AP Chemistry Unit 3 – Stoichiometry Day 4: Stoichiometric Calculation Practice

Warm Up TAKE OUT: Limiting Reactant Guided Inquiry SHOW ME: 1-9 for full credit

Agenda Determining The Molecular Formula Limiting Reactant and Percent Yield Review Guided Inquiry Assignment: Limiting Reactant Unit 3 Quiz: Study Guide Next Classes Monday: Unit 3 Quiz, Chapter 3 WebAssign, Reading: Chapter 4.1-4.7 Due Formal Lab Report Due: November 7th

Determining a Molecular Formula Remember, the number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula. If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula.

Determining a Molecular Formula: Example Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of 200.04 g/mol STEP 1: Molar mass of Empirical Formula: C = 12.01 g/mol x 1 = 12.01 F = 19.00 g/mol x 2 = 38.00 TOTAL  50.01 g/mol

Determining a Molecular Formula: Example Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of 200.04 g/mol STEP 2: Molar masses of multiples to find a match… Multiple Formula Mass x2 C2F4 100.02 x3 C3F6 150.03 x4 C4F8 200.04

Determining a Molecular Formula: Example Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of 200.04 g/mol Is there a simpler way? YES!!!!

Determining a Molecular Formula: Example Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of 200.04 g/mol Molar mass of molecular formula Molar mass of empirical formula

Determining a Molecular Formula: Example Determine the molecular formula of a compound with an empirical formula of CF2 and a molar mass of 200.04 g/mol Molar mass of molecular formula = 200.04 = 4 Molar mass of empirical formula 50.01 50.01 x 4 200.04 CF2 x 4 C4F8

Ribose has a molecular weight of 150 grams per mole and the empirical formula CH2O. The molecular formula of ribose is a. C5H10O5 b. C4H8O4 c. C6H14O4 d. C6H12O6 Answer: a

Limiting Reactant A bicycle manufacturing company has 4802 wheels, 2299 frames, and 2249 handlebars. How many bicycles can be manufactured using these parts? 2250 How many parts of each kind are left over? 304 wheels, 50 frames, 0 handlebars Which part limits the production of bicycles? Handlebars!

Quantitative Relationships The coefficients in the balanced equation show relative numbers of molecules of reactants and products. relative numbers of moles of reactants and products, which can be converted to mass.

Stoichiometric Calculations We have already seen in this chapter how to convert from grams to moles or moles to grams. The NEW calculation is how to compare two DIFFERENT materials, using the MOLE RATIO from the balanced equation!

An Example of a Stoichiometric Calculation How many grams of water can be produced from 1.00 g of glucose? C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) There is 1.00 g of glucose to start. STEP 1: Convert it to moles.

An Example of a Stoichiometric Calculation STEP 2: Convert moles of one substance in the equation to moles of another substance. This is what where the MOLE RATIO comes in from the balanced equation.

An Example of a Stoichiometric Calculation STEP 3: Convert from moles to grams (18.00 g of water for every 1 mole of water). Through canceling units and calculating through you can our answer in grams of water!

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).

Limiting Reactants In the example below, the O2 would be the excess reagent.

Limiting Reactants The limiting reactant is used in all stoichiometry calculations to determine amounts of products and amounts of any other reactant(s) used in a reaction.

Theoretical Yield The theoretical yield is the maximum amount of product that can be made. In other words, the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

Percent Yield Percent yield = × 100 actual yield theoretical yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield): Percent yield = × 100 actual yield theoretical yield

Steps to Stoichiometric Calculation

Approach 1: Find the limiting reagent by looking at the number of moles of each reactant. Determine the balanced chemical equation for the chemical reaction. Convert all given information into moles (most likely, through the use of molar mass as a conversion factor). Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio. Use the amount of limiting reactant to calculate the amount of product produced. If necessary, calculate how much is left in excess of the non-limiting reagent.

Approach 2: Find the limiting reagent by calculating and comparing the amount of product each reactant will produce. Balance the chemical equation for the chemical reaction. Convert the given information into moles. Use stoichiometry for each individual reactant to find the mass of product produced. The reactant that produces a lesser amount of product is the limiting reagent. The reactant that produces a larger amount of product is the excess reagent. To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.