## Presentation on theme: "Chocolate Chip Cookies!!"— Presentation transcript:

1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen 2 Eggs How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? 1 Cup 6 Eggs

Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them chemical equations Furthermore, instead of using cups and teaspoons, we use moles to measure amounts Lastly, instead of eggs, butter, sugar, etc… we use chemical compounds as ingredients

The Math of Chemical Equations
Stoichiometry The Math of Chemical Equations

Just what is stoichiometry?
The word stoichiometry is derived from two Greek words: stoicheion (meaning element) and metron (meaning measure).

Jeremias Benjamin Richter ( ) was the first to lay down the principles of stoichiometry. In 1792 he wrote: “Stoichiometry is the science of measuring the quantitative proportions or mass ratios in which chemical elements stand to one another.” This was verified by Antoine Lavoisier with the Law of Conservation of Mass

When do we use stoichiometry?
Whenever chemists want to know how much of a desired product will be produced. Chemists always start with a balanced equation.

How does it work? The chemist uses the balanced equation to guide them in calculating the maximum amount of products. Since stoichiometry is used to predict things in a “perfect world”, we can expect that the reaction in the lab won’t create the maximum amount of product that was calculated.

How do I use the chemical equation?
Looking at a chemical equation tells us how much of something you need to react with something else to get a product (like the cookie recipe) The chemical equation is written like a mathematical expression. When balanced, the coefficients tell how many moles of each reactant or product is needed or expected.

The important thing is that the equation must be BALANCED!!!!

Example: 2H2 + O2 2H2O One way to understand the information conveyed by a chemical equation is to convert the equation into an English sentence.

2H2(g) + O2(g) 2H2O(l) 2 “parts” hydrogen gas reacts with one “part” oxygen gas to form 2 “parts” liquid water. Now, we can get a bunch of relationships – the “parts” could be: molecules, atoms, moles, or masses (the coefficients tell how many moles, atoms or molecules of each chemical are needed in the “recipe”) The most important relationship for stoichiometry is the MOLE relationship!

What is a mole ratio? Mole Ratios:
The mole ratio is based on the balanced chemical equation. You will use the coefficients to get the mole ratio between two DIFFERENT substances These ratios will be used as conversion factors for stoichiometry problems.

How do you use a mole ratio?
The mole ratio is an equivalent which means that you can arrange the ratio in any way needed. The mole ratio is the KEY to calculations based upon a chemical equation. With it, you can calculate the amount of any other reactant in the equation and the maximum amount of product you can obtain.

Molar Ratios The expressing the ratio of moles in an equation using the coefficients Ex. 2H2 + O2  2H2O What is the mole ratio between H2 and O2? What other mole ratios can we come up with?

Warning: The coefficients of a reaction only give the ratio in which substances react and form. They DO NOT directly tell you, HOW MUCH MASS is reacting. You will need to convert from moles to mass using molar mass as the conversion factor.

Example of a mole ratio problem:
N2 + 3H2  2NH3 Write the mole ratios for N2 and H2 Write the mole ratios for NH3 and H2

Stoichiometry has 5 basic steps:
Step 1: Write and balance the equation

Step 2: Write down all information that you’re given (Make sure to identify what you are trying to find!)

Step 3: Convert everything given into moles
Step 3: Convert everything given into moles. (Use molar mass as the conversion factor if you have to.)

Step 4: Use a mole ratio to solve for what you are trying to find
Step 4: Use a mole ratio to solve for what you are trying to find. (Use the coefficients from the balanced chemical equation to get the correct mole ratio.)

Step 5: Convert, if needed, everything into correct units
Step 5: Convert, if needed, everything into correct units. (IF you need to get from moles to mass use the molar mass of the substance as the conversion factor.)

Mole-to-Mole Stoichiometry
One way to change ion ore, Fe2O3, into metallic iron is to heat it together with hydrogen: Fe2O3 + 3H2  2Fe + 3H2O How many moles of iron are made from 25.0 moles of Fe2O3 reacting with excess hydrogen?

Mole-to-Mole K: 25 moles Fe2O3 UK: ? moles Fe
Use Dimensional Analysis : 25 moles Fe2O3 = 50. mol Fe

More examples of mole-to-mole stoichiometry problems:
2H2 + O2  2H2O  1. How many moles of H2O are produced when 5.00 moles of oxygen are burned in excess hydrogen gas? 10.0 mol H2O

Examples of a mole-mole stoichiometric problem continued:
2H2 + O2  2H2O 2. If 3.00 moles of H2O are produced, how many moles of oxygen must be consumed in a reaction with excess hydrogen gas? 1.50 mol O2

Examples of a mole-mole stoichiometric problem continued:
2H2 + O2  2H2O 3. How many moles of hydrogen gas must be used, when reacting with 0.25 mol oxygen gas to create water. 0.50 mol H2

Mole-to-Mass Stoichiometry
Most of the time in chemistry, the amounts are given in grams instead of moles We STILL go through moles and use the mole ratio, but now we also use molar mass (as a conversion factor) to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2  2 NaCl 5.00 moles Na = 177g Cl2

Practice Calculate the mass, in grams, of iodine required to react completely with 0.50 moles of aluminum to make aluminum iodide. 2 Al + 3 I2  2 AlI3

Mass-to-Mole Stoichiometry
We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest EX: Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2  4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = mol C2H6

Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide 4 Al + 3 O2  2 Al2O3

Mass-to-Mass Stoichiometry
Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)

We must go from grams to moles by using molar mass of the compound given
Go from moles of compound given to moles of compound needed using the mole ratio from the coefficients Go back to grams of compound we are interested in by using molar mass of the compound you are looking for

Mass-to-Mass Stoichiometry Problems
Ex: Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2  2 NH3

Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? 3 Ca + N2  Ca3N2

Gas Stoichiometry (Volume-to-Volume) @ STP
Figure out volume of gas given. Convert to moles using the fact that 22.4 L=1 STP Change from moles of compound given to moles of new compound using molar ratio from the coefficients. Change moles of new compound to volume using 22.4 L = 1 mole

Gases and Stoichiometry
2 H2O2 (g)  2 H2O (g) + O2 (g) 3.5 L of H2O2 was decomposed in a flask. What is the volume of O2 made at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

What if we are not just looking for a volume to volume calculation?
See what the question is asking us and use appropriate conversion factors. For example, going back to: 2 H2O2 (g)  2 H2O (g) + O2 (g) What if you wanted to know how many moles of water were formed from the decomposition of 0.25 L of H2O2 at STP?

Another Example Find the number of grams of oxygen necessary for the combustion of 6.7 L of magnesium, assuming the reaction occurs at STP. 2 Mg + O2  2 MgO Answer: 4.8 g O2

LIMITING REACTANT

1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

Limiting Reactant: Example
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product, in grams, is produced? 2 Al + 3 Cl2  2 AlCl3 Start with Al: Now Cl2: 10.0 g Al 1 mol Al mol AlCl g AlCl3 27.0 g Al mol Al mol AlCl3 = 49.4g AlCl3 35.0g Cl mol Cl mol AlCl g AlCl3 71.0 g Cl mol Cl mol AlCl3 = 43.9g AlCl3

LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete

Limiting Reactant Practice
15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. 2K + I2  2KI

From the previous question, how much of the excess reactant is left over?

Percentage Yield = Actual Yield
Percent Yield Theoretical Yield: The maximum quantity of product that can be obtained, according to the reaction stoichiometry, from a given quantity of reactant Actual Yield: The quantity of product actually obtained from experimentation Percentage Yield = Actual Yield Theoretical yield X 100%

What if I get an answer over 100%?
Percent yield can NEVER be over 100% This would conflict with the Law of Conservation of Matter If you obtain a yield over 100% you either: A. Did your calculation incorrectly by putting theoretical yield over percent yield OR B. Made a mistake in the experiment

Percent yield problems:
We calculated that 19.6g of KI forms from 15.0g K and 15.0g of I2. If we ran the experiment and only obtained 18.5g of KI what is the percent yield?

Limiting Reactant & Percent Yield: Recap
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. Convert ALL of the reactants to the SAME product (pick any product you choose.) The lowest answer is the correct answer. The reactant that gave you the lowest answer is the LIMITING REACTANT. The other reactant(s) are in EXCESS. To find the percent yield, divide the mass of what you get from the experiment from mass of the product from the limiting reactant and multiply by 100%.