Alief ISD CHEMISTRY STAAR Review Reporting Category 3: Bonding and Chemical Reactions Naming Compounds

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1 C.7.A Name ionic compounds main group or transition metals, covalent compounds, acids and bases, using International Union of Pure and Applied Chemistry (IUPAC) nomenclature rules. Alief ISD CHEMISTRY STAAR Review Reporting Category 3: Bonding and Chemical Reactions Naming Compounds How are ionic compounds named using IUPAC rules? Background information: An ion forms when an atom or a group of atoms loses or gains electrons. If the ion has a positive charge, it is a cation. If it has a negative charge, it is an anion. When ionic bonds from between cations and anions, an ionic compound is formed. Metals tend to lose electrons making them a cation ex. Na + Nonmetals tend to gain electrons making them an anion.ex. Cl - Rules for naming ionic compounds: 1. Binary ionic compound (metal-nonmetal): First, name cation using its element name; for a cation with more than one possible charge (like transition metal Fe 2+ or Fe 3+ ), add a Roman numeral for the cation s charge in parentheses Second, name anion with ide ending EXAMPLES: Formula Names CaCl 2 calcium chloride (cation only has one charge) BaO barium oxide (cation only has one charge) Fe 2 S 3 iron(iii) sulfide (cation has more than one charge) FeS iron(ii) sulfide (cation has more than one charge) 2. Polyatomic* ionic compound (metal-polyatomic ion): First, name cation using same rules as binary ionic compound; remember only the cations with more than one charge needs a Roman numeral in parentheses. Second, name the polyatomic anion (NEVER change their names!) EXAMPLES: Formula Names Na 3 PO 4 sodium phosphate (cation only has one charge) Al(NO 2 ) 3 aluminum nitrate (cation only has one charge) Pb(CO 3 ) 2 lead(iv) carbonate (cation has more than one charge) *See STAAR formula chart for polyatomic ion names

2 How are covalent compounds named using the IUPAC rules? Background information: Atoms in covalent compound are held together by covalent bond bonds formed by the sharing of electrons between atoms. A covalent bond forms between two atoms with high electronegativity, usually nonmetals. Because their attractions for a given electron are about equally strong, they share a pair of electrons. This sharing pulls the atoms into close contact with each other. Rules for naming covalent compounds: 3. Binary covalent compound (nonmetal-nonmetal): First, name the element with the lower electronegativity (which is the first element written in the formula) by given it its original name. Second, name element with higher electronegativity (which is the second element written in the formula) with ide ending. For both elements in the formula, add the correct Greek prefix(es) for the number of atoms (mono-is only used for the higher electronegativity element which is the second element in the formula). GREEK PREFIXES: (1) mono-; (2) di-; (3) tri-; (4) tetra-; (5) penta-; ( 6) hexa-; (7) hepta-; (8) octa-; (9) nona-; (10) deca- EXAMPLES: Covalent compound CO CO 2 NO 2 N 2 O N 2 O 3 BF 3 Names carbon monoxide carbon dioxide nitrogen dioxide dinitrogen oxide dinitrogen trioxide boron trifluoride How are acids and bases named using the IUPAC rules? Background information: Acids are compounds in which H + is bonded to an anion (nonmetal) in aqueous solution. Acids are named based on the suffix of the anion. Bases are compounds that produce hydroxide ions [OH 1- ] when dissolved in water. Bases are named in the same way as other ionic compounds. Rules for naming acids: 4. Naming acids that do not contain oxygen (most are binary acids) Use the prefix hydro- followed by the nonmetal s stem name with ic as the suffix (ending) and then the word acid. So, it would look like this: hydro-stem name of nonmetal-ic acid. EXAMPLES: Acid Formulas without oxygen HF HCl H 2 S HCN Names hydrofluoric acid hydrochloric acid hydrosulfuric acid hydrocyanic acid *Notice that the anion does NOT contain oxygen.

3 5. Naming acids that contain oxygen (tertiary acids; H + bonded with a polyatomic anion) NEVER use hydro- to name these! For acids that contain oxygen, look at the polyatomic anion s name first. If the polyatomic anion s name ends in ite then the acid becomes an ous acid If the polyatomic anion s name ends in ate then the acid becomes an ic acid. EXAMPLES: Acid formulas Names with oxygen HNO 2 nitrous acid [(NO 2 ) 1- nitrite so it becomes an -ous acid] HNO 3 nitric acid [(NO 3 ) 1- nitrate so it becomes a ic acid] H 2 SO 3 sulfurous acid H 2 SO 4 sulfuric acid *Notice that the polyatomic anion contains oxygen and hydro- is NEVER used! Rules for naming bases: 6. Naming compounds that contain the hydroxide ion [OH 1- ] Use the same rules as naming polyatomic ionic compounds. That is, name the cation first and then name the polyatomic anion. EXAMPLES: C.7.B Write the chemical formula of common polyatomic ions, ionic compounds containing main group or transition metals, covalent compounds, acid, and bases. Bases NaOH LiOH Ca(OH) 2 KOH NH 3 Name sodium hydroxide lithium hydroxide calcium hydroxide potassium hydroxide ammonia Writing Chemical Formulas How do you write the chemical formulas of common polyatomic ions? Many ionic compounds are made of polyatomic ions. A polyatomic ion is a tightly-bound group of atoms that has a positive or negative charge and behaves as a unit. There is no simple set of rules for naming most polyatomic ions, so you should memorize the chemical formulas for some of the most common ones, see you STAAR chart! Common Chemical Formulas for Some Common Polyatomic Ions Name Formula Ammonium (NH 4 ) 1+ Hydroxide (OH) 1- Nitrite (NO 2 ) 1- Nitrate (NO 3 ) 1- Carbonate (CO 3 ) 2- Phosphate (PO 4 ) 3- Sulfite (SO 3 ) 2- Sulfate (SO 4 ) 2-

4 How do you write chemical formulas of ionic compounds containing main group or transition metals? First you must write the symbol of the cation and then the anion. Next, look up the oxidation number (charges) for the cation and anion. If the charges DO NOT equal zero, then criss-cross the numbers to make them subscripts (do not bring down the charges; just the number). This will balance the positive and negative charges to make the compound neutral (zero charge). If the charges EQUAL zero, then the formula is already correct. EXAMPLES: Name of compound Write symbols and check charges Criss-cross if charges DO NOT equal zero Magnesium oxide Mg 2+ O 2- Charges = 0 So formula correct Lithium sulfide Li 1+ S 2- Charges 0 So criss-cross # s Barium nitride Ba 2+ N 3- Charges 0 So criss-cross # s Iron(III) oxide Fe 3+ O 2- Charges 0 (Roman numeral-charge of iron) So criss-cross # s Lead(IV) sulfate (Roman numeral-charge of lead) Pb 4+ (SO 4 ) 2- Charges 0 So criss-cross # s Calcium hydroxide * Ca 2+ (OH) 1- Charges 0 So criss-cross # s Ammonium (NH 4 ) 1+ (PO 4 ) 3- Charges 0 phosphate So criss-cross # s Ammonium nitrite (NH 4 ) 1+ (NO 2 ) 1- Charges = 0 So formula correct Write the correct formula MgO Li 2 S Ba 3 N 2 Fe 2 O 3 Pb 2 (SO 4 ) 4 reduce it! Pb(SO 4 ) 2 *Ca(OH) 2 *(NH 4 ) 3 PO 4 NH 4 NO 2 *Keep parenthesis around the polyatomic ions only if a number bigger than ONE gets criss-crossed for a subscript. How do you write the chemical formulas of covalent compounds? Because the names of covalent compounds usually indicate the amounts of each atom as a prefix, writing the formula is often a matter of taking the prefixes and translating them into subscripts after each element symbol. No need to check the elements charges! EXAMPLES: Name of Covalent Formula Compounds Dinitrogen difluoride Ni 2 F 2 Carbon dioxide CO 2 Carbon tetrachloride CCl 4 Boron trifluoride BF 3

5 How do you write the chemical formulas for acids and bases? If you know the name of an acid, you can write its formula by using the rules for writing the name of an acid, in reverse. You still have to balance the ionic charges just as you would for any ionic compound. That is, check the charges to see if they equal zero. If not, crisscross the numbers. EXAMPLE: First, you need to look at the acid s name. If you see that the acid s name is a hydro-base name-ic acid, you should recognize that that acid has no oxygen in its formula. Almost all of these acids are binary acid (made up of two elements only; one element being hydrogen and the other element coming from the base name (anion w/-ide ending). Hydrochloric acid: This would be H and Cl now check charges to write formula. H 1+ Cl 1- formula: HCl Hydronitric acid: This would be H and N now check charges to write formula. H 1+ N 3-..formula: H 3 N If you see that the acid s name has no hydro- prefix, then the acid has oxygen in its formula. The anion is a polyatomic. The -ous ending comes for an -ite polyatomic anion ending, and -ic comes from an -ate polyatomic anion ending. Sulfurous acid: This would be H and (SO 3 ) now check charges to write formula. H 1+ (SO 3 ) 2-..formula: H 2 SO 3 Sulfuric acid: This would be H and (SO 4 ).now check charges to write formula. H 1+ (SO 4 ) 2- formula H 2 SO 4 C.7.C Construct electron dot formulas to illustrate ionic and covalent bonds. Constructing Electron Dot Formulas How to construct electron dot formulas: Different kinds of bonds often form between different atoms. A simple way to show how electrons are transferred or shared during bond formation is with an electron dot formula. Learning how to draw and interpret these diagrams is an important skill. An electron dot formula (or diagram) consists of a chemical symbol surrounded by one to eight dots representing valence electrons. Valence electrons are the electrons in the outermost energy level of an element s atom. All of the elements within a given group (with the exception of helium) have the same number of electron dots in their structures. Hydrogen has 1 valence electron, so you would write the electron dot formula for hydrogen as: H Fluorine has 7 valence electrons, so you would write the electron dot formula for fluorine as: : F :

6 Ions can be represented in electron dot formulas by depicting the charge and the corresponding number of valence electrons. For example, the ion fluorine (F 1- ), which has gained one electron, has eight valence electrons, shown below. : F : - Magnesium ion (Mg 2+ ), which has lost both of its valence electrons, is illustrated below without any dots. Mg 2+ How to construct electron dot formulas to illustrate ionic bonds: The electrons involved in an ionic bond are removed or transferred from the cation and found on the anion. Electron dot formulas illustrate this by indication oppositely charged ions side by side. For example, you would write the electron dot formula for sodium chloride (NaCl) as shown. Na+ The electron dot formula for aluminum oxide (Al 2 O 3 ) is written as: Al 3+ Al 3+ How to construct electron dot formulas to illustrate covalent bonds: Covalent bonds are formed from two electrons that are shared between two atoms. The shared pair can be illustrated in electron dot formulas as two dots, or one short straight line, such as the bond between two hydrogen atoms. H:H H H Single bonds are represented by single lines. Double bonds are represented by double lines. Triple bonds are represented by triple lines.

7 How to construct electron dot formulas for complex covalent molecules or ions: Use these steps to help you construct the electron dot formulas for more complex covalent molecules or ions. 1. Add the total number of valence electrons for every atom in the molecule. If it is an ion, add or subtract electrons to produce the correct charge. 2. Write the structure of the skeleton. The more electronegative atom usually belongs in the center. Connect atoms with lines (or pairs of dots). 3. Distribute electrons to each of the outer atoms to satisfy the octet rule. 4. Distribute the remaining electrons to the central atom. Is the octet rule satisfied for the central atom? If not, then there is probably at least one multiple bond in the molecule or ion. Move pairs of electrons from neighboring atoms to form double bonds with the central atom until the octet rule is satisfied for all of the atoms in the molecule. For example, you would write the electron dot formula for the following as follows: Sulfur dioxide Oxygen carbon dioxide nitrogen C.7.D Describe the nature of metallic bonding and apply the theory to explain metallic properties such as thermal and electrical conductivity, malleability, and ductility. The Nature of Metallic Bonding How can you describe the nature of metallic bonding? Metallic atoms have few valence electrons and low ionization energies. The bonds holding metallic atoms together in the solid and liquid phases, however, are apparently strong, as metals have fairly high melting and boiling points. A metallic atom may be considered to have a central portion, or core, made up of its nucleus and its nonvalence electrons. The atom s valence electrons surround the core. The cores of the metallic atoms making up a metallic solid are arranged in the fixed positions of a crystalline lattice. The valence electrons of metal atoms can be modeled as a sea of electrons. That is, the valence electrons are mobile and can drift freely from one part of the metal to another. Metallic bonds consist of the attraction of the free-floating valance electrons for the positively charged metal ions. These bonds are the forces of attraction that hold metals together. A representative of this model in shown in below.

8 Sea of electrons Metal cation How can you apply metallic bonding theory to explain metallic properties? The nature of metallic bonding explains many physical properties of metals. For example, most metals are excellent conductors of thermal energy. When a difference in thermal energy is applied across a metal, it is quickly and evenly transmitted throughout. This is because the crystal lattice of cations making up a metal is held together very tightly. Thermal energy makes cations move about more rapidly. Since the lattice is held together so tightly, the energy of motion is quickly distributed throughout the solid. Most metals are also excellent conductors of electrical energy. When a difference in electrical energy is applied across a metal, it is quickly and evenly distributed throughout. This happen because the electrons flow freely. Electrons move in response to the difference in electrical energy, moving from high electric potential energy to low electric potential energy. Metals are also malleable, which means that they can be shaped and hammered into thin sheets. A force, such as the strike of a hammer, applied to the solid reshapes the lattice of cations because the cations can move through the sea of electrons without breaking the metallic bonds. For this same reason, metals are also ductile, which means that they can be drawn into long wires without breaking. Another property of metals is their high boiling points. It takes a great deal of thermal energy to break the bonds holding the ions of a metal together. STUDY TIP: If you remember that metals have a tendency to give up electrons, then it will be easy to remember the nature of metallic bonds as positively charged ions in a sea of electrons.

9 What are alloys? Metallic bonds bind elemental metals, such as pure sodium. However, metallic bonds also bind alloys of two or more elements. Alloys are mixtures composed of two or more elements, at least one of which is a metal. Brass, for example, is an alloy of copper and zinc. Alloys are important because their properties are often superior to those of their component elements. The most important alloys today are steels. The table below lists some common alloys and the different elements that compose them. Composition Common Alloys Common Name Common Uses of Alloy Locks, gears, musical instruments Copper and zinc Brass Sculptures, bearings, springs Copper and tin Bronze Cookware, appliances Iron and chromium Stainless steel Statuettes, figurines Tin and copper Pewter Fine tableware, jewelry Silver and copper or other metal such as zinc Sterling silver

10 C.7.E Predict molecular structure for molecules with linear, trigonal planar, or tetrahedral electron pair geometries using Valence Shell Electron Pair Repulsion (VSEPR) theory. How to predict molecular structures: Molecular Structures Linear Trigonal planar Tetrahedral The simplest way to predict the molecular structures involves the Valence Shell Electron Pair Repulsion theory. The Valence Shell Electron Pair Repulsion (VSEPR) theory states that the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. Valence electrons, like all electrons, repel each other. The repulsions of valence electrons affect the angles of the covalent bonds, giving the overall shape to each molecule. How to use the VSEPR theory to predict geometric shapes: Once you have drawn the electron dot structure of a molecule and identified any lone pairs of electrons, you can use the VSEPR theory to help determine the geometric shape of the molecule. Since they repel each other, the electron pairs are as far away as possible from each other. The shape of a molecule depends on the arrangement of electron pairs. Linear Arrangement There are two electron pairs around the central atom in a molecule with linear molecular geometry, 2 bonding electron pairs and 0 lone pairs. The ideal bond angle is 180. Trigonal Planar Arrangement If there are three electron pairs on an atom in a molecule, they tend to spread out from each other on the same plane. The ideal bond angle is Tetrahedral Arrangement If there are four electron pairs on an atom in a molecule, they tend to spread out from each other into four points that form a pyramid. The ideal bond angle is

11 To understand any molecule, one must first complete a Lewis dot structure. It is then possible to predict the molecular shape using TWO BASIC PRINCIPLES: 1. The shapes of molecules are determined by the repulsion between electron pairs in the outer shell of the central atom. Both bond pairs (electron pairs shared by two atoms) and lone pairs (those located on a central atom but not shared) must be considered. 2. Lone pairs repel more than bond pairs. The application of these principles is best seen by referring to specific examples. We shall start looking only at molecules with single bonds (for simplicity). A. Central atoms with less than octet configurations: The BeCl 2 molecule has a Lewis dot structure as shown. The central Be atom has two bond pairs in its outer shell. Repulsion between these two pairs (first principle above) causes the molecule to be linear (see above). If the molecule were bent in any direction, the two bond pairs would be brought closer together, increasing the repulsion. The molecule BF 3 has a dot symbol as: Here the B atom has three bond pairs in its outer shell. Minimizing the repulsion causes this molecule to have a trigonal planar shape, with the F atoms forming an equilateral triangle about the B atom. The F-B-F bond angles are all 120, and all the atoms are in the same plane. BF 3

12 B. Central atoms with octet (noble gas) configurations: The CH 4 molecule has a dot structure as shown. The shape of this molecule, however, is not planar, as is suggested by the way we draw this dot structure. Carbon has 4 bond pairs. The four H atoms are arranged about the C atom in a tetrahedral shape. This shape minimizes the repulsion between the bond pairs. The angle is the same for all H-C-H bond angles and is called the tetrahedral angle. CH 4 There are many molecules that have four bond pairs and this regular tetrahedral shape; CCl 4, SiF 4, and SnCl 4 are just a few examples. C.8.A Define and use the concept of a mole Defining and Using the Mole How do chemists define a mole? Atoms and molecules are so small that even a small amount of a substance consists of an extremely large number of them. A drop of water, for example, contains trillions of water molecules. In chemistry, the number of particles of a substance is expressed by a unit called a mole, which is the SI unit for the amount of a substance. It is abbreviated as mol. A mole of a substance is 6.02 x representative particles of that substance. The number of representative particles in a mole, 6.02 x 10 23, is called Avogadro s number. When you refer to the particles in a mole of a substance, you are referring to representative particles of that substance. A representative particle is the type of particle that makes up the substance. It is the smallest particle into which the substance can be divided and still be that substance. Representative particles might be atoms, molecules, formula units, or ions depending on the substance. See chart below: Representative Particles of Selected Substances Substance Representative Particle Chemical Formula Number of representative particles per mole Lithium Atom Li 6.02 x Iron Atom Fe 6.02 x Argon gas Atom Ar 6.02 x Bromine gas Molecule Br x Sodium ion Ion Na x Sodium chloride Formula unit NaCl 6.02 x Water Molecule H 2 O 6.02 x How can you determine the representative particle of a given substance? Substance Representative Particle Examples Most elements Atom Cu, Ag, Mg Diatomic molecules and all Molecule H 2 O, C 12 H 22 O 11 molecular compounds Ionic compounds Formula units CaCO 3, MgO

13 How can you determine the representative particle of a given substance? The relationship 1 mol = 6.02 x representative particles is the basis for the following conversion factors that you can use to convert number of representative particles to moles: 1 mol 6.02 x representative particles You can also use the conversion factor below to convert from moles to number of representative particles x representative particles 1 mol If you know the number of moles of a substance in a sample, you can calculate the number of representative particles. Conversely, you can calculate the number of moles of a substance if you know the number of representative particles. These conversions can be useful in calculation involving the mass of a substance. EXAMPLE #1: Suppose a sample consists of 6.85 x atoms of carbon. How can you determine the number of moles? First list what you know: number of atoms: 6.85 x atoms C 1 mol C = 6.02 x atoms C Next, multiply the number of atoms of C by the conversion factor, and this will give you the answer x atoms C x 1 mol C 6.02 x atoms C = 1.14 x 10-3 mol C EXAMPLE #2: Another sample consists of 2.58 moles of water, H 2 O. How can you determine the number of water molecules in the sample? 2.58 mol H 2 0 x 6.02 x molecules H 2 O 1 mol H 2 O = 1.55 x molecules H 2 O QUESTION: Why are the representative particles called atoms in the first example and molecules in the second example? When do you use formula units for the representative particles? C.8.B Use the mole concept to calculate the number of atoms, ions, or molecules in a sample of material. Calculating Atoms, Ions, or Molecules Using Moles You can use the mole concept to calculate how many particles make up the mass of a sample of material. In order to perform the calculation, you need to know the molar mass. Molar mass is the mass of one mole of the material. One mole of atoms of an element has a molar mass in grams that is equal to the atomic mass(given in periodic table in amu) Example: 1 mol C = grams C = 6.02 x atoms C 6 atomic mass C

14 So, 1 mol Al = grams Al = 6.02 x atoms Al; 1 mol Mg = grams Mg = 6.02 x atoms Mg; 1 mol Mg 3 N 2 = grams Mg 3 N 2 = 6.02 x formula units Mg 3 N 2 CALCULATIONS: You will need to use your STAAR periodic table to do these calculations. Example 1: How many atoms are in 2 moles of He? (start out with what is given) 2 mol He x 6.02 x atoms He = 1.20 x atoms He 1 mol He Example 2: What is the mass in grams of 3 mol N 2? (start out with what is given) 3 mol N 2 x grams N 2 = grams N 2 1 mol N 2 Example 3: How many molecules are in 7.1 grams H 2 O? (start out with what is given) 7.1 g H 2 O x 1 mol H 2 O x 6.02 x molecules H 2 O = 2.4 x molecules H 2 O g H 2 O 1 mol H 2 O Example 4: Find the number of chlorine ions (Cl - ) in 5 grams of CaCl 2. (start our with what is given) 5 g CaCl 2 x 1 mol CaCl 2 x 2 mol Cl - ions x 6.02 x 1023 Cl - ions = 5 x Cl - ions 111 g CaCl 2 1 mol CaCl 2 1 mol Cl - ions C.8.C Calculate percent composition and empirical and molecular formulas. Calculating Percent Composition, Empirical Formulas, and Molecular Formulas How to calculate percent composition: % by mass of element = mass of element in mol compound x 100 % molar mass of compound Example 1: Calculate the percent compositions of each element in MgCl 2. Molar mass of Mg in formula: grams x 1 = grams Molar mass of Cl in formula: grams x 2 = grams Molar mass of compound (MgCl 2 ): = grams % Mg = g x 100 = 26% Mg g % Cl = g x 100 = 74% Cl g

15 How to calculate empirical formula: An empirical formula (reduced formula) is a formula with the lowest whole-number ratio of elements in a compound. It is not always the same as the molecular formula, which tells exactly how many atoms of each element are contained in a molecule or a formula unit of a compound. For example, methane s molecular formula is C 2 H 6, and its empirical formula is CH 3. If you know the percent compositions of the elements in a compound, you can calculate the empirical formula. First, find the number of moles of each element by dividing the percent composition of each element by its atomic mass. Next, divide each result by the lowest number of moles of all the elements contained in the compound. If the results are not integers (whole numbers), multiply by the smallest factor that will make them integers. Remember, whatever you do to one element in the formula you must do it to all! EXAMPLE 1: Find the empirical formula for a compound that is 79.8% C and 20.2% H. To make the calculations easier, consider a 100 g sample. First: Change % to grams and find moles for each element 79.8g C x 1 mol C = 6.64 mol C g C 20.2g H x 1 mol H = 20.0 mol H 1.008g H Next: Divide each answer by the smallest mole answer. In this case, the smallest mole answer is 6.64 mol mol C = mol H = Therefore, the ratio of C to H is 1:3. Now, write the empirical formula: CH 3 How to calculate a molecular formula: Subscript multiplier = actual molecular weight empirical formula weight EXAMPLE 1: The empirical formula for a molecule is NH 2. If its molecular weight is 32 amu, what is the compound s molecular formula? First find the empirical formula mass of NH 2 = (2 x 1.008) = amu Next, find the subscript multiplier. actual molecular weight = 32 amu = 2 empirical formula weight amu Now multiple both subscripts of NH 2 by 2 to get the molecular formula: N 2 H 4.

16 C.8.D Use the law of conservation of mass to write and balance chemical equations Balancing Chemical Equations Law of Conservation of Mass: matter is not created or destroyed in an ordinary chemical reaction (atoms are not created or destroyed) Total mass of reactants = Total mass of products Example: If 4 sulfur atoms are in the reactants, then 4 sulfur atoms are in the product. How to use the law of conservation of mass to write and balance chemical equations: Use coefficients: number to left of element or compound in a chemical equation; to determine the number of a particular atom in any term; multiply the term s coefficient by the atom s subscript. Example: 3H 2 O has 6 H and 3 O Balanced chemical equation: gives the number and kind of substances in a reaction; reactants are to the left of the arrow and products are to the right of the arrow. 2H 2 + O 2 2 H 2 O Reactants Products The following steps describe how to write and balance a chemical equation: 1. Write a skeleton equation. Write formulas for reactants on the left and formulas for products on the right. Place an arrow between them. 2. Count the number of atoms of each element in the reactants and of each element in the products. For simplicity, if a polyatomic ion such as NO 3 1-, SO 4 2-, CO 3 2-, or NH 4 +, is present unchanged on both sides of the equation, treat it as a single unit. 3. Use coefficients to make the number of each element and polyatomic ion the same on both sides of the equation. 4. Make sure the coefficients are in the lowest possible ratio that balances. EXAMPLE: Balance the chemical equation: H 2 + N 2 NH 3 1. Write a skeleton equation by checking elements to see if they are diatomic molecules and then check compounds for correct formulas. 2. Count the atoms: There are 2 H on the left and 3 H on the right. There are 2 N on left and 1 N on right. 3. Insert coefficients so that there are 6 H on each side: 3H 2 + N 2 2NH 3 4. Now check the N. There are 2 N on the left and 2 N on the right. It is balanced with a coefficient of 1 on N 2 : balanced equations: 3H 2 + N 2 2NH 3 5. Check again by counting each element per side. Left side: 3H 2 = 6 H and N 2 = 2 N Right side: 2NH 3 = 2 N and 6 H (left = right)

17 C.8.E Perform stoichiometric calculation, including determination of mass relationships between reactants and products, calculation of limiting reagents, and percent yield. Stoichiometric Calculations How to determine mass relationships between reactants and products: The calculation of quantities in chemical reactions is a subject of chemistry called stoichiometry. BALANCED equations are used to perform stoichiometric calculations. A balanced chemical equation shows molar ratios among substances in a reaction (reactants or products); use molar mass values to convert from moles to grams or from grams to moles. EXAMPLE: Suppose 8.75 g of propane (C 3 H 8 ) react with oxygen gas (O 2 ) to produce carbon dioxide (CO 2 ) and water (H 2 O). How many grams of water are produced? 1. First, write a balanced equation for the reaction: C 3 H 8 + 5O 2 3CO 2 + 4H 2 O 2. Find the two substances that will be compared in the equation. The two substances to compare are propane and water since the problem gave grams of propane and asks for grams of water. 3. This will be a grams to grams problem: Start with grams of propane and take it to grams of water. Use the balance equation to get the mole ratio of propane to water g C 3 H 8 x 1 mol C 3 H 8 x 4 mol H 2 O x g H 2 O = 14.3 g H 2 O g C 3 H 8 1 mol C 3 H 8 1 mol H 2 O How to determine the limiting reagent in a reaction: The limiting reagent in a chemical reaction is the reactant that determines the amount of each product that will form. When all of the limiting reagent has been used, the reaction stops. Any other reactants that remain are excess reagents. There are several ways to find the limiting reagent. The one that follows is one of the simplest ways to find the limiting reagent. EXAMPLE: What is the limiting reagent when 36 g of CH 4 reacts with 98 g O 2 to produce CO 2 and H 2 O? 1. Write the balanced equation for the reaction. CH 4 + 2O 2 CO 2 + 2H 2 O 2. Change grams of each reactant to moles (If moles are given for each reactant then skip this step) 36 g CH 4 x 1 mol CH 4 = 2.25 mol CH 4 16 g CH 4 98 g O 2 x 1 mol O 2 = 3.06 mol O 2 32 g CH 4

18 3. Now, look at the balanced equation and divide each mole answer by the coefficient of each reactant. CH 4 + 2O 2 CO 2 + 2H 2 O So, take 2.25 mol CH 4 and divide it by 1. You will get 2.25 for your answer. Then take 3.06 mol O 2 and divide it by 2. You will get 1.53 for your answer. The lowest answer will be the limiting reagent. That means O 2 is the limiting reagent. CH 4 is the excess reagent. How to calculate percent yield: Percent yield: describes the portion of the theoretical yield that is actually recovered percent yield = actual yield (from experiment) x 100 % theoretical yield (from calculation) Suppose, for example, that you calculate a theoretical yield of 55.0 g H 2 O, but the actual yield from your experimentation was 48.8 g H 2 O. The percent yield is: percent yield = 48.8 g x 100% = 88.7 % 55.0 g