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Ethanol, C2H6O NaCl, salt Buckyball, C60 Just to Review… Please convert 12.4 fluid ounces to cm3 1 fl. oz. = 0.0295735 L 1 cm3 = 1 mL Convert 4.5 centuries to seconds. The Carbon-12 Scale The atomic mass of an element indicates how heavy, on average, an atom of an element is when compared to an atom of another element – Unit is the atomic mass unit (amu) The standard scale is based on the carbon-12 isotope • Mass of one 12C atom = 12 amu (exactly) • Note that 12C and C-12 mean the same thing Isotopic Abundance • • • Most elements exist in nature as a mixture of two or more isotopes. To determine the mass of an element, we must know the mass of each isotope and the atom percent of the isotopes (isotopic abundance) The mass spectrometer can determine the isotopic abundance & atomic mass. Figure 3.1 – Mass Spectrometer • A mass spectrometer is used to determine atomic masses Figure 3.1 – Mass Spectrometer Avogadro’s Number • A sample of any element with a mass equal to its atomic mass contains the same number of atoms, NA, regardless of the identity of the element. – NA = 6.022 X 1023 Avogadro’s # • It represents the number of atoms of an element in a sample whose mass in grams is numerically equal to the atomic mass of the element. Amadeo Avogadro & Avogadro’s Number Avogadro’s Number Amadeo Avogadro 1776-1856 6.02214199 x 23 10 (6.022 x 1023 to 4 s.f.) There is Avogadro’s number of particles in a mole of ANY substance. Examples: • 6.022 x 1023 H atoms in 1.008 g atomic mass H = 1.008 amu • 6.022 x 1023 S atoms in 32.07 g atomic mass S = 32.07 amu Mole: A Perspective on Size The Green Pea Analogy: A Dramatic Reading Examples of Mole Quantities 1 mole of stars in the universe = 6.022 x 1023 stars 1 mole of pennies = 6.022 x 1023 pennies (beats the lottery!) 1 mole of glucose molecules = 6.022 x 1023 molecules 1 mole of helium atoms = 6.022 x 1023 atoms 1 mole of potassium ions (K+) = 6.022 x 1023 ions EOS Molecular Mass Molecular mass: the sum of the atomic masses of all atoms in a molecular formula. - (units are amu or u) 1 Oxygen atom Example: water - H2O 2(1.01 amu) + 16.00 amu = 18.02 amu 2 Hydrogen atoms Sample Problem: Glucose Glucose - C6H12O6 = 6(12.01 u) + 12(1.01 u) + 6(16.00 u) = 180.18 u EOS Formula Mass Formula mass is the sum of the masses of the atoms or ions present in a formula unit – the unit for an ionic formula ClNa+ Cl- Na+ Cl- Na+ Cl- Na+ Crystal of sodium chloride One Na+ and one Cl– make a formula unit for sodium chloride The mass of one formula unit is: = 22.99 amu + 35.45 amu = 58.44 amu Sample Problem Example 3.1 Determine the formula or molecular mass for each of the following: CaI2 (NH4)2S Al(NO3)3 C6H12O6 Mole--Definition Chemistry is a quantitative science—we need a “counting unit” aka the MOLE! 1 mole = the amount of substance that contains as many particles (atoms, molecules, formula units) as there are in 12.0 g of 12C. 518 g of Pb, 2.50 mol One-mole Amounts Analogies We can group items and count by grouping: 12 eggs = 1 dozen eggs like 6.02 x 1023 items = 1 mole We can also group items and count by weighing: Grass seed and nails —Purchased by the POUND, not by the item. The molar mass, MM, in grams/mole, is numerically equal to the sum of the masses (in amu) of the atoms in the formula ***Molar mass is the mass of one mole of a particular substance. Molar Masses of Some Substances MOLECULAR WEIGHT VS. MOLAR MASS Molecular weight = sum of the atomic weights of all atoms in the molecule (in amu) Molar mass = molecular weight in grams/mole We will use molar mass in all problems in this chapter!!!! Equivalencies 1 mole of any substance contains Avogadro’s number of particles (and the mass on the periodic table expressed in grams). 1 mol of C = 12.01 g of C= 6.022 x 1023 atoms of C 1 mol of O2 = 32.00 g of O2= 6.022 x 1023 molecules 1 mol of NaCl = 58.44 g of NaCl= 6.022 x 1023 formula units of NaCl The Mole and Reactions Example: consider the formation of carbon dioxide At the molecular level ... Problem: how does one mass out a single carbon atom? Note that the mass in grams is ~2.00 x 10–23 g! Answer: one doesn’t! EOS The Solution ... Use a measurable amount – molar quantities For carbon, mass out: 2.0 × 10–23 g atom–1 × 6.0 × 1023 atoms mol–1 = 12 g C EOS Must memorize elements that exist as diatomic elements!! Remember HON17 !!! Mole Conversions Moles! Use Avogadro’s # (6.02 x 1023) Representative particles (atoms, molecules, formula units, ions) Use molar mass Mole Conversions Mass Practice problems 1. Calcium carbonate, CaCO3, is the principal mineral found in marble and limestone. How many moles are in 188.0 g of CaCO3? Practice problems 2. What is the mass, in grams, of 0.329 mol of spearmint oil, C10H14O? Practice problems 3. Find the mass of a single lead atom. Practice problems 4. How many individual lead atoms are in a 1.000 g sample of this metal? Practice problems 5. (a) Calculate the number of moles of aluminum in a solid cube that measures 3.40 cm on a side. (d=2.70 g/cm3). (b) How many atoms of aluminum are in the same sample? Practice problems 6. (a) How many molecules of oxygen, O2, are in 0.00100 grams of this gas? (b) How many atoms? Percent Composition by Mass Definition: Describes the proportion of elements in a compound using a percent Equal to the mass of each element present in a 100 g sample of compound! Mass of element Percent composition of an element = ×100 Mass of compound Example: Sodium carbonate is a compound used in the manufacture of soap and glass. Determine the percent composition by mass of each element in this compound. Example: Determine the percent by mass of water in Al2(SO4)3∙18H2O. Example: Magnetite, Fe2O3, is one of the principal iron containing ores. How much elemental iron can be obtained from a metric ton (103 kg) of this ore, assuming 100 % recovery? (hint: first find % iron in Fe2O3) Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. PROBLEM: A compound of B and H is 81.10% B. What is its empirical formula? Empirical Formula Calculations Percent to mass Mass to moles Divide by Small Multiply ‘til whole % g moles Divide by smallest mole value empirical formula A compound of B and H is 81.10% B. What is its empirical formula? Percent to mass: (always assume 100 g sample!) A compound of B and H is 81.10% B. What is its empirical formula? Mass to moles: 1 mol 81.10 g B • = 7.502 mol B 10.81 g 1 mol 18.90 g H • = 18.75 mol H 1.008 g A compound of B and H is 81.10% B. What is its empirical formula? Now, recognize that atoms combine in the ratio of small whole numbers. Find the ratio of moles of elements in the compound by “dividing by small” A compound of B and H is 81.10% B. What is its empirical formula? But we need a whole number ratio! Must multiply each ratio by the smallest integer available to obtain whole numbers (multiply ‘til whole) Multiply ‘til Whole Hints Common Possible Endings: .33 x 3 .25 x 4 .67 x 3 .50 x 2 Sample Problem Example: Bicarbonate of soda is used in products like Alka-Seltzer and generally relieves an upset stomach. Determine the empirical formula of this compound based on the following percent composition: 27.36% Na, 1.200% H, 14.30% C, 57.14% O. Sample Problem Example: A 25.00 gram sample of an orange compound contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of oxygen. Find its empirical formula. A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula? Is the molecular formula B2H5, B4H10, B6H15, B8H20, etc.? B 2H 6 B2H6 is one example of this class of compounds. We need to do an EXPERIMENT to find the MOLAR MASS. Here the experiment gives 53.3 g/mol Compare with the mass of B2H5 = 26.66 g/unit Find the ratio of these masses. 53.3 g/mol 2 units of B2H5 = 26.66 g/unit of B2H5 1 mol Multiply all of the subscripts by the ratio and obtain the molecular formula: Molecular formula = B4H10 Sample Problem Example : A certain compound has the empirical formula C2H4O. Its molar mass is about 90 g/mol. What is the molecular formula? Sample Problem Example: A hydrate of magnesium iodide has the formula MgI2 ∙ X H2O. To determine the value of X, a student heats a sample of the hydrate until all the water is gone. A 1.628 g sample of hydrate is heated to constant mass of 1.072 g. What is the value of X? Writing and Balancing Chemical Equations • All chemical reactions have two parts: • Reactants - the substances you start with (on left side of arrow) • Products - the substances you end up with (on right side of arrow) • The reactants turn into the products. Reactants Products In a chemical reaction… • The way atoms are joined is changed. • Atoms aren’t created or destroyed; they just combine together in new ways. • Can be described using sentences, symbols or word equations: Example: Copper reacts with chlorine to form copper (II) chloride. Copper + chlorine copper (II) chloride Cu + Cl2 CuCl2 Symbols Used in Equations • The arrow separates the reactants from the products; means “reacts” or “yields” • The plus sign = “and” • Subscripts are used to describe the number of atoms in a FORMULA. • Coefficients are used to describe the number of molecules or formula units in the REACTION. They are the only things changed when balancing a reaction. States of Matter • Solid – (s) after the formula – Precipitate -- a solid formed in a reaction • Gas--(g) after the formula • Liquid—(l) after the formula • Aqueous – (aq) after the formula dissolved in water. States of Matter used after a product indicates a gas oSame as writing (g) used after a product indicates a solid or precipitate oSame as writing (s) Other Symbols used in Equations • • indicates a reversible reaction heat , show that heat is supplied to the reaction Pt • is used to indicate a catalyst used or supplied, in this case, platinum. Chemical Equations Example: consider the formation of water H2(g) + O2(g) H2O(g) Law of Conservation of Mass must be obeyed … therefore, equations must be balanced. EOS Balancing Equations Chemical “bookkeeping” of atoms involved in the reaction: H2(g) + O2(g) H2O(g) H–2 O–2 Reactants H–2 O–1 Products Note the imbalance in oxygen atoms COEFFICIENTS must be added so reactant atoms EQUAL product atoms! Hints & Tips for Balancing Equations • Take one element at a time, working from left to right except for H and O. Save H for next to last and O for last. • IF EVERYTHING BALANCES EXCEPT FOR O, and there is no way to balance O with a whole number, double all the coefficients and try again. (Because O is a diatomic element) • (Shortcut) polyatomic ions that appear on both sides of the equation can be balanced as independent units! Balancing Equations Practice Balance the following chemical equation using the appropriate coefficients: ____ Al(s) + _____ Br2 (l) _____ Al2Br6 (s) Balancing Equations Practice Balance the following chemical equation using the appropriate coefficients: ____ Na3PO4 + ____ Fe2O3 ____ Na2O + ____ FePO4 Types of Reactions • Synthesis (combination) reaction • Decomposition reaction • Single replacement reaction • Double replacement reaction • Combustion reaction Types of Reactions Synthesis or Combination Equation in Symbols: A + B AB Sample Equation: 2Cu (s) + O2 (g) 2 CuO (s) Predicting Products: Elements Compounds OR Compounds More Complex Compounds Types of Reactions Decomposition Equation in Symbols: AB A + B Sample Equation: 2 CuO (s) 2Cu (s) + O2 (g) Predicting Products: Compounds Elements OR More Complex Compounds Compounds Types of Reactions Single Replacement Equation in Symbols: A + BC AB + C • Metal replacing metal • Nonmetal replacing nonmetal Sample Equation: Mg (s) + CuCl2 (aq) Cu (s) + MgCl2 (aq) Predicting Products: Cations replace cations (can also have anions replacing anions) Types of Reactions Double Replacement – 2 ionic compounds Equation in Symbols: AX + BY BX + AY Sample Equation: 2AgNO3(aq) + CuCl2 (aq) Cu (NO3)2 (aq) + 2AgCl (s) Predicting Products: Cations switch places; solid formed (must be driving force) Types of Reactions Combustion Equation in Symbols: CxHy + O2 CO2 + H2O Sample Equation: CH4(g) + O2 (g) CO2 (g) + H2O (l) Predicting Products: Hydrocarbons react to form CO2 and H2O Stoichiometric Equivalence and Reaction Stoichiometry CS2 + 3O2 CO2 + 2 SO2 Interpretation in terms of moles: 1 mole of CS2 + 3 moles of O2 form: 1 mole of CO2 + 2 moles of SO2 Stoichiometric Equivalence and Reaction Stoichiometry CS2 + 3O2 CO2 + 2 SO2 Conversion factors extracted from balanced equation: 1 mole of CS2 3 moles of O2 3 moles of O2 1 mole of CO2 etc. These ratios are called MOLE RATIOS! Stoichiometric Equivalents Coefficients from a balanced chemical equation show molar equivalents of reactants and products ==> form conversion factors 2H2 + O2 2H2O In the formation of water: 2 mol H2 = 1 mol O2 2 mol H2 = 2 mol H2O 1 mol O2 = 1 mol H2O Example: Using the equation below, determine: CS2 + 3O2 CO2 + 2SO2 • the number of moles of oxygen required to react with 1.38 mol of carbon disulfide • the number of moles of SO2 produced from 1.38 moles of carbon disulfide. Example 3.13 For 2NH3 + H2SO4 (NH4)2SO4 determine: a) the mass of product possible when 1.43 mol of NH3 are reacted with an excess of sulfuric acid. b) the mass of NH3 required to react completely with 35.00 g of sulfuric acid. c) the mass of sulfuric acid required to form 1000 grams of product. Stoichiometry Diagram Volume (liquids) Volume (liquids) Known Unknown Substance A Substance B Mass Volume 1 mole = 22.4 L @ STP Mass Use coefficients from balanced chemical equation Mole Mole 1 mole = 22.4 L @ STP Volume (gases) (gases) Particles Particles Example : How many milliliters of liquid water can be produced by the combustion of 775 mL of octane with oxygen? Assume that the volumes of the octane and the water are measured at 20oC where the densities are 0.7025 g/mL for octane and 0.9982 g/mL for water. C8H18(l) + O2(g) CO2(g) + H2O(l) Reaction Yields Theoretical yield – predicted amount of product formed from the limiting reagent, based only on the stoichiometry of the reaction 2H2(g) + O2(g) 2 H2O(g) If all worked perfectly ... Example: 1 mol H2 will produce 1 mol of water Actual yield – amount of product produced In practice, actual < theoretical: errors, poor technique, etc. ... Limiting Reactants (Reagents) Chemical reactant that is completely consumed in a reaction and therefore limits the quantity of product formed. **Depends on stoichiometry of reaction Excess Reactant = Reactant left over when limiting reactant is used up How many meals can be made from 105 sandwiches, 202 cookies, and 107 oranges? 1 meal 105 sandwiches x 105 meals 1 sandwich 1 meal 202 cookies x 101 meals 2 cookies 107 oranges x 1 meal 107 meals 1orange Cookies limit the total number of whole meals with excess sandwiches and oranges Steps for Determining Limiting Reactant 1. Write a balanced equation. 2. Take first reactant, calculate theoretical yield of desired product (in grams). 3. Repeat #2 for second reactant. 4. Compare results. Whichever reactant gives the LEAST amount of the product is the limiting reactant and determines the theoretical yield. The other is in excess. Sample Problem CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g) Determine the theoretical yield of product (CO2) in grams if one starts with 1.20 mol of CS2 and 3.83 mol O2 Sample Problem Example: For the following reaction, determine the theoretical yield of product (CO2) if one starts with 105 g of CS2 and 145 g of O2 CS2(l) + 3O2(g) CO2(g) + 2SO2(g) Percent Yield Reaction yields are expressed as a ratio in the form of a percentage: actual yield percent yield = x100 theoretical yield EOS Example For the reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine. 2 Sb(s) + 3 I2(s) 2 SbI3(s) Determine theoretical yield first: Example For the reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine. 2 Sb(s) + 3 I2(s) 2 SbI3(s) Theoretical Yield = 3.17 g SbI3 If 3.00 g of product are actually formed, what is the percent yield?