Presentation on theme: "Learning Goal 4: Use molar relationships in a balanced chemical reaction to predict the mass of product produced in a simple chemical reaction that goes."— Presentation transcript:
Learning Goal 4: Use molar relationships in a balanced chemical reaction to predict the mass of product produced in a simple chemical reaction that goes to completion.
SubstanceSymbolUnit Mass Graham CrackerS7.0 g MarshmallowMm7.1 g Chocolate PiecesOr3.3 g S’moreS 2 MmOr 3 __________ g 2S + Mm + 3Or S 2 MmOr 3 The equation above gives the recipe to make one S’more.
Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass to moles. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired molecule. 5.Convert moles to grams or to particles, if necessary.
Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. Identify reactants and products and write the balanced equation. 4Al + 3O 2 2Al 2 O 3 What are the reactants? Every reaction needs a yield sign! What are the products? What are the balanced coefficients?
Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2 2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al g Al4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O x 2 x ÷ ÷ 4 = 12.3 g Al 2 O 3
Calculating Percent Composition of a Compound Like all percent problems: part whole 1)Find the mass of each of the components (the elements), 2)Next, divide by the total mass of the compound; then x 100 x 100 % = percent
Example Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S g Ag 33.3 g total X 100 = 87.1 % Ag 4.30 g S 33.3 g total X 100 = 12.9 % S Total = 100 %
Theoretical Yield The amount of product predicted from the amount of reactant used. The maximum amount of product that could form. Actual Yield The amount of product that is actually produced. Percent Yield Actual Yield X 100% = Percent Yield Theoretical Yield
First write the balanced equation. 2H 2 + CO CH 3 OH Next Calculate moles of reactants. 68,500 g CO 1 Mol CO = 2450 mol CO g CO 8600 g H 2 1 Mol H 2 = 4270 mol H g H 2
Next determine the limiting factor 2,450 mol CO 2 mol H 2 = 4,900 mol H 2 1 mol CO 4270 mol H 2 < 4,900 mol H 2 Hydrogen is the limiting factor. Next determine the maximum amount of methanol that can be produced in this reaction. 4270 mol H 2 1 mol CH 3 OH = 2140 mol CH 3 OH 2 mol H 2
Convert moles to grams. 2140 mol CH 3 OH g CH 3 OH = g CH 3 OH 1 mol CH 3 OH Percent Yield g CH 3 OH x 100 % = 52.0% g CH 3 OH
Formulas molecular formula = (empirical formula) n [ n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the
Limiting Reactant The limiting reactant is the reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. that is consumed first, limiting the amounts of products formed.
Write and balance the equation for the reaction. Convert known masses of reactants to moles. Determine which reactant is limiting. Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. Convert moles to mass (if required).
Write the balanced equation. N 2 + 3H 2 2NH 3 Convert known masses to moles. g N 2 1 mol N 2 = 892 mol N g N 2 5000 g H 2 1 mol H 2 = 2,480 mol H g H 2
Determine the Limiting reactant. 892 mol N 2 3 mol H 2 = 2676 mol H 2 1 mol N 2 2480 mol H 2 < 2676 mol H 2 Hydrogen is the limiting reactant. To get all of the Nitrogen to react you would need 2676 mol of hydrogen gas.
Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. 2480 mol H 2 2 mol NH 3 = 1653 mol NH 3 3 mol H 2
Convert moles to mass. Calculate the molar mass of NH 3 N x 1 = H X 3 = 1653 mol NH g NH 3 = 28,150 g NH 3 1 mol NH 3