Stoichiometry Section 7.2 pg. 286-293

Stoichiometry A method of problem-solving using known quantities of one entity in a chemical reaction to find out unknown quantities of another entity in the same reaction Always requires a balanced chemical equation Series of steps for measurement calculations Always involves a mole ratio to “switch substances” Writing chemical reactions, net ionic equations and dissociation equations is essential Converting grams to moles (and vice versa) is also essential

Gravimetric Stoichiometry Gravimetric Stoichiometry – method used to calculate the masses of reactants or products in a chemical reaction Gravimetric = mass measurement Using gravimetric stoichiometry, we can apply our knowledge of balanced chemical reactions and mass to mol conversions to: Predict the mass of product we will get from a reaction Estimate the amount of reactant we need for a reaction to produce a certain mass of product

Gravimetric Stoichiometry Steps: Step 1: Write a balanced chemical reaction equation List the measured mass (given), the unknown quantity (required) and conversion factors (M = molar mass) Step 2: Convert the mass of the measured substance to moles Step 3: Calculate the moles of the unknown substance using the mole ratio required given Step 4: Convert from moles of the required substance to its mass.

Stoichiometry Calculations (Measured quantity) solids/liquids m  n (Required quantity) mole ratio

The balanced chemical equation is: How many grams of oxygen are required to completely burn 10.0 g of propane (C3H8(g))? The balanced chemical equation is: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) The number of moles of propane reacting is: (MASS to MOLES) 10.0 g x 1 mol = 0.227 mol 44.11 g The number of moles of oxygen produced is: (MOLE RATIO) 0.227 mol x 5 mol = 1.13 mol 1 mol The mass of oxygen produced is: (MOLES TO MASS) 1.1335 mol x 32.00 g = 36.3 g

Practice #2 20.0 g of butane (C4H10(g)) are completely burned in a lighter. How many grams of CO2(g) are produced? Start with a balanced chemical equation: 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O (g) m = 20.0g m = ? M = 58.14 g/mol M = 44.01 g/mol

Practice #2 (Team Unit Analysis) 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O (g) m = 20.0g m = ? M = 58.14 g/mol M = 44.01 g/mol 2) Mass to moles, mole ratio, moles to mass 20.0 g x 1 mol x . 8 mol CO2 x 44.01 g = 60.6 g 58.14 g 2 mol C4H10 1 mol

Practice #3 (Unit Analysis) What mass of iron (III) oxide is required to produce 100.0 g of iron? Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) m = ? m = 100.0g M = 159.70g/mol M = 55.85 g/mol m Fe2O3(s): 100.0 g x 1 mol x 1 mol x 159.70 g = 143.0 g Fe2O3 55.85 g 2 mol 1 mol

Remember to keep the unrounded values in your calculator for further calculation until the final answer is reported. Pg. 290 #8-14 (Answers pg. 785)

Percent Yield for Reactions We can use stoichiometry to test experimental designs, technological skills, purity of chemicals, etc. We evaluate these by calculating a percent yield. This is the ratio of the actual (experimental) quantity of product obtained and the theoretical (predicted) quantity of product obtained from a stoichiometry calculation Percent yield = actual yield x 100 predicted yield Some forms of experimental uncertainties: All measurements (limitations of equipment) Purity of chemical used (80-99.9% purity) Washing a precipitate (some mass is lost through filter paper) Estimation of reaction completion (qualitative judgements i.e. color)

Percent Yield Example #1 Example: In a chemical analysis, 3.00 g of silver nitrate in solution was reacted with excess sodium chromate to produced 2.81 g of filtered, dried precipitate. Predicted value: 2AgNO3(aq) + Na2CrO7(aq) Ag2CrO7(s) +2NaNO3(aq) m = 3.00 g m = M = 169.88g/mol M = 331.74 g/mol 3.00g x 1 mol x 1 mol x 331. 74 g = 2. 93 g 169.88g 2 mol 1 mol Percent yield = actual yield x 100% = 2.81g x 100% = 95.9% predicted yield 2.93g

Testing the Stoichiometric Method Stoichiometry is used to predict the mass of precipitate actually produced in a reaction. Filtration is used to separate the mass of precipitate actually produced in a reaction

Testing the Stoichiometric Method What mass of lead is produced by the reaction of 2.13 g of zinc with an excess of lead(II) nitrate solution? Design: A known mass of zinc is place in a beaker with an excess of lead(II) nitrate solution. The lead is produced in the reaction is separated by filtration and dried. The mass of the lead is determined Prediction: Zn(s) + Pb(NO3)2(aq)  Zn(NO3)2(aq) + Pb(s) m = 2.13 g m = ? M = 65.41 g/mol M = 207.2 g/mol 2.13 g x 1 mol x 1 x 207.2 g = 6.75 g 65.41 g 1 1 mol

Testing the Stoichiometric Method Prediction: 6.75 g of lead will be produced (Stoichiometric calculation) Evidence: In the beaker, crystals of a shiny black solid were produced, and all the zinc disappeared. Mass of filter paper = 0.92 g Mass of dried filter paper plus lead = 7.60g Analysis: mass of lead product = 7.60g – 0.92g = 6.68g According to the evidence collected, 6.68 g of lead precipitated. Evaluation: % difference = (experimental – predicted) x 100 % predicted 6.68 – 6.75 x 100% = 1% 6.75 Given such a small difference, which can readily be accounted for by normal sources of error, the prediction is clearly verified, and so the stoichiometric method for predicting the mass of lead formed in this experiment is judged to be acceptable.

Testing the Stoichiometric Method Prediction: 6.75 g of lead will be produced (Stoichiometric calculation) Analysis: mass of lead product = 7.60g – 0.92g = 6.68g According to the evidence collected, 6.68 g of lead precipitated. Evaluation: Could you also calculate the % yield ? Percent yield = actual yield x 100% = 6.68 g x 100% = 99.0% predicted yield 6.75g

Homework Investigation 7.2 – Guided Lab Report Pg. 293 #1-2, 6-10