Catalyst

Lecture 3.6 – Stoichiometry, Empirical Formulas, and Limiting Reactants

Today’s Learning Targets LT 3.10 – When given the starting amount of grams of particular substance in a chemical reaction, I can convert this to moles, grams, particles, or liters gas of another compound in the same chemical reaction. LT 3.11 – I can calculate the empirical formula for a given compound using experimental data. Furthermore, I can discuss what these results indicate. LT 3.12 – I can identify the limiting reactant for a reaction. Additionally, I can support my claim through stoichiometric calculations.

Stoichiometry Review Recall, we can convert grams of one substance to grams of another substance using stoichiometry. Label the value that you need to use above each of the arrows Grams of Substance A Grams of Substance B Moles of A Moles of B

Stoichiometry Review We also can convert from particles of one substance to particles of another substance. Label the conversion factors above the arrows Moles of A Moles of B Atoms of Substance A Atoms of Substance B

Converting Moles to Liters of Gas Finally, we can convert from liters of one gaseous substance to liters of another gaseous substance. We know that 1 L of any gas occupies 22.4 L Volume of Substance A Volume of Substance B Moles of A Moles of B

Class Example I run the following reaction: N2 (g) +3 H2 (g)  2NH3 (g) If I start with 14 L of N2, then how many L of NH3 should I produce?

C6H12 (l) + O2 (g)  H2C6H8O4 (l) + H2O (g) Table Talk If I run the following reaction: C6H12 (l) + O2 (g)  H2C6H8O4 (l) + H2O (g) I produce 54.8 L of H2O, then how many L of O2 did I begin with? NOTE – You need to balance this reaction before starting

Volume of Substance A Volume of Substance B Moles of A Moles of B Grams of Substance B Grams of Substance A Have students label the conversion factor above each arrow, then have them label the entire set Atoms of Substance A Atoms of Substance B

Volume of Substance A Volume of Substance B Moles of A Moles of B Grams of Substance B Grams of Substance A Have students label the conversion factor above each arrow, then have them label the entire set Atoms of Substance A Atoms of Substance B

Fritz Habor Research Assistant

C6H12O6 CH2O Empirical Formula Molecular Formula Empirical Formula The empirical formula is the smallest whole number ratio of atoms in a compound. The molecular formula is the actual ratio of atoms in a compound C6H12O6 CH2O Molecular Formula Empirical Formula

Class Example Ascorbic Acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula? If the molecular mass of ascorbic acid is determined to be 176.12 g/mol, then what is the molecular formula of vitamin C?

Table Talk A perfume that you are wearing is 70.57% carbon, 5.93% hydrogen, and 23.50% oxygen. What is the empirical formula of this compound? If the molar mass is 136.15 g/mol, then what is the molecular formula?

Combustion Problems Add in? See page 95

Making Bikes You have 5 bike frames, 12 bike wheels, and 4 handle bars. How many bikes can you make? What stops you from making more bikes?

Limiting Reactants Oftentimes, we do not have equal amounts of reactants, therefore the reaction will stop when one of the reactants runs out. The reactant that runs out first is known as the limiting reactant. We determine the limiting reactant by determining which reactant produces the least amount of a specific product.

Class Example You run the Habor’s reaction: N2 (g) + 3 H2 (g)  2NH3 (g) How many moles of NH3 can be produced from 3.0 mol of N2 and 6.0 mol of H2?

Table Talk The reaction: 2 H2 (g) + O2 (g)  2 H2O (g) is used to produce electricity in hydrogen fuel cells. Suppose a fuel cell contains 150. g of H2 and 1500. g of O2. How many grams of water can form?

Model Cards On the handout, provide an answer to the problem that you are working out and the reasoning behind your answer. Complete all the problems, ask your table partners if you need help on the reasoning!

Closing Time 3.4, 3.5, 3.6, and 3.7