More on Chemical Compounds Nomenclature Chapter #6 More on Chemical Compounds Nomenclature

Ions Atoms can form ions by gaining or losing electrons. Metals tend to lose one or more electrons, to become isoelectronic with Noble gasses, to form positive ions called cations and are named by using the name of the parent atom. Nonmetals tend to gain electrons, to become isoelectronic with Noble gasses, to form negative ions called anions and are named by using the root of the atom name followed by the suffix –ide. Atoms become more stable by gaining or losing electrons.

Common Oxidation States

Common Oxidation States 1+ 2+ 3- 2- 1- 3+ 4+

Common Oxidation States By Group Number

Sample Problem An ion with a 3+ charge contains 23 electrons. Which ion is it?

Sample Problem An ion with a 3+ charge contains 23 electrons. Which ion is it? Fe3+ A certain ion X+ contains 54 electrons and 78 neutrons. What is the mass number of this ion?

Sample Problem An ion with a 3+ charge contains 23 electrons. Which ion is it? Fe3+ A certain ion X+ contains 54 electrons and 78 neutrons. What is the mass number of this ion?

Sample Problem An ion with a 3+ charge contains 23 electrons. Which ion is it? Fe3+ A certain ion X+ contains 54 electrons and 78 neutrons. What is the mass number of this ion? 133

Ionic Compounds Ions combine to form ionic compounds. The simplest ratio between the ions is called the empirical formula or a formula unit. Properties of ionic compounds High melting points Conduct electricity If melted If dissolved in water Ionic compounds are electrically neutral. The charges on the anions and cations in the compound must sum to zero.

Ionic Compounds Crystal Lattice of NaCl Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites. Ions combine in the ratio that results in zero charge to form ionic compounds. Which ions are the smaller ones? Crystal Lattice of NaCl

Formula Writing Rules Examples Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers (O.N. = P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus

Formula Writing Rules Examples Na O Na N Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O Na N Ca O Ca P

Formula Writing Rules Examples Na O Na N Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O 1+ 3- Na N Ca O Ca P

Formula Writing Rules Examples Na O Na N Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O 1+ 3- Na N 2+ 2- Ca O Ca P

Formula Writing Rules Examples Na O Na N Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O 1+ 3- Na N 2+ 2- Ca O 2+ Ca P 3-

Formula Writing Rules Examples Na O Na N Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O 1+ 3- Na N 2+ 2- Ca O 2+ Ca P 3-

Formula Writing Rules Examples Na O Na O Na N Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O Na O ≡ 2 1+ 3- Na N 2+ 2- Ca O 2+ Ca P 3-

Formula Writing Rules Examples Na O Na O Na N Na N Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O Na O ≡ 2 1+ 3- Na N Na N ≡ 3 2+ 2- Ca O 2+ Ca P 3-

Formula Writing Rules Examples Na O Na O Na N Na N Ca O Ca O Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O Na O ≡ 2 1+ 3- Na N Na ≡ N 3 2+ 2- Ca O Ca O Ca O ≡ 2 2 reduced 2+ Ca P 3-

Formula Writing Rules Examples Na O Na O Na N Na N Ca O Ca O Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O Na O ≡ 2 1+ 3- Na N Na N ≡ 3 2+ 2- Ca O Ca O Ca O ≡ 2 2 reduced 2+ Ca P 3-

Formula Writing Rules Examples Na O Na O Na N Na N Ca O Ca O Ca O Ca P Step 1 Write the symbols of the elements. Step 2 Assign oxidation numbers O.N. = (P – e) Step 3 Slide with Clyde! (number only) Step 4 Reduce if the compound is ionic Examples Write formulas for the following two elements 1+ Sodium and oxygen Sodium and nitrogen Calcium and oxygen Calcium and phosphorus 2- Na O Na O ≡ 2 1+ 3- Na N Na N ≡ 3 2+ 2- Ca O Ca O Ca ≡ O 2 2 reduced 2+ Ca P 3- Ca P ≡ 3 2

NOMENCLATURE I. Binary Ionic compounds Binary means two different elements Ionic means metal and nonmetal Step 1 First give the name of the metal, followed by the nonmetal name using the “ide” suffix. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to to describe the charge of the metal. Except Ag, Zn,and Al Examples NaCl Al2O3 FeCl2 FeCl3

NOMENCLATURE I. Binary Ionic compounds Binary means two different elements Ionic means metal and nonmetal Step 1 First give the name of the metal, followed by the nonmetal name using the “ide” suffix. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to to describe the charge of the metal. Except Ag, Zn,and Al Examples NaCl Sodium chloride Al2O3 FeCl2 FeCl3

NOMENCLATURE I. Binary Ionic compounds Binary means two different elements Ionic means metal and nonmetal Step 1 First give the name of the metal, followed by the nonmetal name using the “ide” suffix. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to to describe the charge of the metal. Except Ag, Zn,and Al Examples NaCl Sodium chloride Al2O3 Aluminum oxide FeCl2 FeCl3

NOMENCLATURE I. Binary Ionic compounds Binary means two different elements Ionic means metal and nonmetal Step 1 First give the name of the metal, followed by the nonmetal name using the “ide” suffix. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to to describe the charge of the metal. Except Ag, Zn,and Al Examples NaCl Sodium chloride Al2O3 Aluminum oxide FeCl2 Iron(II) chloride FeCl3

NOMENCLATURE I. Binary Ionic compounds Binary means two different elements Ionic means metal and nonmetal Step 1 First give the name of the metal, followed by the nonmetal name using the “ide” suffix. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to to describe the charge of the metal. Except Ag, Zn,and Al Examples NaCl Sodium chloride Al2O3 Aluminum oxide FeCl2 Iron(II) chloride FeCl3 iron(III) chloride

Nonbinary means more than two different elements II. Nonbinary Ionic compounds Nonbinary means more than two different elements Step 1 First give the name of the metal, followed by the memorized polyatomic ion name. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to describe the charge of the metal. Except Ag, Zn, and Al. Examples NaOH Fe(NO3)3 Zn(C2H3O2)2 Fe(SO4)2

Memorized Polyatomic Ion List Formula Name NH4+ Ammonium O22- Peroxide C2H3O2- Acetate NO3- Nitrate CO32- Carbonate NO2- Nitrite HCO31- Hydorgen carbonate SO42- Sulfate ClO4- Perchlorate SO32- Sulfite ClO3- Chlorate PO43- Phosphate ClO2- Chlorite PO33- Phosphite ClO- Hypochlorite CrO42- Chromate CN- Cyanide Cr2O72- Dichromate OH- Hydroxide

Nonbinary means more than two different elements II. Nonbinary Ionic compounds Nonbinary means more than two different elements Step 1 First give the name of the metal, followed by the memorized polyatomic ion name. Step 2 CO If the metal is to the right of group IIA, then a Roman numeral is used after the metal to describe the charge of the metal. Except Ag, Zn, and Al. Examples NaOH Sodium hydroxide Fe(NO3)3 Zn(C2H3O2)2 Fe(SO4)2

Nonbinary means more than two different elements II. Nonbinary Ionic compounds Nonbinary means more than two different elements Step 1 First give the name of the metal, followed by the memorized polyatomic ion name. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to describe the charge of the metal. Except Ag, Zn, and Al. Examples NaOH Sodium hydroxide Fe(NO3)3 Iron(III) nitrate Zn(C2H3O2)2 FeSO4

Nonbinary means more than two different elements II. Nonbinary Ionic compounds Nonbinary means more than two different elements Step 1 First give the name of the metal, followed by the memorized polyatomic ion name. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to describe the charge of the metal. Except Ag, Zn, and Al. Examples NaOH Sodium hydroxide Fe(NO3)3 Iron(III) nitrate Zn(C2H3O2)2 FeSO4 Iron(II) sulfate

Nonbinary means more than two different elements II. Nonbinary Ionic compounds Nonbinary means more than two different elements Step 1 First give the name of the metal, followed by the memorized polyatomic ion name. Step 2 If the metal is to the right of group IIA, then a Roman numeral is used after the metal to describe the charge of the metal. Except Ag, Zn, and Al. Examples NaOH Sodium hydroxide Fe(NO3)3 Iron(III) nitrate Zn(C2H3O2)2 Fe(SO4)2 Iron(II) sulfate Iron(II) sulfate

III. Binary molecular Compounds Molecular means nonmetals Step 1 First give the name of the first nonmetal, followed by the nonmetal name using the “ide” suffix. Step 2 Give each nonmetal a Latin prefix describing the number of atoms present in the compound. Examples CO CO2 P2O5 CCl4

You will need to learn the Greek numerical prefixes (Table 4.6): Number Prefix 1 Mono-* 2 Di- 3 Tri- 4 Tetra- 5 Penta- 6 Hexa- 7 Hepta- 8 Octa- 9 Nona- 10 Deca- *Note 1 Compound names never start with mono Note 2 When adding a prefix two vowls cannot next to each other

III. Binary Molecular Compounds Molecular means nonmetals Step 1 First give the name of the first nonmetal, followed by the nonmetal name using the “ide” suffix. Step 2 Give each nonmetal a Latin prefix describing the number of atoms present in the compound. Examples CO Carbon monoxide CO2 P2O5 CCl4

III. Binary Molecular Compounds Molecular means nonmetals Step 1 First give the name of the first nonmetal, followed by the nonmetal name using the “ide” suffix. Step 2 Give each nonmetal a Latin prefix describing the number of atoms present in the compound. Examples CO Carbon monoxide CO2 Carbon dioxide P2O5 CCl4

III. Binary Molecular Compounds Molecular means nonmetals Step 1 First give the name of the first nonmetal, followed by the nonmetal name using the “ide” suffix. Step 2 Give each nonmetal a Latin prefix describing the number of atoms present in the compound. Examples CO Carbon monoxide CO2 Carbon dioxide P2O5 CCl4

II. Binary Ionic molecular Molecular means combination of nonmetals Step 1 First give the name of the first nonmetal, followed by the nonmetal name using the “ide” suffix. Step 2 Give each nonmetal a Latin prefix describing the number of atoms present in the compound. Examples CO Carbon monoxide CO2 Carbon dioxide P2O5 Diphosphorus Pentoxide CCl4

II. Binary Ionic molecular Molecular means combination of nonmetals Step 1 First give the name of the first nonmetal, followed by the nonmetal name using the “ide” suffix. Step 2 Give each nonmetal a Latin prefix describing the number of atoms present in the compound. Examples CO Carbon monoxide CO2 Carbon dioxide P2O5 Diphosphorus Pentoxide CCl4 Carbon tetrachloride

III. Nonbinary Molecular Compounds Note: Do not use Latin prefixes Step 1 Write down the memorized polyatomic ions present in the compound. Step 2 Look to see if any monatomic ions are present. If so, then cations use the normal name. If it is an anion, then its name comes last with the “ide” suffix. Examples NH4Cl NH4OH

III. Nonbinary Molecular Compounds Note: Do not use Latin prefixes Step 1 Write down the memorized polyatomic ions present in the compound. Step 2 Look to see if any monatomic ions are present. If so, then cations use the normal name. If it is an anion, then its name comes last with the “ide” suffix. Examples NH4Cl Ammonium chloride NH4OH

III. Nonbinary Molecular Compounds Note: Do not use Latin prefixes Step 1 Write down the memorized polyatomic ions present in the compound. Step 2 Look to see if any monatomic ions are present. If so, then cations use the normal name. If it is an anion, then its name comes last with the “ide” suffix. Examples NH4Cl Ammonium chloride NH4OH Ammonium hydroxide

Compounds that Start with Hydrogen Case 1 If the anion ends in “ide” and it is aqueous, then use the prefix hydro and suffix “ic acid” Case 2 If the anion ends in “ate” then drop it and add the suffix “ic acid” Case 3 If the anion ends in “ite” then drop it and add the suffix “ous acid” Case 4 If the anion ends in” ide” and is a gas, or liquid, then leave the name and do not use Latin prefiex

Compounds that Start with Hydrogen Examples HCl (aq) HNO3 HNO2 H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride HNO3 HNO2 H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid HNO3 HNO2 H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid hydrochloric acid HNO3 HNO2 H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) Chloride chloric acid hydrochloric acid HNO3 nitrate HNO2 H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) Chloride chloric acid hydrochloric acid HNO3 nitrate nitric acid HNO2 H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid hydrochloric acid HNO3 nitrate nitric acid HNO2 nitrite H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid hydrochloric acid HNO3 nitrate nitric acid HNO2 nitrite nitrous acid H2O (l) HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid hydrochloric acid HNO3 nitrate nitric acid HNO2 nitrite nitrous acid H2O (l) oxide HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid hydrochloric acid HNO3 nitrate nitric acid HNO2 nitrite nitrous acid H2O (l) oxide hydrogen oxide HCl (g)

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid hydrochloric acid HNO3 nitrate nitric acid HNO2 nitrite nitrous acid H2O (l) oxide hydrogen oxide HCl (g) chloride

Compounds that Start with Hydrogen Examples HCl (aq) chloride chloric acid hydrochloric acid HNO3 nitrate nitric acid HNO2 nitrite nitrous acid H2O (l) oxide hydrogen oxide HCl (g) chloride hydrogenchloride

Formula Weight Calculation To calculate the molar mass of a compound we sum together the atomic weights of the atoms that make up the formula of the compound. This is called the formula weight (MW, M). Formula weights are the sum of atomic weights of atoms Making up the formula. The following outlines how to find the formula weight of water symbol weight number H 1.01 X 2 = 2.02 O 16.0 X 1 = 16.0 18.0 g/mole

Percent Composition Find the formula weight and the percent composition of glucose (C6H12O6) symbol weight number 72.0 C 12.0 x 6 = H 1.01 x 12 = 12.12 O 16.0 x 6 = 96.0 180.1 g/mole 72.0 %C = X = 40.0 %C 180.1 12.12 %H = X = 6.73 %H 180.1 96.0 X = 53.3 %O %O = 180.1

“the mass of one mole of that substance.” Formula Weight We can define the molar mass or molecular weight of a substance can as: “the mass of one mole of that substance.” We give molar mass the symbol M and it has units g/mol- For an atom the molar mass is equal to the atomic weight we find on the period table.

Its All About Moles For a molecule the molecular weight is equal to the sum of the atomic weights of the atoms that make it up. The mathematical relationship between mass and moles is: m = nM This can be summarized by the following diagram: moles

Mole Concepts The key equation to remember for mole calculations is: m = nM Where: M = molecular weight (gmol-1) n = number of moles (mol) m = mass of sample (g)

Mole Concepts For a mole of a molecule the number of moles of each atom is determined by how many of that atom are in each molecule. e.g. One mole of H2O contains: One mole of oxygen atoms Two moles of hydrogen atoms In 5 moles of H2SO4 how many moles of oxygen atoms is there? 20 moles of O atoms.

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 mole H2SO4 = 98.0g of H2SO4

Mole Conversions In 50.0g of H2SO4 how many moles of oxygen atoms are there? 50.0g of H2SO4 mole H2SO4 4mole O = 2.04 mole O mole H2SO4 98.0g of H2SO4

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present?

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 =

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 4 mole O mole H2SO4

Mole Conversions In 5 moles of H2SO4 how many atoms of oxygen are present? 5 moles H2SO4 4 mole O 6.02 x 1023 atoms O = mole H2SO4 mole O 1.20 x 1025 atoms

Empirical Formulas Empirical formula is the smallest whole number ratio between atoms and can be calculated from the percent composition. Molecular formulas happen to be the exact number of atoms making up a molecule, and may or may no be the simplest whole number ratio. Molecular formulas are whole number multiples of the empirical formula.

Empirical Formula Steps Assume 100 g of compound. Convert percent to a mass number. Convert the mass to moles. Divide each mole number by the smallest mole number. Rounding: If the decimal is ≤ 0.1, then drop the decimals If the decimal is ≥0.9, then round up. All other decimal need to be multiplied by a whole number until roundable.

Empirical Formula Example A compound is composed of 75.0% C and 25.0% H. Find its empirical formula. Step #1 Assume 100 g of compound 75.0 g C 25.0 g H

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #2 Convert grams to moles. Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #3 Divide each mole number by the smallest. Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1.00 = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1.00 = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 3.98 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 4 H 6.225 6.225

Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up Mole C 75.0 g C = 6.225 mole C 12.01 g 25.0 g H Mole H = 24.802 mole H 1.008 g H 6.225 24.802 = 1 C = 4 H 6.225 6.225 Empirical Formula = CH4

Molecular Formulas Empirical formula, is the smallest ratio between atoms in a molecular or formula unit. Molecular formula, is the exact number of atoms in a molecule; a whole number multiple of an empirical formula

Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C3H5O C3H5O C3H5O

6-9 Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C6H10O2 C3H5O C3H5O C3H5O

Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C6H10O2 C9H15O3 C3H5O C3H5O C3H5O

Possible Molecular Formulas Assume an empirical formula of C3H5O Empirical formula Integer Molecular Formula C3H5O C3H5O 1 2 3 4 5 C3H5O C6H10O2 C9H15O3 C3H5O C3H5O C12H20O4 C3H5O C15H25O5

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #1 Assume 100g of compound 83.6 g C 16.3 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C mole 12.01 g C 16.3 g H mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C mole = 6.961 mole 12.01 g C 16.3 g H mole = 16.17 mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 83.6 g C mole = 6.961 mole 12.01 g C 16.3 g H mole = 16.17 mole 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 6.961 83.6 g C mole = 1.00 = 6.961 mole 6.961 12.01 g C 16.3 g H mole = 16.17 mole = 2.32 1.008 g H

Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #4 Round if---Not Roundable 6.961 83.6 g C mole = 6.961 mole = 1.00 6.961 12.01 g C 16.3 g H mole = 16.17 mole = 2.32 1.008 g H Step #4, Multiply by an integer until roundable 1.00 X 3 = 3 2.32 X 3 = 7 Empirical formula C3H7

Molecular Formula Integer Divide empirical weight into molecular weight 3x12 + 7x1 =43 2 43 86 Now multiply the empirical formula by 2

Molecular Formula Integer Divide empirical weight into molecular weight 3x12 + 7x1 =43 2 43 86 Now multiply the empirical formula by 2 Molecular Formula is C6 H14

The End