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Chapter 3
Mole Day is on Monday October 23. Bring your moles/food/mole day items for extra credit.
3 new Webassigns are up. They are balancing equations, Stoichiometry, Limiting reactant.
On your second day of class this week there will be a lab. Dress appropriately
Period 1 Lab is 10/11
Period 3 Lab is 10/13
Period 11 Lab is 10/12
Load the pretest and give it a try now!!!
Remember that reading the textbook is important. If you are not at the point you want to be, then read, ask questions and read.
Some notes:
1 mol = 6.02 x 1023 atoms/molecules = atomic/molecular mass in grams.
Problem type 1: direct conversions. To convert from one to another, use the factor label method. See page 72-73 in your text for specific examples.
Ignore section 3.4 on the mass spectrometer.
Problem type 2: % composition or empirical formula. See sample 3.8 and 3.9 (page 78 and 79 in your text) Remember, for empirical formula, remember
percent to mass, mass to moles, divide by smallest, multiply till whole
For molecular formulas, divide molar mass by empirical formula mass to get a whole number. Multiply each subscript by that number. For example if your molecular mass is 56.16 g/mol, and your empirical formula is CH2:
Multiply each subscript by this number. C's subscript is 1, so 1x4 = 4. H's subscript in the empirical formula is 2, so 2 x 4 is 8. Therefor the molecular formula is C4H8.
Balancing chemical equations. This skill is obtined by practice. Some helpful hints are on page 85 in the text.
Stoichiometry: This allows us to predict how much of each substance is required to react with other substances. It is the same as straightforward conversions, except you must use a mole ratio when converting the amount of one substance to another. The mole ratio is given by the coefficients of the balanced equations. This is why balancing is important. Study example 3.13 and 3.14 on page 89-90 to review stiochiometry.
Limiting reagents. Sometime we add more than we need of a given reagent. If this occurrs then some of that reagent will be left over at the end of the reaction. The reagent that gets fully consumed is the one that determines how much product is formed. This is called the limiting reagent. You can tell a problem is a limiting reagent problem if amounts of more than one reactant are given. It is important to convert each of these to moles product. The one that forms less is the limiting reagent and is fully consumed. All calculations must be done using the limiting reagent. See example 3.15 on page 91 for more help.
Yield. When chemists run reactions, they rarely obtain the stiochiometric amount. (This is the maximum amount formed based on the limiting reagent). There are often competing reactions that cause other undesired products to form. The yield is an important calculation because it esentially tells us how well the reaction works. It is extremely important is optimize reaction conditions to increase yield, and chemists work on this, because more useful product means more dollars for the company. The % yield is the acutal amount formed divided by the stoichiometric (theoretical) yield. See page 94 and 95 for examples of yield calculations.
