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Delve into the fascinating world of chemistry with this comprehensive exploration of the mole concept. We’ll learn how to calculate moles and molecules for various compounds, analyze the quantities of nitrate ions in different scenarios, and even engage in light-hearted scientific banter. From examining Avogadro’s law to comparing bromine content in different compounds, we cover essential calculations that are vital for understanding chemical reactions. Join us for a fun and educational journey through molecular conversions!
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Warmup (5 minutes) 1.0 x10-14 • (1.0 x 109)/(1.0 x 1023) = 2) How many moles are in 40.96 grams of nitrate ions? ( mole NO3-) 1 40.96 g NO3- ( g NO3-) 62.01 Take out your PT and Molar Highway = 0.6605 mole NO3-
a tiny molelike creature with wings which drinks the blood of anyone who doesn't remember when Mole Day is the continual reverence of moles a word describing the Mole Day songs which are played over the loudspeaker moleskito moleism moleodic What do you get when you have a bunch of moles acting like idiots? What kept Avogadro in bed for two months? A bunch of Moleasses Moleonucleosis
Ex 1. How many molecules of CO2are there in 1.50 moles of CO2 ? 1.50 mole CO2 = 9.03x 1023molecules of CO2 (molecules CO2 ) 6.02 x 1023 1 (mole CO2 )
EX 2: How many moles of Na are there in 4.50 x 105 atoms of Na? 1 mole Na 4.50 x105 atoms Na ( ) ( ) = 7.48x10-19 moles Na 6.02 x 1023 atoms Na
The Molar Highway Finding Various Quantities using The Mole Mass compound in sample # molecules compound in sample 1 mole of a compound Need: molar mass of compound Need: 1 mole = 6.02 x 1023 particles
3. Two scientists engage in a weight-lifting competition to see who is stronger (HOT!!!) Avagadro can bench press 7.85 x 1025 atoms carbon Einstein can bench press 6.78 x 1024 atoms einsteinium. Who can bench press the greatest mass? ( g C) ( mole C) 1 12.01 7.85 x 1025 atoms C ( atoms C) ( mole C) 1 6.02 x 1023 = 1570 or 1.57 x 103 g C ( mole Es) 1 ( g Es) 252 6.78 x 1024 atoms Es ( atoms Es) 6.02 x 1023 ( mole Es) 1 = 2840 or 2.84 x 103 g Es
The Molar Highway Finding Various Quantities using The Mole # atoms or ions in sample Need: # of that atom or ion in compound # molecules compound in sample Mass compound in sample 1 mole of a compound Need: molar mass of compound Need: 1 mole = 6.02 x 1023 particles
4. Are there more nitrate ions in 4.00 x 1014 molecules Al(NO3)3OR 2.00 x 1014 molecules of KNO3? ( NO3-ions) 3 4.00 x 1014 molecules Al(NO3)3 molecule Al(NO3)3 1 = 1.20 x 1015 NO3-ions ( NO3-ion) 1 2.00 x 1014 molecules of KNO3 1 molecule KNO3 = 2.00 x 1014 NO3-ions
The Molar Highway Finding Various Quantities using The Mole # atoms or ions in sample moles of element or ion in the compound Need: # of that atom or ion in compound Need: # of that atom or ion in compound # molecules compound in sample Mass compound in sample 1 mole of a compound Need: molar mass of compound Need: 1 mole = 6.02 x 1023 particles
5. A sample of 16.2 grams of MgBr2and a sample of 21.0 grams of AgBr are arguing over who contains more moles of the element bromine. Who is correct? ( mole Br) ( mole MgBr2) 1 2 16.2 g MgBr2 ( mole MgBr2) 1 184.11 ( g MgBr2) = 0.176 mole Br ( mole AgBr ) 1 ( mole Br) 1 21.0 g AgBr ( g AgBr) ( mole AgBr) 1 187.77 = 0.112 mole Br
6. A sample of 3.9 x 1019 strontium chloride molecules contains how many milligrams of chlorine? 3.9 x 1019molec. SrCl2 (2atoms Cl ) (1mole Cl) ( 35.45g Cl) ( 103mg Cl) (1molecSrCl2) (6.02 x 1023atoms Cl) (1mole Cl)( 1g Cl) 4.6 mg chlorine
7. The chemical tryptoline is a serotonin reuptake inhibitor. 76.71%C 7.02%H, 16.27%N. a. Determine the empirical formula, which is also the molecular formula. Answer (work shown on board): C11H12N2 b. How many atoms of hydrogen would be found in 4.5 x 10-6g of tryptoline? (done on board) answer: 1.9 x 1017 atoms H