General Chemistry 1 Reviewer 2024-2025 PDF

General Chemistry 1 Reviewer CHEMISTRY 1st Quarter | 1st Semester | 2024 - 2025 MEASURABLE PROPERTIES OF MATTER:...

General Chemistry 1 Reviewer CHEMISTRY 1st Quarter | 1st Semester | 2024 - 2025 MEASURABLE PROPERTIES OF MATTER: MATTER INTENSIVE PROPERTY - does NOT depend on how much matter is ATOMS - smallest unit of matter. It consists of a nucleus made of being considered. (e.g. boiling point, luster, odor, pressure, protons and neutrons. density) MOLECULE - a group of two or more atoms bonded together. DUCTILITY - the ability of substance to be stretched into ION - an atom or a molecule that has gained or lost one or more wire electrons. It is an atom or a molecule that has CHARGE. MALLEABILITY - the ability of substance to be hammered or transformed into thin sheet STATES OF MATTER: EXTENSIVE PROPERTY - depends on how much matter is being considered. (e.g. volume, weight, energy, mass) SOLID - definite shape and definite volume LIQUID - has a distinct volume independent of its container. SEPARATION TECHNIQUES Assumes the shape of the container it occupies. GAS - no fixed shape or volume FILTRATION - process of separating solid from liquid in which it is suspended PHASE TRANSITIONS: DECANTATION - the process of separation of liquid from solid and other immiscible (non-mixing) liquids, by removing the liquid layer FREEZING - Solid to Liquid at the top from the layer of solid or liquid below. MELTING - Liquid to Solid BOILING/EVAPORATION - Liquid to Gas MAGNETIC SEPARATION/MECHANICAL SEPARATION - involves the CONDENSATION - Gas to Liquid use of forceps, sieves, magnet and other similar tools to separate SUBLIMATION - Solid to Gas the components of mixtures (solids are mostly involved). DEPOSITION - Gas to Solid CENTRIFUGATION - technique which involves application of CLASSIFICATION OF MATTER: centrifugal force to separate particles from a solution according to size, shape, density, viscosity of the medium. DISTILLATION - a physical separation based on the vaporization of the different components of the mixture to be separated. CHROMATOGRAPHY - the separation of a mixture by passing it in solution or suspension or as a vapor through a medium that moves at different rates. ELECTROLYSIS - a chemical decomposition or breakdown produced by passing an electric current through a liquid or solution containing ions. CRYSTALLIZATION - process by which a solid forms where the PROPERTIES OF MATTER: atoms/molecules are highly organized into a crystal. It can be done by freezing, precipitating or deposition. PHYSICAL PROPERTY - can be measured or observed without changing the composition or identity of a substance. CHEMICAL PROPERTY - the ability of a substance of matter to undergo change in composition under a certain condition. CHEMISTRY General Chemistry 1 Reviewer 1  General Chemistry 1 Reviewer CHEMISTRY 1st Quarter | 1st Semester | 2024 - 2025 CONVERSION FACTOR SIGNIFICANT FIGURES 𝐺𝑖𝑣𝑒𝑛 𝑈𝑛𝑖𝑡 ( 𝐷𝑒𝑠𝑖𝑟𝑒𝑑 𝑈𝑛𝑖𝑡 𝐺𝑖𝑣𝑒𝑛 𝑈𝑛𝑖𝑡 ) = 𝐷𝑒𝑠𝑖𝑟𝑒𝑑 𝑈𝑛𝑖𝑡 General Rules in Significant Figures 1. Non-zero digits are ALWAYS significant e.g. 51.26 (4 significant figures) Common Conversion Factors in Chemistry: 2. Any zero contained between two non-zero numbers is significant e.g. 10.5200104 (9 significant figures) 3. Leading zeroes are never significant e.g. 0.52 - 2 significant figures (5 and 2) 0.2006 - 4 significant figures (2, 0, 0, and 6) 4. Final or trailing zeroes are significant only after a decimal point e.g. 0.0025 - 2 significant figures (2 and 5) 0.06800 - 4 significant figures (6, 8, 0, and 0) Addition and Subtraction Rules in Significant Figures: 1. Add the number of significant figures to the right of the decimal part of each number used in the calculation. 1. Calculate as usual 2. The answer must not contain more significant figures to the right of the decimal point than the fewest of any of 1 m = 39.37 in = 1.094 yd the figures 1.00 in = 2.54 cm 3. So for example, if you are adding together two numbers 1.00 lb = 453.5 g 1.000 kg = 2.205 lb with three and four significant figures to the right of the 1.000 L = 1.057 qt (0.946 L = 1.00 qt) decimal point, the answer cannot have more than three significant figures to the right of the point. 1 kilo = 10³ base unit 1 deci = 10⁻¹ base unit (10 deci = 1 base unit) Examples: 1 centi = 10⁻² base unit (10² centi = 1 base unit) 1.13 + 2.2 = 3.33 1 milli = 10⁻³ base unit (10³ milli = 1 base unit) = 3.3 1 micro* = 10⁻⁶ base unit (10⁶ micro = 1 base unit) 3.2 - 1.55 = 1.65 *1 micro = 1μ = 1 mc = 1.7 10³ micro = 1 milli (1 micro = 10-3 milli) 1 cm³ = 1 mL = 1 cc Multiplication and Division Rules in Significant Figures: 1. Calculate the number of significant figures in each 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 14.7 psi = 101,325 number Pa 2. Calculate as usual 1 mmHg = 1 torr = 133.322 Pa 3. The answer must contain the same number of significant figures as the smallest total of them in the initial 1 kPa = 1,000 Pa numbers. 1 bar = 100 kPa = 100,000 Pa Examples: Celsius to Kelvin: K = °C + 273.15 3.10 x 3.50 = 10.85 Kelvin to Celsius: °C = K - 273.15 = 10.9 3.100 x 3.500 = 10.85 Fahrenheit to Celcius: °C = (°F-32) (5/9) = 10.85 Celsius to Fahrenheit: °F = °C (9/5) + 32 Fahrenheit to Kelvin: K = (°F-32) (5/9) + 273.15 CHEMISTRY General Chemistry 1 Reviewer 2  General Chemistry 1 Reviewer CHEMISTRY 1st Quarter | 1st Semester | 2024 - 2025 SCIENTIFIC NOTATION RELATIVE ATOMIC MASS CALCULATION Rules in Writing Scientific Notation: Relative atomic mass (Ar) is the weighted average mass of the atoms of an element, measured relative to 1/12th the mass of a 1. The base should always be 10 carbon-12 atom. It takes into account the masses and natural 2. The exponent (n) must be a non-zero integer, positive or abundances of all the isotopes of the element. It is a negative dimensionless quantity, meaning it has no units, and it is typically 3. The absolute value of the coefficient (a) is greater than expressed on a scale where the mass of a carbon-12 atom is or equal to 1, but it should be less than 10 (1 ≤ a < 10) exactly 12 units. 4. The coefficient (a) can be positive or negative numbers, including whole numbers and decimal numbers How to Calculate Relative Atomic Mass: 5. The mantissa (number that occurs before the 10ⁿ part) contains the remaining significant digits of the number ∑(𝐼𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑀𝑎𝑠𝑠) 𝑥 (𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐴𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒) 𝐴𝑟 = 100 Examples: Numbers in Standard Form) Numbers in Scientific Example: Notation 1,000,000,000 (1 billion) 1 x 10⁹ 24327 2.4327 x 10⁴ 0.053 5.3 x 10⁻² 0.00049386 4.9386 × 10⁻⁴ 10,000,000 (10 million) 1 x 10⁷ However, pwede rin ganito. Just remember lang to put a percent sign in your calculator or convert into decimal: 31,000,000,000 (31 billion) 31 x 10⁹ 𝐴𝑟 = ∑(𝐼𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑀𝑎𝑠𝑠) 𝑥 (𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐴𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒 𝑖𝑛 %) 7,63,000 7.63 × 10⁵ Example: 𝐴𝑟 = (54 · 0. 0595) + (56 · 0. 9188) + (57 · 0. 0217) 𝐴𝑟 = 55. 90 CHEMISTRY General Chemistry 1 Reviewer 3  General Chemistry 1 Reviewer CHEMISTRY 1st Quarter | 1st Semester | 2024 - 2025 MOLE CONCEPT / MOLAR MASS 2. Moles to Atoms How many atoms of oxygen are in 0.45 mol 𝐵𝑎𝑆𝑂4? 23 MOLE - the mole is a fundamental (SI) unit used to measure the 0. 45 𝑚𝑜𝑙 𝐵𝑎𝑆𝑂4 × 4 𝑚𝑜𝑙 𝑂 × 6.022 ×10 𝑎𝑡𝑜𝑚𝑠 𝑂 1 𝑚𝑜𝑙 𝐵𝑎𝑆𝑂4 1 𝑚𝑜𝑙 𝑂 amount of substance. 24 * A mole of a substance ALWAYS contains the same number of = 1. 084 × 10 𝑎𝑡𝑜𝑚𝑠 𝑂 entities whatever the substance may be. Ex: 3. Mass to molecules 23 68.3 grams of KOH are equal to how many molecules of 1 𝑚𝑜𝑙 𝐴𝑟 = 6. 022 × 10 𝑎𝑡𝑜𝑚𝑠 23 KOH? 1 𝑚𝑜𝑙 𝐻20 = 6. 022 × 10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 Avogadro’s Constant: 1 𝑚𝑜𝑙 𝐾𝑂𝐻 23 6.022 ×10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 23 68. 3 𝑔 𝐾𝑂𝐻 × 56.105 𝑔 𝐾𝑂𝐻 × 1 𝑚𝑜𝑙 𝐾𝑂𝐻 𝑁𝐴 = 6. 022 × 10 𝑒𝑛𝑡𝑖𝑡𝑖𝑒𝑠 24 = 7.33 × 10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐾𝑂𝐻 MOLAR MASS - the molar mass of a substance is the mass in 4. Molecules to moles grams of one mole of the compound. Molar mass SI unit is g/mol. 24 *Molar mass = ∑(Number of atoms of each element × Atomic How many moles are in 3. 01 × 10 molecules of 𝐻20? 23 mass of each element) = in g/mol 24 6.022 ×10 𝑚𝑜𝑙 𝐻20 3. 01 × 10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐻20 × 1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐻20 MOLECULAR MASS - the molecular mass is the mass of a single = 5 mol 𝐻20 molecule of a compound and is primarily applied to covalent compounds, where distinct molecules exist. It is measured in 5. Moles to Mass atomic mass units (amu) and is calculated by summing the atomic How many grams of silver are there in 1.34 moles? masses of all the atoms in a single molecule. 107.87 𝑔 𝐴𝑔 1. 34 𝑚𝑜𝑙 𝐴𝑔 × 1 𝑚𝑜𝑙 𝐴𝑔 FORMULA MASS - the formula mass is the mass of the simplest ratio of ions in an ionic compound, calculated by adding the = 144.55 g Ag atomic masses of the atoms in its empirical formula. It is measured in atomic mass units (amu) and applies to ionic compounds, 6. Molecules to Mass 24 which do not form distinct molecules. How many grams are there in 4. 5 × 10 molecules of * Molecular/Formula Mass = ∑(Number of atoms of each element 𝐶𝑂2? × Atomic mass of each element) = in amu 24 1 𝑚𝑜𝑙 𝐶𝑂2 4. 5 × 10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑂2 × 23 × 6.022 ×10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑂2 44.01 𝑔 𝐶𝑂2 Relationship Between Mass & Number of Moles and Between 1 𝑚𝑜𝑙 𝐶𝑂2 Number of Moles and Number of Molecules/Atoms = 328.87 g 𝐶𝑂2 Note: To find the number of molecules or atoms from mass, always convert the mass of the substance to moles first. Likewise, when determining mass from molecules or atoms, convert the molecules or atoms to moles before calculating the mass. MA - MO - MO 1.) Mass — Moles — Molecules/Atoms 2.) Mass — Molar Mass — Moles Examples: 1. Mass to Moles Calculate the number of moles in 20 grams of water (H₂O). 1 𝑚𝑜𝑙 𝐻20 20𝑔 𝐻20 × 18.015𝑔 𝐻20 = 1.11 or 1 mol 𝐻20 CHEMISTRY General Chemistry 1 Reviewer 4  General Chemistry 1 Reviewer CHEMISTRY 1st Quarter | 1st Semester | 2024 - 2025 PERCENT COMPOSITION EMPIRICAL FORMULA - Simplified or reduced ratio of atoms of each element The percentage composition of a given compound is defined as are in the compound the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. (from % composition) 1. Assume 100g of compound General Formula: 2. Convert to moles % by mass = mass of element / mass of the compound × 100% 3. Divide the each moles by the smallest PROBLEM #1 - What is the percent composition by mass of !! remember Hydrogen in H2O Assume MA-MO-MO = Empirical Fomula - From mass percent assume 100 g, therefore multiply Solution: 100 to each element’s % composition H = 1.01 g/mol - When you have the grams of each element, divide it by O = 15.999 g/mol its molar mass to get the moles of each element - When you have the moles, divide it by the smallest Calculating the mass of the compound: number of moles to get the mole ratio of each element = H20 - The mole ratio will be used as the subscript in the = 2(1.01) + 1(15.999) empirical formula = 18.02 g/mol Example: 52.14% C; 13.13% H; 34.73% O Calculate the percent composition of H: % by mass = mass of element / mass of the compound × 100% Assume 100 g of compound % by mass of H = 2(1.01) / 18.02 × 100% 52.14 g C % by mass of H = 11.2% 13.13 g H 34.73 g O PROBLEM #2 - Calculate the percent by mass of Iron in FeCl3 —- Get the moles of each element Solution: 1 𝑚𝑜𝑙 𝐶 Fe = 55.85 g/mol 𝑚𝑜𝑙 𝐶 = 52. 14 𝑔 𝐶 × 12.011 𝑔 𝐶 = 4. 34102 𝑚𝑜𝑙 𝐶 1 𝑚𝑜𝑙 𝐻 Cl = 35.45 g/mol 𝑚𝑜𝑙 𝐻 = 13. 13 𝑔 𝐻 × 1.008 𝑔 𝐻 = 13. 02579 𝑚𝑜𝑙 𝐻 1 𝑚𝑜𝑙 𝑂 𝑚𝑜𝑙 𝑂 = 34. 73 𝑔 𝑂 × 15.999 𝑔 𝑂 = 2. 17076 𝑚𝑜𝑙 𝑂 Calculating the mass of the compound: (note: include atleast 5 decimal places if values cannot be = FeCl3 stored in your calculator) = 1(55.85) + 3(35.45) —- = 162.2 g/mol Divide by the smallest mole 4.34102 C = 2.17076 = 1. 999 ≈ 2 Calculate the percent composition of Fe 13.02579 H= 2.17076 = 6. 005 ≈ 6 % by mass = mass of element / mass of the compound × 100% 2.17076 % by mass of Fe = 1(55.85) / 162.2 × 100% O= 2.17076 = 1 % by mass of Fe = 34.43% —- Answer C₂H₆O If the mole ratio has the following decimal values, multiply each by: -.25 x 4 -.33 x 3 -.4 x 5 -.66 x 3 -.5x2 -.75 x CHEMISTRY General Chemistry 1 Reviewer 5  General Chemistry 1 Reviewer CHEMISTRY 1st Quarter | 1st Semester | 2024 - 2025 MOLECULAR FORMULA - How many atoms of each element are in a compound (from % composition) -can only be computed when molecular mass is given Get the empirical formula 1. Assume 100g of compound 2. Convert to moles 3. Divide the each moles by the smallest From empirical to molecular 4. Get the molecular mass of empirical formula. Add all product of the molar mass of each element to their number of atoms. 5. Divide to the molecular mass given 6. Multiply the quotient to the empirical formula Example: An organic compound has a molar mass of 256.26 g/mol. 56.25% C; 6.29% H; 37.46% O Assume 100g of compound 56.26 g C 6.29 g H 37.46 g O —-- Get the moles of each element 1 𝑚𝑜𝑙 𝐶 𝑚𝑜𝑙 𝐶 = 56. 26 𝑔 𝐶 × 12.011 𝑔 𝐶 = 4. 68403 𝑚𝑜𝑙 𝐶 1 𝑚𝑜𝑙 𝐻 𝑚𝑜𝑙 𝐻 = 6. 29 𝑔 𝐻 × 1.008 𝑔 𝐻 = 6. 24008 𝑚𝑜𝑙 𝐻 1 𝑚𝑜𝑙 𝑂 𝑚𝑜𝑙 𝑂 = 37. 46 𝑔 𝑂 × 15.999 𝑔 𝑂 = 2. 34140 𝑚𝑜𝑙 𝑂 (note: include atleast 5 decimal places if values cannot be stored in your calculator) —-- Divide by the smallest mole 4.68403 C = 2.34140 = 2. 000 × 3 ≈ 6 6.24008 H= 2.34140 = 2. 665 × 3 = 7. 995 ≈ 8 2.34140 O= 2.34140 =1 × 3 = 3 —-- Empirical formula C₆H₈O₃ —-- Get the molacular mass of empirical formula 𝑔/𝑚𝑜𝑙 𝐶6𝐻8𝑂3 = (12. 011 × 6) + (1. 008 × 8) + (15. 999 × 3) 128.127 g/mol C6H8O3 —- Divide molecular mass of empirical to molecular mass given 256.26 𝑔/𝑚𝑜𝑙 𝑟𝑎𝑡𝑖𝑜 = 128.127 𝑔/𝑚𝑜𝑙 ≈ 2 —- Multiply ratio to empirical formula 2(C₆H₈O₃) = C₁₂H₁₆O₆ CHEMISTRY General Chemistry 1 Reviewer 6