Download presentation

Presentation is loading. Please wait.

Published byRoger Ismay Modified over 2 years ago

1
Stoichiometry Percent Yield

2
Important Terms Yield: the amount of product Theoretical yield: the maximum amount of product expected, based on stoichiometric calculations Actual yield: amount of product from a procedure or experiment (this is often given in a question) Percent yield: (actual yield ÷ theoretical yield) × 100

3
Real reactions usually produce less than the “ideal” or theoretical yield. Side Reactions (aka: competing reactions): reactant products Reaction does not go to completion: reactant product Loss of product (vaporizing, spillage, spattering) Reactant impurities due to varying grades of chemicals: Reagent = The highest quality commercially available for this chemical. Practical = chemicals of good quality where there are no official standards. Lab grade = Suitable for histology methods and general laboratory applications. USP = Chemicals manufactured under current Good Manufacturing Practices and which meet the requirements of the US Pharmacopeia. Technical = A grade suitable for general industrial use and non critical laboratory tasks. Fisher Scientific SIGMA-ALDRICH Fisher Scientific SIGMA-ALDRICH More details here: Fisher Scientific or SIGMA-ALDRICHFisher Scientific SIGMA-ALDRICH Why is this so?

4
% Yield Comes from a measured experimental value or is given in the question or problem. Comes from the stoichiometry calculation. It is the “ideal” or “perfect” yield.

5
Finding % Yield Using the equation above, if 39.9 g of water are produced when 26.0 g of ammonia are reacted, what is the % yield of the reaction? 4NH 3 + 5O 2 6H 2 O + 4NO Step 1We find the stoichiometric amount of water that would be produced from 26.0 g of ammonia if we were to obtain a 100% yield.

6
Finding % Yield Using the equation above, if 39.9 g of water are produced when 26.0 g of ammonia are reacted, what is the % yield of the reaction? 4NH 3 + 5O 2 6H 2 O + 4NO Step 2Knowing that the stoichiometric amount of water produced is 41.3 g H 2 O, we can now substitute values into the % Yield equation.

7
Using % Yield to find a product amount 22.0 g of oxygen gas are reacted with excess ammonia as shown above. If the reaction yields 97.0%, what mass of NO will actually be produced? 4NH 3 + 5O 2 6H 2 O + 4NO Step 1We find the stoichiometric amount of NO that would be produced from 22.0 g of oxygen gas if we were to obtain a 100% yield.

8
Using % Yield to find a product amount 22.0 g of oxygen gas are reacted with excess ammonia as shown above. If the reaction yields 97.0%, what mass of NO will actually be produced? 4NH 3 + 5O 2 6H 2 O + 4NO Step 2With the stoichiometric amount of NO determined, we can now reduce it down by 97.0% to the actual yield.

9
Using % Yield to find a reactant amount If the reaction above yields 95.0% and 44.0 g of H 2 O are actually produced, what amount of ammonia gas is required? 4NH 3 + 5O 2 6H 2 O + 4NO Step 1The 44.0 g of H 2 O are already reduced down to 95.0% of the theoretical yield. In order to go from the mass of H 2 O back to the mass of NH 3, we will need to first find what the 100% amount was.

10
Using % Yield to find a reactant amount If the reaction above yields 95.0% and 44.0 g of H 2 O are actually produced, what amount of ammonia gas is required? 4NH 3 + 5O 2 6H 2 O + 4NO Step 2With the 100% yield (stoichiometric amount) of water known, it can now be used to calculate the amount of NH 3 reactant needed.

11
Summary Finding % Yield ► Given actual yield ► Find stoichiometric yield ► Use % yield formula Using % Yield to find amount of product ► Given % yield ► Find stoichiometric yield of product ► Multiply % yield and stoichiometric yield Using % Yield to find amount of reactant ► Given % yield ► Divide actual yield by % yield to get theoretical yield ► Use theoretical yield of product to find reactant amount

12
Have we learned it yet? Try these on your own: Given: 4NH 3 + 5O 2 6H 2 O + 4NO a) What is the % yield of H 2 O if 7.50 g are actually produced using 5.00 g NH 3 ? b) If 9.50 g of O 2 is used to make NO at a 92.5% yield, what mass of NO is actually produced? c) How many grams of NH 3 are needed to produce an actual yield of 33.3 g of H 2 O representing a 94.3% yield?

13
4NH 3 + 5O 2 6H 2 O + 4NO a) b) c) Answers ?g H 2 O=5.00 g NH 3 7.94 g H 2 O = 4 mol NH 3 6 mol H 2 O 17.03 g NH 3 1 mol NH 3 1 mol H 2 O 18.02 g H 2 O % Yield= Theoretical g H 2 O Actual g H 2 O X 100 % Yield= 7.94 g H 2 O 7.50 g H 2 O X 100 94.5% yield = ?g NO=9.50 g O 2 5 mol O 2 4 mol NO 32.00 g O 2 1 mol O 2 1 mol NO 30.01 g NO 100 92.56.59 g NO = ?g NH 3 =33.3 g H 2 O 6 mol H 2 O 4 mol NH 3 18.02 g H 2 O 1 mol H 2 O 1 mol NH 3 17.03 g NH 3 94.3 10022.2 g NH 3 =

Similar presentations

OK

How far away you are from a theoretical or “actual” value.

How far away you are from a theoretical or “actual” value.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on best practices in hr Ppt on total internal reflection fluorescence Ppt on water softening techniques of integration Ppt on social networking website project Ppt on review writing books Ppt on my sweet home Ppt on 2-stroke petrol engine Ppt on fibonacci numbers and the golden Ppt on diffusion taking place in our daily life Ppt on odisha cultures