* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Stoichiometry: Calculations with Chemical Formulas and Equations
Rigid rotor wikipedia , lookup
Process chemistry wikipedia , lookup
Analytical chemistry wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Physical organic chemistry wikipedia , lookup
History of molecular theory wikipedia , lookup
Transition state theory wikipedia , lookup
Molecular dynamics wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Thermometric titration wikipedia , lookup
Host–guest chemistry wikipedia , lookup
Isotopic labeling wikipedia , lookup
IUPAC nomenclature of inorganic chemistry 2005 wikipedia , lookup
Determination of equilibrium constants wikipedia , lookup
Atomic theory wikipedia , lookup
Computational chemistry wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Rate equation wikipedia , lookup
Geometrical frustration wikipedia , lookup
Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Stoichiometry 2006, Prentice-Hall Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789 Stoichiometry Chemical Equations Concise representations of chemical reactions Stoichiometry Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Stoichiometry Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) Reactants appear on the left side of the equation. CO2 (g) + 2 H2O (g) Stoichiometry Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) Products appear on the right side of the equation. CO2 (g) + 2 H2O (g) Stoichiometry Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) The states of the reactants and products are written in parentheses to the right of each compound. Stoichiometry Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) Coefficients are inserted to balance the equation. CO2 (g) + 2 H2O (g) Stoichiometry Subscripts and Coefficients Give Different Information • Subscripts tell the number of atoms of each element in a molecule Stoichiometry Subscripts and Coefficients Give Different Information • Subscripts tell the number of atoms of each element in a molecule • Coefficients tell the number of molecules Stoichiometry Reaction Types Stoichiometry Combination Reactions • Two or more substances react to form one product • Examples: N2 (g) + 3 H2 (g) 2 NH3 (g) C3H6 (g) + Br2 (l) C3H6Br2 (l) 2 Mg (s) + O2 (g) 2 MgO (s) Stoichiometry 2 Mg (s) + O2 (g) 2 MgO (s) Stoichiometry Decomposition Reactions • One substance breaks down into two or more substances • Examples: CaCO3 (s) CaO (s) + CO2 (g) 2 KClO3 (s) 2 KCl (s) + O2 (g) 2 NaN3 (s) 2 Na (s) + 3 N2 (g) Stoichiometry Combustion Reactions • Rapid reactions that produce a flame • Most often involve hydrocarbons reacting with oxygen in the air • Examples: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) Stoichiometry Formula Weights Stoichiometry Formula Weight (FW) • Sum of the atomic weights for the atoms in a chemical formula • So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu • These are generally reported for ionic compounds Stoichiometry Molecular Weight (MW) • Sum of the atomic weights of the atoms in a molecule • For the molecule ethane, C2H6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu Stoichiometry Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: (number of atoms)(atomic weight) % element = (FW of the compound) x 100 Stoichiometry Percent Composition So the percentage of carbon in ethane is… (2)(12.0 amu) %C = (30.0 amu) 24.0 amu x 100 = 30.0 amu = 80.0% Stoichiometry Moles Stoichiometry Avogadro’s Number • 6.02 x 1023 • 1 mole of 12C has a mass of 12 g Stoichiometry Molar Mass • By definition, these are the mass of 1 mol of a substance (i.e., g/mol) – The molar mass of an element is the mass number for the element that we find on the periodic table – The formula weight (in amu’s) will be the same number as the molar mass (in g/mol) Stoichiometry Using Moles Moles provide a bridge from the molecular scale to the real-world scale Stoichiometry Mole Relationships • One mole of atoms, ions, or molecules contains Avogadro’s number of those particles • One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound Stoichiometry Finding Empirical Formulas Stoichiometry Calculating Empirical Formulas One can calculate the empirical formula from the percent composition Stoichiometry Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. Stoichiometry Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C: H: N: O: 1 mol 12.01 g 1 mol 5.14 g x 1.01 g 1 mol 10.21 g x 14.01 g 1 mol 23.33 g x 16.00 g 61.31 g x = 5.105 mol C = 5.09 mol H = 0.7288 mol N = 1.456 mol O Stoichiometry Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: 5.105 mol 0.7288 mol = 7.005 7 H: 5.09 mol 0.7288 mol = 6.984 7 N: 0.7288 mol 0.7288 mol = 1.000 O: 1.458 mol 0.7288 mol = 2.001 2 Stoichiometry Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2 Stoichiometry Empirical to molecular • Since the empirical formula is the lowest ratio, the actual molecule would weigh more. –By a whole number multiple. • Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula Stoichiometry Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – C is determined from the mass of CO2 produced – H is determined from the mass of H2O produced – O is determined by difference after the C and H have been determined Stoichiometry Combustion Analysis p. 94 • Isopropyl alcohol, a substance sold as rubbing alcohol is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g CO2 and 0.306 g H2O. Determine the empirical formula of isopropyl alcohol. We must calculate the grams of C from CO2 And the grams of H from H2O. To calculate O we will subtract grams C and grams O from the total. Stoichiometry Combustion Analysis 0.561 g CO2 1 mol CO2 44.01 g CO2 1 mol C 12.01 g C = 1 mol C 1 mol CO2 0.153 g C 0.306 g H2O 1 mol H2O 18.02 g H2O 2 mol H 1 mol H2O 12.01 g H 1 mol H = 0.0343 g H Mass of O = mass of sample – (mass of C + mass of H) 0.255 g – (0.153 g + 0.0343 g) = 0.068 g O Stoichiometry Combustion Analysis 0.153 g C 1 mol C Moles C = = 0.0128 mol C 12.01 g C 0.0343 g H 1 mol H Moles H = = 0.0340 mol H 1.01 g H 0.068 g O 1 mol O Moles O = = 0.0043 mol O 16.00 g O Stoichiometry Combustion Analysis C: 0.0128 mol 0.0043 mol = 2.977 3 H: 0.0340 mol 0.0043 mol = 7.907 8 O: 0.0043 mol 0.0043 mol = 1.000 empirical formula = C3H8O Stoichiometry Combustion Analysis • Other examples: p.107 #47 a. Combustion analysis of toluene, a common organic solvent gives 5.86 g CO2 and 1.37 g of H2O. If the compound contains only carbon and hydrogen, what is the empirical formula? C7H8 Stoichiometry Combustion Analysis b. Menthol, the substance we smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula for menthol? emp. form. = C10H20O If the compound has a molar mass of 156 g/mol, what is its molecular formula? mol. form. = C10H20O Stoichiometry Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O Stoichiometry Hydrates • Hydrates are solids with water trapped in the solid in a repeating pattern. • When a hydrate is heated, the water will escape leaving an anhydrate (solid without water). –Copper(II) sulfate is a hydrate. –Hydrated copper(II) sulfate is deep blue in color. Stoichiometry How are Names of Hydrates Written? • BaCl2•2H2O • barium chloride dihydrate • FeSO4•6H2O • iron(II) sulfate hexahydrate • Na2CO3•10H2O • sodium carbonate decahydrate • CuSO4•5H2O • copper(II) sulfate pentahydrate Stoichiometry Hydrates Stoichiometry Hydrates Stoichiometry Hydrates Stoichiometry Stoichiometry Greek for “measuring elements” Stoichiometry Quantitative Information from Balanced Equations • Balanced chemical equation gives number of molecules that react to form products. • Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product. • These ratios are called stoichiometric ratios. NB: Stoichiometric ratios are ideal proportions • Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles). Stoichiometry Prentice Hall © 2003 Chapter 3 Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products Stoichiometry Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant) Stoichiometry Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams Stoichiometry Let’s ride the Mole Train! grams of known moles of known molar mass of known moles of unknown mole to mole ratio grams of unknown molar mass of unknown Stoichiometry Stoichiometric Calculations • p. 107 #55 Na2SiO3 + 8 HF H2SiF6 + 2 NaF + 3H2O a. How many moles of HF are needed to react with 0.300 mol of Na2SiO3? 2.40 mol HF b. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3? 5.25 g NaF c. How many grams of Na2SiO3 can react with 0.800 g of HF? 0.610 g Na2SiO3 Stoichiometry Stoichiometric Calculations • Additional examples #57 and #59 Stoichiometry Limiting Reactants Stoichiometry How Many Cookies Can I Make? • You can make cookies until you run out of one of the ingredients • Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat) Stoichiometry How Many Cookies Can I Make? • In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make Stoichiometry Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount Stoichiometry Limiting Reactants • The limiting reactant is the reactant present in the smallest stoichiometric amount – In other words, it’s the reactant you’ll run out of first (in this case, the H2) Stoichiometry Limiting Reactants In the example below, the O2 would be the excess reagent Stoichiometry Limiting Reactants • The Limiting Reactant: – The reactant that is used up first in a reaction. – It controls the amounts of the other reactants that are used. – It controls the amount of product produced (maximum amount of product). – It produces less product than the other reactants possibly could produce. (***Use this for problem solving). Stoichiometry Theoretical Yield • The theoretical yield is the amount of product that can be made – In other words it’s the amount of product possible as calculated through the stoichiometry problem • This is different from the actual yield, the amount one actually produces and measures Stoichiometry Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Percent Yield = Theoretical Yield x 100 Stoichiometry Limiting Reactants • p. 109 #69 • Sodium hydroxide reacts with carbon dioxide as follows: 2 NaOH + CO2 Na2CO3 + H2O • Which reagent is the limiting reactant when 1.70 mol NaOH and 1.00 mol CO2 are allowed to react? • How many moles of Na2CO3 can be produced? • How many moles of the excess reactant remain after the completion of the reaction? Stoichiometry • To calculate the limiting reactant 1) Convert to moles 2) Compare – pick 1 reactant, solve for the amount needed of the other and compare. 1.00 mol CO2 2 mol NaOH = 2.00 mol NaOH required (given = 1.70 mol) 1 mol CO2 or 1.70 mol NaOH 1 mol CO2 = 0.85 mol CO2 required 2 mol NaOH (given = 1.00 mol) So: NaOH is the limiting reactant Stoichiometry How many moles of Na2CO3 can be produced? 1.70 mol NaOH 1 mol Na2CO3 = 0.85 mol Na2CO3 produced 2 mol NaOH How many moles of the excess reactant remain after the completion of the reaction? 1) use L.R. to determine quantity of excess reactant used. 2)subtract to find quantity that remains. 1.70 mol NaOH 1 mol CO2 2 mol NaOH = 0.85 mol CO2 used Stoichiometry 1.00 mol given – 0.85 mol used = 0.15 mol CO2 remains