cations and vice-versaMade from combining metals (which form cations) with non-metals (which form anions)No individual molecular units, only formula unitsCan contain polyatomic ions2+ atoms attached together by covalent bonds with an overall charge on the entire ion*Table salt containsan array of Na+ ionsand Cl- ions

*= Halogens1A2A3A4A5A6A7A8A

OSSeTe+1+2-1-2-3N= Alkali earth metalsAl+3Ions with predictable charges

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Compounds that Contain IonsA neutral compound must have no total charge, therefore the number of cations and anions multiplied by their charges, must balance to get 0 charge

If Na+ is combined with S2, you will need 2 Na+ ions for every S2 ion to balance the charges, therefore the formula must be Na2S *

Classify the Following as Either an Atomic Element, Molecular Element, Molecular Compound, or Ionic Compound Aluminum, AlAluminum chloride, AlCl3Chlorine, Cl2Acetone, C3H6OCarbon monoxide, COCobalt, Co

*atomic elementionic compoundmolecular elementmolecular compoundmolecular compoundatomic element

Ex 3.4: Write a formula for ionic compound that forms between calcium and oxygen*Ca2+ O2CaOcations: +2anions: 2The charges cancel

Write the symbol for the metal cation and the nonmetal anion. Obtain the ions charges from the elements group number on the periodic table and write it as a superscript on the symbol.Adjust the number of cations and anions using subscripts to balance the overall charge.Check that the sum of the charges of the cations equals the sum of the charges of the anions.

Example 3.3: Write the formula of a compound made from aluminum ions and oxide ionsWrite the symbol for the metal cation and its chargeWrite the symbol for the nonmetal anion and its chargeCharge (without sign) becomes subscript for other ionReduce subscripts to smallest whole number ratioCheck that the total charge of the cations cancels the total charge of the anions Al3+ column 3AO2 column 6AAl+3 O2Al2O3Al = (2)(+3) = +6O = (3)(2) = 6*no change required

Write the formulas for compounds made from the following ionsPotassium ion with a nitride ionK+ with N3Calcium ion with a bromide ionCa2+ with Br1Aluminum ion with a sulfide ionAl3+ with S2

*K3NCaBr2Al2S3

Rules for Naming Ionic Compounds*Some compounds have common names that can only be learned by memorizationNaCl = table saltNaHCO3 = baking soda

Others have a systematic name based on naming the component ions

Naming Binary Ionic Compounds forMetals with Invariant Charge*First, use the metals elemental name for the cationNext, add the nonmetal anions nameassign the proper negative charge based its group number on the Periodic Tablethe nonmetal names suffix changes to -ide

CsF: A Binary Ionic Compound with a Metal having Invariant Charge*1.Identify cation and anionCs = Cs+ because it is Group 1AF = F because it is Group 7A2.Name the cation Cs+ = cesium3.Name the anion F = fluoride4.Write the cation name first, then the anion namecesium fluoride

Naming Binary Ionic Compounds forMetals with Variable ChargeAgain, use the elemental metal name for the metal cationDetermine the anions charge cations chargeA (Roman Numeral) follows the metal ion to indicate its chargeNext add the nonmetal anions namethe nonmetal names suffix changes to -ide*

CuF2: A Binary Ionic Compound with a Metal having Variable Charge*Identify cation and anion F = F because it is Group 7ACu = Cu2+ to balance the two () charges from 2 F2.Name the cationCu2+ = copper(II)3.Name the anionF = fluoride4.Write the cation name first, then the anion namecopper(II) fluoride

Name the compound composed of chlorine and potassium: KCl*potassium (I) chloridechlorine potassiumidepotassium chloridepotassium chlorine

Name the compound composed of magnesium and bromine: MgBr2*magnesium bromidemagnesium (II) dibromidebromine magnesidebromine magnesiumidemagnesium dibromine

Name the compound composed of aluminum and sulfur: Al2S3*trisulfur dialuminidealuminum (III) trisulfidesulfur magnesiumidealuminum sulfuridealuminum sulfide

Determining the charge on a metal cation having variable charge in a compound composed of gold and sulfur: Au2S3*Determine the invariant charge on the anionAu2S3 : the anion is S = Group 6A, its charge is 2Determine the total negative charge3 S in the formula, the total negative charge is 3 x 2 = 6Set the total positive charge = total negative chargetotal negative charge = 6, total positive charge = 6+Divide by the number of cations in the formulawith 2 Au in the formula and a total positive charge of 6+, each Au must have a 3+ charge The compound name is Gold (III) Sulfide

Find the charge on the cationCrO3

Fe3N2*3 O = 63 Fe = 6+each O = 2 2 N = 6 Cr = 6+ Fe = 2+each N = 3 Name the compoundchromium (VI) oxideName the compoundiron (II) nitride

Find the charge on the cation in TiCl4*-1-4+1+4

Find the charge on the cationTiCl44 Cl = 4* Ti = 4+each Cl = 1-Name the compoundtitanium(IV) chloride

Name the compoundPbBr2

Fe2S3lead(II) bromide

iron(III) sulfide*each Br = -12 x -1= -2 Pb 2+each S = -23 x -2= -6-6/2 Fe = -3 Fe 3+

To Write the formula for a binary ionic compound containing metal with variable chargemanganese(IV) sulfide1.Write the symbol for the cation and its charge2.Write the symbol for the anion and its charge3.The charge (without the sign) becomes subscript for other ion4.Reduce subscripts to smallest whole number ratio5.Check that the total charge of the cations cancels the total charge of the anions Mn4+ S2-Mn4+ S2Mn2S4Mn = (1)(4+) = +4S = (2)(2) = 4MnS2*

What are the formulas for compounds made from copper (II) nitride, Cu2+ with N3*Cu2N3Cu3N2N3Cu2N2Cu3

What are the formulas for compounds made from iron(III) with bromide: Fe3+ with Br- *FeBr3Fe3BrFeBrBr3FeBrFe3

Compounds Containing Polyatomic Anions*Name the cation first, then the polyatomic anionThe name and the charge of the polyatomic ion do not changeIf more than one polyatomic ion is present, the polyatomic ion is surrounded by parenthesesa subscript tells how many polyatomic ions there are.

NO3- (NO3-)2 (NO3-)3

Periodic Pattern of Polyatomic Ions Elements in the same column form similar polyatomic ions - ate groups*

Patterns for Polyatomic Ions*-ate ion chlorate = ClO3add 1 oxygenwith same chargeperchlorate = ClO4remove 2 oxygenwith same chargehypochlorite = ClOremove 1 oxygenwith same chargechlorite = ClO2

Some Common Polyatomic Ions*If the polyatomic ion has a H in the front, add hydrogen- as a prefix and adjust the overall charge by +1

Example: Naming ionic compounds containing a polyatomic ion - Na2SO4*1.Identify the ions2Na = 2Na+ because in Group 1ASO4 = SO42 a polyatomic ion2.Name the cationNa+ = sodium, metal with invariant charge3.Name the anionSO42 = sulfate4.Write the name of the cation followed by the name of the anionsodium sulfate

Example: Naming ionic compounds containing a polyatomic ion Fe(NO3)3*1.Identify the ionsNO3 = NO3 a polyatomic ionFe = Fe3+ to balance the charge of the 3 NO32.Name the cationFe3+ = iron(III), a metal with variable charge3.Name the anionNO3 = nitrate4.Write the name of the cation followed by the name of the anioniron(III) nitrate

Name the Following Compounds1.Ca(C2H3O2)22.Cu(NO3)2calcium acetatecopper(II) nitrate*1.NH4Clammonium chloridePolyatomic Ammonium Cations NH4+

Example Writing formula for ionic compounds containing polyatomic ionIron(III) phosphate1.Write the symbol for the cation and its charge2.Write the symbol for the anion and its charge3.Charge (without sign) becomes subscript for other ion4.Reduce subscripts to smallest whole number ratio5.Check that the total charge of the cations cancels the total charge of the anions Fe3+ PO43Fe3+ PO43Fe3(PO4)3Fe = (1)(3+) = +3PO4 = (1)(3) = 3FePO4*

Practice What are the formulas for compounds made from the following ions?aluminum ion with a sulfate ionchromium(II) with hydrogen carbonate Al3+ with SO42Cr2+ with HCO3*Al2(SO4)3Cr(HCO3)2

Hydrates*Hydrates are ionic compounds containing a specific number of waters in each formula unitWater of hydration can be driven off by heatingIn formula, attached waters written as CoCl2 6H2ONamed with hydrate suffix. The # of attached waters is indicated by the prefixes to the rightCoCl2 6H2O = cobalt(II) chloride hexahydrateCaSO4 H2O = calcium sulfate hemihydrate

Cobalt(II) chloride hexahydrate*

What is the formula of magnesium sulfate heptahydrate?What is the name of NiCl26H2O?Mg2+ + SO42MgSO4MgSO47H2O2Cl Ni2+nickel(II) chloridenickel(II) chloride hexahydrate*

Classifying CompoundsMolecular compounds compounds made of only nonmetalscovalent compounds composed of individual molecule units where atoms of different elements are chemically attached by covalent bonds*Propane contains individual C3H8 molecules

Naming Binary Molecular Compounds of 2 nonmetals*1.Write the name of the first element in the formulaa)Which is the element furthest left and down on the Periodic Table2.Write the second elements name in the formula with an -ide suffixas if it were an anion, but these compounds do not contain ions!3.Use a prefix in front of each name to indicate the number of atoms (slide 59)a)never use the prefix mono- for the first element

Subscript PrefixesDrop the final a if the elements name begins with a vowel*1 = mono- 2 = di-3 = tri-4 = tetra-5 = penta- 6 = hexa- 7 = hepta- 8 = octa- 9 = nona-10 = deca-

Naming binary molecular BF3*1.Name the first element: boron2.Name the second element with an idefluorine fluorideAdd a prefix to the 2nd nonmetals name to indicate the subscript a) Do not use the prefix mono on the first elementmonoboron trifluoride4.Combine the first elements name without the mono- prefix with the second elements name with a prefixboron trifluoride

Name the FollowingNO2PCl5I2F7nitrogen dioxidephosphorus pentachloridediiodine heptafluoride*

Binary Molecular Compounddinitrogen pentoxide*Identify the symbols of the elementsnitrogen = Noxide = oxygen = OWrite the formula using the prefix numbers as subscriptsdi = 2, penta = 5N2O5

Write Formulas for the Followingdinitrogen tetroxidesulfur hexafluoridediarsenic trisulfideN2O4SF6As2S3*

Formula Mass = the sum of the mass of all atoms in a single molecule or formula unit*Formula mass, aka molecular mass, aka molecular weight (MW), aka molar massthe formula mass of 1 H2O molecule, which is made up of 2 hydrogen and 1 oxygen = 2x(1.01 amu/H) + 1x(16.00 amu/O) = 18.02 amu/H2OMolar mass can be used as a conversion factor to convert grams to moles and moles to grams

How many moles are in 50.0 g of PbO2? (Pb = 207.2 amu, O = 16.00 amu)because the given amount is less than 239.2 g, the moles being < 1 makes sense

1 mol PbO2 = 239.2 g50.0 g mol PbO2moles PbO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:*Calc molar mass of PbO2Remember, adding decimals use the leastnumber of decimal places for sig. figs.207.2 amu

Find the number of CO2 molecules in 10.8 g of dry icebecause the given amount is much less than 1 mol CO2, the number makes sense

1 mol CO2 = 44.01 g, 1 mol = 6.022 x 102310.8 g CO2molecules CO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:*

How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2)because the given amount is less than 1 mol PbO2, the number makes sense

1 mol PbO2 = 239.2 g,1 mol = 6.022 x 102350.0 g PbO2formula units PbO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:*

What is the mass of 4.78 x 1024 NO2 molecules?because the given amount is more than Avogadros number, the mass > 46 g makes sense

1 mol NO2 = 46.01 g, 1 mol = 6.022 x 10234.78 x 1024 NO2 moleculesg NO2Check:Solution:Conceptual Plan:

Relationships:Given:Find:*

Percent Composition*Mass percent of each element in a compoundCan be determined from the formula of the compound, or the experimental mass analysis of the compoundThe percentages may not always total to 100% due to rounding

Example 3.13: Find the mass percent of Cl in C2Cl4F2because the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes senseC2Cl4F2% Cl by massCheck:Solution:Conceptual Plan:

Relationships:Given:Find:*

Determine % Ca and % Cl in CaCl2*Molar mass Ca = 40.08 g/mol

Mass Percent as a Conversion FactorThe mass percent tells you the mass of a constituent element in 100 g of the compoundthe fact that CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains 58.64 g ClThis can be used as a conversion factor100 g CCl2F2 : 58.64 g Cl*

Given that there are 39 g Na in 100 g NaCl and you have 2.4 g Na. to find the number of grams of NaCl, you have, which conversion factor would you use?*39 g Na / 100 g NaCl100 g NaCl / 39 g Na

Example 3.14: Find the mass of table salt containing 2.4 g of Nabecause the mass of NaCl is more than 2x the mass of Na, the number makes sense

100 g NaCl : 39 g Na2.4 g Na, 39% Nag NaClCheck:Solution:Conceptual Plan:

Relationships:Given:Find:*

Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?because the mass of benzaldehyde is more than the mass of C, the number makes sense

100 g benzaldehyde : 79.2 g C19.8 g C, 79.2% Cg benzaldehydeCheck:Solution:Conceptual Plan:

Relationships:Given:Find:*

Conversion Factors in Chemical FormulasChemical formulas show inherent relationships between the number of atoms and moleculesor between the moles of atoms and moleculesThese relationships can be used to convert between amounts of constituent elements and moleculeslike percent composition*

Example 3.15: Find the mass of hydrogen in 1.00 gal of waterbecause 1 gallon weighs about 3800 g, and H is light, the number makes sense

3.785 L = 1 gal, 1 L = 1000 mL, 1.00 g H2O = 1 mL, 1 mol H2O = 18.02 g, 1 mol H = 1.008 g, 2 mol H : 1 mol H2O1.00 gal H2O, dH2O = 1.00 g/mlg HCheck:Solution:Conceptual Plan:

Relationships:Given:Find:*

1 mol NaCl = 58.44 g, 1 mol Na = 22.99 g, 1 mol Na : 1 mol NaClFind the mass of sodium in 6.2 g of NaClbecause the amount of Na is less than the amount of NaCl, the answer makes sense 6.2 g NaClg NaCheck:Solution:Conceptual Plan:

Relationships:Given:Find:*-END LECTURING

*Empirical Formulas via elemental analysisIf given the percentages of each element, convert them to grams by assuming you have 100 g of total compoundIf given grams, proceed to next step.Example 3.17: A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.C = 60.00% H = 4.48% O = 35.53% CONVERT TO A BASIS OF 100 TOTAL GRAMS

C = 60.00 g H = 4.48 g O = 35.53 gDetermined by combustion analysis or % composition

How much aspirin do you assume you have at the start?*50 g1 mol100 g1/3 mole

Example: Find the empirical formula of aspirin with the given mass percent compositionWrite down the quantity to find and/or its units Find: the empirical formula,

InformationGiven:60.00 g C, 4.48 g H, 35.53 g O*

Example:Find the empirical formula of aspirin with the given mass percent compositionApply the conceptual plancalculate the moles of each element*InformationGiven:60.00 g C, 4.48 g H, 35.53 g OFind: empirical Formula, CxHyOzCP: g C,H,O mol C,H,O pseudo form. mol ratio emp. form.Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g

Given 60.00 g C / 4.48 g H / 35.53 g O; convert the grams of each element into moles *1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O5.00 mol C / 4.44 mol H / 2.22 mol O.200 mol C / 0.225 mol H / 0.450 mol O721 mol C / 4.52 mol H / 568 mol O

Given 4.996 mol C, 4.44 mol H, 2.220 mol O; what is the pseudo formula *C4.996H4.44O2.220C5H4O2

Example:Find the empirical formula of aspirin with the given mass percent compositionApply the conceptual planwrite the pseudo formulaC4.996H4.44O2.220*InformationGiven:60.00 g C, 4.48 g H, 35.53 g OFind: empirical formula, CxHyOzCP: g C,H,O mol C,H,O pseudo form. mol ratio emp. form.Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g

Example:Find the empirical formula of aspirin with the given mass percent compositionApply the concept planfind the mole ratio by dividing by the smallest number of moles (i.e. normalize the smallest value to 1)InformationGiven:60.00 g C, 4.48 g H, 35.53 g OFind: empirical formula, CxHyOzCP: g C,H,O mol C,H,O pseudo form. mol ratio emp. form.Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g*

Example:Find the empirical formula of aspirin with the given mass percent compositionApply the conceptual planmultiply subscripts by factor to give whole number{C2.25H2O1} x 4C9H8O4InformationGiven:60.00 g C, 4.48 g H, 35.53 g OFind: empirical formula, CxHyOzCP: g C,H,O mol C,H,O pseudo form. mol ratio emp. form.Rel: 1 mol C = 12.01 g; 1 mol H = 1.008 g; 1 mol O = 16.00 g*

Example:Find the empirical formula of aspirin with the given mass percent compositionWrite a conceptual plan InformationGiven:60.00 g C, 4.48 g H, 35.53 g OFind: empirical formula, CxHyOzCP: g C,H,O mol C,H,O pseudo form. mol ratio emp. form.

g Cmol Cg Hmol Hpseudo-formulaempiricalformulamoleratiowholenumberratiog Omol O*Collect needed relationships: 1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O

*Finding an Empirical FormulaConvert grams to moles using each elements molar mass60.00 g C x 1 mol C/12.01 g C = 4.996 mol C4.48 g H x 1 mol H/1.008 g H = 4.44 mol H35.53 g O x 1 mole O/16.00 g O = 2.220 mol O

Write a pseudo-formula using moles as subscriptsCxHyOz C4.996H4.44O2.220

Finding an Empirical Formula*5.Divide all subscripts by the smallest number of molesif result is within 0.1 of whole number, round to the whole numberfind the mole ratio by dividing by the smallest number of moles (i.e. normalize the smallest value to 1)

Finding an Empirical Formula*6.Multiply all mole ratios by a common factor to make sure the subscripts are all whole numbersa)if the mole ratio is 0.5, multiply all by 2; if the ratio is 0.33 or 0.67, multiply by 3; if the ratio is 0.25 or 0.75, multiply by 4; etc. skip this step if the subscripts are already whole numbers

{C2.25H2O1} x 4C9H8O4

Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00)Given: 75.7% Sn, F must be (100 75.3) = 24.3% in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g FFind: the empirical formula SnxFyRelationships: 1 mol Sn = 118.70 g; 1 mol F = 19.00 gConceptual plan:g Snmol Sng Fmol Fpseudo-formulaempiricalformulamoleratiowholenumberratio*

Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00)Apply the conceptual planSn0.638F1.28SnF2*

Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)Given: 72.4% Fe, and O must be (100 72.4) = 27.6% in 100 g magnetite there are 72.4 g Fe and 27.6 g OFind: FexOyRelationships: 1 mol Fe = 55.85 g; 1 mol O = 16.00 g

Conceptual plan:g Femol Feg Omol Opseudo-formulaempiricalformulamoleratiowholenumberratio*

Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)Apply the conceptual planFe1.30O1.73*

Molecular Formulas*The molecular formula is a multiple of the empirical formulaTo determine the molecular formula you need to know the empirical formula and the molar mass of the compound

3.18 Given the empirical formula, C2H3O, Find the molecular formula of butanedionethe molar mass of the calculated formula is in agreement with the given molar mass

empirical formula = C2H3Obutanedione MM = 86.03 g/molmolecular formulaCheck:Solution:Conceptual Plan: and Relationships:Given:

Find:*

Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01)C5 =5(12.01 g) = 60.05 gH3 =3(1.01 g) = 3.03 gC5H3 = 63.08 gMolecular formula = {C5H3} x 4 = C20H12*

Combustion AnalysisA common technique for analyzing compounds

burn a known mass of the compoundweigh the quantity of products madeused for organic compounds containing C,H,O Carbon CO2, Hydrogen H2O, Oxygen = original mass (C + H)

Knowing the masses of all the constituent elements in the original compound, the empirical formula can be found*

*Combustion Analysis

Example 3.20Combustion of a 0.8233 g sample of an organic compound containing only carbon, hydrogen, and oxygen produced the following products: CO2 = 2.445 gH2O = 0.6003 gDetermine the empirical formula of the compound

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Example 3.19: Find the empirical formula of a compound with the given amounts of combustion productsWrite down the given quantity and its units Given: combustion of 0.8233 g of a compound that contains only C,H,O; yields products:2.445 g of CO20.6003 g of H2O*

Example: Find the empirical formula of a compound with the given amounts of combustion products2. Write down the quantity to find and/or its units Find: empirical formula, CxHyOz

1. Write down the given information with units: 0.8233 g compound, 2.445 g CO2 0.6003 g H2O*3. Write a conceptual plan gCO2, H2OmolC, H, O

Example: Find the empirical formula of a compound with the given amounts of combustion productsCollect needed relationships 1 mole CO2 = 44.01 g CO21 mole H2O = 18.02 g H2O1 mole C = 12.01 g C1 mole H = 1.008 g H1 mole O = 16.00 g O1 mole CO2 = 1 mole C1 mole H2O = 2 mole H

InformationGiven: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2OFind: empirical formula, CxHyOzCP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formula*

Example: Find the empirical formula of a compound with the given amounts of combustion productsApply the conceptual plancalculate the moles of C and HInformationGiven: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2OFind: empirical formula, CxHyOzCP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formulaRel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound*

Example: Find the empirical formula of a compound with the given amounts of combustion productsApply the conceptual plancalculate the grams of C and HInformationGiven: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2OFind: empirical formula, CxHyOzCP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formulaRel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound*

Apply the conceptual plancalculate the grams and moles of OExample: Find the empirical formula of a compound with the given amounts of combustion productsInformationGiven: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2OFind: empirical formula, CxHyOzCP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formulaRel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound*11am

Apply the conceptual planwrite a pseudoformulaC0.05556H0.06662O0.00556Example: Find the empirical formula of a compound with the given amounts of combustion products InformationGiven: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol OFind: empirical formula, CxHyOzCP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formulaRel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound*

Apply the conceptual planfind the mole ratio by dividing by the smallest number of molesExample: Find the empirical formula of a compound with the given amounts of combustion productsInformationGiven: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol OFind: empirical formula, CxHyOzCP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formulaRel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound*

Apply the conceptual planmultiply subscripts by factor to give whole number, if necessarywrite the empirical formulaExample: Find the empirical formula of a compound with the given amounts of combustion productsInformationGiven: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol OFind: empirical formula, CxHyOzCP: g CO2 & H2O mol CO2 & H2O mol C & H g C & H g O mol O mol ratio empirical formulaRel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound*

The smell of dirty gym socks is caused by the compound caproic acid. Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula of caproic acid?(MM C = 12.01, H = 1.008, O = 16.00) *

molesg0.01450.08700.04360.2320.08770.524OHCC0.0436H0.0870O0.0145*

Molecular formula = {C3H6O} x 2 = C6H12O2*

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