Presentation on theme: "Stoichiometry! The math of chemistry. Stoichiometry The study of quantifiable (measurable) ratios/relationships that exist in chemical formulas and reactions."— Presentation transcript:
Stoichiometry The study of quantifiable (measurable) ratios/relationships that exist in chemical formulas and reactions. In many ways, stoichiometry is the backbone of the most practical part of chemistry; it helps us relate actual quantities (measured by mass or volume) of reactants to products in a chemical reaction.
Stoichiometry – a lot like baking cookies! You have to put the ingredients together in the proper proportions to make a cookie that tastes good! Consider the following balanced equation: 2Na + Cl 2 2NaCl
Consider the following balanced equation: 2Na + Cl 2 2NaCl We can look at this equation two different ways: 1. at the atomic/molecular scale: 2 Na atoms react with 1 Chlorine molecule to produce 2 formula units of NaCl 2. at the real-world, measurable scale: 2 mol of Na react with 1 mol of Cl 2 to produce 2 mol of NaCl (you could measure out those amounts)
Dimensional Analysis Can be used to relate the moles of reactants and products in a chemical reaction to one another. It is essentially a matter of determining the ratio of reactants and products by looking at the coefficients in the balanced chemical equation. In the reaction above, the ratio of sodium to chlorine is 2:1 – The ratio of sodium to sodium chloride is 2 Na:2 NaCl – The ratio of chlorine to sodium chloride is 1 Cl 2 :2 NaCl
Dimensional Analysis If you’re starting out with 0.650 moles of sodium and you want to know how many moles of chlorine you need to fully react with the sodium, you can use dimensional analysis: 0.650 mol Na x 1 mol Cl 2 = 0.325 Mol Cl 2 2 mol Na Mole Ratio
Dimensional Analysis If you want to produce 6.4 moles of sodium chloride, how many moles of chlorine do you need to use? 6.4 mol NaCl x 1 mol Cl 2 = 3.2 mol Cl 2 2 mol NaCl Because we can literally measure out the moles of sodium by weighing it (1 mol = 22.99 g) and we can literally measure out the moles of chlorine by volume (1 mol = 22.4 L), we can literally put the reactants together in the proper proportions. Mole Ratio
When solving a stoichiometry problem… You’ll always be given a quantity of either a product or reactant that you have to relate to a quantity of another product or reactant. To do this, you’ll use the balanced chemical equation as well as the mole equivalencies you’ve been using in mole conversions.
These equivalencies are: 1 mole = 6.02 x 10 23 particles – (molecules, formula units, atoms, etc.) 1 mole = 22.4 L of gas at STP – (Standard Temperature 0° C & Pressure 1 atm) 1 mole = mass in grams – (from periodic table, add masses of elements for compounds)
The equivalencies (conversion factors) you’ll use depend on the specific problem, but the logic is always the same: Quantity of given (measurable) moles of given moles of unknown quantity of unknown (measurable)
New and Improved Mole Map! moles of substance B moles of substance A liters of substance B particles of substance A grams of substance A liters of substance A particles of substance B grams of substance B This is your Mole Ratio (from balanced eq’n)
Given the following equation: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) What mass of oxygen will react with 36.0 g of methane? Plan: grams CH 4 mol CH 4 mol O 2 grams O 2 Calculations: 36.0 g CH 4 x 1 mol CH 4 x 2 mol O 2 x 32.00 g O 2 = 143.64 g O 2 = 144 g O 2 16.04 g CH 4 1 mol CH 4 1 mol O 2
Given the following equation: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) What mass of water would be produced when oxygen reacts with 36.0 g of methane? Plan: grams CH 4 mol CH 4 mol H 2 O grams H 2 O Calculations: 36.0 g CH 4 x 1 mol CH 4 x 2 mol H 2 O x 18.02 g H 2 O = 80.9 g H 2 O 16.04 g CH 4 1 mol CH 4 1 mol H 2 O
Given the following equation: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) How many Liters of carbon dioxide would be produced from the reaction if 44.0 g of oxygen are consumed? Plan: grams O 2 mol O 2 mol CO 2 L CO 2 Calculations: 44.0 g O 2 x 1 mol O 2 x 1 mol CO 2 x L CO 2 = ?? L CO 2 g O2 2 mol O 2 1 mol CO 2