Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Unit 7 Quantity in Chemistry Unit 7 Learning Goals • • • • • • • • • • • • Define & differentiate between atom, molecule and formula unit Define & differentiate between atomic mass and formula mass Determine the molar mass when given symbol or formula Define mole and Avogadro’s number Convert moles to mass and mass to moles Convert particles to moles and moles to particles Define molar volume and convert liters to moles & moles to liters Determine % composition when given formula or name for compound Determine empirical formula when given % composition Define & differentiate between molecular formula and empirical formula Determine empirical formula when given molecular formula Determine molecular formula when given empirical formula and formula mass Key Terms – Part One Microscopic View of Matter • Atom • Atomic mass • Atomic element • Diatomic element • Molecule • Formula unit • Formula mass Macroscopic View of Matter • Mole • Molar mass • Molar volume • Avogadro’s Number Learning Goals • Define & differentiate between atom, molecule and formula unit • Identify the elements that are diatomic and describe difference between diatomic and atomic elements • Define & differentiate between atomic mass and formula mass Atom and Atomic Mass • Atom is the simplest form of an element that represents that element • Each element has a unique symbol (ex: Fe) • Atomic mass is the weighted average of the mass numbers of all the naturally‐occurring isotopes of that element Atomic Element and Diatomic Element • Atomic element is an element that exists in nature as a single atom – Fe, He, S, U, Cu... • Diatomic element is an element that exists in nature as two atoms covalently bonded together – There are only 7 (GEN – U – INE) – H2, N2, O2, F2, Cl2, Br2, I2 Molecule and Formula Unit: Calculating Formula Mass • One unit of a covalent compound or diatomic element is called a molecule of that substance (ex: H2O; Br2) • One unit of an ionic compound is called a formula unit of that compound (ex: NaCl) • The formula mass of a molecule or formula unit is the sum of all the atomic masses of the elements of which it is comprised • One molecule of H2O = 2xH + 1xO = (2x1) + 16 = 18amu • One formula unit of NaCl = 23 + 35.5 = 58.5amu • One molecule of Br2 = 80 + 80 = 160amu diatomic element element ‐ + ‐ + ‐ Ionic compound + Learning Goals • Determine the molar mass when given symbol or formula • Define mole and Avogadro’s number • Convert moles to mass and mass to moles • Convert particles to moles and moles to particles • Define molar volume and convert liters to moles & moles to liters Mole • Mole is the “chemists’ serving” used to count only the extremely small particles that make up matter • The particles that make up a mole could be: – atoms (Fe) – molecules (H2O or O2) – formula units (NaCl) Avogadro’s Number • Number of particles in one mole • It is exactly 6.02 x 1023 • The number of atoms in one mole of atoms • The number of molecules in one mole of molecules • The number of formula units in one mole of formula units From amu to grams: what you need to know 1 amu = mass of 1 proton 1 amu = 1.67x10‐24 grams atomic mass (aka gram atomic mass) = mass of 1 mole of an element in grams 1 mole = 6.02x1023 particles Atomic mass = (mass in amu) x (# of atoms in 1 mole) x (mass of 1amu in grams) Fe = . . _ = 56g/mole How to do calculation From amu to grams: what you need to know 1 amu = mass of 1 proton 1 amu = 1.67x10‐24 grams molar mass = mass of 1 mole of any substance in grams 1 mole = 6.02x1023 particles Molar mass = (mass in amu) x (# of particles in 1 mole) x (mass of 1amu in grams) H2O = . . _ = 18g/mole How to do calculation Molar Mass and Molar Volume • Molar mass is the mass, in grams, of one mole of any substance – 1 mole Fe = 56 g – 1 mole H20 = 18 g – 1 mole NaCl = 58.5 g • Molar volume is the volume, in liters, of one mole of any gas at STP (std. temp & pressure) – 1 mole He gas = 22.4 L – 1 mole N2 gas = 22.4 L – 1 mole CO2 gas = 22.4 L Molar Mass of Selected Substances Substance Type of Substance Atomic or Molar Mass Formula Mass of (mass of one one particle mole) Calcium (Ca) Atomic element 40.1 amu 40.1 grams Carbon (C0 Atomic element 12.0 amu 12.0 grams Hydrogen (H2) Diatomic element 2.0 amu 2.0 grams Hydrogen peroxide (H2O2) Covalent compound 34.0 amu 34.0 grams Sodium chloride (NaCl) Ionic compound 58.5 amu 58.5 grams Calcium chloride (CaCl2) Ionic compound 111.0 amu 111.0 grams Mass of One Mole of Selected Substances Substance Representative Particle Number of Particles in 1 Mole Mass of 1 mole Neon (Ne) Atom 6.02 x 1023 20.2 grams Zinc (Zn) Atom 6.02 x 1023 65.4 grams Oxygen (O2) Molecule 6.02 x 1023 32.0 grams Water (H2O) Molecule 6.02 x 1023 18.0 grams Magnesium oxide Formula unit (MgO) 6.02 x 1023 40.3 grams Calcium fluoride (CaF2) 6.02 x 1023 78.1 grams Formula unit Multiplying and Dividing with Exponents • Positive vs negative exponents – 3.2 x 102 = 320 – 3.2 x 10‐2 = 0.0320 • (6.02 x 1023) x 3 = (18.06 x 1023) = 1.806 x 1024 • (6.02 x 1023) ÷ 4 = (1.505 x 1023) • (8.74 x 1023) ÷ (6.02 x 1023) = 1.45 • (7.9 x 1024) ÷ (6.02 x 1023) = (7.9 x 1024 ) ÷ (0.602 x 1024 ) = 13.2 How do you do mole conversions using ratios? 1) You will create a ratio that reflects the information and “ask” in the problem. How many moles in 150g of NaCl? 2) Create an equivalent ratio using either molar mass, molar volume or Avogadro’s number 3) Solve for X X = 2.56 mol NaCl EXAMPLES: Solving Using Ratios 1) How many moles in 150g of NaCl? . X = 2.56 mol 2) How many molecules are in 5.5 moles of CO2? . . X =3.31 x1024 molecules 3) How many moles are in 50L of N2? . X = 2.23 moles 4) How many grams are in 3.8 moles of Al2(CO3)3? . X =889.2 grams More Practice • Determine the following: – molar mass Mg(OH)2 – molar volume CO2 (gas) – molecules of CH4 in 5 moles – moles of C6H12O6 in 360 grams – moles of N2 gas in 11.2 liters ACTIVITY: % Composition of M&Ms • Count the total number of M&Ms • Determine the mass of total of all M&Ms • Count the total number of each color in sample • Determine the mass of each color (all) • Calculate the % by number of each color • Calculate the % by mass of each color Analysis: Is the % by number the same as the % by mass? Why or why not? Key Terms – Round 2 • Percent composition – Of formula – By mass – By volume • Hydrated compounds • Empirical Formula • Molecular Formula Learning Goals • Define percent composition of elements in compound based on molar mass • Use percent composition to determine actual mass of an element in a known quantity of a compound • Determine percent composition when given actual amount of each element in compound • Determine the percent water in a hydrated compound Percent Composition The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100 Mass Element Mass Compound X 100 Calculating: Percent Composition by mass Calculate: Percentage composition of Fe in Fe2O3 Mass of Fe2 Mass of Fe2O3 X 100 (56g) x 2 [(56g) x 2] + [(16g) x 3] X 100 Answer: Percentage composition of Fe = 70% Calculating: Percent Composition by mass Calculate: Percentage composition of O in Fe2O3 Mass of O3 Mass of Fe2O3 X 100 (16g) x 3 [(56g) x 2] + [(16g) x 3] X 100 Answer: Percentage composition of O = 30% Calculating: Mass of element in compound Given: % composition and total mass of compound Calculate: Grams of O and Fe in 15.2 grams of Fe2O3 30% Oxygen X 15.2 grams of Fe2O3 = 4.56 grams O 70% Iron X 15.2 grams of Fe2O3 = 10.64 grams Fe Calculating: Mass of element in compound Given: mass of each element in a compound Calculate: 5.7g of Ca combine completely with 2.1g of Cl to form a compound. What is the percent composition of the elements in this compound? Total mass of compound = 5.7g + 2.1g = 7.8g Mass of Ca ÷ Mass of Compound = (5.7 ÷ 7.8) x 100 = 73% Mass of Cl ÷ Mass of Compound (2.1 ÷ 7.8) x 100 = 27% Hydrated Compounds • Ionic compounds with a specific amount of water • Examples: MgSO4 • 5H2O = Magnesium sulfate pentahydrate KNO3 • 3H2O = Potassium nitrate trihydrate Cu3PO4 • 7H2O = Copper (I) phosphate heptahydrate Calculating: Percent of Water in Hydrated Compounds • MgSO4 • 5H2O Magnesium sulfate pentahydrate Calculate: Percentage composition of H2O in MgSO4 • 5H2O Mass of 5H2O Mass of MgSO4 • 5H2O (18g) x 5 [(24g) x 1] + [(32g) x 1] + [(16g) x 4] + [(18g) x 5] Answer: Percentage composition of H2O = 43% x 100 Learning Goals Determine empirical formula when given % composition Define and identify empirical formulas What is a chemical formula? • A qualitative description of what is in a compound CH2O 1 mol C 2 mol H 1 mol O • A quantitative description of how much of a component is in a compound • When compound formulas are described as mole ratios of atoms or ions they can be connected to mass easier C6H12O6 6 mol C 12 mol H 6 mol O • Subscripts represent mole ratios of atoms or ions Empirical Formula • An empirical formula is the simplest formula for a compound • It gives the simplest whole‐number ratio of the atoms of the elements • The empirical formula is based on the percent composition or atom to atom ratio of the elements in that compound • Helps define the mass ratios of atoms or ions, and hence identify the presence of a substance Calculation of Empirical Formula from Data • Percent to mass Change the % unit to gram by replacing it (we presume 100 grams of sample) • Mass to Mole Convert each gram amount to the mole amount (mole bridge) • Divide by small Divide each mole amount by the smallest one obtained • Multiply ‘til whole If the previous step did not produce whole numbers (± 0.1) find a multiplier that can convert all numbers to whole, or “near” (± 0.1) whole numbers Determining Empirical Formula Ex: A compound is 31.9% potassium, 28.9% chlorine and 39.2% oxygen. what is empirical formula? – 31.9g ÷ 39g/mole K = .82 mol K ÷ .81 = 1 – 28.9g ÷ 35.5g/mole Cl = .81 mol Cl ÷ .81 = 1 – 39.2g ÷ 16g/mole O = 2.45 mol O ÷ .81 = 3 Atom to atom ratio is 1:1:3 Empirical formula is KClO3 Determining Empirical Formula Ex: A compound is 50.05% sulfur and 49.95% oxygen. What is empirical formula? _____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____ _____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____ Atom to atom ratio is _________ Empirical formula is ____________ http://firstyear.chem.usyd.edu.au/calculators/empirical_formula.shtml Determining Empirical Formula Ex: A compound is 64.8% carbon, 13.62% hydrogen and 21.58% oxygen. What is empirical formula? _____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____ _____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____ _____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____ Atom to atom ratio is _________ Empirical formula is ____________ Determining Empirical Formula Ex: A compound is 31.42% sulfur, 31.35% oxygen and 37.23% fluorine. What is empirical formula? Atom to atom ratio is _________ Empirical formula is ____________ Determining Empirical Formula Ex: A compound is found to contain 23.3% magnesium, 30.7% sulfur and 46.0% oxygen. What is the empirical formula of this compound? _____g ÷ _____g/mole Mg = ____ mol Mg ÷ ____ = ____ _____g ÷ _____g/mole S = _____ mol S ÷ _____ = ____ _____g ÷ _____g/mole O = _____ mol O ÷ _____ = ____ Atom to atom ratio is _________ Empirical formula is ____________ Try This: NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet. _____g ÷ ___g/mole C = ____ mol C ÷ ____ =___ X ___ =___ _____g ÷ ___g/mole H = ____ mol H ÷ ____ =___ X ___ =___ _____g ÷ ___g/mole N = ____ mol N ÷ ____ =___ X ___ =___ _____g ÷ ___g/mole O = ____ mol O ÷ ____ =___ X ___ =___ http://firstyear.chem.usyd.edu.au/calculators/empirical_formula.shtml Learning Goals • Define and differentiate between empirical formula and molecular formula • Determine empirical formula when given molecular formula • Determine molecular formula when given empirical formula and molar mass of molecular formula. Molecular Formula • A molecular formula is the actual ratio of atoms in a compound • It is either the same as, or a multiple of, the empirical formula For ionic compounds, it is ALWAYS the same as the empirical formula Examples: lead (II) nitrate‐ Pb(NO3)2 carbon dioxide‐ CO2 sodium chloride – NaCl glucose – C6H12O6 Molecular Formula to Empirical Formula Problems Ex: Glucose is C6H12O6 what is empirical formula? – atom to atom ratio is 1:2:1 – empirical formula is CH2O Comparing Molecular and Empirical Formulas Compound Molecular Formula Empirical Formula Glucose C6H12O6 CH2O Ibuprofen C13H18O2 Caffeine C8H10N4O2 Ethylene glycol C2H6O2 Acetaminophen C8H9NO2 Comparing Molecular and Empirical Formulas Compound Molecular Formula acetylene C2H2 benzene C 6 H6 formaldehyde CH2O acetic acid C2H4O2 glucose C6H12O6 Empirical Formula Calculating Molecular Formula from Empirical Formula Data • Determine a common factor by comparing the molar mass of the unknown compound with the formula mass of the empirical formula (efm) • Apply that factor to the empirical formula as a multiplier • The result is the molecular formula Calculating Molecular Formula from Empirical Formula Data • Empirical Mass Determine the empirical formula mass • Divide for whole Divide the molar mass by the empirical mass • Multiply for mole Multiply the whole number by each atom in the compound to get molecular formula Calculating Molecular Formula An empirical formula is CH4 and the molar mass is 48 g/mol. What is molecular formula? 1. Determine the efm (empirical formula mass) 2. Divide the molar mass by the (efm) empirical formula mass to achieve a whole # 3. Use this whole # as a multiplier for every type of atom in the empirical formula to get the molecular formula Calculating Molecular Formula What is the molecular formula of a compound that has a molar mass of 118.0 g/mol and an empirical formula of C3H7O 1. Determine the efm (empirical formula mass) 2. Divide the molar mass by the (efm) empirical formula mass to achieve a whole # 3. Use this whole # as a multiplier for every type of atom in the empirical formula to get the molecular formula Calculating Molecular Formula What is the molecular formula of a compound that has a molar mass of 60.0 g/mol and an empirical formula of CH4N. 1. Determine the efm (empirical formula mass) 2. Divide the molar mass by the (efm) empirical formula mass to achieve a whole # 3. Use this whole # as a multiplier for every type of atom in the empirical formula to get the molecular formula Empirical/Molecular Formula Problems Ex: An empirical formula is CH4 and the formula mass is 48amu. What is molecular formula? – 48amu ÷ mass of CH4 = 48 ÷ 16 = 3 (this is multiplier) – molecular formula is C3H12 (3 x CH4 ) Ex: A compound is 63% Mn, 37% O. (Hint: presume 100g and change % to grams) – 63g ÷ 55g/mole Mn = 1.1 mol Mn – 37g ÷ 16g/mole O = 2.3 mol O – Atom to atom ratio is 1:2 – Empirical formula is MnO2 Empirical/Molecular Formula Problems Ex: The molecular formula for hydrogen peroxide is H2O2. What is the empirical formula? – Divide formula by least common denominator – empirical formula is HO (H2O2 ÷ 2 ) Ex: A compound contains 56g Nitrogen and 160g Oxygen. What is the empirical formula? What is the molecular formula? – 56g ÷ 14g/mole N = 4 mol N – 160g ÷ 16g/mole O = 10 mol O – Atom to atom ratio is 2:5 – Empirical formula is N2O5 – Molecular formula is N4O10