H O H C C O H H
Molecular Compounds A molecular compound is made up of discrete units called molecules, which typically consist of a small number of non-metal atoms held together by covalent bonds. Molecular compounds are represented by chemical formulas, symbolic representations that, at minimum, indicate: the elements present and the relative number of atoms of each element. The 2 elements
H2O
Example:
Two H atoms per
Lack of subscript means 1 atoms of O per molecule
Empirical Formula The empirical formula is the simplest formula for a compound where its subscripts are reduced to their simplest whole number ratios. Generally, the empirical formula does not tell one a significant about of information about a given compound. Acetic acid (C2H4O2) and glucose (C6H12O6) both have the empirical formula CH2O. Molecular Formula The molecular formula is based on an actual molecule of a compound. In some cases, the empirical and molecular formulas are identical, such as formaldehyde (CH2O). In other cases, the molecular formula is a multiple of the empirical formula. Both the empirical and molecular formulas provide information on the combining rations of the atoms in the compounds, however, they show nothing on how the atoms are attached relative to each other. Structural Formula The structural formula shows the order in which atoms are bonded together in a molecule and by what types of bonds. For example acetic acid:
H
H
O
C
C
O
H
H The covalent bonds in the structural formula are represented by lines or dashes and a double bond is represented by a double dash.
Condensed Structural Formula This formula is a representation of the molecular structure of a compound which is written in one line and is an alternate way of showing how the atoms of a molecule are connected. For example with acetic acid: CH3COOH or CH3CO2H. It can also be used to show how a group of atoms is attached to another atom. For example, methylpropane (C4H10).
CH3 CH3 CH CH3
can be written as
CH3CH (CH3) CH3
or as
CH (CH3)3
CH3 The parentheses indicate that that ligand is connected to the molecule at that given point. Line angle Formula The line angle formula contains lines that represent chemical bonds. For example, the male hormone molecule, testosterone.
OH
H3C H3 C O Ionic Compounds An ionic compound is made up of positive and negative ions joined together by electrostatic forces of attraction. The atoms of metallic elements tend to lose one or more electrons when combining with non-metal atoms. The non-metal atoms, in return, gains one or more electrons in the process. When the metal loses an electron during transfer, it becomes a cation (positive ion) and the non-metal atom becomes an anion (negative ion). Consider NaCl. An ionic crystal of NaCl shows that one Na+ ion is surrounded by six Cl- ions. We cannot pick any one of these Cl- ions and exclusively associate it with a given Na+, however, since the ratio of chloride ions to sodium ions is 1:1, and so a
combination of one sodium ion and one chloride ion as a formula unit. The formula unit of an ionic compound is the smallest electrically neutral collection of ions. The ratio of atoms in the formula unit is essentially the same as in the chemical formula. Since a formula unit is only a fragment of a vast network of ion (crystal), a formula unit cannot be distinguished as distinct. Therefore, it is incorrect to label a formula unit as a molecule. Ions such as Na+, Mg2+ and Cl- are monatomic, meaning that each consists of a single ionized atom. Alternately, and ion made up of two or more atoms are called polyatomic ions (e.g. NO3-). (Study polyatomic ions) The Mole Concept and Chemical Compounds Formula mass: the mass of a formula unit in atomic mass units. It is appropriate to use the term formula mass, but for molecular compounds, the formula unit is actually a molecule, so the term molecular mass is used. The molecular mass is the mass of a molecule in atomic mass units (amu). Mole of a Compound The molar mass is the mass of one mole of compound – one mole of molecules of a molecular compound and one mole of formula units of an ionic compound. Problem: An analytical balance can detect a mass of 0.1 mg. What is the total number of ions present in this minimally detectable quantity of MgCl2? Solution: Use molar mass to convert from mass to number of moles of MgCl2. Then use Avogadro constant as a conversion factor to convert moles to number of formula units (fu). Finally, there are three ions (1 Mg2+ and two Cl-) per formula unit of magnesium chloride. ? ions = [0.1mg MgCl2 x (1 g MgCl2 / 1000 mg MgCl2) x (1 mol MgCl2 / 95 g MgCl2) x (6.0 x 10²³ fu MgCl2 / 1 mol MgCl2) x (3 ions / 1 fu MgCl2)] = 2 x 1018 ions
Problem: The volatile liquid ethyl mercaptan, C2H6S, is one of the most odoriferous substances known. It is sometimes added to natural gas to make gas leaks detectable. How many C2H6S molecules are contained in a 1.0 µL sample? (d=0.84 g/mL) Solution: ? molecules C2H6S =
[1.0 µL x (1 x 10-6 / 1µL) x (1000mL / 1L) x (8.4 g C2H6S / 1 mL) x (1 mol C2H6S / 62.1 g C2H6S) x (6.022 x 10²³ molecule C2H6S / 1 mol C2H6S] = 8.1 x 1018 molecules C2H6S
Some Familiar Molecular Formulas H2, O2, N2, F2, Cl2, Br2, I2, P4, S8 Composition of Chemical Compounds Consider Halothane C2HBrClF3
Mole ratio: nc / nhalothane
Mass ratio: mc / mhalothane
M(C2HBrClF3)= 2MC + MH + MBR + MCl + 3MF =(2 x 12.01) + 1.01 + 79.90 + 35.45 + (3 X 19.00) = 197.38 g / mol Problem: How many moles of F atoms are in a 75.0 mL sample of halothane (d = 1.871 g / mL)? Solution: ? mol F = 75.0 mL C2HBrClF3 x (1.871g C2HBrClF3 / 1 mL C2HBrClF3) x (1 mol C2HBrClF3 / 197.4g C2HBrClF3) x (3 mol F / 1 mol C2HBrClF3) Calculating Percent Composition From a Chemical Formula 1. Determine the molar mass of the compound. This is the denominator in the equation. 2. Determine the contribution of the given element to the molar mass. This product of the formula subscript and the molar mass of the element appears in the numerator of the equation. 3. Formulate the ratio of the mass of the given element to the mass of the compound as a whole. This is the ratio of the numerator from step 2 to the denominator in step 1. 4. Multiply this ratio by 100% to obtain the mass percent of the element. Mass % element = [(# of atoms of element per formula unit) x (molar mass of element) / molar mass of compound] x 100%
Problem: What is the mass percent composition of halothane, C2HBrClF3? Solution:
%C = [(2 mol C) x (12.01 g C / 1 mol C) / 197.38g C2HBrClF3] x 100% = 12.17% C %H= [(1 mol H) x (1.01 g H / 1 mol H) / 197.38g C2HBrClF3] x 100% = 0.51% H %C = [(1 mol Br) x (79.70 g Br / 1 mol Br) / 197.38g C2HBrClF3] x 100% = 40.48% Br %C = [(1 mol Cl) x (35.45 g Cl / 1 mol Cl) / 197.38g C2HBrClF3] x 100% = 17.96% Cl %C = [(3 mol F) x (19.00 g CF/ 1 mol F) / 197.38g C2HBrClF3] x 100% = 28.88% F
Establishing Formulas from Experimentally Determined Percent Consider the following question: Determine a formula form the experimentally determined percent composition of the compound 2-deoxyribose, which is 44.77% C, 7.52% H and 47.71% O. 1. Choose an arbitrary sample size (100g): 44.77 g C, 7.52 g H and 47.71 g O 2. Convert masses to amounts in moles ? mol C = [44.77g C x (1 mol C / 12.011 g C)] = 3.727 mol C ? mol H = [7.52g H x (1 mol H / 1.008 g H)] = 7.46 mol H ? mol O = [47.71g O x (1 mol O / 15.999 g O)] = 2.982 mol O 3. Write a formula: C3.727H7.46O2.982
4. Convert formula to small whole numbers: C(3.727/
H(7.46/ 2.982)O(2.982/2.982) = C(1.25)H(2.50)O
2.982)
5. Multiply all subscripts by a small whole integer to make the subscripts integral:
C(1.25x4)H(2.50x4)O(1x4) = C5H10O4 Combustion Analysis
Before Combustion
After Combustion
CxHyOz
xCO2
and
and
O2
y/2 H2O
Combustion Problem Vitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of these carbon-hydrogen-oxygen compound yields 0.2998 g CO2 and 0.00819 g H2O. What are the percent composition and the empirical formula of vitamin C? Solution: First, determine the mass of carbon in sample by converting to mol C... ? mol C = [0.2998 g CO2 x (1 mol CO2 / 44.010 g CO2) x ( 1 mol C / 1 mol CO2)] = 0.006812 mol C ...and then to g C... ? g C = [0.006182 mol C x (12.011 g C / 1 mol C)] = 0.08182 g C
...Proceed in similar method for H2O ? mol H = [0.0819 g H2O x (1 mol H2O / 18.02 g H2O) x ( 2 mol H / 1 mol H2O)] = 0.00909 mol C ...and... ? g H = [0.00909 mol H x (1.008 g H / 1 mol H)] = 0.00916 g H ...obtain the mass of O in the sample as a difference... ? g O = 0.2000 g sample – 0.08182 g C – 0-00916 g H = 0.1090 g O ...Finally, multiply the mass fractions of the three elements by 100% to obtain mass percentages. %C = (0.08182 g C / 0.2000 g sample) x 100% = 40.91% C %H = (0.00916 g H / 0.2000 g sample) x 100% = 4.58% H %O= (0.1090 g O / 0.2000 g sample) x 100% = 54.50% O From here, we must obtain the empirical formula by calculating the number of oxygen. ? mol O = 0.1090 g O x (1 mol O / 15.9994 g O) = 0.006813 mol O
From the number of moles of each element, we now have a tentative empirical
C0.006812H0.00909O0.006813
formula:
Next divide each subscript by the smallest (0.006812) to obtain:
CH1.33O Multiply all subscripts by three to get:
C3H4O3
Oxidation States Metals tend to lose electrons. Na becomes Na+ + e-. This is known as oxidation. Non-metals, however, tend to gain electrons. Cl + e- become Cl-. This is known as reduction. Rules for Oxidation States 1. The oxidation state (OS) of an individual atom in a free element is 0. 2. The total of the OS in all atoms in: i. Neutral species is 0. ii. Ionic species is equal to the charge on the ion. 3. In their compounds, the alkali metals and the alkaline earths have OS of +1 and +2 respectively. 4. In compounds the OS of fluorine is ALWAYS -1. 5. In compounds, the OS of hydrogen is USUALLY +1. 6. In compounds, the OS of oxygen is USUALLY -2.
7. In binary (two element) compounds with metals: i. Halogens have OS of -1, ii. Group 16 have OS of -2 and iii. Group 15 have OS of -3 8. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have it were a monatomic ion. ***Review Oxidation States on Page 83*** Chemical Nomenclature
***STUDY : WILL BE ON TEST and EXAM*** •
Ionic Compounds are often a combination of metal and non-metal. The anion (non-metal), add –ide to the element name.
Examples: BaCl2: barium chloride; K2O: potassium oxide; Mg(OH)2: magnesium hydroxide; KNO3: potassium nitrate •
Transition metal ionic compounds must be indicated with a charge on metal with Roman Numerals:
Example: FeCl2: iron (II) Chloride; FeCl3: iron (III) chloride; Cr2S3: chromium (III) sulphide •
Molecular Compounds
Example: HI: hydrogen iodide; NF3: nitrogen trifluoride; SO2: Sulfur dioxide; N2Cl4: dinitrogen tetrachloride; NO2: nitrogen dioxide; N2O: dinitrogen monoxide •
Naming Oxoacids and Oxoanions Oxoacid Per- -ic acid Representat ive –ic acid
-ous acid
Hypo- ous acid
Oxidation State
Removal of all
Oxoanion Per- -ate
+ O
-ate
+ O
-ite
+ O
Hypo- -ite
Formula of Acid Name of Acid
Cl: +1 HClO hypochlorite
Hypochlorous acid
Cl: +3
Chlorous acid
HClO2 Sodium chlorite
Cl: +5 chlorate
HClO3
Chloric acid
Formula of Salt Name of Salt
NaClO
Sodium NaClO2
NaClO3
Sodium
Cl: +7
HClO4 Perchloric acid Sodium perchlorate
NaClO4
N: +3
HNO2
Nitrous acid
NaNO2
Sodium nitrite
N: +5
HNO3
Nitric acid
NaNO3
Sodium nitrate
S: +4
H2SO3 Sodium sulfite
S: +6 sulphite
H2SO4
Sulfurous acid Sulfuric acid
Na2SO3 Na2SO4
Sodium
H2O
Example:
Two H atoms per
Lack of subscript means 1 atoms of O per molecule
Empirical Formula The empirical formula is the simplest formula for a compound where its subscripts are reduced to their simplest whole number ratios. Generally, the empirical formula does not tell one a significant about of information about a given compound. Acetic acid (C2H4O2) and glucose (C6H12O6) both have the empirical formula CH2O. Molecular Formula The molecular formula is based on an actual molecule of a compound. In some cases, the empirical and molecular formulas are identical, such as formaldehyde (CH2O). In other cases, the molecular formula is a multiple of the empirical formula. Both the empirical and molecular formulas provide information on the combining rations of the atoms in the compounds, however, they show nothing on how the atoms are attached relative to each other. Structural Formula The structural formula shows the order in which atoms are bonded together in a molecule and by what types of bonds. For example acetic acid:
H
H
O
C
C
O
H
H The covalent bonds in the structural formula are represented by lines or dashes and a double bond is represented by a double dash.
Condensed Structural Formula This formula is a representation of the molecular structure of a compound which is written in one line and is an alternate way of showing how the atoms of a molecule are connected. For example with acetic acid: CH3COOH or CH3CO2H. It can also be used to show how a group of atoms is attached to another atom. For example, methylpropane (C4H10).
CH3 CH3 CH CH3
can be written as
CH3CH (CH3) CH3
or as
CH (CH3)3
CH3 The parentheses indicate that that ligand is connected to the molecule at that given point. Line angle Formula The line angle formula contains lines that represent chemical bonds. For example, the male hormone molecule, testosterone.
OH
H3C H3 C O Ionic Compounds An ionic compound is made up of positive and negative ions joined together by electrostatic forces of attraction. The atoms of metallic elements tend to lose one or more electrons when combining with non-metal atoms. The non-metal atoms, in return, gains one or more electrons in the process. When the metal loses an electron during transfer, it becomes a cation (positive ion) and the non-metal atom becomes an anion (negative ion). Consider NaCl. An ionic crystal of NaCl shows that one Na+ ion is surrounded by six Cl- ions. We cannot pick any one of these Cl- ions and exclusively associate it with a given Na+, however, since the ratio of chloride ions to sodium ions is 1:1, and so a
combination of one sodium ion and one chloride ion as a formula unit. The formula unit of an ionic compound is the smallest electrically neutral collection of ions. The ratio of atoms in the formula unit is essentially the same as in the chemical formula. Since a formula unit is only a fragment of a vast network of ion (crystal), a formula unit cannot be distinguished as distinct. Therefore, it is incorrect to label a formula unit as a molecule. Ions such as Na+, Mg2+ and Cl- are monatomic, meaning that each consists of a single ionized atom. Alternately, and ion made up of two or more atoms are called polyatomic ions (e.g. NO3-). (Study polyatomic ions) The Mole Concept and Chemical Compounds Formula mass: the mass of a formula unit in atomic mass units. It is appropriate to use the term formula mass, but for molecular compounds, the formula unit is actually a molecule, so the term molecular mass is used. The molecular mass is the mass of a molecule in atomic mass units (amu). Mole of a Compound The molar mass is the mass of one mole of compound – one mole of molecules of a molecular compound and one mole of formula units of an ionic compound. Problem: An analytical balance can detect a mass of 0.1 mg. What is the total number of ions present in this minimally detectable quantity of MgCl2? Solution: Use molar mass to convert from mass to number of moles of MgCl2. Then use Avogadro constant as a conversion factor to convert moles to number of formula units (fu). Finally, there are three ions (1 Mg2+ and two Cl-) per formula unit of magnesium chloride. ? ions = [0.1mg MgCl2 x (1 g MgCl2 / 1000 mg MgCl2) x (1 mol MgCl2 / 95 g MgCl2) x (6.0 x 10²³ fu MgCl2 / 1 mol MgCl2) x (3 ions / 1 fu MgCl2)] = 2 x 1018 ions
Problem: The volatile liquid ethyl mercaptan, C2H6S, is one of the most odoriferous substances known. It is sometimes added to natural gas to make gas leaks detectable. How many C2H6S molecules are contained in a 1.0 µL sample? (d=0.84 g/mL) Solution: ? molecules C2H6S =
[1.0 µL x (1 x 10-6 / 1µL) x (1000mL / 1L) x (8.4 g C2H6S / 1 mL) x (1 mol C2H6S / 62.1 g C2H6S) x (6.022 x 10²³ molecule C2H6S / 1 mol C2H6S] = 8.1 x 1018 molecules C2H6S
Some Familiar Molecular Formulas H2, O2, N2, F2, Cl2, Br2, I2, P4, S8 Composition of Chemical Compounds Consider Halothane C2HBrClF3
Mole ratio: nc / nhalothane
Mass ratio: mc / mhalothane
M(C2HBrClF3)= 2MC + MH + MBR + MCl + 3MF =(2 x 12.01) + 1.01 + 79.90 + 35.45 + (3 X 19.00) = 197.38 g / mol Problem: How many moles of F atoms are in a 75.0 mL sample of halothane (d = 1.871 g / mL)? Solution: ? mol F = 75.0 mL C2HBrClF3 x (1.871g C2HBrClF3 / 1 mL C2HBrClF3) x (1 mol C2HBrClF3 / 197.4g C2HBrClF3) x (3 mol F / 1 mol C2HBrClF3) Calculating Percent Composition From a Chemical Formula 1. Determine the molar mass of the compound. This is the denominator in the equation. 2. Determine the contribution of the given element to the molar mass. This product of the formula subscript and the molar mass of the element appears in the numerator of the equation. 3. Formulate the ratio of the mass of the given element to the mass of the compound as a whole. This is the ratio of the numerator from step 2 to the denominator in step 1. 4. Multiply this ratio by 100% to obtain the mass percent of the element. Mass % element = [(# of atoms of element per formula unit) x (molar mass of element) / molar mass of compound] x 100%
Problem: What is the mass percent composition of halothane, C2HBrClF3? Solution:
%C = [(2 mol C) x (12.01 g C / 1 mol C) / 197.38g C2HBrClF3] x 100% = 12.17% C %H= [(1 mol H) x (1.01 g H / 1 mol H) / 197.38g C2HBrClF3] x 100% = 0.51% H %C = [(1 mol Br) x (79.70 g Br / 1 mol Br) / 197.38g C2HBrClF3] x 100% = 40.48% Br %C = [(1 mol Cl) x (35.45 g Cl / 1 mol Cl) / 197.38g C2HBrClF3] x 100% = 17.96% Cl %C = [(3 mol F) x (19.00 g CF/ 1 mol F) / 197.38g C2HBrClF3] x 100% = 28.88% F
Establishing Formulas from Experimentally Determined Percent Consider the following question: Determine a formula form the experimentally determined percent composition of the compound 2-deoxyribose, which is 44.77% C, 7.52% H and 47.71% O. 1. Choose an arbitrary sample size (100g): 44.77 g C, 7.52 g H and 47.71 g O 2. Convert masses to amounts in moles ? mol C = [44.77g C x (1 mol C / 12.011 g C)] = 3.727 mol C ? mol H = [7.52g H x (1 mol H / 1.008 g H)] = 7.46 mol H ? mol O = [47.71g O x (1 mol O / 15.999 g O)] = 2.982 mol O 3. Write a formula: C3.727H7.46O2.982
4. Convert formula to small whole numbers: C(3.727/
H(7.46/ 2.982)O(2.982/2.982) = C(1.25)H(2.50)O
2.982)
5. Multiply all subscripts by a small whole integer to make the subscripts integral:
C(1.25x4)H(2.50x4)O(1x4) = C5H10O4 Combustion Analysis
Before Combustion
After Combustion
CxHyOz
xCO2
and
and
O2
y/2 H2O
Combustion Problem Vitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of these carbon-hydrogen-oxygen compound yields 0.2998 g CO2 and 0.00819 g H2O. What are the percent composition and the empirical formula of vitamin C? Solution: First, determine the mass of carbon in sample by converting to mol C... ? mol C = [0.2998 g CO2 x (1 mol CO2 / 44.010 g CO2) x ( 1 mol C / 1 mol CO2)] = 0.006812 mol C ...and then to g C... ? g C = [0.006182 mol C x (12.011 g C / 1 mol C)] = 0.08182 g C
...Proceed in similar method for H2O ? mol H = [0.0819 g H2O x (1 mol H2O / 18.02 g H2O) x ( 2 mol H / 1 mol H2O)] = 0.00909 mol C ...and... ? g H = [0.00909 mol H x (1.008 g H / 1 mol H)] = 0.00916 g H ...obtain the mass of O in the sample as a difference... ? g O = 0.2000 g sample – 0.08182 g C – 0-00916 g H = 0.1090 g O ...Finally, multiply the mass fractions of the three elements by 100% to obtain mass percentages. %C = (0.08182 g C / 0.2000 g sample) x 100% = 40.91% C %H = (0.00916 g H / 0.2000 g sample) x 100% = 4.58% H %O= (0.1090 g O / 0.2000 g sample) x 100% = 54.50% O From here, we must obtain the empirical formula by calculating the number of oxygen. ? mol O = 0.1090 g O x (1 mol O / 15.9994 g O) = 0.006813 mol O
From the number of moles of each element, we now have a tentative empirical
C0.006812H0.00909O0.006813
formula:
Next divide each subscript by the smallest (0.006812) to obtain:
CH1.33O Multiply all subscripts by three to get:
C3H4O3
Oxidation States Metals tend to lose electrons. Na becomes Na+ + e-. This is known as oxidation. Non-metals, however, tend to gain electrons. Cl + e- become Cl-. This is known as reduction. Rules for Oxidation States 1. The oxidation state (OS) of an individual atom in a free element is 0. 2. The total of the OS in all atoms in: i. Neutral species is 0. ii. Ionic species is equal to the charge on the ion. 3. In their compounds, the alkali metals and the alkaline earths have OS of +1 and +2 respectively. 4. In compounds the OS of fluorine is ALWAYS -1. 5. In compounds, the OS of hydrogen is USUALLY +1. 6. In compounds, the OS of oxygen is USUALLY -2.
7. In binary (two element) compounds with metals: i. Halogens have OS of -1, ii. Group 16 have OS of -2 and iii. Group 15 have OS of -3 8. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have it were a monatomic ion. ***Review Oxidation States on Page 83*** Chemical Nomenclature
***STUDY : WILL BE ON TEST and EXAM*** •
Ionic Compounds are often a combination of metal and non-metal. The anion (non-metal), add –ide to the element name.
Examples: BaCl2: barium chloride; K2O: potassium oxide; Mg(OH)2: magnesium hydroxide; KNO3: potassium nitrate •
Transition metal ionic compounds must be indicated with a charge on metal with Roman Numerals:
Example: FeCl2: iron (II) Chloride; FeCl3: iron (III) chloride; Cr2S3: chromium (III) sulphide •
Molecular Compounds
Example: HI: hydrogen iodide; NF3: nitrogen trifluoride; SO2: Sulfur dioxide; N2Cl4: dinitrogen tetrachloride; NO2: nitrogen dioxide; N2O: dinitrogen monoxide •
Naming Oxoacids and Oxoanions Oxoacid Per- -ic acid Representat ive –ic acid
-ous acid
Hypo- ous acid
Oxidation State
Removal of all
Oxoanion Per- -ate
+ O
-ate
+ O
-ite
+ O
Hypo- -ite
Formula of Acid Name of Acid
Cl: +1 HClO hypochlorite
Hypochlorous acid
Cl: +3
Chlorous acid
HClO2 Sodium chlorite
Cl: +5 chlorate
HClO3
Chloric acid
Formula of Salt Name of Salt
NaClO
Sodium NaClO2
NaClO3
Sodium
Cl: +7
HClO4 Perchloric acid Sodium perchlorate
NaClO4
N: +3
HNO2
Nitrous acid
NaNO2
Sodium nitrite
N: +5
HNO3
Nitric acid
NaNO3
Sodium nitrate
S: +4
H2SO3 Sodium sulfite
S: +6 sulphite
H2SO4
Sulfurous acid Sulfuric acid
Na2SO3 Na2SO4
Sodium
