#  The Materials:  Pick up a packet and assessment plan.  Paper, pencil, scientific calculator, periodic table  The Plan:  Learn about 3 definitions.

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 The Materials:  Pick up a packet and assessment plan.  Paper, pencil, scientific calculator, periodic table  The Plan:  Learn about 3 definitions of a mole  Solve dimensional analysis problems  HOMEWORK:  See your calendar!

 The mole is a unit of measurement used in chemistry.  The unit can be defined in multiple ways. (We’ll learn 3 today.)  In its simplest terms, it represents a specific number.  Dozen = what number?  Pair = what number?  Baker’s dozen = what number?  Mole = 6.022 x 10 23

 Just like a dozen means 12 of anything... 6.022x10 23 of anything equals of mole.  Consider the size of 6.022x10 23. Is it large or small?  Would you commonly use 6.022x10 23 with large things or small things?

 In chemistry, we are often dealing with VERY SMALL things.  Atoms are SUBmicroscopic. In order to have an amount large enough with which to really interact, we need quite a few atoms. 6.022x10 23 is the number that Amadeo Avogadro chose in his lab using carbon.

 6.022x10 23  My pet mole is named Avogadro.

 Counting  Weighing  Amount of Space Needed  Let’s represent our 3 mole definitions in a graphic organizer.

 A compound is a collection of atoms.  To calculate the mass of one mole of a compound, you’d need to add up the mass of all the atoms. This is called the MOLAR MASS.  Example: 1 mole CH 4 = ______ g CH 4  1 C = 12.011 g C  4 H = 4(1.0079 g H)  Total = 16.04 g/mol

 Calculate the molar mass of sulfur dioxide, a gas produced when sulfur- containing fuels are burned.  SO 2  S = 32.07 g  O = 2(16.00 g)  Total = 64.07 g/mol  Can also be expressed as 1 mol SO 2 = 64.07 g

POLYVINYL CHLORIDE, CALLED PVC, WHICH IS WIDELY USED FOR FLOOR COVERINGS (“VINYL”) AS WELL AS FOR PLASTIC PIPES IN PLUMBING SYSTEMS, IS MADE FORM A MOLECULE WITH THE FORMULA C 2 H 3 CL. CALCULATE THE MOLAR MASS OF THIS SUBSTANCE. RECORD THE ANSWER TO TWO DECIMAL PLACES.

 Polyvinyl chloride, called PVC, which is widely used for floor coverings (“vinyl”) as well as for plastic pipes in plumbing systems, is made form a molecule with the formula C 2 H 3 Cl. Calculate the molar mass of this substance.  62.49 g/mol

 Using the Mole Concept to Calculate  Page 182-7 EX 6.3, EX 6.4, EX. 6.6

 Aluminum (Al), a metal with a high strength- to-weight ratio and a high resistance to corrosion, is often used for structures such as high-quality bicycle frames. Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.  How many calculations is this prompt asking me to carry out?

 Aluminum (Al), a metal with a high strength- to-weight ratio and a high resistance to corrosion, is often used for structures such as high-quality bicycle frames. Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.  Concentrate on one calculation at a time

A.) 3.36X10 -25 B.) 1.22X10 23 C.) 1.22X10 20 D.) none of these

 Calcium carbonate, CaCO 3 (also called calcite), is the principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals. a) Calculate the molar mass of calcium carbonate. b) A certain sample of calcium carbonate contains 4.86 mol. What is the mass in grams of this sample?

A.) 2.112 B.) 42,600 C.) 2.520 D.) none of these

 If converting between the units of moles and liters of a gas, what conversion factor is needed?  1 mole = ________________  1 mole = 22.4 Liters  Example: During cellular respiration, a cell releases 2.1 mol of O 2 gas. What volume is needed to hold that gas?

 Juglone, a dye known for centuries, is produced from the husks of black walnuts. The formula for juglone is C 10 H 6 O 3.  A sample of 1.56 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent?  How many atoms of carbon are present in the sample?

 On a clean sheet of paper, show your work for 10-2 Practice Problems (1-3, 12-14, 17, 18, 23- 25).

 Last week, you created a poster using chalk.  How many grams of chalk did you use on the poster?  How many moles of chalk did you use?  How many formula units of chalk did you use? How many atoms of Ca did you use? How many atoms of C did you use? How many atoms of O did you use?

 I marked mistakes on your papers, but I did not correct them.  Correct and complete 1-8 tonight for homework.  Let’s solve 9 and 10 together now.

 The relative amounts of each element in a compound are expressed in percent composition. AKA: percent by mass of each element  % of element = grams of element X 100 grams of compound

 Carvone is a substance that occurs in two forms, both of which have the same molecular formula (C 10 H 14 O) and molar mass. One type of carvone give caraway seeds their characteristic smell; the other is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone.

 Individually, calculate the mass percents of the first three compounds on the worksheet.  Let’s check your work.

 Mass % is easily related to chemical formulas  The relationship of moles to chemical formulas requires a little more thought...

 Empirical formula: lowest whole number ratio of the atoms of the elements in a compound  Ionic compounds are criss-crossed and then reduced. Ionic compound formulas are nearly always empirical formulas.  Empirical formula doesn’t have to be the same as the actual molecular formula of the compound.  Remember, “molecule” means covalent compound. Covalent compounds weren’t always in a reduced form.

 CH 2 O  Calculate the mass percent of each element. C = 40% H = 6.7% O = 53.3%  Isn’t this the reduced formula for: C 3 H 6 O 3, C 4 H 8 O 4, C 5 H 10 O 5, & C 6 H 12 O 6  Row 1: Calculate the mass percent of C 3 H 6 O 3.  Row 2: Calculate the mass percent of C 4 H 8 O 4.  Row 3: Calculate the mass percent of C 5 H 10 O 5.  Row 4: Calculate the mass percent of C 6 H 12 O 6.  The percentages (ratio) of the elements is the same with all of these formulas because they share the same reduced form.

 Molecular formula: actual formula for the compound which gives the composition of the molecule  Glucose shares an empirical formula with many compounds, but it has its molecular formula all to itself.  6(CH 2 O) = glucose  C 6 H 12 O 6

 When an unknown compound is found, instruments can tell scientists the mass percent composition of the compound.  Calculations are required to convert that series of percentages into a chemical formula.  We start by converting to the empirical formula.

 White powder found in the hallway  Mass spectroscopy instrument used to analyze the unknown powder  Data from the instrument: 40.9 % Carbon 4.58% Hydrogen 54.5% Oxygen Molar mass of 180 grams/mole  Use these %s, formula knowledge, & mole knowledge to figure out the chemical formula of the powder

 Use a simple rhyme!  % to gram  Gram to mole  Divide by the smallest  Multiply ‘til whole.

 An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound.  % to gram = IS DONE FOR YOU  Gram to mole = dimensional analysis  Let’s carry out the calculation on the board.

 A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g of lead, 0.00672 g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.

 The most common form of nylon is 63.68% carbon, 12.38% nitrogen, 9.80% hydrogen, and 14.4% oxygen. Calculate the empirical formula for nylon.

 Molecular Formula= n(empirical formula) (Remember 6(CH 2 O) = glucose)  We know how to calculate the empirical formula, but how do we know what number to multiply it by? n= molecular formula mass/molar mass of empirical

 A white powder is analyzed and found to have an empirical formula of P 2 O 5. The compound has a molar mass of 283.88 g/mol. What is the compound’s molecular formula?

 A compound used as an additive for gasoline to help percent engine knock shows the following percentage composition:  71.65% Cl  24.27% C  4.07% H The molar mass is known to be 98.96 g. Determine the empirical formula and the molecular formula for this compound.

 Page 208 4-6

 On a clean sheet of paper, work the following problems individually:  Calculate the number of oxygen atoms in 3.5 g aluminum sulfate.  Calculate the number of molecules of O 2 gas in 2.5 L of O 2.  Calculate the number of grams of iron that contain the same number of atoms as 2.24 g of cobalt. (pg 214 #57)  Quiz answers are coming.

 The final product in protein metabolism is urea. Urea contains 20.00% C, 6.73% H, 46.65% N, and 26.64% O. The molar mass of urea is 60.07g/mol. Calculate the empirical formula and molecular formula.

 You’ll have to READ (not skim) the lab to be successful.  Steps 1-5 are probably unnecessary.  Steps 6-11 are vital. Substitute hot plate for Bunsen burner.  Notice the “Observations and Data” area on the back. You’ll need those measurements.  The calculations are described to you in each question. Just follow the directions.  You’ll need to answer the “Questions for Discussion,” too.

1) Subtract. 2) % water lost =[water lost/hydrate mass]100 3) % water in hydrate = [5(water molar mass)/total molar mass] 100 4) Subtract. %water in hydrate - %water lost from your hydrate 5) % error = [#4 answer/#3 answer]100

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