Unit 7b: Chemical Formulas

Unit 7b: Chemical Formulas Student Name: Key Class Period: Website upload 2014 Page 1 of 55

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Unit 7b Vocabulary: 1. Binary compound: A compound that consists of only two elements. 2. Dipole attractions: The attraction of the partially negative (δ - ) end of one polar molecule to the partially positive (δ + ) end of another polar molecule. 3. Dipole moment: An arrow along the line of symmetry in a polar molecule that shows the net direction that electrons are being pulled towards the partially negative (δ - ) end of the molecule. 4. Electronegativity difference: The difference in electronegativity between two elements in a bond. 5. Electrical conductivity (of metals): The ability of a substance to allow electrons to pass from atom to atom through a substance from a source of electricity to an electrical ground. 6. Electrolyte: A solution containing dissolved ions that can conduct electricity within the solution. 7. Empirical formula: The simplest whole-number mole ratio of elements in a compound; used to write the formulas of ionic compounds. 8. Formula mass: The sum of the atomic masses of an element or compound, measured in grams per mole (g/mole). Reported to the nearest tenth (0.1) of a g/mole. 9. Hydrogen bonds: The strong attraction of the H (δ + ) end of one polar molecule to the N, O, or F (δ - ) ends of another molecule. The two molecular ends form temporary covalent bonds. 10. InterMolecular Attractive Force (IMAF): The forces that hold molecules together in the solid and liquid phases. These are the forces that must be overcome to melt or boil a substance. IMAF forces are also called van der Waal s forces. 11. Ionic compound: Compounds consisting of a metal and a nonmetal ionic bonded in a whole-number ratio. 12. London Dispersion Force: The weak attractive forced caused by temporary dipoles in nonpolar molecules. Website upload 2014 Page 3 of 55

13. Metallic bond: A bond formed between metal atoms as they collectively share their conducting electrons evenly between metal kernals. 14. Molecular formula: The actual number of nonmetal atoms in a molecule; a whole-number multiple of the empirical formula. 15. Molecule: A group of nonmetal atoms covalently bonded together to form a distinct particle. 16. Network solid: A crystal lattice formed from covalently bonded nonmetal atoms with no distinct molecules. 17. Nonpolar molecule: A molecule with symmetrical electron distribution resulting in any polar bonds cancelling each other out to yield no partially charged ends. 18. Percent composition: The formula mass of an element divided by the formula mass of the compound containing the element and the divisor then multiplied by 100. 19. Polar molecule: A molecule with asymmetric electron distribution resulting in partially charged ends. 20. Polyatomic ion: An ion formed by atoms bonding together in way that a net charge (positive or negative) is formed. 21. Ternary compound: A compound that consists of three (or more) elements, usually containing a polyatomic ion. Website upload 2014 Page 4 of 55

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Unit 7b Homework Assignments: Assignment: Date: Due: Website upload 2014 Page 6 of 55

Topic: Writing Binary Formulas Objective: How do you write formulas for TWO element compounds? Binary Ionic Compounds: Binary Compounds are compounds that are made of only TWO elements combined in a simple, whole-number ratio. The ratio of the elements in the compounds is dependent on the oxidation numbers (or charges) of the elements that are combining to form the compound. Atoms combine in ratios so that the oxidation numbers add up to ZERO (numbers cancel out). In a chemical formula, the CATION is written first and the ANION is written second. A subscript (#) that represents the number of atoms in the compound is placed after the atom. The number 1 is not written as a subscript for single atoms. Examples would include: NaCl has one sodium atom and one chlorine atom; Li 2 O has two lithium atoms and one oxygen atom. Website upload 2014 Page 8 of 55

Writing Binary Ionic Formulas: *NOTE: X designates a metal ion; Y designates a nonmetal ion Ions to be Bonded How to determine the formula (How to make the + and - charges each cancel out) Formula Example X +1 and Y -1 +1 and -1 cancel each other XY K +1 and F -1 KF X +2 and Y -2 +2 and -2 cancel each other XY Zn +2 and S -2 ZnS X +3 and Y -3 +3 and -3 cancel each other XY Al +3 and N -3 AlN X +2 and Y -1 X +3 and Y -1 Need two -1 ions to cancel out +2 ion Need three -1 ions to cancel out +3 ion XY 2 Ca +2 and 2 x Cl -1 CaCl 2 XY 3 Fe +3 and 3 x Br -1 FeBr 3 X +1 and Y -2 Need two +1 ions to cancel out -2 ion X 2 Y 2 x Li +1 and O -2 Li 2 O X +1 and Y -3 X +2 and Y -3 X +3 and Y -2 X +4 and Y -2 Need three +1 ions to cancel out -3 ion Find COMMON DENOMINATOR of 6: Need three +2 ions to cancel out two -3 ions Find COMMON DENOMINATOR of 6: Need two +3 ions to cancel out three -2 ions Need two -2 ions to cancel out +4 ion X 3 Y 3 x Na +1 and P -3 Na 3 P X 3 Y 2 3 x Cu +2 and 2 x N -3 Cu 3 N 2 X 2 Y 3 2 x Cr +3 and 3 x S -2 Cr 2 S 3 XY 2 Pb +4 and 2 x O -2 PbO 2 Website upload 2014 Page 9 of 55

Topic: Naming Binary Compounds Objective: How do we name compounds containing TWO elements? Naming Binary Ionic compounds: 1. When naming Binary Compounds we place the name of the metal ion first. 2. If the metal ion has more than one listed charge on the Periodic Table, use the Stock system to place a roman numeral in parenthesis after the metal s name representing the charge of the metal ion. 3. Use the first syllable of the nonmetal s name and add the suffix ide to the nonmetal. Roman numeral system: Watch Bozeman Science Naming Compounds video https://www.youtube.com/watch?v=cvkqbhk7vhq Website upload 2014 Page 11 of 55

Naming Binary Ionic Compounds: Use the ion charges as shown here when looking at the samples below. Formula How to determine the name Compound Name KCl ZnO FeBr 2 FeBr 3 K has only one charge listed, so K +1 is named potassium. Chlorine -1 is changed to chloride. Zn has only one charge listed, so Zn +2 is named zinc. Oxygen -2 is changed to oxide. Fe has two charges listed, so use the Br to determine the charge of the Fe. Br has the first listed charge as -1, so two Br -1 ions would be (-1) + (-1) = -2. To make the -2 charge from the two Br -1 ions cancel to zero, the Fe ion MUST have a charge of +2. The Fe +2 cation would be named iron (II), and bromine -1 is changed to bromide. Fe has two charges listed, so use the Br to determine the charge of the Fe. Br has the first listed charge as -1, so three Br -1 ions would be (-1) + (-1) + (-1) = -3. To make the - 3 charge from the three Br -1 ions cancel to zero, the Fe ion MUST have a charge of +3. The Fe +3 cation would be named iron (III), and bromine -1 is changed to bromide. Potassium chloride Zinc oxide Iron(II) bromide Iron(III) bromide Website upload 2014 Page 12 of 55

Formula How to determine the name Ag 2 S Cr 2 O 3 CrO 3 Ca 3 P 2 CuS Cu 2 S Ag has only one charge listed, so Ag +1 is named silver. Two Ag +1 ions are needed to cancel the S -2 ion. Sulfur -2 is changed to sulfide. Cr has three charges listed, so use the O to determine the charge of the Cr. Three oxygen ions each have a -2 charge totaling a -6 charge. The Cr therefore MUST have a +6 charge, so each of the two Cr ions has a +3 charge. The Cr +3 cation would be chromium (III) and the oxygen -2 is changed to oxide. Cr has three charges listed, so use the O to determine the charge of the Cr. Three oxygen ions each have a -2 charge totaling a -6 charge. The Cr therefore MUST have a +6 charge, so the single Cr ion has a +6 charge. The Cr +6 cation would be chromium (VI) and the oxygen -2 is changed to oxide. Ca has only one charge listed, so Ca +2 is named calcium. P has the first listed charge as -3, and two P -3 ions have a charge of -6, so three Ca +2 ions are needed. The Ca +2 cation would be calcium and the phosphorous -3 is changed to phosphide. Cu has two charges listed, so use the S to determine the charge of Cu. S has a -2 charge, so the Cu MUST have a +2 charge. The Cu +2 cation is copper (II) and the sulfur -2 is changed to sulfide. Cu has two charges listed, so use the S to determine the charge of Cu. S has a -2 charge, so the two Cu ions MUST equal a +2 charge. The Cu +1 cation is copper (I) and the sulfur -2 is changed to sulfide. Compound Name Silver sulfide Chromium(III) oxide Chromium(VI) oxide Calcium phosphide Copper(II) sulfide Copper(I) sulfide Website upload 2014 Page 13 of 55

Topic: Writing Formulas Objective: How do we write a formula given the compound name? Writing the formulas when given an Ionic Compound name: If you are given the name of a compound, use the Period Table to confirm the ion charge (oxidation number) of the element to ensure you have the correct number of ions in each compound. If given a metal with a Stock number, use that to determine the number of ions in the compound. Compound Name How to write the formula (using Periodic Table) Potassium sulfide Potassium is always K +1 ; sulfide is S -2 K 2 S Derived Formula Cobalt(II) oxide Oxide is always O -2 ; cobalt is Co +2 CoO Cobalt(III) oxide Oxide is always O -2 ; cobalt is Co +3 Co 2 O 3 Tin(II) sulfide Sulfide is S -2 ; tin is Sn +2 SnS Tin(IV) sulfide Sulfide is S -2 ; tin is Sn +4 SnS 2 Calcium phosphide Calcium is always Ca +2 ; phosphorous is P -3 Ca 3 P 2 Aluminum bromide Aluminum is always Al +3 ; bromide is Br -1 AlBr 3 Silver nitride Silver is always Ag +1 ; nitride is usually N -3 Ag 3 N Watch Bozeman Science Writing Formulas video https://www.youtube.com/watch?v=mrhe4lyqj0a Website upload 2014 Page 14 of 55

Topic: Ternary Compounds Objective: What compounds do polyatomic ions form? Ternary Compound: A Ternary Compound is an ionic compound containing at least one polyatomic ion. Polyatomic ion: A Polyatomic Ion is an ion made from two or more atoms that is treated as one charged ion in chemistry, and not as the individual atoms it is composed of. Polyatomic atoms are listed in Reference Table E. Note these four things about polyatomic ions. Table E: Selected Polyatomic Ions 1. They are made of more than one atom; 2. All but three have a negative charge; 3. Most of them contain oxygen; 4. The polyatomic cations end in -ium ; the polyatomic anions end in -ide, -ite, or -ate. If the ion contains two atoms it ends in -ide ; (CN -, OH -, O 2-2 ) If the ion has few oxygen atoms, it ends in -ite ; (NO 2 - = nitrite) If the ion has many oxygen atoms, it ends in -ate ; (NO 3 - = nitrate) Website upload 2014 Page 15 of 55

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Topic: Writing Ternary Compounds Objective: How do we write a formula with polyatomic ions in it? Writing Ternary Compound Names: When writing names for Ternary Compounds the rules are the same as those for binary compounds, with following exceptions: 1. Parentheses surround the polyatomic ion and a subscript will be used when more than polyatomic ion is in the formula; 2. The metal cation name remains and still uses the Stock system, if needed; Only the ammonium (NH + 4 ) cation retains its name as is. 3. The anion will most likely be the polyatomic ion, and it will keep its given name. Website upload 2014 Page 18 of 55

Writing Ternary Formulas from a given Ternary Compound Name: Compound Name How to write the formula (using Periodic Table E Selected Polyatomic Ions) Derived Formula Ammonium sulfide Ammonium is NH 4 +1 ; sulfide is S -2 (NH 4 ) 2 S Cobalt(II) cyanide Cobalt(II) is Co +2 ; cyanide is CN -1 Co(CN) 2 Cobalt(III) sulfate Cobalt(III) is Co +3 ; sulfate is SO 4-2 Co 2 (SO 4 ) 3 Tin(II) nitrate Tin(II) is Sn +2 ; nitrate is NO 3-1 Sn(NO 3 ) 2 Tin(IV) nitrate Tin(IV) is Sn +4 ; nitrate is NO 3-1 Sn(NO 3 ) 4 Calcium phosphate Calcium is Ca +2 ; phosphate is PO 4-3 Ca 3 (PO 4 ) 2 Aluminum dichromate Aluminum is Al +3 ; dichromate is Cr 2 O 7-2 Al 2 (Cr 2 O 7 ) 3 Silver nitrite Silver is Ag +1 ; nitrate is NO 2-1 AgNO 2 Website upload 2014 Page 19 of 55

Topic: Naming Ternary Compounds Objective: How do we name a formula with polyatomic ions in it? *Note: Use the listed ionic charges on the Periodic Table. Formula How to determine the name Derived Name KNO 3 ZnSO 4 Fe(NO 2 ) 2 Fe(NO 2 ) 3 K is always +1, so K +1 is potassium; NO 3-1 is nitrate. Zn is always +2, so Zn +2 is zinc; SO 4-2 is sulfate. Fe has two listed charges, so use the polyatomic NO 2-1 ion to determine the charge of the Fe cation. There are two NO 2-1 anions, so together they have a total -2 charge. The Fe cation MUST be +2 to cancel the charges. Fe +2 is then iron (II), and NO 2-1 is nitrite. Fe has two listed charges, so use the polyatomic NO 2-1 ion to determine the charge of the Fe cation. There are three NO 2-1 anions, so together they have a total -3 charge. The Fe cation MUST be +3 to cancel the charges. Fe +3 is then iron (III), and NO 2-1 is nitrite. Potassium nitrate Zinc sulfate Iron(II) nitrite Iron(III) nitrite Ag 2 CO 3 Ag is always +1, so Ag +1 is silver; CO 3-2 is carbonate. There needs to be two Ag +1 ions to cancel the charge of the single polyatomic CO 3-2 ion. Silver carbonate Website upload 2014 Page 20 of 55

Formula How to determine the name Derived Name Cr 2 (SO 3 ) 3 Cr(SO 3 ) 3 Ca 3 (PO 4 ) 2 CuCrO 4 Cu 2 CrO 4 Cr has three listed charges, so use the polyatomic SO 3-2 ion to determine the charge of the Cr cation. There are three SO 3-2 ions, totaling -6 charge, so the total charge of cations MUST be +6. There are two Cr ions, so each Cr cation would be a Cr +3 ion. Cr +3 is then chromium (III), and SO 3-2 is sulfite. Cr has three listed charges, so use the polyatomic SO 3-2 ion to determine the charge of the Cr cation. There are three SO 3-2 ions, totaling -6 charge, so the total charge of the cation MUST be +6. There is only one Cr ion, so the Cr cation would be a Cr +6 ion. Cr +6 is then chromium (VI), and SO 3-2 is sulfite. Ca is always +2, so Ca +2 is calcium; PO 4-3 is phosphate. The three Ca +2 cations have a total +6 charge, and the two PO 4-3 polyatomic ions have a total -6 charge. Cu has two listed charges, so use the charge of the polyatomic CrO 4-2 chromate ion. There is only one CrO 4-2 ion, so the single Cu ion MUST be a Cu +2 cation. Cu +2 is then copper (II), and CrO 4-2 is chromate. Cu has two listed charges, so use the charge of the polyatomic CrO 4-2 chromate ion. There is only one CrO 4-2 ion, but the two Cu ions MUST be Cu +1 cations. Cu +1 is then copper (I), and CrO 4-2 is chromate. Chromium(III) sulfite Chromium (VI) sulfite. Calcium phosphate Copper(II) chromate Copper(I) chromate Website upload 2014 Page 21 of 55

Topic: Molecular Compounds Names Objective: How do we name a molecular compound? Molecular compounds may be written two ways: using the Stock System, or using the Prefix System. 1. The Stock System for Molecular Compounds: The Stock System works the same for molecular compounds as with ionic compounds. A Roman numeral is used to show the charge of the first atom written in the formula (which has the LOWER electronegativity). Molecular Formula Name (Stock System) Molecular Formula Name (Stock System) CO 2 Carbon(IV) oxide SO Sulfur(II) oxide CO Carbon(II) oxide SO 2 Sulfur(IV) oxide NO 2 Nitrogen(IV) oxide CCl 4 Carbon(IV) chloride NO 3 Nitrogen(VI) oxide NCl 3 Nitrogen(III) chloride N 2 O 5 Nitrogen(V) oxide P 2 O 3 Phosphorous(III) oxide *Note: NO metals in molecules here! Website upload 2014 Page 22 of 55

2. The Prefix System for Molecular Compounds: The Prefix System uses specific prefixes to describe how many atoms of each element are found in the molecule. 1 atom 2 atoms 3 atoms 4 atoms 5 atoms 6 atoms Mono- (or none) Di- Tri- Tetra- Penta- Hexa- Molecular Formula Name (Prefix System) Molecular Formula Name (Prefix System) CO 2 Carbon dioxide SO Sulfur monoxide CO Carbon monoxide SO 2 Sulfur dioxide NO 2 Nitrogen dioxide CCl 4 Carbon tetrachloride NO 3 Nitrogen trioxide NCl 3 Nitrogen trichloride N 2 O 5 Dinitrogen pentoxide P 2 O 3 Diphosphorous trioxide *NOTE: The Prefix System MAY NOT be used for naming ionic compounds. The Prefix System may ONLY be used to name molecular compounds. *Note: NO metals in molecules here! Website upload 2014 Page 23 of 55

Student name: Key Class Period: _3, 5, & 10_ Please carefully remove this page from your packet to hand in. Naming and Writing Binary and Ternary Formulas Homework: Binary Compounds: Write the formulas for the following binary compounds (4 pts): Name Formula Lead (II) nitride Pb 3 N 2 Potassium oxide K 2 O Iron (III) chloride FeCl 3 Lead (IV) oxide PbO 2 Write the name of the following binary compounds (4 pts): Formula AuCl Cu 2 S CuS FeO Name Gold(I) chloride Copper(I) sulfide Copper(II) sulfide Iron(II) oxide Ternary Compounds: Write the formulas of the following ternary ionic compounds (4 pts): Name Formula Potassium oxalate K 2 C 2 O 4 Lead (II) nitrite Pb(NO 2 ) 2 Lead (IV) hypochlorite Pb(ClO) 4 Ammonium nitrate NH 4 NO 3 Cont d next page Website upload 2014 Page 25 of 55

Name the following ternary compounds (4 pts): Formula Cu 3 PO 4 Cu 3 (PO 4 ) 2 (NH 4 ) 2 O Zn(CH 3 COO) 2 Name Copper(I) phosphate Copper(II) phosphate Ammonium oxide Zinc acetate Both Binary and Ternary Compounds (1 pt ea): Indicate if the formula is binary or ternary, and then write the compound name. Formula Type (Binary or Ternary) Name Na 2 O B Sodium oxide Fe 3 (PO 4 ) 2 T Iron(II) phosphate (NH 4 ) 2 SO 4 T Ammonium sulfate Fe 2 S 3 B Iron(III) sulfide Indicate if the formula is binary or ternary, and then write the compound formula. Name Type (Binary or Ternary) Formula Potassium oxide B K 2 O Sodium phosphate T Na 3 PO 4 Aluminum iodide B AlI 3 Sodium cyanide T NaCN Website upload 2014 Page 26 of 55

Topic: The Mole Concept Objective: How do we know what a chemical formula means in use? The Mole Concept: A Mole is to chemistry as the term dozen applies to doughnuts or eggs. You buy packages of doughnuts or eggs by the dozen, and we use chemical formulas and calculate reactions by the mole. A mole equals 6.023 x 10 23 of ANYTHING. One mole is the number of atoms that have a combined mass of their given atomic mass measured in grams. According to the Periodic Table, hydrogen has an atomic mass number of 1.00794, so one mole of hydrogen gas would have a mass of 1.00794 grams. o Just how big is this number? o One mole of marbles spread over the earth would result in a layer three miles thick. Website upload 2014 Page 27 of 55

Topic: Empirical Formulas Objective: How do we know what a chemical formula means in use? Empirical Formulas: An Empirical Formula is the simplest, whole-number mole ratio of the elements in an ionic compound. If no subscript is shown, then the number of moles of that ion is given as one. When interpreting chemical formulas, if there are parentheses around the written elements then any subscript outside the parentheses multiplies ALL the elements within the parentheses by that subscript. As an example, Ca(NO 3 ) 2 has a 2 as the subscript outside the parentheses, so multiply the number of each element INSIDE the parentheses to find the total number of atoms for that element: (2 x one N = 2 N; 2 x three O = 6 O). As Ca is NOT within the parentheses, it is not affected. There remains only the single Ca in the formula. Website upload 2014 Page 28 of 55

Empirical Formula Examples: Empirical Formula # of ions of each element in the compound # of moles of each element in one mole of the compound Total # of moles of ions in one mole of the compound NaCl 1 x Na, 1 x Cl 1 x Na, 1 x Cl 1 + 1 = 2 CaCl 2 1 x Ca, 2 x Cl 1 x Ca, 2 x Cl 1 + 2 = 3 Al 2 O 3 2 x Al, 3 x O 2 x Al, 3 x O 2 + 3 = 5 KNO 3 1 x K, 1x N, 3 x O 1 x K, 1 x N, 3 x O 1 + 1 + 3 = 5 K 2 SO 4 2 x K, 1 x S, 4 x O 2 x K, 1 x S, 4 x O 2 + 1 + 4 = 7 Al(NO 3 ) 3 1 x Al, 3 x N, 9 x O 1 x Al, 3 x N, 9 x O 1 + 3 + 9 = 13 Al 2 (SO 4 ) 3 2 x Al, 3 x S, 12 x O 2 x Al, 3 x S, 12 x O 2 + 3 + 12 = 17 Website upload 2014 Page 29 of 55

Topic: Interpreting Formulas Objective: How do we know what a chemical formula means in use? Coefficients: A coefficient is a number in FRONT of given formula that tells you how many moles of the compound you are working with. To find the total of moles of atoms you have, multiply the number of a particular element by the coefficient given. If no coefficient is given, then it is assumed as 1. Coefficient Examples: Empirical Formula # of moles of each atom in one mole the compound Formula with coefficient Total # of moles each of atom in given number of moles of the compound KBr 1 K, 1 Br 2 KBr 2 x (1 K, 1 Br) = 2 K, 2 Br MgF 2 1 Mg, 2 F 3 MgF 2 3 x (1 Mg, 2 F) = 3 Mg, 6 F CaSO 4 1 Ca, 1 S, 4 O 4 CaSO 4 4 x (1 Ca, 1 S, 4 O) = 4 Ca, 4 S, 16 O Au(NO 2 ) 3 1 Au, 3 N, 6 O 5 Au(NO 2 ) 3 5 x (1 Au, 3 N, 6 O) = 5 Au, 15 N, 30 O Co(OH) 3 1 Co, 3 O, 3 H 4 Co(OH) 3 4 x (1 Co, 3 O, 3 H) = 4 Co, 12 O, 12 H Mg 3 (PO 4 ) 2 3 Mg, 2 P, 8 O 3 Mg 3 (PO 4 ) 2 3 x (3 Mg, 2 P, 8 O) = 9 Mg, 6 P, 24 O Website upload 2014 Page 30 of 55

Topic: Molecular Formulas Objective: How do we know what a molecular formula means in use? Molecular Formulas: A Molecular Formula gives the total number of atoms of each element needed to form the molecule. Molecules are particles formed from the covalent bonding of two or more types of nonmetal atoms. Molecular Formula examples: i. A molecule of methane has one carbon atom and four hydrogen atoms, giving methane the molecular formula of CH 4. ii. A molecule of benzene has six carbon atoms and six hydrogen atoms, giving benzene a molecular formula of C 6 H 6. Website upload 2014 Page 31 of 55

Topic: Empirical Formulas Objective: How do we know what an empirical formula means? Empirical Formulas: Empirical Formulas show the simplest whole-number ratio in which atoms may combine to form a compound. Empirical formulas are used to represent ionic compounds which form crystals of alternating + and - charges rather than separate molecules. Calcium chloride is ionic, and the empirical formula CaCl 2 represents a ratio of one Ca +2 cation to every two Cl -1 anions. i. If a sample of calcium chloride contains 3000 Ca +2 ions, then it must contain 6000 Cl -1 ions. ii. If a sample of calcium chloride contains 1.5 moles of Ca +2 ions, then it must contain 3.0 moles of Cl -1 ions. ALL ionic compound formulas are empirical formulas. o KCl has a 1:1 ratio, CaCl 2 has a 1:2 ratio, and Al 2 (SO 4 ) 3 has a 2:3:12 ratio. None of these may be further simplified. Molecular Compound Empirical formulas: o Certain molecular compounds such as CH 4 (1:4), NO 2 (1:2), and H 2 O (2:1) cannot be simplified further. For these molecular compounds, the molecular formula is also the empirical formula. o Other molecular compounds may have their molecular formula simplified to a new empirical formula. Website upload 2014 Page 32 of 55

Benzene, C 6 H 6 (6:6), may be simplified to a 1:1 ratio, giving CH as the empirical formula for benzene. Hydrogen peroxide, H 2 O 2 (2:2), may be simplified to a 1:1 ratio, giving HO as the empirical formula for hydrogen peroxide. Isobutene, C 4 H 8 (4:8), may be simplified to a 1:2 ratio, giving isobutene the empirical formula of CH 2. Website upload 2014 Page 33 of 55

Topic: Stoichiometry Objective: Define the process of stoichiometry in chemistry. Stoichiometry: The study of quantitative relationships as derived from chemical formulas and equations is known as stoichiometry. When working with chemical formulas, stoichiometry involves the relationships of the ratios between the numbers of moles of atoms of different compounds that comprise the various compounds. i. The mass of one mole of atoms of an element is equal to that element s atomic mass in grams. This is known as the gram atomic mass of the element. The gram atomic mass is written above left of the atomic symbol on the Periodic Table, and may be rounded to the nearest tenth of a gram for the purposes of this course. ii. The mass of one mole of molecules or formula units is equal to the sum of the gram atomic masses of the atoms that make up a mole of a particular molecule or formula unit. This number represents the gram formula mass, and is also known as just the formula mass. Website upload 2014 Page 34 of 55

Topic: Gram Formula Mass Objective: How do we calculate the Molar Mass of a compound? Determining the Gram Formula Mass of a Compound: You will find below some atomic symbols and information as found on the Periodic Table. Before working on the gram formula mass calculations for the examples given, we will round each gram atomic mass as given to the nearest tenth (written below each atomic symbol). KCl: (1 mole of K @ 39.1 g/mol = 39.1 g of K) + (1 mole of Cl @ 35.5 g/mol = 35.5 g of Cl) = 74.6 g/mol each mole of KCl Ca(NO 3 ) 2 : (1 mole of Ca @ 40.1 g/mol = 40.1 g of Ca) + (2 moles of N @ 14.0 g/mol each = 28.0 g of N) + (6 moles of O @ 16.0 g/mol each = 96.0 g of O) = 164.1 g/mol each mole of Ca(NO 3 ) 2 Al 2 (SO 4 ) 3 : (2 moles of Al @ 27.0 g/mol each = 54.0 g of Al) + (3 moles of S @ 32.1 g/mol each = 96.3 g of S) + (12 moles of O @ 16.0 g/mol each = 192.0 g of O) = 342.3 g/mol each mole of Al 2 (SO 4 ) 3 Website upload 2014 Page 35 of 55

Topic: Chemistry recipes Objective: What does chemistry and cooking have in common? Think about using a recipe to create some new food. You need to buy certain ingredients, measure them, follow directions, apply energy, and hopefully you will end up with what was in the picture! Chemistry is the same idea; we measure, follow directions, apply energy, and hopefully you will end up with what you originally planned to make. The recipe to be followed in chemistry is called a CHEMICAL REACTION. To make a triple batch of chicken wings for a party, triple the recipe ingredients. In chemistry, we can triple a reaction by tripling the reactants (ingredients). For chemistry, our recipes call for measurements in moles, but we can t measure moles. We need to measure the amount of ingredients (reactants) in grams; therefore we have a need to convert moles to grams, and also grams to moles. Website upload 2014 Page 36 of 55

Topic: Converting Grams to Moles Objective: How do we convert grams to moles? Determining the number of moles in a measured mass of a substance: To convert grams to moles, divide the given mass of your substance by the formula mass of that substance: given mass (g) formula mass ( g mole ) = # of moles Examples: 1. How many moles of KCl are in 182.9 grams of KCl? Step #1: Find the formula mass for KCl: (1 mole of K @ 39.1 g/mole = 39.1 g of KCl) + (1 mole of Cl @ 35.5 g/mole = 35.5 g of Cl) = 74.6 g/mole of KCl Step #2: given mass (g) formula mass ( g mole ) = # of moles 182.9 g of KCl / 74.6 g/mole of KCl = 2.45 moles of KCl Website upload 2014 Page 37 of 55

2. How many mole of Ca(NO 3 ) 2 are in 45.5 grams of Ca(NO 3 ) 2? Step #1: Find the formula mass for Ca(NO 3 ) 2 : (1 mole of Ca @ 40.1 g/mole = 40.1 g of Ca) + (2 moles of N @ 14.0 g/mole each = 28.0 g of N) + ( 6 moles of O @ 16.0 g/mole each = 96.0 g of O) = 164.1 g/mole of Ca(NO 3 ) 2 Step #2: given mass (g) formula mass ( g mole ) = # of moles 45.5 g of Ca(NO 3 ) 2 / 164.1 g/mole of Ca(NO 3 ) 2 = 0.277 moles of Ca(NO 3 ) 2 Website upload 2014 Page 38 of 55

Topic: Converting Moles to Grams Objective: How do we convert moles to grams? Determining the mass (in g) of a given number of moles of substance: To convert moles to grams, multiply the given number of moles of your substance by the formula mass for that substance: [given # of moles x formula mass (g/mole)] = mass in grams Examples: 3. What will be the mass of a 2.45 mole sample of KCl? Step #1: Find the formula mass of KCl: (1 mole of K @ 39.1 g/mole = 39.1 g of KCl) + (1 mole of Cl @ 35.5 g/mole = 35.5 g of Cl) = 74.6 g/mole of KCl Step #2: [given # of moles x formula mass (g/mole)] = mass in grams 2.45 moles of KCl x 74.6 g/mole of KCl = 183 g of KCl Website upload 2014 Page 39 of 55

4. What will be the mass of a 0.554 mole sample of Ca(NO 3 ) 2? Step #1: Find the formula mass of Ca(NO 3 ) 2 : (1 mole of Ca @ 40.1 g/mole = 40.1 g of Ca) + (2 moles of N @ 14.0 g/mole each = 28.0 g of N) + ( 6 moles of O @ 16.0 g/mole each = 96.0 g of O) = 164.1 g/mole of Ca(NO 3 ) 2 Step #2: [given # of moles x formula mass (g/mole)] = mass in grams 0.554 moles of Ca(NO 3 ) 2 x 164.1 g/mole of Ca(NO 3 ) 2 = 91.0 g of Ca(NO 3 ) 2 Watch Bozeman Science Mole conversions video https://www.youtube.com/watch?v=xpdqex_wmjo Website upload 2014 Page 40 of 55

Student name: Key Class Period: _3, 5, & 10_ Please carefully remove this page from your packet to hand in. Formula Mass homework (1 pt ea.) Determine the number of moles of an element in one mole of these compounds: Formula Number of moles: Number of moles: Number of moles (if any): CaBr 2 Ca: 1 Br: 2 XXXXXXXXXXXXXXXXX K 2 SO 4 K: 2 S: 1 O: 4 SO 3 S: 1 O: 3 XXXXXXXXXXXXXXXXX CH 3 COOH C: 2 H: 4 O: 2 Determine the total number of moles of atoms in the following formulas: Formula Total # of moles of atoms Formula Total # of moles of atoms Formula Total # of moles of atoms NaCl 2 2 NaNO 3 10 3 K 2 Cr 2 O 7 33 Calculate the formula masses in g/mole for the compounds listed below. Determine the gram atomic masses to the nearest tenth. Formula Work space (Show ALL work!) Gram Formula Mass Na 2 SO 4 (2 x 23.0)+(1 x 32.1)+(4 x 16.0) 46.0+32.1+64.0 = 142.1 g/mol NaOH (1 x 23.0)+(1 x 16.0)+(1 x 1.0) 23.0+16.0+1.0 = 40. g/mol H 2 O (2 x 1.0)+(1 x 16.0) 2.0+16.0 = 18. g/mol C 6 H 12 O 6 (6 x 12.0)+(12 x 1.0)+(6 x 16.0) 72.0+12.0+96.0 = 180. g/mol N 2 O 4 (2 x 14.0)+(4 x 16.0) 28.0+64.0 = 92.0 g/mol Ba(NO 3 ) 2 (1 x 137.3)+(2 x 14.0)+(6 x 16.0) 137.3+28.0+96.0 = 261.3 g/mol C 2 H 6 (2 x 12.0)+(6 x 1.0) 24.0+6.0 = 30. g/mol Website upload 2014 Page 41 of 55

Identify each of the following formulas as either empirical or molecular. If it is a molecular formula, write the simplified empirical formula. Formula Empirical or Molecular? Simplified Empirical Formula Empirical or Molecular? Simplified Empirical C 2 H 6 M CH 3 N 2 O 4 M NO 2 Ba(NO 3 ) 2 E --------------- C 6 H 12 O 6 M CH 2 O Determine how many moles of each substance is represented by the given mass. Mass and formula Work space (Show ALL work!) Moles 36. g of H 2 O 36 g H 2 O x 1mol H 2 O / 18.0 g/mol of H 2 O = 2 2 60. g of NaOH 60. g NaOH x 1 mol NaOH / 40.0 g/mol of NaOH = 1.5 1.5 71. g of Na 2 SO 4 71 g Na 2 SO 4 x 1 mol Na 2 SO 4 / 142.1 g mol of Na 2 SO 4 = 0.5 0.5 Website upload 2014 Page 42 of 55

Topic: Percent Composition Objective: How do we find the percentage of each element? Percent Composition: The Percent Composition of a compound is the proportion by mass of all the elements that are in a compound. Determination of Percent Composition: Qualitative analysis tells us WHAT elements are in a compound, but we need quantitative analysis to determine how much of each element is in a compound. One of the ways this is accomplished is to weigh a sample of the compound, then take the compound and decompose it into its individual elements. After decomposition, weigh each element alone and then calculate what percent by mass each element was of the whole compound. Watch Percent Mass, Composition, Empirical Formulas & Molecular Formulas video https://www.youtube.com/watch?v=mdnydmoq6as Website upload 2014 Page 43 of 55

Experimental Determination of Percent Composition: given mass of an element in a sample given mass of the total sample x 100 Example: A sample of compound containing only nitrogen and oxygen has a mass of 80.0 grams. Experiments show that the sample is made up of 56.0 grams of oxygen and 24.0 grams of nitrogen. What is the percent composition by mass of each element in the compound? o Percent of oxygen: (56.0 g of O/80.0 g of sample) x 100 = 70.0% O o Percent of nitrogen: (24.0g of N/80.0 g of sample) x 100 = 30.0% N Determination by given formula (need periodic table): given mass of an element in a sample given mass of the total sample x 100 Examples: 1. What is the percent composition by mass of the elements in SiO 2? Formula mass of SiO 2 : (1 Si @ 28.1 g/mole) + (2 O @ 16.0 g/mole) = 60.1 g/mole of SiO 2 o Percent of Si: (28.1/60.1) x 100 = 46.8% Si in SiO 2 o Percent of O: (32.0/60.1) x 100 = 53.2% O in SiO 2 Website upload 2014 Page 44 of 55

2. What is the percent composition by mass of the elements in H 2 O? Formula mass of H 2 O: (2 H @ 1.0 g/mole) + (1 O @ 16.0 g/mole) = 18.0 g/mole of H 2 O o Percent of H: (2.0/18.0) x 100 = 11% H in H 2 O o Percent of O: (16.0/18.0) x 100 = 88.9% O in H 2 O 3. What is the percent composition by mass of each element in H 2 SO 4? Formula mass of H 2 SO 4 : (2 H @ 1.0 g/mole) + (1 S @ 32.1 g/mole) + (4 O @ 16.0 g/mole) = 98.1 g/mole of H 2 SO 4 o Percent of H: (2.0/98.1) x 100 = 2.0% H in H 2 SO 4 o Percent of S: (32.1/98.1) x 100 = 32.7% S in H 2 SO 4 o Percent of O: (64.0/98.1) x 100 = 65.2% O in H 2 SO 4 Website upload 2014 Page 45 of 55

Topic: Hydrated Compounds Objective: What is a hydrate, and how does it affect chemistry? Hydrates: i. A hydrate is an ionic solid crystal that has water trapped within the crystalline structure. Different hydrates may have differing amounts of water within the crystal structure. ii. A hydrate is represented by n H 2 0, where the n is the number of water molecules within each hydrate crystal. iii. The water molecules are not a part of the compound; the water molecule(s) are trapped between the crystals ions. iv. The water molecules within the hydrate may be removed by heating, leaving the anhydrous salt behind. Hydrate examples: 5 water molecules locked in 10 water molecules locked in 7 water molecules locked in 2 water molecules locked in 2 water molecules locked in Watch Hydrates video https://www.youtube.com/watch?v=4hkxmttuzhq Website upload 2014 Page 46 of 55

Topic: Hydrate Percentage Objective: How may we determine the ratio of water in a hydrate? Determining the Percentage of Water in a Hydrate Experimentally: given mass of the water in a hydrate compound given mass of the hydrate compound x 100 i. To determine the mass of water that was removed from the hydrate: a. Subtract the mass of the (dried) anhydride from the original hydrate sample mass. ii. To determine the percent of water in the original hydrate sample: a. Take the mass of removed water (found above) and divide it by the Example: original hydrate sample mass and multiply by 100. A 10.40 gram sample of hydrate crystal is heated to an anhydride with a final mass of 8.72 grams. (Assumption is that ALL water has been lost) 1. Calculate the mass of water removed from the original hydrate sample. (mass of sample - mass of anhydride) = 10.40 g - 8.72 g = 1.68 g H 2 O lost 2. Calculate the percent by mass of the water in the hydrate sample. (mass of H 2 O lost / mass of hydrate) x 100 = (1.68 g/10.40 g) x 100 = 16.2% water by mass Website upload 2014 Page 47 of 55

Determining the Percentage of Water in a Hydrate using the formula: formula mass of the water in a hydrate compound formula mass of the hydrate compound x 100 Formula mass of H 2 O: (2 H @ 1.0 g/mole) + (1 O @ 16.0 g/mole) = 18.0 g/mole of H 2 O o If the hydrate coefficient (n) is 2, then (2 x 18.0 g/mole) = 36.0 g/mole H 2 O in the hydrate o If the hydrate coefficient (n) is 4, then (4 x 18.0 g/mole) = 72.0 g/mole H 2 O in the hydrate To determine the formula mass of the hydrate: 1. Find the formula mass of the crystal; 2. Find the formula mass of the number of moles (n) in the hydrate; 3. Add the two numbers together. 4. To determine the percentage of water in the hydrate: 5. Divide the formula mass of the water in the hydrate by the total formula mass and multiply by 100. Website upload 2014 Page 48 of 55

Examples: 1. What is the percent composition by mass of each element in CuCO 3 2 H 2 O? Note that the 2 H 2 O means this is a hydrate, in this case containing 2 molecules of water. i. Find the formula mass for CuCO 3 2 H 2 O: (1 Cu @ 63.5 g/mole) + (1 C @ 12.0 g/mole) + (5 O @ 16.0 g/mole) + (4 H @ 1.0 g/mole) = 159.5 g/mole CuCO 3 2 H 2 O Note the three O in the crystal, and a total of two O in the hydrate. Note also the hydrate has a total of four H (2 x 2). % Cu: (63.5 g/mole Cu / 159.5 g/mole CuCO 3 2 H 2 O) x 100 = 39.8% Cu % O: (80.0 g/mole O / 159.5 g/mole CuCO 3 2 H 2 O) x 100 = 50.2% O % C: (12.0 / 159.5 g/mole CuCO 3 2 H 2 O) x 100 = 7.52% C % H: (4.0 / 159.5 g/mole CuCO 3 2 H 2 O) x 100 = 2.5% H 2. What is the percent composition by mass of only the water in CuCO 3 2 H 2 O? i. Formula mass of H 2 O: (2 H @ 1.0 g/mole) + (1 O @ 16.0 g/mole) = 18.0 g/mole of H 2 O ii. Find the formula mass for CuCO 3 2 H 2 O: (1 Cu @ 63.5 g/mole) + (1 C @ 12.0 g/mole) + (5 O @ 16.0 g/mole) + (4 H @ 1.0 g/mole) = 159.5 g/mole CuCO 3 2 H 2 O % H2O: (18.0 g/mole H 2 O / 159.5 g/mole CuCO 3 2 H 2 O) x 100 = 22.6% H 2 O in CuCO 3 2 H 2 O Website upload 2014 Page 49 of 55

Topic: Molecular Formulas Objective: How do we calculate a molecular formula? A molecular formula shows the true number of atoms of each element in a chemical formula. An empirical formula only states the ratio of the different types of elements in a chemical formula. If we have the empirical formula and a molecular mass we can calculate the molecular formula. To determine the molecular formula of a compound given the molecular mass and empirical formula: i. Determine the formula mass of the empirical formula; ii. Divide the molecular mass by the empirical mass to give a wholenumber multiple; iii. Multiply the whole number above by the empirical formula. Website upload 2014 Page 50 of 55

Example: 1. What is the molecular formula of a compound with an empirical formula of C 2 H 3 and a molecular mass of 54.0 g/mole? i. Determine the formula mass of the empirical formula: C 2 H 3 : (2 C @ 12.0 g/mole) + (3 H @1.0 g/mole) = 27.0 g/mole C 2 H 3 ii. Divide the molecular mass by the empirical mass, giving the wholenumber multiple: (54.0 g/mole) / (27.0 g/mole) = 2 The 2 means the molecular formula is TWICE the empirical formula iii. Multiply the calculated whole-number multiple by the empirical formula: 2 x C 2 H 3 = C 4 H 6 The molecular formula for the compound in this problem is C 4 H 6 Website upload 2014 Page 51 of 55

Student name: Key Class Period: _3, 5, & 10_ Please carefully remove this page from your packet to hand in. Percent Composition and Formulas homework Determine the percentage by mass of each element in the following compounds. Show ALL work, and make sure that your answers are properly rounded. (3 pts each) CO: % C = _42.9%_ % O = _57.1%_ 42.9% + 57.1% = 100% C = 12.0 g/mol O = 16.0 g/mol % C = (12.0 g/mol of C) / (28.0 g/mol of CO) x 100 = 42.9% C in CO % O = (16.0 g/mol of O) / (28.0 g/mol of CO) x 100 = 57.1% O in CO CO = (12.0 g/mol of C) + (16.0 g/mol of O) = 28.0 g/mol of CO (NH 4 ) 2 SO 3 : % N = 24.1% % O = 41.3% % S = 27.6% % H = 6.9% N = 14.0 g/mol % N = (28.0 g of N) / {116.1 g/mol of (NH 4 ) 2 SO 3 } x 100 = 24.1% N in (NH 4 ) 2 SO 3 O = 16.0 g/mol % O = (48.0 g of O) / {116.1 g/mol of (NH 4 ) 2 SO 3 } x 100 = 41.3% O in (NH 4 ) 2 SO 3 H = 1.0 g/mol % H = (8.0 g of H) / {116.1 g/mol of (NH 4 ) 2 SO 3 } x 100 = 6.9% H in (NH 4 ) 2 SO 3 S = 32.1 g/mol % S = (32.1 g of S) / {116.1 g/mol of (NH 4 ) 2 SO 3 } x 100 = 27.6% S in (NH 4 ) 2 SO 3 (NH 4 ) 2 SO 3 = 2 x {14.0 + (4 x 1.0)} + 32.1 + (3 x 16.0) = 116.1 g/mol of (NH 4 ) 2 SO 3 CuSO 4 5 H 2 O : % Cu = 25.4% % O = 57.7% % S = 12.9% % H = 4.0% O = 16.0 g/mol H = 1.0 g/mol S = 32.1 g/mol Cu = 63.5 g/mol % O = (144.0 g of O) / (249.6 g/mol of CuSO 4 5 H 2O) x 100 = 57.7% O in CuSO 4 5 H 2O % H = (10.0 g of H) / (249.6 g/mol of CuSO 4 5 H 2O) x 100 = 4.0 % H in CuSO 4 5 H 2O % S = (32.1 g/mol of S) / (249.6 g/mol of CuSO 4 5 H 2O) x 100 = 12.9 % S in CuSO 4 5 H 2O % Cu = (63.5 g/mol of Cu) / (249.6 g/mol of CuSO4 5 H2O) x 100 = 25.4 % Cu in CuSO4 5 H2O CuSO 4 5 H 2O = (63.5) + (32.1) + (9 x 16.0) + (10 x 1.0) = 249.6 g/mol of CuSO 4 5 H 2O Cont d next page Website upload 2014 Page 53 of 55

Determine the percent by mass of water in the hydrate below. Show ALL work, and correct labels. (3 pts) Formula mass of H 2 O: (2 x 1.0 g/mol of H) + 16.0 g/mol of O = 18.0 g/mol for H 2 O Formula mass of CuSO 4 5 H 2 O: 249.6 g/mol of CuSO 4 5 H 2 O Formula mass of _5_ H 2 O: (5 x 18.0 g/mol H 2 0) = 90.0 g of H 2 O / mol of CuSO 4 5 H 2 O Calculate the % of water in CuSO 4 5 H 2 O: (90.0 g/mol of H 2 O in CuSO 4 5 H 2 O) / (249.6 g/mol of CuSO 4 5 H 2 O) x 100 = 36.1% H 2 O in CuSO 4 5 H 2 O Determine the molecular formulas for each question below. Show ALL work. (3 pts ea.) 1. The empirical formula of a compound is found to be CH, and the molecular mass is determined to be 78.0 g/mole. Write the molecular formula for this compound. CH = (12.0 g/mole of C) + (1.0 g/mol of H) = 13.0 g/mol for CH (78.0 g/mol molecular mass) / (13.0 g/mol for CH) = 6 (6 x 1 C) + (6 x 1 H) = C 6 H 6 molecular formula 2. The empirical formula of a compound is found to be CH 2 O, and the molecular mass is determined to be 180.0 g/mole. Write the molecular formula for this compound. CH 2 O = (12.0 g/mole of C) + (2 x 1.0 g/mol of H) + (16.0 g/mol of O) = 30. g/mol for CH 2 O (180.0 g/mol molecular mass) / (30. g/mol for CH 2 O) = 6 (6 x 1 C) + (6 x 2 C) + (6 x 1 O) = C 6 H 12 O 6 Website upload 2014 Page 54 of 55