Stoichiometry Chapter/Unit 3

Significant Figures Review (1.5) Rules for counting Significant Figures: 1. Nonzero integers always count! 2. Zeroes (3 classes): a. Leading zeroes do not count (place holders) b. Captive zeroes do count c. Trailing zeroes are if there is a decimal point 3. Exact numbers (counting or from definitions)

Sig Fig Calculation Review (1.5) Rules for Significant Figures in calculations: 1. For multiplication or division: the resultant is the same as the LEAST precise number in the calculation. 2. For addition or subtraction: the resultant is the same number of decimal places as the least precise measurement in the calculation. 3. Rounding: DO NOT round until all calculations are completed. Use only the first number to the right of the last significant figure.

Stoichiometry Chemical Stoichiometry: The quantities of materials consumed and those produced in chemical reactions. How much of A How much of B How much of AB To understand Stoichiometry it is necessary to understand how average atomic mass is calculated.

Average Atomic Mass http://www.tutorvista.com/content/chemistry/chemistry-iii/organic-compounds/molecular-mass-determination.php

Average Atomic Mass Average Atomic Mass of Carbon 12C = 98.89 % at 12 amu 13C = 1.11 % at 13.0034 amu 14C = Negligibly small at this level http://harmonyscienceacademy.web.officelive.com/AtomicTheoryandStructure2.aspx 98.89%*(12 amu) + 1.11%*(13.0034 amu) = 12.01 amu

Average Atomic Mass What does knowing that carbons average atomic mass is 12.01 amu do for us? Even though natural carbon does not contain a single atom with a mass of 12.01, for Stoichiometry purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01 amu. So to obtain 1000 atoms of Carbon, with an average atomic mass of 12.01, we can weigh out 12,010 atomic mass units. (this would be a mixture of 12C and 13C)

The Mole Mole: the number equal to the number of carbon atoms in exactly 12 grams of pure 12C (abbreviated as mol). Turns out that number is 6.022 x 1023 which is referred to Avogrado’s number. One Mole of something equals 6.022 x 1023 units of that something. Ex: 1 mol of students contains 6.022 x 1023 students Ex: 1 mol of Carbon has 6.022 x 1023 atoms and has an amu of 12.01.

The Mole The mole is defined such that a sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms. 1 mol of Helium = 6.022 x 1023 atoms = 1.00794 grams 1 mol of Oxygen = 6.022 x 1023 atoms =15.9994 grams 1 mol of Zirconium = 6.022 x 1023 atoms =91.224 grams (6.022 x 1023 atoms)(12 amu) = 12 grams (atom) 6.022 x 1023 amu = 1 g or 6.022 x 1023 amu 1g

The Mole Determining the mass of a sample of Atoms Remember that 1 atom of Carbon = 12.01 amu If we have 6 atoms of Carbon what is the mass of that sample? 6 atoms x 12.01 amu x _____1g_____ atom 6.022x1023 amu = 1.197 x 10-22 g

The Mole Determining the moles of a sample of atoms Aluminum has mass of 26.98 amu If we have 10.0 grams of Al how many moles is in the sample? 10.0 grams Al x 1 mol Al __ 26.98 g Al 6.022x1023 atoms x 0.371 mol Al 1 mol Al = 0.371 mol Al Atoms = 2.23 x 1023 Atoms

The Mole Calculating the numbers of Atoms in a sample Silicon has an amu of 28.09 A silicon ship used in an integrated circuit has a mass of 5.68 mg. How many silicon atoms are present in the chip? 5.68 mg Si x 1 g Si __ 1000 mg Si = 5.68 x 10-3 g Si 5.68 x 10-3 g Si x 1 mol Si 28.09 g Si = 2.02 x 10-4 mol Si 2.02 x 10-4 mol Si x 6.022 x 1023 atoms 1 mol Si = 1.22 x 1020 atoms of Si

The Mole Calculating the numbers of Moles and Mass Cobalt has an amu of 58.93 Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample? 5.00 x 1020 atoms Co x _____1 mol Co_____ 6.022x 1023 atoms Co = 8.30 x 10-4 mol Co 8.30 x 10-4 mol Co x ____58.93 grams Co 1 mol Co = 4.89 x 10-2 grams Co

The Mole Calculating Molar Mass Molar Mass: the mass in grams of one mole of the compound Calculating the molar mass of a molecule requires that we sum the molar masses of the individual elements. Example: The molar mass of methane (CH4)? Mass of 1 mol of C = Mass of 1 mol of H = 12.011 g 4 x 1.00794 g Mass of 1 mol of CH4 = 16.043 g

The Mole Calculating Molar Mass Example: Juglone, C10H6O3, is a natural herbicide. Calculate the molar mass and how many moles are present in 1.56 x10-2 g of the pure compound. Mass of 1 mol of C (x10) = Mass of 1 mol of H (x6)= Mass of 1 mol of O (x3)= 10 x 12.011 g 6 x 1.00794 g 3 x 15.9994 g Mass of 1 mol of C10H6O3 = 174.156 g 1.56 x 10-2 g C10H6O3 x _1 mol of C10H6O3__ 174.156 g = 8.96x 10-5 mols of C10H6O3

The Mole Calculating Molar Mass Example: Bees release 1 x10-6 grams of Isopentyl acetate (C7H14O2) when they sting. How many molecules are present per sting? Mass of 1 mol of C (x7) = Mass of 1 mol of H (x14)= Mass of 1 mol of O (x2)= 7 x 12.011 g 14 x 1.00794 g 2x 15.9994 g Mass of 1 mol of C7H14O2 = 130.187 g 1 x 10-6 g C7H14O2 x _1 mol of C7H14O2__ 130.187 g = 8 x 10-9 mols of C7H14O2 8 x 10-9 mol C7H14O2 x _6.022 x 1023 molecules__ 1 mol of C7H14O2 = 5 x 1015 molecules C7H14O2

The Mole Calculating Percent Composition Percent composition: comparison of the mass of individual elements to its parent molecules mass. Example: What percent mass is each element in ethanol, C2H5OH? 2x12.011 g = 24.022 g C 6 x1.00794 g = 6.04764 g H 1x15.9994 g = 15.9994 g O Mass of 1 mol of C (x2) = Mass of 1 mol of H (x6)= Mass of 1 mol of O (x1)= 2x 12.011 g 6 x 1.00794 g 1x 15.9994 g Mass of 1 mol of C2H5OH = 46.069 g ____24.022g C___ 46.069 g C2H5OH = 52.14 % Carbon X 100% ____6.04674 g H___ 46.069 g C2H5OH X 100% = 13.13 % Hydrogen = 34.73 % Oxygen

The Mole Calculating Percent Composition Example: What percent mass is each element in Penacillin C14H20N2SO4? Mass of 1 mol of C (x14) = Mass of 1 mol of H (x20)= Mass of 1 mol of O (x4)= Mass of 1 mol of N (x2)= Mass of 1 mol of S (x1)= 14x 12.011 g 20x 1.00794 g 4x 15.9994 g 2x 14.00674 g 1x 32.066 g = 168.154 g C = 20.1588 g H = 63.9976 g O = 28.01348 g N = 32.066 g S Mass of 1 mol of C14H20N2SO4 = 312.38988 g ____168.154 g C___ 312.38988 g C14H20N2SO4 = 53.83 % Carbon = 20.49 % Oxygen X 100% = 10.27 % Sulfur ____20.1588 g H __ 312.38988 g C14H20N2SO4 X 100% = 6.45 % Hydrogen = 8.96 % Nitrogen

The Mole Calculating empirical and Molecular formula Example: A compound contains 71.65% Cl, 24.27% C and 4.07% H. The molar mass is known to be 98.95916 g/mol. What is the empirical and molecular formula? 1.) Convert the mass percents to masses in grams (use 100 grams as your conversion) 71.65 g Cl x 1 mol Cl__ 34.4527 g = 2.079662842 mol Cl 24.27 g C x 1 mol C__ 12.011 g = 2.02064774 mol C 4.07 g H x 1 mol H__ 1.00794 g = 4.037938766 mol H

The Mole Calculating empirical and Molecular formula Example: A compound contains 71.65% Cl, 24.27% C and 4.07% H. The molar mass is known to be 98.95916 g/mol. What is the empirical and molecular formula? 2.) Divide each mole value by the smallest mole number present 2.079662842 Cl 2.02064774 = 1.029206032 Chlorines 2.02064774__C 2.02064774 = 1 Carbons Empirical Formula = ClCH2 4.037938766_H 2.02064774 = 1.998338793 Hydrogens

The Mole Calculating empirical and Molecular formula Example: A compound contains 71.65% Cl, 24.27% C and 4.07% H. The molar mass is known to be 98.95916 g/mol. What is the empirical and molecular formula? 3.) Determine the molar mass of the empirical formula and compare to the given molar mass. Empirical Formula = ClCH2 Empirical Formula molar mass = 49.47958 g/mol 98.95916 g/mol 49.47958 g/mol = 2 Molecular Formula = (ClCH2)2 Molecular Formula = Cl2C2H4

Physical vs. Chemical Changes in physical properties Melting Boiling Condensation No change occurs in the identity of the substance Example: Ice , rain, and steam are all water

Physical vs. Chemical In a chemical Reaction: Atoms in the reactants are rearranged to form one or more different substances. Old bonds are broken; new bonds form Examples: Fe and O2 form rust (Fe2O3) Ag and S form tarnish (Ag2S)

Chemical Reactions http://www.mikeblaber.org/oldwine/chm1045/notes/Stoich/Equation/Stoich01.htm

Balancing Chemical Reactions States within a chemical reaction: Solid: symbolized using a (s) Liquid: symbolized using a (l) Gas: symbolized using a (g) Dissolved in water: symbolized using a (aq) for aqueous HCl(aq) + NaHCO3(s) CO2(g) + H2O(l) +NaCl(aq)

Balancing Chemical Reactions Rules for balancing chemical equations: 1. Determine what reaction is occurring. What are the reactants and what are the products? What are the physical states involved? 2. Write the unbalanced equation that summarizes the reaction. 3. Balance the equation by inspection, starting with the most complicated molecules. Determine the coefficients necessary. DO NOT change the formulas of the reactants or products.

Balancing Chemical Reactions Balancing by inspection C2H5OH(l) + O2(g) CO2(g) + H2O(g) Make a list of what atoms are present on each side: Reactants Products C x 2 = 2C H x 6 = 6H O x 3 = 3O C x 1 = 1C H x 2 = 2H O x 3 = 3O Start with the most complicated molecule Since C2H5OH contains 2 carbons then we need to have 2 carbons on the reactant side. We place a coefficient of 2 in front of carbon dioxide C2H5OH(l) + O2(g) 2CO2(g) + H2O(g)

Balancing Chemical Reactions Balancing by inspection C2H5OH(l) + O2(g) 2CO2(g) + H2O(g) Change the list of what atoms are present on each side after the change: Reactants Products C x 2 = 2C H x 6 = 6H O x 3 = 3O C x 2 = 2C H x 2 = 2H O x 5 = 5O Since C2H5OH contains 6 H’s we can place a coefficient of 3 in front of water. Which will give us a change in the number of H’s and O’s C2H5OH(l) + O2(g) 2CO2(g) + 3H2O(g)

Balancing Chemical Reactions Balancing by inspection C2H5OH(l) + O2(g) 2CO2(g) + 3H2O(g) Reactants Products C x 2 = 2C H x 6 = 6H O x 3 = 3O C x 2 = 2C H x 6 = 6H O x 7 = 7O All that’s left is to balance the oxygens. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) Reactants Products C x 2 = 2C H x 6 = 6H O x 7 = 7O C x 2 = 2C H x 6 = 6H O x 7 = 7O

Chemical Stoichiometry Rules for calculating masses of reactants and products in chemical equations: 1. Write the equation and balance 2. Convert the known mass of the reactant or product to moles of that substance 3. Use the balanced equation to set up the appropriate mole ratios 4. Use the mole ratio to calculate the number of moles of the desired reactant/product 5. Convert from moles back to grams if the desired by question

Chemical Stoichiometry Example: Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide forming solid lithium carbonate and liquid water. What mass of carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? Step 1: Write the equation and balance 2 LiOH(s) + CO2(g) Li2CO3(s) + H2O(l) Step 2: Convert the mass of LiOH to moles 1.00 kg LiOH 1 1000 g LiOH 1 mol LiOH 23.948 g LiOH x = 41.8 moles of LiOH

Chemical Stoichiometry Example: Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide forming solid lithium carbonate and liquid water. What mass of carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? Step 3: Write the appropriate mole ratio 1 mol CO2_ 2 mol LiOH Step 4: Calculate the moles of CO2 needed to react with the moles of LiOH 41.8 mol LiOH 1 1 mol CO2_ 2 mol LiOH x = 20.9 moles of CO2

Chemical Stoichiometry Example: Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide forming solid lithium carbonate and liquid water. What mass of carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? Step 5: Convert the moles of the desired product to mass using the molar mass of the product 20.9 mol CO2 1 44.010 g CO2_ 1 mol CO2 x = 9.20 x 102 g of CO2

Limiting Reactants Limiting Reactant: the reactant that is consumed first and therefore limits the amount of products that can be formed. The rules for determining Limiting reactants are the same as for mass to mass calculations, but you are given the masses of both reactants.

Limiting Reactants Example: Nitrogen gas can be prepared by passing gaseous ammonia over solid Copper (II) Oxide at high temperatures producing solid copper and water vapor. If a sample containing 18.1 grams of NH3 is reacted with 90.4 grams of Copper (II) Oxide, which is the limiting reactant? How many grams of Nitrogen gas are produced? Step 1: Write the equation and balance 3 CuO(s) + NH3(g) N2(g) + Cu(s) + H2O(l) 2 3 3

Limiting Reactants CuO(s) + NH3(g) N2(g) + Cu(s) + H2O(l) 2 3 Step 2: Convert the mass of CuO and NH3 to moles 18.1 g NH3 1 1 mol NH3 17.03056 g NH3 x = 1.062795 moles of NH3 90.4 g CuO 1 1 mol CuO 79.5454 g CuO x = 1.136457 moles of CuO

Limiting Reactants CuO(s) + NH3(g) N2(g) + Cu(s) + H2O(l) 2 3 Step 3: Write the a mole ratio for the two reactants 3 mol CuO_ 2 mol NH3 Required = 3/2 = 1.5 1.14 mol CuO_ 1.06 mol NH3 Actual = 1.14/1.06 = 1.08 Since the actual ratio is smaller than the required CuO will be limiting

Limiting Reactants CuO(s) + NH3(g) N2(g) + Cu(s) + H2O(l) 2 3 Step 4: Calculate the number of moles that are produced for your target product using a your limiting reactant mole ratio. In this case the ratio between N2 and CuO. 1 mol N2_ 3 mol CuO 1.14 moles of CuO 1 x = 0.380 moles of N2 Step 5: Convert the moles of the desired product to mass using the molar mass of the product 0.380 moles of N2 1 x 28.01348 g N2_ 1 mol N2 = 10.6 grams of N2