Chemistry Unit 3

Introduction to Nomenclature and Note-Taking

Making questions based on subtitles from textbooks is a useful note-taking method.
Questions vary between students, enhancing personal understanding.
Additionally, post-reading statements or questions can help reinforce material.

Types of Compounds and Their Formulas

Types of Bonds

Covalent Bonds: These bonds form when atoms share electrons. This type of bonding typically occurs between non-metal atoms.
Ionic Bonds: These bonds form when there is a transfer of electrons from one atom to another, resulting in the formation of ions (charged atoms). This usually occurs between a metal and a non-metal.

Molecular Compounds

Definition: Compounds that are formed when non-metal atoms share electrons, forming discrete units called molecules.
Characteristics:
- Made of individual, distinct molecules.
- Represented by chemical formulas, which provide the following information:
- The elements present, indicated by their chemical symbols.
- The relative number of atoms of each element using subscripts. For example, in water, H_2O, there are two hydrogen atoms and one oxygen atom.
Example: Glucose
- Chemical formula: C6H{12}O_6
- Elements present: Carbon (C), Hydrogen (H), and Oxygen (O).
- Atom counts: 6 Carbon atoms, 12 Hydrogen atoms, 6 Oxygen atoms.

Representing Compounds: Chemical Formulas & Molecular Models

Types of Chemical Formulas

Empirical Formula
- The simplest type of chemical formula, providing the atoms present and their relative quantities in the simplest whole-number ratio.
- Example:
  - The empirical formula for P4O{10} (tetraphosphorus decoxide) is P2O5. This is because the ratio of phosphorus to oxygen atoms (4:10) simplifies to 2:5.
  - Similarly, for glucose (C6H{12}O6) and formaldehyde (CH2O), their empirical formula is the same: CH_2O. In glucose, the 6:12:6 ratio simplifies to 1:2:1. For formaldehyde, it's already in the simplest 1:2:1 ratio.
Molecular Formula
- This formula shows the actual number of atoms of each element present in one molecule of the compound.
- It can be the same as the empirical formula or a whole-number multiple of it.
- Example: The empirical formula of glucose is CH2O. Its molecular formula, C6H{12}O6, is six times the empirical formula (each subscript is multiplied by 6).
- Important Note: A molecular formula does not show how atoms are connected in space. For example, glucose and fructose both have the molecular formula C6H{12}O_6 but have different structural arrangements.
Structural Formula
- This formula illustrates the order in which atoms are bonded together within a molecule and indicates the types of bonds (single, double, or triple).
- Example: For acetic acid, the structural formula clearly shows the connectivity of atoms and bonds:
\text{One carbon atom is double-bonded to an oxygen atom and single-bonded to another oxygen atom (which is then single-bonded to a hydrogen) and a methyl (CH_3) group.}

Condensed Structural Formula

A simplified, linear representation of a structural formula that groups together atoms connected to a central atom.
Example: For acetic acid, the condensed structural formula can be written as CH3COOH or CH3CO2H. Here, CH3 indicates a carbon with three hydrogens, and COOH (or CO_2H) represents the carboxylic acid functional group, showing a carbon double-bonded to one oxygen and single-bonded to another oxygen, which is attached to a hydrogen.

Molecular Models

Ball and Stick Model:
- Atoms are represented by spheres (balls) and chemical bonds by rods (sticks).
- These models are excellent for visualizing bond lengths, bond angles, and the overall three-dimensional geometry of molecules.
Space-Filling Model:
- These models represent atoms as spheres proportioned to their actual relative sizes, depicting the space occupied by each atom in a molecule.
- They show the electron cloud of atoms in a molecule as if they are in actual contact with one another, giving a more accurate representation of molecular volume.
- Constructed to scale to reflect covalent radii.

Ionic Compounds

Definition

Compounds formed by ionic bonds, typically occurring between a metal and a non-metal where electrons are transferred.

Bonding Process

The metal atom loses one or more electrons to form positively charged ions called cations.
The non-metal atom gains one or more electrons to form negatively charged ions called anions.
These oppositely charged ions are then attracted to each other, forming a stable ionic compound.
Example: Calcium chloride
- Calcium (Ca), a metal, donates two electrons to form a Ca^{2+} cation.
- Chlorine (Cl), a non-metal, accepts one electron to form a Cl^- anion.
- To maintain electrical neutrality, one Ca^{2+} ion combines with two Cl^- ions, resulting in the chemical formula CaCl_2.

Crystal Structure of Ionic Compounds

Ionic compounds form crystal lattices, where each ion is surrounded by multiple ions of opposite charge.
In calcium chloride, for instance, each Ca^{2+} ion is generally surrounded by multiple Cl^- ions, and each Cl^- is surrounded by multiple Ca^{2+} ions, maintaining an overall 1:2 ratio of Ca^{2+} to Cl^-. It's not a discrete molecule, but a continuous network.
Formula Unit: The smallest electrically neutral collection of ions that represents the composition of an ionic compound. For CaCl_2, one Ca^{2+} ion and two Cl^- ions constitute one formula unit.
Monoatomic Ions: Ions consisting of a single ionized atom (e.g., Mg^{2+}, Br^-, Al^{3+}).
Polyatomic Ions: Ions composed of two or more atoms covalently bonded together, but carrying an overall net charge (e.g., NO_3^-, the nitrate ion).

Naming Compounds

Organic vs. Inorganic Compounds

Organic Compounds: These are compounds primarily composed of carbon and hydrogen, often with oxygen, nitrogen, or other elements. (Detailed nomenclature for organic compounds is typically covered in organic chemistry courses like CHEM 101/102).
Inorganic Compounds: These encompass all compounds that do not contain carbon and hydrogen together (though they may contain carbon or hydrogen individually, like CO2 or H2O).

Binary Compounds

Definition: Compounds formed between exactly two different elements.

Type I Binary Ionic Compounds

Formed between a metal that forms only one type of cation (fixed charge) and a non-metal.
Common Type I metals include Group 1 alkali metals (Li^+, Na^+, K^+, Cs^+), Group 2 alkaline earth metals (Mg^{2+}, Ca^{2+}, Sr^{2+}, Ba^{2+}), aluminum (Al^{3+}), zinc (Zn^{2+}), cadmium (Cd^{2+}), and silver (Ag^+).
Rules for Naming:
- The cation (metal) is named first, using its unmodified element name.
- The anion (non-metal) is named second. Its element name is modified by dropping the ending and adding the suffix "-ide".
Examples:
- NaBr: Sodium bromide (Sodium is the metal, bromine becomes bromide).
- CaBr_2: Calcium bromide (Calcium is the metal, bromine becomes bromide. The subscript "2" for bromine is accounted for by calcium's fixed +2 charge balancing with two Br^- ions; it's not explicitly named in a Type I compound name).
- AlBr_3: Aluminum bromide (Aluminum is the metal, bromine becomes bromide. Similar to calcium bromide, the subscript "3" for bromine is balanced by aluminum’s fixed +3 charge).

Chart of Common Monoatomic Ions

Cations (Positive Ions):
- +1: H^+(hydrogen), Li^+(lithium), Na^+(sodium), K^+(potassium), Cs^+(cesium), Ag^+(silver)
- +2: Mg^{2+}(magnesium), Ca^{2+}(calcium), Sr^{2+}(strontium), Ba^{2+}(barium), Zn^{2+}(zinc), Cd^{2+}(cadmium)
- +3: Al^{3+}(aluminum)
Anions (Negative Ions):
- -1: H^- (hydride), F^- (fluoride), Cl^- (chloride), Br^- (bromide), I^- (iodide)
- -2: O^{2-}(oxide), S^{2-}(sulfide)
- -3: N^{3-}(nitride)

Type II Binary Ionic Compounds

Formed between a metal that can form more than one type of cation (variable charge, typically transition metals) and a non-metal.
Rules for Naming:
- The cation (metal) is named first, followed by a Roman numeral in parentheses to indicate its charge.
- The anion (non-metal) is named second, with its ending changed to "-ide".
Example: Cobalt can form Co^{2+} and Co^{3+} ions.
- CoCl_2: To determine cobalt's charge, we know Cl is -1. Since there are two Cl^- ions (2 \times -1 = -2), cobalt must be +2 to balance the charge. Thus, the name is Cobalt(II) chloride.
- CoCl_3: Here, three Cl^- ions mean cobalt must be +3. The name is Cobalt(III) chloride.

Polyatomic Ions

Definition: Ions composed of two or more atoms covalently bonded together, but carrying an overall net electrical charge. They behave as a single unit in chemical reactions.
Common Examples (Memorization is key!):
- NH_4^+: ammonium ion
- CO_3^{2-}: carbonate ion
- SO_4^{2-}: sulfate ion
- NO_3^-: nitrate ion
- OH^-: hydroxide ion
- PO_4^{3-}: phosphate ion
- MnO_4^-: permanganate ion
Naming Rules for Oxoanions (polyatomic anions containing oxygen):
- Many polyatomic ions have similar structures but differ in the number of oxygen atoms.
- The most common form usually ends in "-ate" (e.g., SO4^{2-} is sulfate, NO3^- is nitrate).
- An ion with one less oxygen atom than the "-ate" form receives the "-ite" suffix (e.g., SO3^{2-} is sulfite, NO2^- is nitrite).
- For elements forming more than two oxoanions, prefixes are used:
- "hypo-" (least oxygen, e.g., ClO^- , hypochlorite)
- "per-" (most oxygen, e.g., ClO_4^- , perchlorate)

Acid Nomenclature

Acids are compounds that produce H^+ (hydrogen ions) when dissolved in water. Their naming depends on whether they contain oxygen.

Binary Acids (Acids that do not contain oxygen):
- These are typically formed when hydrogen combines with a non-metal from Group 17 (halogens) or Group 16 (chalcogens).
- Naming Convention: Use the prefix "hydro-" + the root name of the non-metal + the suffix "-ic" + the word "acid".
- Examples:
  - HCl: hydrogen chloride becomes hydrochloric acid (non-metal root is "chlor-").
  - HF: hydrogen fluoride becomes hydrofluoric acid.
  - H_2S: hydrogen sulfide becomes hydrosulfuric acid.
Oxoacids (Acids that contain oxygen):
- These acids are formed from oxoanions (polyatomic ions containing oxygen).
- Naming Convention: Based on the name of the polyatomic oxoanion, drop the "ion" and change the suffix:
  - If the oxoanion ends in "-ate", the acid name ends in "-ic".
  - If the oxoanion ends in "-ite", the acid name ends in "-ous".
- Examples:
  - H2SO4: Derived from the sulfate ion (SO_4^{2-}). Since sulfate ends in "-ate", the acid is sulfuric acid.
  - HNO3: Derived from the nitrate ion (NO3^-). "-ate" becomes "-ic", so it's nitric acid.
  - HNO2: Derived from the nitrite ion (NO2^-). "-ite" becomes "-ous", so it's nitrous acid.
  - H3PO4: Derived from the phosphate ion (PO_4^{3-}). "-ate" becomes "-ic", so it's phosphoric acid.

Formula Mass & Mole Concept for Compounds

Definitions

Molecular Mass (or Molecular Weight): The sum of the atomic masses of all atoms present in a molecule of a molecular compound. It is expressed in atomic mass units (amu). For example, the molecular mass of H_2O is (2 \times 1.008 \text{ amu}) + (1 \times 15.999 \text{ amu}) = 18.015 \text{ amu}.
Formula Mass (or Formula Weight): The sum of the atomic masses of all atoms present in one formula unit of any compound (ionic or molecular). For ionic compounds, this term is preferred because they do not form discrete molecules. It is also expressed in amu. For example, the formula mass of CaCl_2 is (1 \times 40.078 \text{ amu}) + (2 \times 35.453 \text{ amu}) = 110.984 \text{ amu}.
Molar Mass: The mass of 1 mole of a pure substance (atoms, molecules, or formula units), expressed in grams per mole (g/mol). Numerically, the molar mass in g/mol is equal to the atomic, molecular, or formula mass in amu. For example, the molar mass of H_2O is 18.015 \text{ g/mol}.

Density, Molar Mass, and Avogadro’s Constant: Conversion Relationships

Understanding these relationships is crucial for solving stoichiometry problems:
- Mass (g) \rightleftharpoons Moles (mol): Use Molar Mass (\text{g/mol}).
  - To convert moles to mass: \text{mass} = \text{moles} \times \text{molar mass}
  - To convert mass to moles: \text{moles} = \frac{\text{mass}}{\text{molar mass}}
- Moles (mol) \rightleftharpoons Number of Elementary Entities (atoms, molecules, ions, formula units):
  - Use Avogadro’s Constant: 1 \text{ mole} = 6.022 \times 10^{23} \text{ entities}.
  - To convert moles to entities: \text{entities} = \text{moles} \times (6.022 \times 10^{23} \text{ entities/mol})
  - To convert entities to moles: \text{moles} = \frac{\text{entities}}{6.022 \times 10^{23} \text{ entities/mol}}
- Volume (L) of a gas at STP \rightleftharpoons Moles (mol): Use the Molar Volume of gas at STP: 1 \text{ mole} = 22.4 \text{ L}.
- Density (\text{g/mL} or \text{g/cm}^3): Relates mass and volume for liquids and solids.
  - \text{density} = \frac{\text{mass}}{\text{volume}} so \text{mass} = \text{density} \times \text{volume} and \text{volume} = \frac{\text{mass}}{\text{density}}

Example Calculations (Mass-Mole-Number Conversions)

Example 1: Calculating the mass of oxygen in a sample of aluminum sulfate.
Determine the mass of oxygen in 250.0 \text{ g} of Al2(SO4)_3.
- Find the Molar Mass of Al2(SO4)_3:
  - Al: 2 \times 26.98 \text{ g/mol} = 53.96 \text{ g/mol}
  - S: 3 \times 32.07 \text{ g/mol} = 96.21 \text{ g/mol}
  - O: 12 \times 16.00 \text{ g/mol} = 192.00 \text{ g/mol} (Note: 3 \times 4 = 12 oxygen atoms)
  - Molar Mass of Al2(SO4)_3 = 53.96 + 96.21 + 192.00 = 342.17 \text{ g/mol}
- Calculate the Percentage by Mass of Oxygen in Al2(SO4)_3:
  - \text{Mass % of O} = \frac{\text{total mass of O}}{\text{molar mass of } Al2(SO4)_3} \times 100\%
  - \text{Mass % of O} = \frac{192.00 \text{ g/mol}}{342.17 \text{ g/mol}} \times 100\% = 56.11\%
- Calculate the Mass of Oxygen in the Sample:
  - \text{Mass of O} = 250.0 \text{ g } Al2(SO4)3 \times \frac{56.11 \text{ g O}}{100 \text{ g } Al2(SO4)3} = 140.3 \text{ g O}

Composition of Chemical Compounds

Mass Percent Calculation

The mass percent of an element in a compound is the mass of that element present in one mole of the compound, divided by the molar mass of the compound, multiplied by 100\%.
Formula for calculating mass percent of an element X in a compound:
- \text{mass percent of } X = \frac{\text{number of atoms of } X \times \text{atomic mass of } X}{\text{molar mass of compound}} \times 100\%
Example: Calculate the mass percent of Carbon in Glucose (C6H{12}O_6).
- Find the Molar Mass of Glucose:
  - C: 6 \times 12.01 \text{ g/mol} = 72.06 \text{ g/mol}
  - H: 12 \times 1.008 \text{ g/mol} = 12.096 \text{ g/mol}
  - O: 6 \times 16.00 \text{ g/mol} = 96.00 \text{ g/mol}
  - Molar Mass of C6H{12}O_6 = 72.06 + 12.096 + 96.00 = 180.156 \text{ g/mol}
- Mass Percent of Carbon:
  - \text{mass percent of C} = \frac{6 \times 12.01 \text{ g/mol}}{180.156 \text{ g/mol}} \times 100\% = \frac{72.06 \text{ g/mol}}{180.156 \text{ g/mol}} \times 100\% = 39.99\%

Determining a Chemical Formula from Experimental Data

Process Overview

Empirical formulas can be determined from the experimentally measured percent composition of a compound. The steps involve converting mass percentages to a mole ratio, which then simplifies to the smallest whole-number ratio for the empirical formula.
Steps to Determine the Empirical Formula:
1. Assume a 100 \text{ g} sample: This allows you to directly convert mass percentages into grams for each element.
2. Convert grams to moles: Use the atomic mass of each element to convert the mass (in grams) of each element to moles.
3. Find the simplest mole ratio: Divide the number of moles of each element by the smallest number of moles calculated in step 2. This will give you the tentative subscripts for the empirical formula.
4. Convert to whole numbers: If the ratios from step 3 are not whole numbers, multiply all ratios by the smallest integer that converts them into whole numbers (e.g., if you have 1.5, multiply by 2; if 0.33, multiply by 3). These whole numbers are the subscripts for the empirical formula.

Practical Applications: Determining Empirical Formula from Combustion Data

Combustion analysis is a common experimental technique used to determine the empirical formula of organic compounds. When a compound containing carbon, hydrogen, and sometimes oxygen is burned in excess oxygen, all the carbon is converted to carbon dioxide (CO2) and all the hydrogen to water (H2O).
Example: Determining the empirical formula of an unknown compound.
A 0.250 \text{ g} sample of a compound containing only carbon, hydrogen, and oxygen is completely combusted. The combustion produces 0.366 \text{ g} of CO2 and 0.150 \text{ g} of H2O. Determine the empirical formula of the compound.
- Calculate the mass of C and H in the sample:
  - Mass of C in CO2: 0.366 \text{ g } CO2 \times \frac{1 \text{ mol } CO2}{44.01 \text{ g } CO2} \times \frac{1 \text{ mol C}}{1 \text{ mol } CO_2} \times \frac{12.01 \text{ g C}}{1 \text{ mol C}} = 0.100 \text{ g C}
  - Mass of H in H2O: 0.150 \text{ g } H2O \times \frac{1 \text{ mol } H2O}{18.02 \text{ g } H2O} \times \frac{2 \text{ mol H}}{1 \text{ mol } H_2O} \times \frac{1.008 \text{ g H}}{1 \text{ mol H}} = 0.0168 \text{ g H}
- Calculate the mass of O in the sample:
  - Mass of O = Total sample mass - (mass of C + mass of H)
  - Mass of O = 0.250 \text{ g} - (0.100 \text{ g} + 0.0168 \text{ g}) = 0.1332 \text{ g O}
- Convert masses to moles:
  - Moles of C: 0.100 \text{ g C} \times \frac{1 \text{ mol C}}{12.01 \text{ g C}} = 0.00833 \text{ mol C}
  - Moles of H: 0.0168 \text{ g H} \times \frac{1 \text{ mol H}}{1.008 \text{ g H}} = 0.0167 \text{ mol H}
  - Moles of O: 0.1332 \text{ g O} \times \frac{1 \text{ mol O}}{16.00 \text{ g O}} = 0.00833 \text{ mol O}
- Find the simplest mole ratio (divide by the smallest number of moles):
  - Ratio C: \frac{0.00833}{0.00833} = 1.00
  - Ratio H: \frac{0.0167}{0.00833} = 2.00
  - Ratio O: \frac{0.00833}{0.00833} = 1.00
- Empirical Formula: The simplest whole-number ratio is 1:2:1. Thus, the empirical formula is CH_2O.

Multistep Examples and Problem Solving

Example: Calculating Molecular Formula from Empirical Formula and Molar Mass.
If the molar mass of the compound in the previous combustion analysis example (CH_2O) is experimentally determined to be 180.16 \text{ g/mol}, what is its molecular formula?
- Calculate the empirical formula mass:
  - Empirical formula: CH_2O
  - Empirical formula mass = (1 \times 12.01) + (2 \times 1.008) + (1 \times 16.00) = 30.026 \text{ g/mol}
- Determine the whole-number multiple (n):
  - n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}
  - n = \frac{180.16 \text{ g/mol}}{30.026 \text{ g/mol}} \text{ (approximately)} = 6
- Calculate the molecular formula: Multiply the subscripts in the empirical formula by n
  - Molecular formula = (CH2O)6 = C6H{12}O_6
- This compound is glucose.