Chem Skills Tests
Know the difference between a compound or element (both are pure substances)
Element:
A pure substance made of only one type of atom.
Cannot be broken down into anything simpler.
Example: Oxygen (O₂), Gold (Au).
Compound:
A pure substance made of two or more different types of atoms bonded together.
Can be broken down into simpler substances (its elements).
Example: Water (H₂O), Salt (NaCl).
Summary:
Element: Only one kind of atom.
Compound: Two or more kinds of atoms combined.
Be able to convert between prefixes of milli, centi, and kilo, from a base unit. For example, can you convert 10 mL to L?
Summary of Conversion Process:
Convert to the base unit by multiplying or dividing based on the prefix.
Convert from the base unit to the desired prefix, if necessary.
Be able to write full and noble gas shorthand electron configurations, or identify the correct element when an electron configuration is given to you.
Be able to draw orbital diagrams or determine what element it is when given an orbital diagram.
Be able to identify the types of reactions below, as well as be able to predict what the products would be:
Combustion
Decomposition
Synthesis
Single replacement
Double replacement
1. Combustion Reaction:
General Form: A + O₂ → CO₂ + H₂O (if A is a hydrocarbon)
Description: In a combustion reaction, a substance (usually a hydrocarbon) reacts with oxygen (O₂), producing carbon dioxide (CO₂) and water (H₂O).
Example:
Reactants: CH₄ (methane) + 2 O₂
Products: CO₂ + 2 H₂O
Type: Combustion
2. Decomposition Reaction:
General Form: AB → A + B
Description: A single compound breaks down into two or more simpler substances.
Example:
Reactants: 2 H₂O (water)
Products: 2 H₂ + O₂
Type: Decomposition
3. Synthesis Reaction:
General Form: A + B → AB
Description: Two or more simple substances combine to form a more complex compound.
Example:
Reactants: 2 H₂ + O₂
Products: 2 H₂O
Type: Synthesis
4. Single Replacement Reaction:
General Form: A + BC → AC + B
Description: One element replaces another in a compound. The more reactive element displaces the less reactive one.
Example:
Reactants: Zn + 2 HCl (hydrochloric acid)
Products: ZnCl₂ + H₂
Type: Single Replacement
5. Double Replacement Reaction:
General Form: AB + CD → AD + CB
Description: The ions in two compounds exchange partners to form two new compounds.
Example:
Reactants: NaCl (sodium chloride) + AgNO₃ (silver nitrate)
Products: NaNO₃ + AgCl (silver chloride)
Type: Double Replacement
How to Predict Products:
Combustion: If the reactant is a hydrocarbon (like methane, CH₄), the products will always be CO₂ and H₂O.
Decomposition: Break down the compound into its elements or simpler compounds.
Synthesis: Combine the reactants directly into one compound.
Single Replacement: Replace one element in a compound with a more reactive element.
Double Replacement: Swap the cations or anions between two compounds to form new products.
By identifying the type of reaction, you can predict the products based on the patterns provided.
Write chemical names/formulas for:
Ionic compounds (binary or polyatomic)
Covalent compounds (the ones that use the prefixes, so you need to memorize the prefixes 1-10)
1. Naming Ionic Compounds
Ionic compounds are formed between metals and nonmetals, where electrons are transferred from the metal to the nonmetal.
Rules for Naming Ionic Compounds:
Name the Cation (Metal) First:
Use the full name of the metal.
If the metal can have more than one charge (transition metals), specify the charge using Roman numerals in parentheses.
Name the Anion (Nonmetal) Second:
Change the ending of the nonmetal's name to "-ide."
For Polyatomic Ions:
Use the name of the polyatomic ion without changing its ending (e.g., sulfate, nitrate).
Examples:
NaCl:
Name: Sodium Chloride
Explanation: "Sodium" is the metal, and "chlorine" becomes "chloride."
FeCl₃:
Name: Iron(III) Chloride
Explanation: "Iron" can have multiple charges; here, it's +3, so it's "Iron(III)," and "chlorine" becomes "chloride."
Ca(NO₃)₂:
Name: Calcium Nitrate
Explanation: "Calcium" is the metal, and "nitrate" is a polyatomic ion.
2. Naming Covalent Compounds
Covalent compounds are formed between nonmetals, where atoms share electrons.
Rules for Naming Covalent Compounds:
Use Prefixes to Indicate the Number of Atoms:
Prefixes show the number of each type of atom (e.g., mono-, di-, tri-, etc.).
If there’s only one atom of the first element, the prefix "mono-" is often omitted.
Name the First Element:
Use the full name of the first element.
Name the Second Element with the Suffix "-ide":
The second element's name ends with "-ide."
Apply the Prefix to the Second Element:
Always use the appropriate prefix for the number of atoms, even if it's just one (mono-).
Examples:
CO₂:
Name: Carbon Dioxide
Explanation: "Carbon" for C, "Di-" for two oxygen atoms, and "oxide" for O.
P₄O₁₀:
Name: Tetraphosphorus Decaoxide
Explanation: "Tetra-" for four phosphorus atoms, "Deca-" for ten oxygen atoms, and "oxide" for O.
N₂O₅:
Name: Dinitrogen Pentoxide
Explanation: "Di-" for two nitrogen atoms, "Penta-" for five oxygen atoms, and "oxide" for O.
Summary of Differences:
Ionic Compounds: Metal + Nonmetal (or polyatomic ions), with Roman numerals for transition metals, and "-ide" for nonmetals.
Covalent Compounds: Nonmetal + Nonmetal, using prefixes to indicate the number of atoms, with "-ide" at the end of the second element's name.
Define the difference between an empirical formula and a molecular formula (remember, they could actually be the same thing)
Sure! Let's break it down further with some examples:
1. Empirical Formula:
The empirical formula shows the simplest whole-number ratio of atoms in a compound. It does not necessarily reflect the actual number of atoms in a single molecule, just the relative proportions.
Example 1: Glucose
Molecular Formula: C₆H₁₂O₆
This tells us that one molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
Empirical Formula: CH₂O
This is the simplest ratio of the elements in glucose, obtained by dividing each subscript in the molecular formula by 6. It shows that for every 1 carbon atom, there are 2 hydrogen atoms and 1 oxygen atom.
Example 2: Hydrogen Peroxide
Molecular Formula: H₂O₂
This tells us that one molecule of hydrogen peroxide contains 2 hydrogen atoms and 2 oxygen atoms.
Empirical Formula: HO
The simplest ratio of hydrogen to oxygen in hydrogen peroxide is 1:1, so the empirical formula is HO.
2. Molecular Formula:
The molecular formula shows the exact number of each type of atom in a single molecule of the compound. It gives the full composition of the molecule.
Example 1: Ethylene (C₂H₄)
Molecular Formula: C₂H₄
This indicates that each molecule of ethylene contains 2 carbon atoms and 4 hydrogen atoms.
Empirical Formula: CH₂
The ratio of carbon to hydrogen is 1:2, so the empirical formula is CH₂.
Example 2: Benzene (C₆H₆)
Molecular Formula: C₆H₆
This tells us that each molecule of benzene contains 6 carbon atoms and 6 hydrogen atoms.
Empirical Formula: CH
The simplest ratio of carbon to hydrogen is 1:1, so the empirical formula is CH.
When They Are the Same:
Sometimes, the empirical formula and the molecular formula are identical because the simplest ratio is the actual number of atoms in the molecule.
Example 1: Water (H₂O)
Molecular Formula: H₂O
Empirical Formula: H₂O
The ratio of hydrogen to oxygen is 2:1, which is the same as the actual number of atoms in a water molecule.
The empirical formula gives the simplest ratio of elements.
The molecular formula gives the actual number of atoms in a molecule.
Sometimes, these formulas are the same (e.g., water, methane), but they can also differ when the molecular formula represents a multiple of the empirical formula (e.g., glucose, hydrogen peroxide).
When given the mass of the molecular formula and what the empirical formula is, be able to calculate what the molecular formula would be.
Steps to Determine the Molecular Formula:
Find the Empirical Formula Mass:
Calculate the mass of the empirical formula by adding up the atomic masses of all the atoms in the empirical formula.
Divide the Molar Mass by the Empirical Formula Mass:
Divide the given molar mass (mass of the molecular formula) by the empirical formula mass. This will give you a ratio (n).
Multiply the Empirical Formula by the Ratio:
Multiply each subscript in the empirical formula by the ratio nnn to get the molecular formula.
Example:
Given:
Empirical Formula: CH₂O
Molar Mass of Molecular Formula: 180 g/mol
Step 1: Calculate the Empirical Formula Mass:
C: 12.01 g/mol
H₂: 2(1.01) = 2.02 g/mol
O: 16.00 g/mol
Total: 12.01 + 2.02 + 16.00 = 30.03 g/mol
Step 2: Divide the Molar Mass by the Empirical Formula Mass:
n=180 g/mol30.03 g/mol≈6n = \frac{180 \text{ g/mol}}{30.03 \text{ g/mol}} \approx 6n=30.03 g/mol180 g/mol≈6
Step 3: Multiply the Empirical Formula by the Ratio:
Multiply each subscript in CH₂O by 6:
C: 1×6=61 \times 6 = 61×6=6
H: 2×6=122 \times 6 = 122×6=12
O: 1×6=61 \times 6 = 61×6=6
Molecular Formula: C₆H₁₂O₆
Summary:
Calculate the empirical formula mass.
Divide the molar mass by the empirical formula mass to get the ratio.
Multiply the subscripts in the empirical formula by this ratio to get the molecular formula.