Lab #6

Key Terms  Atomic Weight  Molecular Weight  Gram Atomic & Molecular Weight  Avogardro’s Number  Moles  Formula Weight  Simplest Empirical Formula  Law of Definite Composition  Stoichiometery

What is Atomic Weight?  This is the mass of an atom.  Each element’s atom has a different mass.  How is it measured?  In Amus (atomic mass units).  How do we determine the mass of an atom?  We use the periodic table.

Atomic Mass Units  Example # 1 What is the mass of one Iron atom? 55.8 amus. What is the of a Copper atom? amus

Ionic Vs. Molecular Weight  Ionic Weight is the mass of all of the components in an ionic compound.  Molecular Weight is the mass of all of the components in a molecular compound.  Both are determined the same way.  For now on if both Molecular Wt refers to both ionic and molecular compounds.

What is a Mole?  6.02X10 23 parts.  We use the mole concept to talk about how many parts not mass.  If have a mole of computers we have 6.02X10 23 computers.  This number is used just like how we say we have 12 parts in a dozen.  A mole of cars has a different mass then a mole of potatoes, but each mole contains 6.02X10 23 respective parts.

Grams Vs. Amu  To determine how many grams are in one mole of any ELEMENT all we have to do is look at the periodic table.  55.8 grams = 1 mole of iron.  So when we are talking about moles the units are grams, and when we are talking about atoms the units are in amus.  Remember the reason for using moles is to make the numbers easier to use.

Example (1)  1 mole of Na +1 ions reacts with 1 mole of Cl -1 ions to form one mole of Sodium Chloride.  This is easier then saying 6.02X sodium ions reacted with 6.02X chloride ions to form 6.02X atoms of sodium chloride

We use the formula:

Determine the number of moles in.64g grams of NaCl?

Formula Weight  If the formula is wrong then the mass will be wrong. Per means 1 more oxygen then the base acid.  Base= HClO oxygen ANS : HClO 4  H= 1.0 g/mol  Cl = g/mol  O= g/mol Answer: g/mol

What is the percent composition of all of the elements in HIO?  Add the masses up for a grand total of g/mol  1.0g/mol/ X100 = 0.69%  / X 100 = 88.0%  16.0g/mol/ X 100 = 11.0%

 Simplest Empirical  This tells us the number of different elements present in a compound and the lowest possible ratio of these elements.  True Molecular Formula  True molecular formula is always a true multiple of the simplest formula. Types of Formulas

1.We need a balanced equation 2.Start with the number of grams given and convert to to moles 3.Use your mole ratio and number of grams produced to determine the number of moles asked for. 4.Convert this number of moles back to grams. Stoichiometery Rules:  Given 8.0g of CH 4 how many grams of H 2 O can be produced? CH 4 + 2O 2 →CO 2 +2H 2 O

(1) Balance Equation

(2) Convert given grams to moles Given 8.0g of CH 4 how many grams of H 2 O can be produced?

(3) Use mole to mole ratio CH 4 + 2O 2 →CO 2 +2H 2 O mole ratio 1mole of CH 4 reacts with 2 moles O 2 to produce 1 mole of CO 2 and 2 moles of H 2 O

 From our problem: Given 8.0g of CH 4 how many grams of H 2 O can be produced? CH 4 + 2O 2 →CO 2 +2H 2 O We know 1 mole of CH 4 will produce 2 moles of H 2 O

Now Do the math:.5X 2 = 1 mole of H 2 0 can be produced

(4) Convert moles back to grams

Lab # 6  Obtain a sample of salt and determine its composition.  This will be accomplished by determining the % oxygen in your sample.  Review how to use a crucible.

Table A  Wash, dry, cool and then determine the mass of your empty crucible.  Add the salt and record the mass of salt and crucible.  Once the weight is determined add a pinch of the catalyst to your crucible.  Heat crucible according to directions in lab.  Allow crucible to cool and determine the final mass

 For KClO 2  Mass of each K = 39g, Cl = 36.g and O =16 and 2 present so 2 X16 = 32g  Total mass: 39g+ 36g +32g = 107g. Table B Determine the percent oxygen in the salts listed.

Table C

Due Next Week  Complete Lab: pgs  Complete homework and exercise for lab # 6: 46-47pgs  Study for quiz # 2