Chapter 3 Molecules, Compounds, and Chemical Equations

Elements and Compounds Elements combine together to make an almost unlimited number of compounds The properties of a compound are totally different from its constituent elements

Formation of Water from Its Elements

Chemical Bonds Compounds are made of atoms held together by chemical bonds attractive forces between each atoms’ protons and the bonding electrons

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

Formulas Describe Compounds Each element is represented by its letter symbol The number of atoms of each element is written to the right of the element’s symbol as a subscript A compound is a distinct substance composed of atoms from two or more elements the compound is defined by the number and type of each atom compounds can be a neutral molecule or an ion

An element is made up of Multiple types of atoms Compounds A single type of atom

Representing Compounds with Chemical Formulas Compounds are generally represented with a chemical formula All chemical formulas tell what elements are in the compound through use of the letter symbol of the element All formulas and models convey a limited amount of information – none are perfect representations

Types of Formulas: Empirical Formula A compound’s empirical formula gives the relative number of atoms of each element reduced to the lowest possible denominator it doesn’t describe the # atoms, the order of attachment, or the shape formulas of ionic compounds are empirical Fluorspar (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 formula e.g. the molecular formula for oxalic acid is C2H2O4 but the empirical formula is CHO2

Types of Formula: Molecular Formula A molecular formula gives the actual number of atoms of each element in a molecule it does not describe the order of attachment, or the shape The molecular formula C2H2O4. tells you that there are 2 carbon atoms, 2 hydrogen atoms, and 4 oxygen atoms Doesn’t tell you how the carbon, oxygen and hydrogen atoms are attached together

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

Representing Compounds Molecular Models Molecular models show the 3-d structure of the molecule plus all the information given by the structural formula Ball-and-stick models use balls to represent the atoms and sticks to represent the bonds between them Space-filling models use interconnected spheres to show the electron clouds of atoms connecting together

Structural Formula of Oxalic Acid Space filling Model Ball and Stick Model

The following formulas represent: / C2O4H2 / CO2H Empirical / Molecular / Structural Molecular / Empirical / Structural Structural / Molecular / Empirical Empirical / Structural / Molecular Molecular / Structural / Empirical Structural / Empirical / Molecular

Write the empirical formula for each of the following The ionic compound that has two aluminum ions for every three oxide ions arabinose, C5H10O5 pyrimidine ethylene glycol Al2O3 CH2O C2H2N CH3O

Classification of Elements and Compounds

Classifying Elements more than 2 atoms (e.g. P4, S8, Se8) Atomic elements Single atoms Most elements are atomic elements Diatomic elements 2 atoms (e.g. H2) Polyatomic elements more than 2 atoms (e.g. P4, S8, Se8)

Molecular Elements N7 O8 F9 H1 Cl17 Br35 I53 7 elements occur as diatomic molecules (2 atoms) H1 Cl17 Br35 I53 5A 6A 7A N7 O8 F9 1A shaped like a 7 on the periodic table… start with element 7, N – group V, go across the table to group VIIA, F, then go down group VIIA to I Hydrogen stands alone Other elements occur as polyatomic molecules P4, S8, Se8

Molecular Elements

Classifying Compounds Ionic compounds a 3-D array of anions surrounded by cations and vice-versa Made from combining metals (which form cations) with non-metals (which form anions) No individual molecular units, only formula units Can contain polyatomic ions 2+ atoms attached together by covalent bonds with an overall charge on the entire ion Table salt – contains an array of Na+ ions and Cl- ions

Ions with predictable charges = Alkali metals = Alkali earth metals = Halogens Ions with predictable charges -3 1A 2A 7A 8A +3 3A 4A 5A 6A +1 +2 Cs Rb K Na Li N O I Br Cl F -1 Ba Sr Ca Mg Al S Se add pictures of elements from text Te -2

Compounds that Contain Ions A 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

atomic element ionic compound molecular element molecular compound Classify the Following as Either an Atomic Element, Molecular Element, Molecular Compound, or Ionic Compound atomic element ionic compound molecular element molecular compound Aluminum, Al Aluminum chloride, AlCl3 Chlorine, Cl2 Acetone, C3H6O Carbon monoxide, CO Cobalt, Co

Ex 3.4: Write a formula for ionic compound that forms between calcium and oxygen Write the symbol for the metal cation and the nonmetal anion. Obtain the ion’s charges from the element’s 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. Ca2+ O2− CaO cations: +2 anions: −2 The charges cancel

Example 3.3: Write the formula of a compound made from aluminum ions and oxide ions Write the symbol for the metal cation and its charge Write the symbol for the nonmetal anion and its charge Charge (without sign) becomes subscript for other ion Reduce subscripts to smallest whole number ratio Check that the total charge of the cations cancels the total charge of the anions Al3+ column 3A O2− column 6A Al+3 O2− Al2O3 no change required Al = (2)∙(+3) = +6 O = (3)∙(−2) = −6

Write the formulas for compounds made from the following ions Potassium ion with a nitride ion K+ with N3− Calcium ion with a bromide ion Ca2+ with Br1− Aluminum ion with a sulfide ion Al3+ with S2− K3N CaBr2 Al2S3

Rules for Naming Ionic Compounds Some compounds have common names that can only be learned by memorization NaCl = table salt NaHCO3 = baking soda Others have a systematic name based on naming the component ions

Naming Binary Ionic Compounds for Metals with Invariant Charge First, use the metal’s elemental name for the cation Next, add the nonmetal anion’s name assign the proper negative charge based its group number on the Periodic Table the nonmetal name’s suffix changes to -ide

CsF: A Binary Ionic Compound with a Metal having Invariant Charge 1. Identify cation and anion Cs = Cs+ because it is Group 1A F = F− because it is Group 7A 2. Name the cation Cs+ = cesium 3. Name the anion F− = fluoride 4. Write the cation name first, then the anion name cesium fluoride

Naming Binary Ionic Compounds for Metals with Variable Charge Again, use the elemental metal name for the metal cation Determine the anion’s charge  cation’s charge A (Roman Numeral) follows the metal ion to indicate its charge Next add the nonmetal anion’s name the nonmetal name’s 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 7A Cu = Cu2+ to balance the two (−) charges from 2 F− 2. Name the cation Cu2+ = copper(II) 3. Name the anion F− = fluoride 4. Write the cation name first, then the anion name copper(II) fluoride

Name the compound composed of chlorine and potassium: KCl potassium (I) chloride chlorine potassiumide potassium chloride potassium chlorine

Name the compound composed of magnesium and bromine: MgBr2 magnesium bromide magnesium (II) dibromide bromine magneside bromine magnesiumide magnesium dibromine

Name the compound composed of aluminum and sulfur: Al2S3 trisulfur dialuminide aluminum (III) trisulfide sulfur magnesiumide aluminum sulfuride aluminum 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 anion Au2S3 : the anion is S = Group 6A, its charge is 2− Determine the total negative charge 3 S in the formula, the total negative charge is 3 x 2 − = 6− Set the total positive charge = total negative charge total negative charge = 6−, total positive charge = 6+ Divide by the number of cations in the formula with 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 cation CrO3 Fe3N2 each O = 2− Name the compound 3 O = 6− chromium (VI) oxide  Cr = 6+ each N = 3− Name the compound 2 N = 6− iron (II) nitride 3 Fe = 6+  Fe = 2+

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

Find the charge on the cation TiCl4 each Cl = 1- 4 Cl = 4−  Ti = 4+ Name the compound titanium(IV) chloride

Name the compound PbBr2 Fe2S3 each Br = -1 2 x -1= -2  Pb 2+ lead(II) bromide iron(III) sulfide each S = -2 3 x -2= -6 -6/2 Fe = -3  Fe 3+

Mn4+ S2- Mn4+ S2− Mn2S4 MnS2 Mn = (1)∙(4+) = +4 S = (2)∙(2−) = −4 To Write the formula for a binary ionic compound containing metal with variable charge manganese(IV) sulfide 1. Write the symbol for the cation and its charge 2. Write the symbol for the anion and its charge 3. The charge (without the sign) becomes subscript for other ion 4. Reduce subscripts to smallest whole number ratio 5. Check that the total charge of the cations cancels the total charge of the anions Mn4+ S2- Mn4+ S2− Mn2S4 MnS2 Mn = (1)∙(4+) = +4 S = (2)∙(2−) = −4

What are the formulas for compounds made from copper (II) nitride, Cu2+ with N3− Cu2N3 Cu3N2 N3Cu2 N2Cu3

What are the formulas for compounds made from iron(III) with bromide: Fe3+ with Br- FeBr3 Fe3Br FeBr Br3Fe BrFe3

Compounds Containing Polyatomic Anions Name the cation first, then the polyatomic anion The name and the charge of the polyatomic ion do not change If more than one polyatomic ion is present, the polyatomic ion is surrounded by parentheses a 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 = ClO3− remove 2 oxygen with same charge hypochlorite = ClO− add 1 oxygen with same charge perchlorate = ClO4− remove 1 oxygen with same charge chlorite = 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 ions 2Na = 2Na+ because in Group 1A SO4 = SO42− a polyatomic ion 2. Name the cation Na+ = sodium, metal with invariant charge 3. Name the anion SO42− = sulfate 4. Write the name of the cation followed by the name of the anion sodium sulfate

Example: Naming ionic compounds containing a polyatomic ion Fe(NO3)3 1. Identify the ions NO3 = NO3− a polyatomic ion Fe = Fe3+ to balance the charge of the 3 NO3− 2. Name the cation Fe3+ = iron(III), a metal with variable charge 3. Name the anion NO3− = nitrate 4. Write the name of the cation followed by the name of the anion iron(III) nitrate

Name the Following Compounds 1. Ca(C2H3O2)2 2. Cu(NO3)2 calcium acetate copper(II) nitrate Polyatomic Ammonium Cations – NH4+ 1. NH4Cl ammonium chloride

Example – Writing formula for ionic compounds containing polyatomic ion Iron(III) phosphate 1. Write the symbol for the cation and its charge 2. Write the symbol for the anion and its charge 3. Charge (without sign) becomes subscript for other ion 4. Reduce subscripts to smallest whole number ratio 5. Check that the total charge of the cations cancels the total charge of the anions Fe3+ PO43− Fe3+ PO43− Fe3(PO4)3 FePO4 Fe = (1)∙(3+) = +3 PO4 = (1)∙(3−) = −3

Practice — What are the formulas for compounds made from the following ions? aluminum ion with a sulfate ion chromium(II) with hydrogen carbonate Al3+ with SO42− Al2(SO4)3 Cr2+ with HCO3− Cr(HCO3)2

Hydrates Hydrates are ionic compounds containing a specific number of waters in each formula unit Water of hydration can be driven off by heating In formula, attached waters written as CoCl2 ∙ 6H2O Named with hydrate suffix. The # of attached waters is indicated by the prefixes to the right CoCl2 ∙ 6H2O = cobalt(II) chloride hexahydrate CaSO4 ∙ ½H2O = calcium sulfate hemihydrate

Cobalt(II) chloride hexahydrate

nickel(II) chloride hexahydrate What is the formula of magnesium sulfate heptahydrate? What is the name of NiCl2•6H2O? Mg2+ + SO42− MgSO4 MgSO47H2O 2Cl−  Ni2+ nickel(II) chloride nickel(II) chloride hexahydrate

Classifying Compounds Molecular compounds compounds made of only nonmetals covalent 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 formula a) Which is the element furthest left and down on the Periodic Table 2. Write the second element’s name in the formula with an -ide suffix as 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

Drop the final “a” if the element’s name begins with a vowel Subscript – Prefixes 1 = mono- 2 = di- 3 = tri- 4 = tetra- 5 = penta- 6 = hexa- 7 = hepta- 8 = octa- 9 = nona- 10 = deca- Drop the final “a” if the element’s name begins with a vowel

Naming binary molecular BF3 1. Name the first element: boron 2. Name the second element with an –ide fluorine  fluoride Add a prefix to the 2nd nonmetal’s name to indicate the subscript a) Do not use the prefix mono on the first element monoboron trifluoride 4. Combine the first element’s name without the mono- prefix with the second element’s name with a prefix boron trifluoride

Name the Following NO2 PCl5 I2F7 nitrogen dioxide phosphorus pentachloride diiodine heptafluoride

Binary Molecular Compound dinitrogen pentoxide Identify the symbols of the elements nitrogen = N oxide = oxygen = O Write the formula using the prefix numbers as subscripts di = 2, penta = 5 N2O5

Write Formulas for the Following dinitrogen tetroxide sulfur hexafluoride diarsenic trisulfide N2O4 SF6 As2S3

2x(1.01 amu/H) + 1x(16.00 amu/O) = 18.02 amu/H2O 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 mass the 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/H2O Molar 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) Given: Find: 50.0 g mol PbO2 moles PbO2 Conceptual Plan: Relationships: 1 mol PbO2 = 239.2 g g PbO2 mol PbO2 Calc molar mass of PbO2 207.2 amu Solution: Remember, adding decimals use the least number of decimal places for sig. figs. Check: because the given amount is less than 239.2 g, the moles being < 1 makes sense

Find the number of CO2 molecules in 10.8 g of dry ice Given: Find: 10.8 g CO2 molecules CO2 g CO2 mol CO2 molecules CO2 Conceptual Plan: Relationships: 1 mol CO2 = 44.01 g, 1 mol = 6.022 x 1023 Solution: Check: because the given amount is much less than 1 mol CO2, the number makes sense

How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2) Given: Find: 50.0 g PbO2 formula units PbO2 Conceptual Plan: Relationships: 1 mol PbO2 = 239.2 g,1 mol = 6.022 x 1023 g PbO2 mol PbO2 units PbO2 Solution: Check: because the given amount is less than 1 mol PbO2, the number makes sense

What is the mass of 4.78 x 1024 NO2 molecules? Given: Find: 4.78 x 1024 NO2 molecules g NO2 Conceptual Plan: Relationships: 1 mol NO2 = 46.01 g, 1 mol = 6.022 x 1023 molecules mol NO2 g NO2 Solution: Check: because the given amount is more than Avogadro’s number, the mass > 46 g makes sense

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

Example 3.13: Find the mass percent of Cl in C2Cl4F2 Given: Find: C2Cl4F2 % Cl by mass Conceptual Plan: Relationships: Solution: Check: because the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense

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

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 CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains 58.64 g Cl This can be used as a conversion factor 100 g CCl2F2 : 58.64 g Cl

Given that there are 39 g Na in 100 g NaCl and you have 2. 4 g Na… 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 NaCl 100 g NaCl / 39 g Na

Example 3.14: Find the mass of table salt containing 2.4 g of Na Given: Find: 2.4 g Na, 39% Na g NaCl Conceptual Plan: Relationships: 100 g NaCl : 39 g Na g Na g NaCl Solution: Check: because the mass of NaCl is more than 2x the mass of Na, the number makes sense

Benzaldehyde is 79. 2% carbon. What mass of benzaldehyde contains 19 Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C? Given: Find: 19.8 g C, 79.2% C g benzaldehyde Conceptual Plan: Relationships: 100 g benzaldehyde : 79.2 g C g C g benzaldehyde Solution: Check: because the mass of benzaldehyde is more than the mass of C, the number makes sense

Conversion Factors in Chemical Formulas Chemical formulas show inherent relationships between the number of atoms and molecules or between the moles of atoms and molecules These relationships can be used to convert between amounts of constituent elements and molecules like percent composition

Example 3.15: Find the mass of hydrogen in 1.00 gal of water Given: Find: 1.00 gal H2O, dH2O = 1.00 g/ml g H Conceptual Plan: Relationships: 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 H2O gal H2O L H2O mL H2O g H2O g H2O mol H2O moL H g H Solution: Check: because 1 gallon weighs about 3800 g, and H is light, the number makes sense

Find the mass of sodium in 6.2 g of NaCl Given: Find: 6.2 g NaCl g Na Conceptual Plan: Relationships: 1 mol NaCl = 58.44 g, 1 mol Na = 22.99 g, 1 mol Na : 1 mol NaCl g NaCl mol NaCl mol Na g Na Solution: Check: because the amount of Na is less than the amount of NaCl, the answer makes sense

Empirical Formulas via elemental analysis Determined by combustion analysis or % composition If given the percentages of each element, convert them to grams by assuming you have 100 g of total compound If 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 g

How much aspirin do you assume you have at the start? 50 g 1 mol 100 g 1/3 mole 15

Find: the empirical formula, Example: Find the empirical formula of aspirin with the given mass percent composition Information Given: 60.00 g C, 4.48 g H, 35.53 g O Write down the quantity to find and/or its units Find: the empirical formula,

Apply the conceptual plan Example: Find the empirical formula of aspirin with the given mass percent composition Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical Formula, CxHyOz CP: 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 Apply the conceptual plan calculate the moles of each element

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 O 5.00 mol C / 4.44 mol H / 2.22 mol O .200 mol C / 0.225 mol H / 0.450 mol O 721 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 C4.996H4.44O2.220 C5H4O2

Example: Find the empirical formula of aspirin with the given mass percent composition Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, CxHyOz CP: 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 Apply the conceptual plan write the pseudo formula C4.996H4.44O2.220

Example: Find the empirical formula of aspirin with the given mass percent composition Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, CxHyOz CP: 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 Apply the concept plan find the mole ratio by dividing by the smallest number of moles (i.e. normalize the smallest value to 1)

Apply the conceptual plan Example: Find the empirical formula of aspirin with the given mass percent composition Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, CxHyOz CP: 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 Apply the conceptual plan multiply subscripts by factor to give whole number {C2.25H2O1} x 4 C9H8O4

Write a conceptual plan Example: Find the empirical formula of aspirin with the given mass percent composition Information Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, CxHyOz CP: g C,H,O  mol C,H,O  pseudo form.  mol ratio  emp. form. Write a conceptual plan g C mol C whole number ratio mole ratio empirical formula pseudo- formula g H mol H g O mol 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 Formula Convert grams to moles using each element’s molar mass 60.00 g C x 1 mol C/12.01 g C = 4.996 mol C 4.48 g H x 1 mol H/1.008 g H = 4.44 mol H 35.53 g O x 1 mole O/16.00 g O = 2.220 mol O Write a pseudo-formula using moles as subscripts CxHyOz  C4.996H4.44O2.220

Finding an Empirical Formula 5. Divide all subscripts by the smallest number of moles if result is within 0.1 of whole number, round to the whole number find 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 numbers a) 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 4 C9H8O4 

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 F Find: the empirical formula SnxFy Relationships: 1 mol Sn = 118.70 g; 1 mol F = 19.00 g Conceptual plan: whole number ratio g Sn mol Sn mole ratio pseudo- formula empirical formula g F mol F

Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Apply the conceptual plan Sn0.638F1.28 SnF2

Determine the empirical formula of magnetite, which contains 72 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 O Find: FexOy Relationships: 1 mol Fe = 55.85 g; 1 mol O = 16.00 g Conceptual plan: whole number ratio g Fe mol Fe mole ratio pseudo- formula empirical formula g O mol O

Determine the empirical formula of magnetite, which contains 72 Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Apply the conceptual plan  Fe1.30O1.73

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

3.18 – Given the empirical formula, C2H3O, Find the molecular formula of butanedione butanedione MM = 86.03 g/mol molecular formula Conceptual Plan: and Relationships: Solution: Check: the molar mass of the calculated formula is in agreement with the given molar mass

Molecular formula = {C5H3} x 4 = C20H12 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 g H3 = 3(1.01 g) = 3.03 g C5H3 = 63.08 g Molecular formula = {C5H3} x 4 = C20H12

Combustion Analysis Hydrogen  H2O, Oxygen = original mass – (C + H) A common technique for analyzing compounds burn a known mass of the compound weigh the quantity of products made used 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 3.20 Determine the empirical formula of the compound Combustion of a 0.8233 g sample of an organic compound containing only carbon, hydrogen, and oxygen produced the following products: CO2 = 2.445 g H2O = 0.6003 g Determine the empirical formula of the compound

Example 3.19: Find the empirical formula of a compound with the given amounts of combustion products Write 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 CO2 0.6003 g of H2O

Find: empirical formula, CxHyOz Example: Find the empirical formula of a compound with the given amounts of combustion products 1. Write down the given information with units: 0.8233 g compound, 2.445 g CO2 0.6003 g H2O 2. Write down the quantity to find and/or its units Find: empirical formula, CxHyOz 3. Write a conceptual plan g CO2, H2O mol CO2, H2O mol C, H g C, H g O mol O mol C, H, O pseudo formula mol ratio empirical formula

Example: Find the empirical formula of a compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Collect needed relationships 1 mole CO2 = 44.01 g CO2 1 mole H2O = 18.02 g H2O 1 mole C = 12.01 g C 1 mole H = 1.008 g H 1 mole O = 16.00 g O 1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H

Apply the conceptual plan calculate the moles of C and H Example: Find the empirical formula of a compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan calculate the moles of C and H

Apply the conceptual plan calculate the grams of C and H Example: Find the empirical formula of a compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan calculate the grams of C and H

11am Apply the conceptual plan calculate the grams and moles of O Example: Find the empirical formula of a compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan calculate the grams and moles of O 11am

Example: Find the empirical formula of a compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan write a pseudoformula C0.05556H0.06662O0.00556

Apply the conceptual plan Example: Find the empirical formula of a compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan find the mole ratio by dividing by the smallest number of moles

Example: Find the empirical formula of a compound with the given amounts of combustion products Information Given: 0.8233 g compound, 2.445 g CO2, 0.6003 g H2O 0.05556 mol C, 0.6673 g C, 0.06662 mol H, 0.06715 g H, 0.0889 g O, 0.00556 mol O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan multiply subscripts by factor to give whole number, if necessary write the empirical formula 1 12 10 O H C

The smell of dirty gym socks is caused by the compound caproic acid 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)

C H O g 0.524 0.0877 0.232 moles 0.0436 0.0870 0.0145 C0.0436H0.0870O0.0145

Molecular formula = {C3H6O} x 2 = C6H12O2