Tro: Chapter 3 - Molecules and Compounds
Chapter 3: Molecules and Compounds
Compounds Are Not Their Elements
Chemical compounds possess distinct chemical and physical properties that are entirely unrelated to the properties of the individual elements from which they are formed.
Example: Water (H_2O)
Hydrogen (H_2): Boiling Point = -253 \text{ °C}, State at Room Temperature = Gas, Flammability = Explosive.
Oxygen (O_2): Boiling Point = -183 \text{ °C}, State at Room Temperature = Gas, Flammability = Necessary for combustion.
Water (H_2O): Boiling Point = 100 \text{ °C}, State at Room Temperature = Liquid, Flammability = Used to extinguish flame.
Kinds of Compounds
Molecular Compounds (Covalent Compounds)
Composed of atoms held together by covalent bonds, where pairs of electrons are "shared" between atoms.
Typically formed between nonmetals bonded to each other.
Exist as discrete, individual units called molecules.
Example: Water (H_2O) is a molecular compound.
Ionic Compounds
Composed of positively charged ions (cations) and negatively charged ions (anions).
Held together by electrostatic attraction (ionic bonds).
Usually formed between a metal (cation) and a nonmetal (anion).
Do not exist as discrete molecules but as a repeating, ordered arrangement called a crystal lattice.
Example: Sodium chloride (NaCl) is an ionic compound.
Terms Related to Chemical Formulas
Molecular formula: Represents the exact number and type of atoms present in one molecule of a compound.
Used specifically for molecular compounds only.
Example: C6H{12}O_6 for glucose.
Empirical formula: Represents the simplest whole-number ratio of elements in a compound.
Can be used for both molecular and ionic compounds, but for ionic compounds, it's generally the only formula given.
Example: CH2O for C6H{12}O6 (glucose).
Formula unit: The smallest electrically neutral unit within the crystal lattice of an ionic compound.
Example: NaCl represents one formula unit of sodium chloride (one Na^+ and one Cl^-).
Representing Chemical Formulas
Chemical formulas use elemental symbols to represent atoms and subscripts (to the right of the symbol) to indicate the number of atoms (a subscript of 1 is omitted).
Methane (CH_4) Example:
Molecular formula: CH_4
Structural formula: Shows all atoms and covalent bonds.
Ball-and-stick model: Represents atoms as spheres and bonds as sticks, showing molecular geometry.
Space-filling model: Shows the relative sizes of atoms and their electron clouds, depicting the overall shape and volume of the molecule.
Atomic-Level Classification of Elements and Compounds
Pure Substances
Elements:
Atomic Elements: Exist as individual atoms. Example: Neon (Ne).
Molecular Elements: Exist as diatomic or polyatomic molecules. Example: Oxygen gas (O_2).
Compounds:
Molecular Compounds: Composed of discrete molecules. Example: Water (H_2O).
Ionic Compounds: Composed of a crystal lattice of ions, represented by a formula unit. Example: Sodium chloride (NaCl).
Molecular Elements
Elements that exist as molecules rather than individual atoms in their standard state.
Diatomic Molecular Elements: Consist of two atoms of the same element bonded together. There are seven total.
H2, N2, O2, F2, Cl2, Br2, I_2.
Mnemonic: "Horses Need Oats For Clear Brown I's" or their locations on the periodic table form a "7" shape (starting from Nitrogen).
Polyatomic Molecular Elements: Consist of more than two atoms of the same element bonded together.
P4 (white phosphorus), S8 (rhombic sulfur), Se_8 (selenium, less common).
Naming Binary Ionic Compounds
Compounds formed between a metal and a nonmetal.
General Rule: Name the cation (metal) first, then the anion (nonmetal) with an "-ide" ending.
\text{name of cation (metal) + base name of anion (nonmetal) + -ide}
Examples:
NaCl: Sodium chloride.
MgI_2: Magnesium iodide.
Al2O3: Aluminum oxide.
Metals Forming Multiple Cations (Transition Metals & some Main Group Metals):
If the metal can form cations with different charges, a Roman numeral in parentheses is added after the metal's name to indicate the charge of the cation.
Exceptions: Sc^{3+}, Zn^{2+}, and Ag^+ usually form only one common ion, so Roman numerals are not typically used for them.
\text{name of cation (metal) + (roman numeral for charge) + base name of anion (nonmetal) + -ide}
Examples:
FeCl_2: Iron(II) chloride (since Cl^- is -1, two Cl total -2, so Fe must be +2).
FeCl_3: Iron(III) chloride (since Cl^- is -1, three Cl total -3, so Fe must be +3).
Formulas of Binary Ionic Compounds
Procedure:
Write the two ions with their charges.
The net charge on any stable compound must be zero.
Determine the smallest whole number of each ion that will sum to a zero net charge. These values become the subscripts.
Example: Titanium(IV) oxide
Ions: Ti^{4+} and O^{2-}.
To balance charges (+4 and -2), we need one Ti^{4+} and two O^{2-}.
Formula: TiO_2
Practice: Naming/Formulas of Binary Ionic Compounds
CaF_2
CrCl_3
Zinc nitride
Copper(I) oxide
K_3P
Polyatomic Ions
Definition: A group of covalently bonded atoms that, as a unit, carry a net charge (either positive or negative).
Naming: Use the ion's name as part of the compound name.
Parentheses in Formulas: Parentheses are used around the polyatomic ion in the chemical formula if more than one such ion is present to ensure the subscript applies to the entire polyatomic ion.
Example: Ca(OH)_2 indicates one calcium ion and two hydroxide ions.
Common Polyatomic Ions (to be known):
Ammonium: NH_4^+
Acetate: C2H3O_2^-
Carbonate: CO_3^{2-}
Hydroxide: OH^-
Nitrate: NO_3^-
Phosphate: PO_4^{3-}
Sulfate: SO_4^{2-}
Oxyanion Name Conventions
Oxyanion: An anion that contains oxygen and at least one other element.
Series of Two Ions:
The ion with more oxygen atoms takes the -ate suffix.
The ion with less oxygen atoms takes the -ite suffix.
Example: Nitrogen oxyanions
NO_3^-: Nitrate (more oxygen)
NO_2^-: Nitrite (less oxygen)
Series of Four Ions (Halogens, e.g., Chlorine):
Per- + base name + -ate: Most oxygen (one more than -ate). Example: ClO_4^- (Perchlorate).
Base name + -ate: More oxygen. Example: ClO_3^- (Chlorate).
Base name + -ite: Less oxygen. Example: ClO_2^- (Chlorite).
Hypo- + base name + -ite: Least oxygen (one less than -ite). Example: ClO^- (Hypochlorite).
Hydrated Ionic Compounds
Hydrates: Ionic compounds that contain a specific number of "waters of hydration" incorporated into each formula unit.
These water molecules are chemically associated with the ionic compound but can usually be removed by heating.
In the chemical formula, the waters of hydration are separated from the ionic compound by a dot ($\cdot$).
Naming: Add "prefix-hydrate" to the name of the ionic compound.
Prefixes for Number of Water Molecules:
hemi = 1/2
mono = 1
di = 2
tri = 3
tetra = 4
penta = 5
hexa = 6
hepta = 7
octa = 8
Example: CoCl2 \cdot 6H2O is cobalt(II) chloride hexahydrate.
The anhydrous form would be CoCl_2 (cobalt(II) chloride).
Practice: Naming/Formulas with Polyatomic Ions and Hydrates
Cr(C2H3O2)3
NH4NO3
Lithium hydroxide
Calcium phosphate
MgSO4 \cdot 7H2O
Naming Binary Molecular Compounds
Compounds formed from two nonmetals.
Rules:
The elements are named in the order they appear in the formula.
The name of the first element is used as is, with a prefix indicating the number of atoms (if greater than one).
The name of the second element ends in "-ide", with a prefix indicating the number of atoms.
The prefix "mono-" is only used for the second element, never for the first element.
\text{prefix + name of 1st element + prefix + base name of 2nd element + -ide}
Prefixes:
mono = 1
di = 2
tri = 3
tetra = 4
penta = 5
hexa = 6
hepta = 7
octa = 8
nona = 9
deca = 10
Examples:
CO_2: Carbon dioxide.
CO: Carbon monoxide (not "monocarbon monoxide").
Practice: Naming/Formulas of Binary Molecular Compounds
Tetraphosphorus decoxide
CCl_4
P2N5
Sulfur trioxide
N_2O
Naming Binary Acids
Definition of an Acid: Compounds that generate protons (H^+) when dissolved in water (aqueous solution, denoted by (aq)).
Binary Acids: Contain hydrogen and one other nonmetal atom (usually a halogen).
Naming Rules:
Use the prefix "hydro-".
Use the base name of the nonmetal.
Add the suffix "-ic".
Follow with the word "acid".
\text{hydro- + base name of nonmetal + -ic + acid}
Example: HCl(aq) is hydrochloric acid.
Naming Oxyacids
Oxyacids: Acids that contain hydrogen and an oxyanion (hence, they contain oxygen).
Naming Rules: Based on the name of the oxyanion.
If the oxyanion name ends in "-ate", change the suffix to "-ic acid".
Example: NO3^- (nitrate) $\rightarrow$ HNO3(aq) (nitric acid).
\text{base name of oxyanion + -ic + acid}
If the oxyanion name ends in "-ite", change the suffix to "-ous acid".
Example: SO3^{2-} (sulfite) $\rightarrow$ H2SO_3(aq) (sulfurous acid).
\text{base name of oxyanion + -ous + acid}
Mnemonic: "I ate something icky, and Sprite is delicious!" (-ate $\rightarrow$ -ic, -ite $\rightarrow$ -ous).
Summary of Inorganic Nomenclature Flowchart
Start: Is it IONIC or MOLECULAR or an ACID?
IONIC (Metal and nonmetal):
Metal forms one type of ion only: Name of cation + base name of anion + "-ide". Example: CaI_2 (calcium iodide).
Metal forms more than one type of ion: Name of cation + (Roman numeral for charge) + base name of anion + "-ide". Example: FeCl_3 (iron(III) chloride).
MOLECULAR (Nonmetals only):
Prefix of 1st element + name of 1st element + prefix of 2nd element + base name of 2nd element + "-ide". Example: P2O5 (diphosphorus pentoxide).
ACIDS* (H and one or more nonmetals): *Acids must be in aqueous solution.
Binary acids (Two-element): hydro- + base name of nonmetal + "-ic acid". Example: HCl (hydrochloric acid).
Oxyacids (Contain oxygen):
From -ate oxyanion: base name of oxyanion + "-ic acid". Example: H3PO4 (phosphoric acid).
From -ite oxyanion: base name of oxyanion + "-ous acid". Example: H2SO3 (sulfurous acid).
Review Practice
PCl_3
Sodium sulfate
CoCl_3
H2CO3(aq)
Iron(II) phosphide
Terminology for Mass
Molecular mass (or Molecular weight):
The mass of one molecule of a molecular compound.
Measured in atomic mass units (amu).
Formula mass:
The mass of one formula unit of a compound.
Used for both molecular and ionic compounds.
Measured in atomic mass units (amu).
Molar mass (M):
The mass of one mole of the particles that comprise a substance.
Expressed in grams per mole (g/mol).
Relates "particle" scale masses (amu) to macroscopic quantities (grams).
Calculation of Molar Mass
To calculate the molar mass of a compound:
Count the number of each type of atom in the chemical formula.
Multiply the count of each atom by its atomic mass (from the periodic table).
Sum these products.
Example: Calculating Molar Mass of CO_2
1 Carbon atom (1 \times 12.01 \text{ g/mol})
2 Oxygen atoms (2 \times 16.00 \text{ g/mol})
Molar Mass (CO_2) = 12.01 \text{ g/mol} + (2 \times 16.00 \text{ g/mol}) = 44.01 \text{ g/mol}
Significance: Molar mass serves as a crucial conversion factor between the mass of a substance and the moles of that substance.
Percent Composition
Mass Percent Composition: The mass percentage of an element in a compound.
Formula:
\text{Mass % of element} = \frac{\text{mass of element in compound}}{\text{molar mass of compound}} \times 100 \%Example: Calculate the mass % O in CO_2
Molar mass of CO_2 = 44.01 \text{ g/mol}
Mass of O in CO_2 = 2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol} (from two oxygen atoms)
\text{Mass % O} = \frac{32.00 \text{ g/mol}}{(12.01 + 2 \times 16.00) \text{ g/mol}} \times 100 \% = \frac{32.00}{44.01} \times 100 \% = 72.71 \% O
Conversion Factors from Chemical Formulas
The subscripts in chemical formulas provide conversion factors relating moles of elements or ions to moles of the compound (or "molecules" / "formula units").
This applies to both molecular and ionic compounds.
Example: For CCl2F2
1 \text{ mol } CCl2F2 contains 2 \text{ mol } Cl (and 1 \text{ mol } C and 2 \text{ mol } F).
Possible conversion factors:
\frac{2 \text{ mol } Cl}{1 \text{ mol } CCl2F2}
\frac{1 \text{ mol } CCl2F2}{2 \text{ mol } Cl}
Example: Mass of Molecules
Question: What is the mass of 1.06 \times 10^{24} molecules of carbon tetrachloride (CCl_4)?
Calculation involves converting molecules to moles using Avogadro's number, then moles to mass using molar mass.
Molar mass of CCl_4 = 12.01 + 4(35.45) = 153.81 \text{ g/mol}
1.06 \times 10^{24} \text{ molecules } CCl4 \times \frac{1 \text{ mol } CCl4}{6.022 \times 10^{23} \text{ molecules }} \times \frac{153.81 \text{ g } CCl4}{1 \text{ mol } CCl4} = 270.9 \text{ g } CCl_4
Example: Atoms in a Given Mass
Question: How many atoms of carbon are present in a teaspoon of table sugar (C{12}H{22}O_{11}) with a mass of 4.16 \text{ g}?
Calculation involves converting mass to moles of sugar, then moles of sugar to moles of carbon, then moles of carbon to atoms of carbon.
Molar mass of C{12}H{22}O_{11} = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 \text{ g/mol}
4.16 \text{ g } C{12}H{22}O{11} \times \frac{1 \text{ mol } C{12}H{22}O{11}}{342.30 \text{ g } C{12}H{22}O{11}} \times \frac{12 \text{ mol } C}{1 \text{ mol } C{12}H{22}O{11}} \times \frac{6.022 \times 10^{23} \text{ atoms } C}{1 \text{ mol } C} = 8.76 \times 10^{22} \text{ atoms } C
Practice: Moles of Atoms from Mass
Question: How many moles of hydrogen atoms are present in 50.0 \text{ g} of methane, CH_4?
Answer choices: A. 12.5 \text{ mol}, B. 1.88 \times 10^{24} \text{ mol}, C. 3.12 \text{ mol}, D. 7.51 \times 10^{24} \text{ mol}
Practice: Atoms from Moles of Compound
Question: How many oxygen atoms are present in 5.32 \text{ moles} of chalk (CaCO_3)?
Answer choices: A. 1.07 \times 10^{24}, B. 9.65 \times 10^{23}, C. 3.20 \times 10^{24}, D. 4.89 \times 10^{23}, E. 9.61 \times 10^{24}
Percent Composition and Empirical Formulas
Procedure to Determine Empirical Formula from Mass Percent Composition:
Assume a 100 \text{ g} sample: This makes the percentage values directly equal to the number of grams of each element.
Convert grams to moles: Use the atomic mass of each element to convert its mass into moles.
Write a "tentative formula": Use the calculated mole values as subscripts (they will likely not be whole numbers).
Divide by the smallest mole value: Divide all mole values by the smallest mole value determined in step 2. This aims to get at least one subscript to 1.
Convert fractions to whole numbers: If any values are still not whole numbers (e.g., 1.5, 2.33, 2.5), multiply all the mole values by the smallest integer needed to convert all of them into whole numbers (e.g., multiply by 2 for 0.5, by 3 for 0.33 or 0.66, by 4 for 0.25 or 0.75).
Verify whole numbers and no common factors: Ensure all subscripts are whole numbers and share no common divisor (other than 1). This is the empirical formula.
Note: Significant figures are less critical here; keep extra digits during intermediate calculations to avoid rounding errors that could lead to an incorrect empirical formula.
Example: Empirical Formula from Mass % (Dibutyl Succinate)
Dibutyl succinate composition: 62.58 \% C, 9.63 \% H, and 27.79 \% O
Mass of each element in 100 \text{ g} sample:
C: 62.58 \text{ g}
H: 9.63 \text{ g}
O: 27.79 \text{ g}
Convert masses to moles:
n_C = 62.58 \text{ g } C \times \frac{1 \text{ mol } C}{12.011 \text{ g } C} = 5.210 \text{ mol } C
n_H = 9.63 \text{ g } H \times \frac{1 \text{ mol } H}{1.008 \text{ g } H} = 9.554 \text{ mol } H
n_O = 27.79 \text{ g } O \times \frac{1 \text{ mol } O}{16.00 \text{ g } O} = 1.737 \text{ mol } O
Tentative formula: C{5.210}H{9.554}O_{1.737}
Divide by the smallest mole value (which is 1.737 for Oxygen):
C{\frac{5.210}{1.737}} H{\frac{9.554}{1.737}} O_{\frac{1.737}{1.737}}
C{3.00}H{5.50}O_1
Convert to small whole number ratio: The 5.50 indicates a fraction of 1/2. Multiply all subscripts by 2.
[C{3.00}H{5.50}O1] \times 2 = C{6.00}H{11.00}O2
Empirical formula: C6H{11}O_2
Practice: Empirical Formula from Mass %
Question: A mineral containing magnesium, silicon, hydrogen, and oxygen has the following composition: 28.03 \% Mg, 21.60 \% Si, 1.16 \% H, & remaining % O. What is the mineral's empirical formula?
First, calculate the % O: 100 - 28.03 - 21.60 - 1.16 = 49.21 \% O
Answer choices: A. Mg2SiH4O2, B. Mg3Si4H2O{12}, C. Mg3Si2H3O8, D. Mg{24}Si{19}HO{42}, E. MgSiHO_4
Determining the Molecular Formula
If the molecular mass of a compound is known (e.g., from a mass spectrometer), the molecular formula can be determined from the empirical formula.
The molecular formula will always be a whole-number multiple of the empirical formula.
Procedure:
Calculate the molar mass of the empirical formula unit.
Divide the experimental molecular mass by the empirical formula molar mass. This will give a whole number (n).
Multiply all the subscripts in the empirical formula by this whole number (n) to get the molecular formula.
Example: Empirical formula CH_2O and molecular mass 180.2 \text{ amu}
Empirical formula molar mass of CH_2O = 12.01 + (2 \times 1.008) + 16.00 = 30.03 \text{ g/mol}
n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Molar Mass}} = \frac{180.2 \text{ amu}}{30.03 \text{ amu}} = 6.00 \approx 6
Molecular formula = (CH2O)6 = C6H{12}O_6
Practice: Determining the Molecular Formula
Question: A compound has an empirical formula of C3H7O and its molecular mass was determined to be approximately 118 \text{ amu}. What is the compound's molecular formula?
Answer choices: A. C4H6O4, B. C6H{14}O2, C. C3H7O, D. C9H{21}O_3, E. Not enough information given.
Combustion Analysis
Purpose: A common experimental technique used to determine the empirical formula of organic compounds (containing C, H, and sometimes O, N, S, etc.).
Method: An unknown compound (CxHy or CxHyO_z) is burned completely in excess oxygen.
CxHy + \text{excess } O2 \rightarrow x CO2(g) + \frac{y}{2} H_2O(l)
The gaseous products, carbon dioxide (CO2) and water (H2O), are isolated and weighed using specialized absorbing tubes.
The masses of carbon and hydrogen in the original sample can be determined directly from the mass of CO2 and H2O produced, respectively.
If other elements (like oxygen) are present, their mass is determined by subtracting the calculated masses of C and H (and any other measured elements) from the total mass of the original sample.
Empirical Formula from Combustion Data
Procedure:
Calculate mass and moles of C: Use the mass of CO2 produced. (grams CO2 \rightarrow moles CO_2 \rightarrow moles C \rightarrow grams C).
Calculate mass and moles of H: Use the mass of H2O produced. (grams H2O \rightarrow moles H_2O \rightarrow moles H \rightarrow grams H).
Calculate mass and moles of "other" element (e.g., O): If the compound contains an element other than C and H (e.g., Oxygen), subtract the calculated masses of C and H from the original mass of the sample to find the mass of the "other" element. Then convert this mass to moles.
Write a "tentative formula": Use the calculated mole values as subscripts.
Divide by the smallest mole value: Divide all mole values by the smallest one found.
Convert fractions to whole numbers: Multiply all mole values by the smallest integer needed to achieve whole numbers.
Verify whole numbers and no common factors: Ensure the simplest whole-number ratio for the empirical formula.
Note: Again, keep extra digits during intermediate calculations to minimize rounding errors.
Example: Empirical Formula from Combustion Data (Dimethylhydrazine)
Question: Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned completely, a 0.312 \text{ g} sample yields 0.458 \text{ g } CO2 and 0.374 \text{ g } H2O. What is the empirical formula of dimethylhydrazine?
This example requires calculating C and H moles from CO2 and H2O, then finding N by difference from the original sample mass, then determining the empirical formula ratio for C:H:N.
Moles of C: 0.458 \text{ g } CO2 \times \frac{1 \text{ mol } CO2}{44.01 \text{ g } CO2} \times \frac{1 \text{ mol } C}{1 \text{ mol } CO2} = 0.010406 \text{ mol } C
Mass C = 0.010406 \text{ mol } C \times 12.011 \text{ g/mol } C = 0.12499 \text{ g } C
Moles of H: 0.374 \text{ g } H2O \times \frac{1 \text{ mol } H2O}{18.015 \text{ g } H2O} \times \frac{2 \text{ mol } H}{1 \text{ mol } H2O} = 0.04152 \text{ mol } H
Mass H = 0.04152 \text{ mol } H \times 1.008 \text{ g/mol } H = 0.04185 \text{ g } H
Mass and Moles of N:
Mass N = Total sample mass - mass C - mass H = 0.312 \text{ g} - 0.12499 \text{ g} - 0.04185 \text{ g} = 0.14516 \text{ g } N
Moles N = 0.14516 \text{ g } N \times \frac{1 \text{ mol } N}{14.007 \text{ g } N} = 0.010363 \text{ mol } N
Tentative formula: C{0.010406}H{0.04152}N_{0.010363}
Divide by smallest (N): C{\frac{0.010406}{0.010363}} H{\frac{0.04152}{0.010363}} N_{\frac{0.010363}{0.010363}}
C1H4N_1
Empirical formula: CH_4N