General Chemistry 1 - Past Paper PDF

LESSON 1: Use Properties of Matter to Identify Substances and to Separate Them GENERAL CHEMISTRY 1  SHORT RECALL… IDENTIFY THE TERMS BASED ON ITS DEFINITION 1. Fixed volume and shape SOLID 2. Fixed volume, indefinite shape LIQUID 3. Indefinite volume and shape...

LESSON 1: Use Properties of Matter to Identify Substances and to Separate Them GENERAL CHEMISTRY 1  SHORT RECALL… IDENTIFY THE TERMS BASED ON ITS DEFINITION 1. Fixed volume and shape SOLID 2. Fixed volume, indefinite shape LIQUID 3. Indefinite volume and shape GAS 4. Anything that has mass and occupies space MATTER 5. Substance composed only of one component PURE SUBSTANCE 6. Substance composed of several components CHEMICAL COMPOUND 7. Pure substance composed of identical atoms ELEMENT 8. Pure substance composed of atoms chemically bonded together CHEMICAL COMPOUND 9. Mixture with indistinguishable components HOMOGENEOUS MIXTURE 10.Mixture with distinguishable components HETEROGENEOUS MIXTURE Read the situation carefully about Lino’s Chemistry project. Lino’s Chemistry Project Lino was given a vial containing an unknown substance that he must identify as ONE of three given options: table salt, table sugar, and iron filings. First, Lino observed that the substance was a (1) fine, white powder. Using a magnifying glass, he observed further that the substance was (2)shiny and had angular edges so he inferred that it was crystalline. Read the following situation carefully about Lino’s Chemistry project. Lino’s Chemistry Project Second, he got a small amount and added it to two separate test tubes— one containing water, and the other containing oil. He shook the test tubes and noticed that the powder disappeared in water and was there visible in oil. He inferred that the substance is (3) soluble in water, but insoluble in oil. Third, he brought the substance near a magnet and observed that it did not get attracted to the magnet. He thus inferred that the substance is (4)not magnetic. Read the following situation carefully about Lino’s Chemistry project. Lino’s Chemistry Project Fourth, he heated the substance under high flame. He noticed that the substance began to melt into a (5) clear, colorless liquid. After some time, the liquid began to (6) bubble and change into a brown color while emitting a sweet aroma. Afterwards, (7) black solids began to appear at the edges and a burnt smell was observed. Lino then inferred that the substance decomposes at high temperatures. Using all these data, Lino now had an answer and confidently submitted his laboratory notebook with his answer to his teacher. He got a perfect score 1. If you were Lino, what substance did you receive? 2. Why do you think this is the correct answer? Point out the observations that Lino made that support this answer. 3. What evidence made you decide that the two other options are not correct? 4. Which of the properties observed by Lino (indicated in bold) are observed without changing the substance’s composition? Write the number only. 5. Which properties are observed by changing the composition of the original substance? 6. What do you call the properties indicated in number 3? How about in number 4? LESSON 1.1: PHYSICAL AND CHEMICAL PROPERTIES OF MATTER Scenario: Lino examined an unknown substance for its various properties. Observations (PHYSICAL PROPERTIES) 1. Visual Examination Naked Eye: Fine, white powder. With Magnifying Glass: Shiny particles with angular edges.  Scenario: Lino examined an unknown substance for its various properties. 2. Solubility Soluble in water. Insoluble in oil. 3. Magnetic Property Not attracted to a magnet. 4. Behavior when Heated Melted, changing state from solid to liquid. WHAT IS PHYSICAL PROPERTIES? Properties that can be observed or measured without changing the identity or composition of the substance. Examples of Physical Properties: Size Shape Color Solubility – ability of solid, liquid or gaseous chemical substance to dissolve in solvent and form a solution. Magnetic character Density – measures how compact the mass is. Volatility – substance tendency to vaporize or turn into gas. Scenario: After melting, Lino observed additional changes in the substance. Observations Indicating Chemical Change: 1.Color change. 2.Emission of a sweet aroma. 3.Formation of black solids. 4.Smell of burning. WHAT IS CHEMICAL PROPERTIES? Properties measured with an accompanying chemical change in a substance. WHAT IS CHEMICAL PROPERTIES? Examples: Flammability (the measure of how easily a material can catch fire and burn) Decomposition at high temperatures Reactivity with acids, bases, oxidizing agents, or reducing agents  INTENSIVE PHYSICAL PROPERTIES These are properties that do not depend on the amount of matter in a substance.  INTENSIVE PHYSICAL PROPERTIES Examples: Density Boiling point Melting point Color Hardness Temperature  EXTENSIVE PHYSICAL PROPERTIES These are properties that do depend on the amount of matter in a substance.  EXTENSIVE PHYSICAL PROPERTIES Examples: Mass Volume Length Energy (e.g., heat capacity) Weight  Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 1. Coal is used to produce electricity by burning them to power the generator.  Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 2. Water freezes at 0 degree Celcius.  Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 3. Salt and refined sugar are both white in color.  Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 4. 5 kg. rice is as heavy as 5 kg. newly harvested cotton.  Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 5. Silver is usually tarnished when exposed to air forming silver oxide. Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 6. Red folded paper  Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 7. Combustibility of plastic Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 8. 12-meter rope.  Categorize the following properties of matter as intensive physical property, extensive physical property or chemical property. 9. Gallium starts to melt at 29.76°C. Lesson 2: Chemical Formula of Common Chemical Substances GENERAL CHEMISTRY 1  Food, medicine, and other product labels often include chemical formulas. Theseformulas can be confusing due to unfamiliarity. Why is chemical Many assume chemical formulas formulas are for scientists important? only.  Why should you care? Society is rapidly changing with increasing information. Grasping chemical formulas helps in making informed decisions about: – Food safety – Medicine ingredients – Everyday products Promotes awareness  H H O water H2O (DIHYDROGEN OXIDE) WHAT IS A CHEMICAL FORMULA? If we are to study the meaning of chemical formula, two words are involved namely chemical and formula.  CHEMICAL refers to a compound or substance FORMULA refers to the symbolic expression  CHEMICAL FORMULA is a symbolic expression of a compound or substance. It is also described as shorthand of expressing the types and the number of atoms in a substance.  CHEMICAL SYMBOL An abbreviation that represents the name of an element.  SUBSCRIPT A small number written at the lower right corner of a chemical symbol. the number after the letter refers to its number of atoms.  Central atom The central atom is usually written first in the chemical formula. Central atom C O O Carbon Dioxide (CO2) C = 1 Carbon atom (central atom). O₂ = 2 Oxygen atoms (bonded to Carbon). Therefore, -The first symbol in the formula is the central atom. -A subscript after a symbol indicates the number of atoms. -If no subscript, it is presumed to be 1. Identify: Central atom. Number of hydrogen atoms. Total number of atoms. Question: What does the subscript "4" mean?  Answers: 1. Central Atom: The central atom is Carbon (C) because it forms bonds with all four hydrogen atoms. 2. Number of Hydrogen Atoms: There are 4 hydrogen atoms, as indicated by the subscript "4" in the formula. 3. Total Number of Atoms: The molecule contains 5 atoms in total: 1. 1 carbon atom 2. 4 hydrogen atoms 4. What does the subscript "4" mean? The subscript "4" indicates that there are 4 hydrogen atoms bonded to the central carbon atom.  Central atom Exception to the Rule: HYDROGEN atom cannot be a central atom because it can only form 1 bond. Example: The central atom of H₂O (water) is oxygen (O). Why? The oxygen atom is the central atom because it is the atom that other atoms (hydrogen) bond to in the molecule.  Complete each statement with a word/ concept in the blank of the item. 1. The chemical symbol of oxygen is _____. 2. There is/are ____ hydrogen atom/s present in this chemical formula C6H12O6. 3. _____ is shorthand of expressing the types and the number of atoms present in a substance. 4. C is the chemical symbol for ______. 5. There is/are _____ oxygen atom/s in H2O. Determine the types and number of atoms in the following chemical formulas. 6. BF3 7. C3H8 8. K2CO3 9. CH4 10. CF4  Chemical Formula of Common Chemical Substances 1. NaCl Common Name: Table Salt Type of Atoms: Sodium (Na), Chlorine (Cl) Number of Each Atom: Sodium: 1 Chlorine: 1  Chemical Formula of Common Chemical Substances 2. NH₃ Common Name: Ammonia Type of Atoms: Nitrogen (N), Hydrogen (H) Number of Each Atom: Nitrogen: 1 Hydrogen: 3  Chemical Formula of Common Chemical Substances 3. C₁₂H₂₂O₁₁ Common Name: Sucrose (Table Sugar) Type of Atoms: Carbon (C), Hydrogen (H), Oxygen (O) Number of Each Atom: Carbon: 12 Hydrogen: 22 Oxygen: 11  Chemical Formula of Common Chemical Substances 4. CH₄ Common Name: Methane Type of Atoms: Carbon (C), Hydrogen (H) Number of Each Atom: Carbon: 1 Hydrogen: 4  Chemical Formula of Common Chemical Substances 5. C₉H₈O₄ Common Name: Aspirin (Acetylsalicylic Acid) Type of Atoms: Carbon (C), Hydrogen (H), Oxygen (O) Number of Each Atom: Carbon: 9 Hydrogen: 8 Oxygen: 4  Chemical Formula of Common Chemical Substances 6. NaOH Common Name: Sodium Hydroxide (Lye or Caustic Soda) Type of Atoms: Sodium (Na), Oxygen (O), Hydrogen (H) Number of Each Atom: Sodium: 1 Oxygen: 1 Hydrogen: 1  LESSON 3: CONSUMER PRODUCTS GENERAL CHEMISTRY 1 CONSUMER PRODUCTS are something that we cannot live without. Every activity we do such as taking a bath, cooking our favorite dishes, driving our cars or monitoring and maintaining our good health condition, there will always be a consumer product that is involved.  Those consumer products were made possible because of the interactions of Science, Technology and Society.  PICTURE ANALYSIS 1. Which problem is being addressed by the invention of the bath soap? 2. What substances are needed to make bath soap? 1. Bath soap was invented to address the problem of personal hygiene by removing dirt, oils, and bacteria from the skin. 2. Fats or oils (e.g., coconut oil, palm oil, olive oil) Alkali (e.g., sodium hydroxide for solid soap or potassium hydroxide for liquid soap) Water Optional: fragrances, colorants, essential oils, and additives like glycerin or herbs. CONSUMER PRODUCT Consumer product is the final good that is bought by individuals or households for personal use. In other words, consumer products are goods that are bought for consumption by the average consumer. The Science, Technology and Society are interrelated to one another because any change in one aspect could trigger also a change to other aspects. Meaning if Knowledge (Science) will improve due to new or latest discoveries made by scientists or even an ordinary person (Society), there will be also a corresponding effect or improvement to the consumer products (Technology) we use every day. CONSUMER PRODUCT The Knowledge (Science) is being used to come up with the different consumer products (Technology) that address problems of the community (Society). Consumer products are widely available because of their usefulness. However, some products can pose dangers despite their benefits.  When Can Consumer Products Cause Harm? 1. Misuse or Abuse: Example: Solvent adhesives like Rugby are intended for use as adhesive materials. Danger: Misused as inhalants by drug dependents.  When Can Consumer Products Cause Harm? 1. Misuse or Abuse: Example: Bleach, ammonia, drain cleaners, oven cleaners. Danger: Direct inhalation or ingestion can cause burns, poisoning, or respiratory failure.  When Can Consumer Products Cause Harm? 1. Misuse or Abuse: Example: Gasoline and Fuels Danger: Inhalation of fumes can cause intoxication, organ damage, or explosion risks.  When Can Consumer Products Cause Harm? 1. Misuse or Abuse: Example: Energy Drinks (Red Bull, Monster, other caffeinated beverages.) Danger: Excessive consumption leads to caffeine overdose, causing heart palpitations, seizures, or even cardiac arrest.  When Can Consumer Products Cause Harm? 1. Misuse or Abuse: Example: Smartphones, gaming consoles. Danger: Excessive use leads to addiction, sleep disruption, and mental health issues.  When Can Consumer Products Cause Harm? 2. Intended Harm: Some products are explicitly designed to cause harm. Examples: Gunpowder, poisons.  When Can Consumer Products Cause Harm? 2. Intended Harm: Example: Tobacco / e-cigars (Nicotine, Tar, Carbon Monoxide, Formaldehyde, Ammonia, Propylene Glycol and Glycerin, Diacetyl)  Lesson 4: Methods of Separating Components of Mixtures and Compounds GENERAL CHEMISTRY 1 Situational Analysis. Analyze the scenario below then answer the corresponding guide questions. Your plane suddenly crushed in an island and luckily you survived. The island is full of fruit bearing plants but no source of fresh water. Situational Analysis. Analyze the scenario below then answer the corresponding guide questions. Guide Questions 1. What would you do to convert the ocean water into potable water? How? 2. Which separating method could be used to separate water and salt from sea water? IMPORTANT VOCABULARY WORDS 2. Separate- 1. Component- A Cause to move, part or element segregate, or of a larger whole. keep apart. IMPORTANT VOCABULARY WORDS 3. Method - is 4. Mixture- a substance describe as produced when two or procedure, more pure substances process, way or are put together in any technique proportion Thus, method of separating components of mixtures is a procedure or process of segregating or moving the constituent or part of a substance which is made up of two or more pure substances. Example of mixture: 1. FILTRATION It is the process of separating the insoluble solid from the liquid substance by allowing the liquid to pass through a porous material called filtering agent. 1. FILTRATION Residue- The solid particles which are retained in the filter paper Filtrate - the clear liquid which passes through a filtering agent such as filter paper  Laboratory apparatus needed: Glass funnel 1. FILTRATION Filter Paper Beaker Stirring Rod 2. DECANTATION (liquid-solid) This is a method used to separate the liquid layer known as supernatant and a heavy insoluble solid known as precipitate by allowing the said solid to settle at the bottom of the container then the liquid above the solid is poured off carefully into another container. 2. DECANTATION (liquid-liquid) Basis of Separation: Differences in density Method: The denser component settles at the bottom and the liquid above is poured off. Example: Separating oil from water. 3. MECHANICAL SEPARATION This involves the use of forceps, sieves, magnet and other similar tools to separate the components of mixtures 3. MECHANICAL SEPARATION the involved substances are mostly solids that can be separated manually 4. CENTRIFUGATION It is the process that uses a motor device known as centrifuge that speeds up the settling of the precipitate using centrifugal or rotating motion. 4. CENTRIFUGATION The heavier component settles at the bottom and the less dense component settles at the upper portion 5. DISTILLATION This is a process that involves the evaporation and condensation of a liquid solvent from a mixture or solution. 3. DISTILLATION Basis of Separation: Differences in boiling points Method: The mixture is heated to boil one component into vapor, which is then condensed back into a liquid. Example: Separation of alcohol and water 6. CHROMATOGRAPHY Basis of Separation: Differences in solubility and adsorption Method: Components of a mixture move at different speeds on a stationary phase (e.g., paper or filter) when a solvent is applied. Example: Separating pigments in ink. 7. ELECTROLYSIS This is a chemical decomposition or breakdown produced by passing an electric current through a liquid or solution containing ions. 7. ELECTROLYSIS Example of this is the separation of oxygen atom to the hydrogen atoms from the water molecules.  8. EVAPORATION Basis of Separation: Volatility (ability to vaporize) Method: The liquid is evaporated, leaving the solid behind. Example: Obtaining salt from seawater. 9. FLOTATION Basis of Separation: Differences in density Method: Lighter materials float while denser materials sink in a liquid. Example: Separating sawdust from water. 10. DISSOLVING Basis of Separation: Difference in solubility: Components have varying solubilities in a solvent. Method: Dissolving mixture in a solvent: Choose a solvent that selectively dissolves one or more components. Examples: Separating salt and sand: Dissolve salt in water, filter, and evaporate water to recover salt.  Let’s Try… Identify the following statements / pictures of methods for separating mixtures or compounds. FILTRATION DECANTATION MECHANICAL (SEIVING)  A) Distillation Which method B) Filtration separates components based C) Decantation on particle size? D) Centrifugation  A) Density What basis of B) Solubility separation does distillation use? C) Boiling points D) Magnetism  A) Centrifugation Which process B) Decantation separates magnetic materials? C) Mechanical D) Filtration  A) Decantation Which method B) Filtration separates salt from water? C) Evaporation D) Centrifugation  LESSON 5: ISOTOPES AND THEIR USES GENERAL CHEMISTRY  What is an ATOM? Atoms are the smallest unit of matter that retains the properties of an element. Example:  ELEMENT An element is a pure substance consisting of only one type of atom, identified by its atomic number. Example: Gold (Au), Carbon (C). STRUCTURE OF AN ATOM  STRUCTURE OF AN ATOM NUCLEUS - Center of the atom containing protons and neutrons. SUBATOMIC PARTICLES: PROTONS - Positively charged particles in the nucleus. NEUTRONS - Neutral particles in the nucleus. ELECTRONS - Negatively charged particles orbiting the nucleus.  PERIODIC TABLE Elements are organized based on their atomic number. Groups (vertical columns) and periods (horizontal rows) categorize elements with similar properties. ATOMIC NUMBER Number of protons in the nucleus of an atom Mass Number Sum of protons and neutrons in the nucleus. Formula: Mass Number =Pro tons +Neutrons  ISOTOPES derived from Greek words isos and topos which means “the same place”. Frederick Soddy.  ISOTOPES are atoms of the same element that have the same number of protons but different in the number of neutrons To identify a specific isotope of an element, write the name of the element followed by a hyphen and the mass number of the isotope.  Due to the variations of the mass number of the isotopes like Oxygen-16, Oxygen-17 and Oxygen-18, the relative atomic mass should be computed as indicated in the periodic table.  To calculate the atomic mass of an element, you have to multiply the atomic mass number of each isotope by its percentage abundance in decimal form. Then add these amounts together to find the relative atomic mass.  For example, Chlorine-35 makes up 75.53 percent of all the chlorine in nature, and Chlorine-37 makes up the other 24.47 percent. The atomic mass unit of the isotopes are 34.969 and 36.966. The relative atomic mass of chlorine is calculated as follows; PRACTICE EXERCISE : Problem Solving Boron has two isotopes, Boron-10 and Boron-11, whose percentage abundances are 19.8% and 80.2% respectively. The atomic masses of Boron-10 and Boron-11 are 10.0129 amu and 11.0093 amu respectively. Write the symbols for the two isotopes of boron and determine the relative atomic mass. Boron- 10 10 10.0129 19.8% 10.812 Boron- 11 11 11.0093 80.2% Problem Solving Silicon has three isotopes, Silicon-28, Silicon- 29 and Silicon-30, whose percentage abundances are 92.21%, 4.70% and 3.09% respectively. The atomic masses are 27.977 amu, 28.976 amu and 29.974amu respectively. Write the symbols for the three isotopes of silicon and determine the relative atomic mass. Silicon-28 28 27.97 92.21% 28.079 Silicon-29 29 28.976 4.70% Silicon-30 30 29.974 3.09%  QUIZ Naturally occurring Argon consists of three isotopes, the atoms of which occur in the following abundances: Argon 36 = 0.34% Argon 38 = 0.07% Argon 40 = 99.59% Calculate the relative atomic mass of argon using the atomic mass unit below: Argon 36 = 35.9676 Argon 38 = 37.9627 Argon 40 = 39.9624  L E S S O N 6 : W RI T I NG A ND NA M I NG T H E C H E M I C A L F O RM UL A O F C O M P O UND GENERAL CHEMISTRY 1 CHEMICAL FORMULA symbolic expression of a compound or substance. It is also defined as shorthand of expressing the types and the number of atoms in a substance. CHEMICAL NAME the scientific name given to a chemical in accordance with the nomenclature system developed by the International Union of Pure and Applied Chemistry (IUPAC).  STRUCTURAL FORMULA a graphical representation of the molecular structure showing how the atoms are possibly arranged in the real three-dimensional space COMP OUND This is a substance which is made up of two or more atoms joined together by a chemical bond. TWO CLASSIFICATIONS OF COMPOUNDS IONIC COVALENT COMPOUNDS COMPOUNDS  1. IONIC COMPOUND is formed between a metal atom and a nonmetal atom, and the type of intramolecular bond exist between atoms is ionic bond due to the presence of ionic charges  2. COVALENT COMPOUND is formed between two or more nonmetal atoms, and the type of intramolecular bond that exist between atoms is covalent bond.  H OW T O W RI T E C H E MI C A L NA ME , A N D F O RM U L A O F I O NI C C O MP O UND S ? STEPS IN WRITING CHEMICAL FORMULA OF IONIC COMPOUND (binary compound) Example: Sodium Chloride Step 1: Identify the atoms involved. Sodium Chloride Atoms: Sodium and Chlorine Step 2: Write the chemical symbols of each atom Sodium Chloride Symbols: Na and Cl Step 3: Identify and write the charges of each atom on the upper right of the symbol Sodium Chloride Symbols: and Step 3: Identify and write the charges of each atom on the upper right of the symbol The Na has a charged of +1 because it tends to give 1 valence electron to chlorine atom to become stable. Step 3: Identify and write the charges of each atom on the upper right of the symbol Most metal atoms whose valence electrons are ranging from 1-3, they tend to give valence electron to become stable thus becomes positive ion known as Cation. Step 3: Identify and write the charges of each atom on the upper right of the symbol Since Na has 1 valence electron, it will give its valence electron to chlorine forming sodium ion. Step 3: Identify and write the charges of each atom on the upper right of the symbol While Cl has a charged of -1 because it tends to accept 1 valence electron from sodium atom to become stable. Step 3: Identify and write the charges of each atom on the upper right of the symbol Most nonmetal atoms whose valence electrons are ranging from 5-7, they tend to accept valence electron to become stable thus becomes negative ion known as Anion.  CATION AND ANION Cation: A positively charged ion formed when an atom loses electrons. (Example: Na+) Anion: A negatively charged ion formed when an atom gains electrons. (Example: Cl−) Step 4: Criss-cross multiply the charges Sodium Chloride Step 5: The charges will become subscript; remove the charge signs Sodium Chloride Chemical Formula: NaCl Example #2 Calcium Sulfide 1. Calcium and Sulfur 2. Ca and S (Ca + S) 3.Ca+2 + S-2 4. Ca+2 + S-2 5. CaS S T E P S W RI T I NG CH E MI CA L F O RM UL A W I T H P O LYATO M I C I O NS 1. Identify the atoms/ions and their charges Magnesium Hydroxide Magnesium (Mg+2) Hydroxide (OH-) 2. Criss-cross multiply the charges Magnesium Hydroxide Mg+2 OH- 3. Drop the charge sign, the charges will become subscript Magnesium Hydroxide Mg OH2 4. Use parentheses for polyatomic ions if necessary. Magnesium Hydroxide Mg(OH)2 Example #2 Aluminum Sulfate 1.Aluminum (Al+3) Sulfate (SO4-2) 2.Al+3 SO4-2 3.Al2 SO4 4.Al2 (SO4 ) 3  STEPS WRITING CHEMICAL FORMULA FOR IONIC COMPOUND USING TRANSITION METALS EXAMPLE: Iron (III) Chloride 1. Identify the cation (transition metal) and its oxidation state, and anion and its charge Iron (III) Chloride Cation: Anion: Iron (Fe3+) Chlorine (Cl-1) 2. Criss-cross multiply the charges Iron (III) Chloride Fe3+ Cl-1 3. Drop the charge signs and the charges will become subscripts Iron (III) Chloride FeCl3 Example #2 Chromium (III) Oxide 1. Chromium (Cr+3) and Oxygen (O-2) 2. Cr+3 O-2 3. Cr2O3  S T E P S W RI T I N G C H E M I C A L F O RM U L A W I T H T RA N S I T I O N M E TA L S A N D P O L YA T O M I C I O N S 1. Identify the atoms/ions and their charges Cobalt(II) nitrate. Cobalt (Co+2) Nitrate (NO3-1) 2. Criss-cross multiply the charges Cobalt(II) nitrate. Co+2 NO3-1 3. Drop the charge sign, the charges will become subscript Cobalt(II) nitrate. Co NO3 2 4. Use parentheses for polyatomic ions if necessary. Cobalt Nitrate Co(NO3)2 Example #2 Copper(II) Sulfate. CuSO4 Write the chemical formula of the following ionic compound: 1. Potassium bromide 2. Calcium fluoride 3. Potassium permanganate 4. Zinc (II) nitrate 5. Lead (II) chromate  NA MI NG OF I ONI C COMP UND S  S T E P S I N NA MI NG I ONI C COMP OUND (B I NA RY) Example: NaCl Step 1: Name the cation (metal) first; Name the anion (nonmetal) Example: NaCl Sodium (metal) Chlorine (nonmetal) Step 2: Use the root name of the nonmetal and add the suffix " -ide. Example: NaCl Sodium (metal) Chlorine (nonmetal) – Chlor+ide= Chloride Step 2: Use the root name of the nonmetal and add the suffix " -ide. Example: NaCl Sodium chloride Examples: 1.MgO 2.Al₂S₃ 3.MgBr₂ 4.AlF₃ 5.CaO Examples: 1.MgO – Magnesium oxide 2.Al₂S₃ - Aluminum sulfide 3.MgBr₂ - Magnesium bromide 4.AlF₃ - Aluminum fluoride 5.CaO – Calcium oxide S T E P S I N NA MI NG I O NI C C O MP O UND W I T H P O LYATO MI C I O NS Example: NaNO₃ Step 1: Name the cation (metal) first; Name the polyatomic ion Example: NaNO₃ Sodium (metal) Nitrate (polyatomic ion) Step 2: Combine the name of cation and polyatomic ion directly Example: NaNO₃ Sodium nitrate Examples: 1.Ca(OH)₂ 2.NH₄Cl 3.K₂SO₄ Examples: 1.Ca(OH)₂ - Calcium hydroxide 2.NH₄Cl – Ammonium chloride 3.K₂SO₄ - Potassium sulfate S T E P S I N NA MI NG I O NI C C O MP O UND S W I T H T RA NS I T I O N ME TA L S ( B I NA RY A ND P O LYATO MI C I O NS ) Example: FeCl₂ Step 1: Name the transition metal cation first; Name the anion or polyatomic ion next. Example: FeCl₂ Iron Chlorine Step 2:Identify the charge of the transition metal and use its name followed by a Roman numeral in parentheses to indicate its charge Example: FeCl₂ Iron (II) Chlorine Important Note: Determine the charge of the transition metal based on the total charge of the compound. Step 3: For monatomic anions, use the "-ide" suffix; For polyatomic ions, use their given names Example: FeCl₂ Iron (II) Chlorine –ide (Chloride) Examples: 1.FeCl₃ 2.MnO₂ 3.Cu(NO₃)₂ 4.SnO 5.SnO₂ Examples: 1.FeCl₃ - Iron(III) chloride 2.MnO₂ - Manganese(IV) oxide 3.Cu(NO₃)₂ - Copper(II) nitrate 4.SnO – Tin (II) oxide 5.SnO₂ - Tin (IV) oxide  W RI T I NG CH E MI CA L F ORMUL A A ND CH E MI CA L NA ME F OR COVA L E NT COMP OUND  STEPS IN WRITING CHEMICAL FORMULA FOR COVALENT COMPOUND (BINARY) Example: Carbon dioxide Step 1: Identify the elements from the compound’s name Example: Carbon dioxide Carbon - C Oxygen - O Step 2: Determine the number of atoms for each element from the prefixes Example: Carbon dioxide Carbon (1) - C Oxygen (2) - O Step 2: Determine the number of atoms for each element from the prefixes Example: Carbon dioxide Mono- (1), Di- (2), Tri- (3), Tetra- (4), Penta- (5), Hexa- (6), Hepta- (7), Octa- (8), Nona- (9), Deca- (10). Step 2: Write the chemical formula: Add subscripts to indicate the number of atoms. Example: Carbon dioxide CO₂ Examples: 1. Phosphorus trichloride 2. Carbon tetrachloride 3. Dinitrogen tetroxide 4. Tetraphosphorus decoxide 5. Tetrasulfur tetranitride Examples: 1. Phosphorus trichloride - PCl₃ 2. Carbon tetrachloride - CCl₄ 3. Dinitrogen tetroxide - N₂O ₄ 4. Tetraphosphorus decoxide - P₄O₁₀ 5. Tetrasulfur tetranitride - S₄N₄  STEPS IN NAMING COVALENT COMPOUNDS Example: PCl₅ Step 1: Identify the elements in the compound Example: PCl₅ Phosphorous Chlorine Step 2: Name the first element: Use the full name of the first element. Add a prefix if there is more than one atom of the first element. Example: PCl₅ Phosphorous Note: Do not use " mono-" if there is only one atom of the first element Step 3: Name the second element: Use the root of the second element’s name and add the suffix " - ide.“ ; use a prefix to indicate the number of atoms of the second element, even if there is only one. Example: PCl₅ Chlorine – Chloride - Pentachloride Step 4: Combine the names: Example: PCl₅ Phosphorous pentachloride Examples: 1.SO₂ 2.N2O5 3.NO 4.SeBr4 5.SiO2 Examples: 1.SO₂ - Sulfur dioxide 2.N2O5 – Dinitrogen pentoxide 3.NO – Nitrogen monoxide 4.SeBr4 – Selenium tetrabromide 5.SiO2 – Silicon Dioxide  LESSON 7.1: FORMULA MASS AND MOLECULAR MASS GENERAL CHEMISTRY 1  FORMULA MASS ◦is used for compounds that are made up of ions and have primarily ionic bonding  FORMULA MASS ◦is the sum of all mass numbers of the atoms that make up the compound. ◦Unit used - atomic mass unit (amu) / u  MOLECULAR MASS ◦known as molar mass is used for compounds that are composed of molecules and have primarily covalent bonding. ◦Unit used – g/mol  MASS NUMBER ◦is the total number of protons and neutrons in the nucleus of an atom, Determine the mass number of the ff elements: ◦Ca – ◦Si – ◦Cu – ◦P - Determine the mass number of the ff elements: ◦Ca – 40.08 u ◦Si – 28.09 u ◦Cu – 63.55 u ◦P – 30.97 u Important Steps in Solving the Formula Mass of a Compound Example: CuSO4 Step 1: List down the atoms involved in the formula which are represented by chemical symbol. Write this in downward manner in the first column. Example: CuSO4 Atoms Cu S O Step 2:. Determine the number of atoms in the formula. Write this in the second column. Consider the subscript which is written outside the parenthesis. Example: CuSO4 Atoms Number of Atoms Cu 1 S 1 O 4 Step 2:. Determine the number of atoms in the formula. Write this in the second column. Consider the subscript which is written outside the parenthesis. Example: CuSO4 Atoms Number of Atoms Cu 1 S 1 O 4 Step 3: Write the mass number of each element in the third column. Example: CuSO4 Atoms Number of Atoms Mass Number Cu 1 63.55 u S 1 32.07 u O 4 16.00 u Step 4: Multiply the number of each atom to its mass number to determine the total mass number of each atom. Write this in the fourth column. Example: CuSO4 Atoms Number of Atoms Mass Number Total mass number of each atom Cu 1 63.55 u 63.55 u S 1 32.07 u 32.07 u O 4 16.00 u 64.00 u Step 4: Get the sum of all total masses to determine the formula mass Example: CuSO4 Atoms Number of Atoms Mass Number Total mass number of each atom Cu 1 63.55 u 63.55 u S 1 32.07 u 32.07 u O 4 16.00 u 64.00 u 159.62 u Determine the formula mass of the following compounds: ◦Al(OH)3 ◦(NH4)2SO4  Example: Al(OH) 3 Atoms Number of Atoms Mass Number Total mass number of each atom Al 1 26.98 u 26.98 u O 3 16.00 u 48.00 u H 3 1.01 u 3.03 u 78.01 u  LESSON 7.2: EMPIRICAL FORMULA FROM PERCENT COMPOSITION AND MOLECULAR FORMULA FROM MOLECULAR MASS GENERAL CHEMISTRY 1 Empirical Formula ◦is the lowest whole- numbered ratio of the elements in a compound  Molecular Formula ◦is the actual composition of a compound indicating the number of atoms per element; number of moles of each kind of atom per mole of compound.  Steps on How to Calculate the Empirical Formula SAMPLE PROBLEM: Determine the empirical formula of a compound whose percentage composition is 50.05% S and 49.95% O by mass Step 1: Use 100 g of the compound as basis of calculation. SAMPLE PROBLEM: Determine the empirical formula of a compound whose percentage composition is 50.05% S and 49.95% O by mass Meaning there will be 50.05 g would be atoms of Sulfur and 49.95 g would be atoms of Oxygen. 50.05% S = 50.05g S 49.95% O = 49.95g O Step 2. Convert the given mass of each atom into their equivalent mole. To do this, convert the given mass of atom in the compound by using the mole of the atom as the conversion factor. SAMPLE PROBLEM: Determine the empirical formula of a compound whose percentage composition is 50.05% S and 49.95% O by mass Step 3. The equivalent value of mole of each atom will be used as the subscript. Therefore, the formula maybe written as S1.56O3.12, but this formula does not have whole-number subscripts. One way to get a formula with whole-number subscript is to divide all the subscripts by the smallest subscript. SAMPLE PROBLEM: Determine the empirical formula of a compound whose percentage composition is 50.05% S and 49.95% O by mass  SAMPLE PROBLEM: Determine the empirical formula of a compound whose percentage composition is 50.05% S and 49.95% O by mass ◦Therefore, the empirical formula is S1O2 or SO2 because it has the lowest possible ratio of whole- numbered subscript. The chemical name of the compound is Sulfur dioxide. Example #2 A compound is found to contain 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen by mass. Determine its empirical formula. Use two decimal places in your calculations. Step 1 ◦Carbon (C): 40.00 g ◦Hydrogen (H): 6.72 g ◦Oxygen (O): 53.28 g Step 2 40.00 gC x 1mol/12.01gC = 3.33 mol 6.72 gH x 1mol/1.01 gH = 6.65 mol 53.28 gO x 1mol/16.00 gO = 3.33 mol C3.33H6.65O3.33 Step 3 C3.33H6.65O3.33 C: 3.33 / 3.33 = 1 H: 6.65 / 3.33 = 2 O: 3.33 / 3.33 = 1 Step 3 C: 3.33 / 3.33 = 1 H: 6.65 / 3.33 = 2 O: 3.33 / 3.33 = 1 Therefore, the empirical formula is C1H2O1 or CH2O. STEPS IN CALCULATING MOLECULAR FORMULA SAMPLE PROBLEM: A sample is found to have 40.0% C, 6.6% H and 53.4% O by mass. The molecular mass of the compound is 180. What its empirical and molecular formulas? Step 1: Get the empirical formula of the compound SAMPLE PROBLEM: A sample is found to have 40.0% C, 6.6% H and 53.4% O by mass. The molecular mass of the compound is 180. What its empirical and molecular formulas? Step 1: Get the empirical formula of the compound SAMPLE PROBLEM: A sample is found to have 40.0% C, 6.6% H and 53.4% O by mass. The molecular mass of the compound is 180. What its empirical and molecular formulas? C H O 3.33 6.6 3.33 Step 1: Get the empirical formula of the compound C H O 3.33 6.6 3.33 Empirical Formula: CH2O Step 2: Calculate the molecular mass of the compound CH2O Atom # of Atoms Mass # Total mass # C 1 12.01 g/mol 12.01 g/mol H 2 1.01 g/mol 2.02 g/mol O 1 16.00 g/mol 16.00 g/mol 30.03 g/mol Step 3: Divide the molecular mass of the compound by its molecular mass by empirical formula SAMPLE PROBLEM: A sample is found to have 40.0% C, 6.6% H and 53.4% O by mass. The molecular mass of the compound is 180. What its empirical and molecular formulas? 180 g/mol = 6 30 g/mol Step 4: Multiply quotient to the empirical formula 180 g/mol = 6 30 g/mol CH2O x 6 = C6H12O6 (Molecular Formula) Example #2: A compound is found to contain 30.4% nitrogen and 69.6% oxygen. If it has a molecular mass of 92.0, what is its molecular formula? LESSON 8: WRITING AND BALANCING CHEMICAL REACTION GENERAL CHEMISTRY 1 Examples of Chemical Reactions daily WORD EQUATION describes a chemical change using the names of the reactants and products. Common Names of some compounds Diatomic Elements/ Molecules Composed of two atoms of the same element. They exist as diatomic molecules naturally rather than single atom because they are more stable in that way. Steps in Writing and Balancing Chemical Equations Step 1. Convert the given problem into a word equation. Example: Nitrogen gas reacts to Hydrogen gas to produce gaseous Ammonia.. Nitrogen(g)+ Hydrogen(g) Ammonia gas(g) Step 2. Convert the word equation into a chemical equation. Example: Nitrogen(g)+ Hydrogen(g) Ammonia gas(g) N2(g) + H2 (g) NH3 (g) Step 3. Balance the given chemical equation written in the step 2 by adding coefficient after the chemical symbol or chemical formula. N2(g) + H2 (g) NH3 (g) N: 2 N: 1 H: 2 H: 3 Step 3. Balance the given chemical equation written in the step 2 by adding coefficient after the chemical symbol or chemical formula. N2(g) + H2 (g) 2NH3 (g) N: 2 N: 1 = 2 H: 2 H: 3 = 6 Step 3. Balance the given chemical equation written in the step 2 by adding coefficient after the chemical symbol or chemical formula. N2(g) + 3H2 (g) 2NH3 (g) N: 2 N: 1 = 2 H: 2 = 6 H: 3 = 6 More Examples… Problem #1: Solid elemental Phosphorus (P4) reacts with Fluorine gas to produce gaseous Phosphorus pentafluoride. Step 1: Phosphorus(s) + Fluorine(g) Phosphorus pentafluoride(g) More Examples… Problem #1: Solid elemental Phosphorus (P4) reacts with Fluorine gas to produce gaseous Phosphorus pentafluoride. Step 1: Phosphorus(s) + Fluorine(g) Phosphorus pentafluoride(g) Step 2: P4(s) + F2(g) PF5(g) Step 3: P4(s) + F2(g) PF5(g) Step 3: P: 4 P: 1 F: 2 F: 5 Step 3: P4(s) + F2(g) 4PF5(g) Step 3: P: 4 P: 1 = 4 F: 2 F: 5 = 20 Step 3: P4(s) + 10F2(g) 4PF5(g) Step 3: P: 4 P: 1 = 4 F: 2 = 20 F: 5 = 20 More Examples… Problem #2: Calcium metal reacts with Nitrogen gas to produce Calcium Nitride. Problem #3: Solid elemental Sulfur (S8) reacts with Oxygen gas to produce Sulfur Trioxide gas. Problem #4: Hydrochloric acid(aq) reacts with an aqueous solution Barium hydroxide to produce Water and aqueous solution Barium chloride. Problem #5: Zinc metal reacts with an aqueous solution of Hydrobromic Acid to produce Hydrogen gas and Zinc bromide(aq). Problem #6: Pentane (C5H12) in liquid form burns in air in a combustion reaction to produce water and Carbon dioxide. Note: Combustion (a.k.a. burning) is a type of chemical reaction in which a substance reacts rapidly with oxygen, releasing heat and light. LESSON 9: CALCULATIONS WITH BALANCED EQUATIONS GENERAL CHEMISTRY 1  LAW OF CONSERVATION OF MATTER also known as the conservation of mass, states that the amount of matter in a closed system never changes. Meaning, the amount of matter before and after the reaction are always equal. CALCULATIONS INVOLVED IN BALANCED EQUATIONS: Mole-to-Mole Conversion Mole-to-Mass Conversion Mass-to-Mass Conversion 1. MOLE-TO-MOLE CONVERSION If the number of moles of one substance is given in the problem, we can determine the number of moles of any of the other substances in the chemical reaction or chemical equation by using a mole conversion factor based on the coefficients in the balanced chemical equation.  STEPS IN MOLE-TO-MOLE CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many moles of O2 are needed to react with 8 moles of NH3? STEP 1: Write the Chemical Equation. NH3 + O2 → N2 + H2O  STEPS IN MOLE-TO-MOLE CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many moles of O2 are needed to react with 8 moles of NH3? STEP 2: Balance the Chemical Equation. 4NH3 + 3O2 → 2N2 + 6H2O  STEPS IN MOLE-TO-MOLE CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many moles of O2 are needed to react with 8 moles of NH3? STEP 3: Analyze the balanced equation and list the number of moles (coefficient) on each substance. 4NH3 + 3O2 → 2N2 + 6H2O No. of Moles Substance 4 NH3 3 O2 2 N2 6 H 2O  STEPS IN MOLE-TO-MOLE CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many moles of O2 are needed to react with 8 moles of NH3? STEP 4: Identify what is asked in the problem. 4NH3 + 3O2 → 2N2 + 6H2O ? Moles of O2 : 8 moles of NH3  STEPS IN MOLE-TO-MOLE CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many moles of O2 are needed to react with 8 moles of NH3? STEP 5: Identify the mole conversion factor based from the balanced chemical equation 4NH3 + 3O2 → 2N2 + 6H2O ? Moles of O2 : 8 moles of NH3 Based from the balanced chemical equation, there are 3 moles of O2 for every 4 moles of NH3. Therefore, the conversion factor would be; 3 mole O2 or 4 mole NH3 4 mole NH3 3 mole O2  STEPS IN MOLE-TO-MOLE CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many moles of O2 are needed to react with 8 moles of NH3? STEP 5: Convert the given mole using the conversion factor 4NH3 + 3O2 → 2N2 + 6H2O ? Moles of O2 : 8 moles of NH3 8 mole NH3 x 3 mole O2 = 6 moles O2 4 mole NH3 2. MOLE-TO-MASS CONVERSION The gram is the most convenient unit for measuring amounts of reactants or products in the laboratory. In order to convert moles to grams or grams to moles, you have to make use of the molar mass of the given compound.  STEPS IN MOLE-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How grams of O2 are needed to react with 10 moles of NH3? STEP 1: Write the Chemical Equation. STEP 2: Balance the Chemical Equation. 4NH3 + 3O2 → 2N2 + 6H2O  STEPS IN MOLE-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How grams of O2 are needed to react with 10 moles of NH3? 4NH3 + 3O2 → 2N2 + 6H2O STEP 3: Identify what is asked in the problem. ? grams of O2 : 10 moles of NH3  STEPS IN MOLE-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How grams of O2 are needed to react with 10 moles of NH3? STEP 4: Identify the mole conversion factor based from the balanced chemical equation 4NH3 + 3O2 → 2N2 + 6H2O ? Moles of O2 : 10 moles of NH3 Based from the balanced chemical equation, there are 3 moles of O2 for every 4 moles of NH3. Therefore, the conversion factor would be; 3 mole O2 or 4 mole NH3 4 mole NH3 3 mole O2  STEPS IN MOLE-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How grams of O2 are needed to react with 10 moles of NH3? STEP 5: Calculate the mole of the substance being asked. 4NH3 + 3O2 → 2N2 + 6H2O ? grams of O2 : 10 moles of NH3 10 mole NH3 x 3 mole O2 = 7.5 mole O2 4 mole NH3  STEPS IN MOLE-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How grams of O2 are needed to react with 10 moles of NH3? STEP 6: Convert the mole into grams using the molar mass (mass number) of O2 as conversion factor. 4NH3 + 3O2 → 2N2 + 6H2O ? grams of O2 : 10 moles of NH3 Mass number of O2 = 16 g/mol x2 atoms = 32 g/mol 7.5 mole O2 x 32 g O2 = 240 g O2 1 mole O2 3. MASS-TO-MASS CONVERSION If the problem asks for the mass in grams of one substance and the mass of another substance is given, do the following steps below. We will use the sample problem below as an example.  STEPS IN MASS-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many grams of O2 are needed to react with 250g of NH3. STEP 1: Write the Chemical Equation. STEP 2: Balance the Chemical Equation. 4NH3 + 3O2 → 2N2 + 6H2O STEP 3: Identify what is asked in the problem. ? grams of O2 : 250 grams of NH3  STEPS IN MASS-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many grams of O2 are needed to react with 250g of NH3. 4NH3 + 3O2 → 2N2 + 6H2O ? grams of O2 : 250 grams of NH3 STEP 4: Convert the given mass to moles using its molar mass (mass number) as conversion factor. Calculate the molar mass of NH3: N (14 g/mole x 1 = 14g/mole) + H (1.01 g/mole x 3 = 3.03g/mole) = 17.03g/mole Convert grams to mole of NH3: 250 g NH3 x 1mole NH3 = 14.68 mole of NH3 17.03 g NH3  STEPS IN MASS-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many grams of O2 are needed to react with 250g of NH3. 4NH3 + 3O2 → 2N2 + 6H2O ? grams of O2 : 250 grams of NH3 STEP 5: Convert the mole of NH3 to mole of O2. (use mole-to-mole conversion steps) 14.68 mole NH3 x 3 mole O2 = 11.01 mole O2 4 mole NH3  STEPS IN MASS-TO-MASS CONVERSION Example: Ammonia gas reacts with Oxygen and produces Nitrogen and Water. How many grams of O2 are needed to react with 250g of NH3. 4NH3 + 3O2 → 2N2 + 6H2O ? grams of O2 : 250 grams of NH3 STEP 6: Convert the mole of O2 to grams of O2 using its molar mass (mass number) 11.01 mole O2 x 32 grams O2 = 352.32 g O2 1 mole O2 Example #2 Nitrogen substance reacts with Hydrogen and produces Ammonia gas. How many grams of Hydrogen are needed to react with 4.5 moles of Nitrogen? How many grams of Nitrogen needed to react with the mass of Hydrogen?