Stoichiometry

Stoichiometry John-Nicholas Furst

Meaning of the Word • • • •

Comes from 2 Greek words stoicheion – element metron – measure Stoichiometry is calculations relating to the masses of reactants and products used in chemical reactions. • Lots of MATH!

What to Expect • Most stoichimetric problems will give you an amount of reactant and ask you to solve for the amount of product formed.

Steps to Solving Stoichimetric Problems 1. 2. 3. 4.

Balance equation Convert into moles Set up molar proportion Convert back into mass

Molar Ratios • The ratio is determined by the coefficients Equation: 2 H2 + O2 ---> 2 H2O Question: What is the molar ratio between H2 and O2? Answer: 2:1 Explanation: This is because the coefficient of H2 is 2 and the coefficient of O2 is 1(we just don’t write it).

Molar Ratio Example #1 Equation: 2 H2 + O2 ---> 2 H2O Question: What is the molar ratio between O2 and H2O? Answer: 1:2 Explanation: This is again because the O2 has an unwritten coefficient of 1 and the H2 has a coefficient of 2. Question: What is the molar ratio between H2 and H2O?

Molar Ratio Example #2 Equation: 2 O3 ---> 3 O2 Question: What is the molar ratio between O3 and O2? Answer: 2:3 Question: What is the molar ratio between O2 and O3? Answer: 3:2 Explanation: It all depends on how the question is worded, the first element is the numerator in the ratio and the second is the denominator

Mole-Mole Stoichimetric Problems • The procedure to solve mole-mole problems involves making two ratios and setting them equal to each other. • One ratio will come from the coefficients of the balanced equation, and the other will be set up from data in the problem and will have one unknown in which you are solving for. • You will then cross-multiply and divide to get the answer.

Mole-Mole Example #1 Equation: N2 + 3 H2 ---> 2 NH3 Question: If we have 2.00 mol of N2 reacting with sufficient H2, how many moles of NH3 will be produced? Solution: The ratio from the problem will have N2 and NH3 in it. When making the two ratios, make sure that the numbers are in the same relative positions. For example, if the value associated with NH3 is in the numerator, then MAKE SURE it is in both numerators. Use the coefficients of the two substances to make the ratio from the equation. Why isn't H2 involved in the problem? Answer: The word "sufficient" removes it from consideration. Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The proportion (setting the two ratios equal) is: Solving by cross-multiplying gives x = 4.00 mol of NH3 produced.

Mole-Mole Example #2 Equation: N2 + 3 H2 ---> 2 NH3 Question: Suppose 6.00 mol of H2 reacted with sufficient nitrogen. How many moles of ammonia would be produced? Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The proportion (setting the two ratios equal) is: Solving by cross-multiplying, 3x = 12, and dividing gives x = 4.00 mol of NH3 produced.

Converting Grams-Moles Steps: 2. Determine how many grams are given in the problem. 3. Calculate the molar mass of the substance. 4. Divide step one by step two.

Grams-Moles Example #1 Question: Convert 25.0 grams of KMnO4 to moles. Solution: (Remember the 3 steps) 3. The problem gives us 25.0 grams. • The molar mass of KMnO4 is 158.034 grams/mole • You divide the grams given by the substance's molar mass: Answer: 0.158 (Remember 3 sig. figs. from the 25.0)

Grams-Moles Example #2 Question: Calculate how many moles are in 17.0 grams of H2O2 Solution: 3. 17.0 grams are given in the text of the problem. 4. The molar mass is 34.0146 grams/mole. 5. You divide the grams given by the substance's molar mass: Answer: 0.500 moles

Converting Moles-Grams Steps: (Exactly the same except you multiply this time) 2. Determine how many moles are given in the problem. 3. Calculate the molar mass of the substance. 4. Multiply step one by step two.

Moles-Grams Example #1 Question: Calculate how many grams are in 0.700 moles of H2O2 Solution: (Again with the 3 steps) • 0.700 moles are given in the problem. • The molar mass of H2O2 is 34.0146 grams/mole. • Step Three: You multiply the moles given by the substance's molar mass: 0.700 mole x 34.0146 grams/mole = 23.8 grams Answer: 23.8 grams

Moles-Grams Example #2 Question: Convert 2.50 moles of KClO3 to grams. Solution: • 2.50 moles is given in the problem. • The molar mass for KClO3 is 122.550 grams/mole. • Step Three: You multiply the moles given by the substance's molar mass: 2.50 moles x 122.550 grams/mole = 306.375 grams Answer: 306.375 grams

Mole-Mass Problems Equation: 2 KClO3 ---> 2 KCl + 3 O2 Question: 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? Solution: Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The proportion (setting the two ratios equal) is: Cross-multiplying, 2x=4.5, and dividing gives x = 2.25 mol of O2 produced.

Mass-Mole Example #1 Equation: 2 KClO3 ---> 2 KCl + 3 O2 Problem: If 80.0 grams of O2 was produced, how many moles of KClO3 decomposed? Solution: Let's use this ratio to set up the proportion: That means the ratio from the equation is: The ratio from the data in the problem will be: The 2.50 mole came from 80.0 g ÷ 32.0 g/mol. The 32.0 g/mol is the molar mass of O2. Be careful to keep in mind that oxygen is O2, not just O. The proportion (setting the two ratios equal) is: Solving by cross-multiplying and dividing gives x = 1.67 mol of KClO3 decomposed.

Mass-Mass Problems • Most common type of Stoichimetric problem Steps: 3. Make sure you are working with a balanced equation. 4. Convert grams to moles. 5. Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem. 6. Convert moles into grams

Mass-Mass Example #1 Question: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl3 Equation: 2 AuCl3 ---> 2 Au + 3 Cl2 Solution: 4. Molar ratio: 5. Convert grams-moles: 6. Set ratios equal: 7. Solve for x and convert moles-grams:

Mass-Mass Example #2 Calculate mass of PbI2 produced by reacting of 30.0 g KI with excess Pb(NO3)2 Equation: 2 KI + Pb(NO3)2 --> PbI2 + 2 KNO3 3. 4. 5. 6.

Molar ratio: Convert grams-moles: Set ratios equal: Solve for x and convert moles-grams:

Limit Reagent •

The Limiting Reagent is simply the substance in a chemical reaction that runs out first. [Ex. #1] Reactant A is a test tube. I have 20 of them. Reactant B is a stopper. I have 30 of them. Product C is a stoppered test tube. The reaction is: A + B ---> C or: test tube plus stopper gives stoppered test tube. • So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up). • Suddenly, we run out of one of the "reactants." • We ran out of test tubes first. We had only had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted. • So, which the test tubes are limiting and the stopper are in excess?

Limiting Reagent Example #2 Equation: 2 Al + 3 I2 ------> 2 AlI3 Question: Determine the limiting reagent of the product if we use: 1.20 mol Al and 2.40 mol iodine. • Solution: Here is how to find out the limiting reagent: take the moles of each substance and divide it by the coefficient of the balanced equation. • For aluminum: 1.20 / 2 = 0.60 For iodine: 2.40 / 3 = 0.80 • The lowest number indicates the limiting reagent. Aluminum will run out first in part

Percent Yield • Percent Yield is found by taking the actual yield and dividing it by the theoretical yield and then multiplying by 100. • The actual yield is the amount formed when the experiment is actually carried out. • The theoretical yield is determined by how much can be produced with the limiting reagent.

 Actual  (100)    Theoretical 

Sources • • • • • • • • • • • • •

Mr. Beran http://members.tripod.com/~EppE/stoictry.htm http://www.shodor.org/UNChem/basic/stoic/index.html http://en.wikipedia.org/wiki/Stoichiometry http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch3/massmolfra http://www.chemtutor.com/mols.htm http://www.science.uwaterloo.ca/~cchieh/cact/c120/stoichio.html http://www.chem4kids.com/files/react_stoichio.html http://members.aol.com/profchm/eq_form.html http://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Stoichiometry.html http://members.aol.com/profchm/eq_form.html http://chemistry.about.com/od/stoichiometry/Stoichiometry.htm http://www.chemcollective.org/tutorials.php