The Mole.


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The Mole The Mole It's not a see, a machine for burrowing burrows, a tunneling creature, or a spot of skin pigmentation - it's - a unit of estimation containing around 6.02 x 10 23 particles. Scientists made the mole as their own particular tallying unit. The Compound Mole 1mole of iotas is
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The Mole

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The Mole

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It\'s not a see, a machine for burrowing passages, a tunneling creature, or a spot of skin pigmentation - it\'s - a unit of estimation containing around 6.02 x 10 23 particles. Scientific experts made the mole as their own particular checking unit. The Chemical Mole

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1mole of molecules is 602,214,199,000,000,000,000,000 particles!

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Avogadro’s Number 6.02 x 10 23 particles/mole

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Just How Big is a Mole? Enough soda pop jars to cover the world\'s surface to a profundity of more than 200 miles. In the event that you had Avogadro\'s number of unpopped popcorn bits, and spread them over the United States of America, the nation would be secured in popcorn to a profundity of more than 9 miles. In the event that we had the capacity tally iotas at the rate of 10 million every second, it would take around 2 billion years to include the particles one mole.

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Moles of Elements 1 mole component = (nuclear mass) in grams of that component 1 mole C = 12.011 g C 1 mole B = 10.81 g B 1 mole Cu = 63.55 g Cu

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Avogadro’s Number as Conversion Factor 6.02 x 10 23 particles 1 mole or 1 mole 6.02 x 10 23 particles Note that a molecule could be an iota OR an atom!

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Equation to discover number of particles! 6.02 x 10 23 particles 1 mole number of moles x = number of particles Example: what number particles of sucrose are in 3.50 moles? 6.02 x 10 23 atoms 1 mole 3.50 mol sucrose x = 2.11 x 10 24 particles of sucrose

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Equation to discover number of moles! number of particles x = number of moles 1 mole 6.02 x 10 23 particles Example: what number moles are contained in 4.50 x 10 24 molecules of zinc? 1 mole Zn 6.02 x 10 23 particles 4.50 x 10 24 iotas of zinc x = 7.48 mol Zn

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Learning Check 1. Number of particles in 0.500 mole of Al a) 500 Al molecules b) 6.02 x 10 23 Al iotas c) 3.01 x 10 23 Al molecules 2.Number of moles of S in 1.8 x 10 24 S particles a) 1.0 mole S iotas b) 3.0 mole S molecules c) 1.1 x 10 48 mole S iotas

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Molar Mass 11.2

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Molar Mass The Mass of 1 mole (in grams ) Molar mass is equivalent to the nuclear mass (look on the occasional table ) 1 mole of C atoms = 12.01 g 1 mole of Au particles = 196.97g 1 mole of Cu molecules = 63.55 g Always round to the hundredths spot.

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Molar Mass of Molecules and Compounds Mass in grams of 1 mole meet numerically to the entirety of the nuclear masses 1 mole of CaCl 2 Ca = 40.08 Cl = 35.453 x 2 70.906 40.08 + 70.906 = 110.99 g/mol

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Molar Mass of Molecules and Compounds 1 mole of N 2 O 4 N = 14.007 O = 15.999 x 2 x 4 28.014 63.996 28.014 + 63.996 = 92.01 g/mol

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Where will you locate the quantity of grams per mole? Mole to Mass Conversion Number of moles x number of grams = mass 1 mole Example: Calculate the mass in grams of 0.0450 moles of chromium. 0.0450 moles Cr x 52.0 g Cr = 2.34 g Cr 1 mol Cr The Periodic Table Look at your Periodic Table

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Where will you locate the quantity of grams per mole? Mass to Mole Conversion Mass x 1 mole = number of moles number of grams Example: what number moles of calcium arrive in 525 grams of calcium? 525 g Ca x 1 mol Ca = 13.1 mol Ca 40.1 g Ca The Periodic Table Look at your Periodic Table

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Mass to Atoms Conversion Two stages: Use the mass to moles transformation Now figure the quantity of iotas utilizing the moles to particles mathematical statement.

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Example: what number molecules of gold are in an immaculate gold chunk having a mass of 25.0 g? Step 1 Mass x 1 mole = number of moles number of grams 25.0 g Au x 1 mol Au = 0.13 mol Au 197.0 g Au Round to the hundredths place!!!!!!

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6.02 x 10 23 particles 1 mole number of moles x = number of particles Step 2 0.13 mol Au x = 7.83 x 10 22 iotas Au 6.02 x 10 23 molecules 1 mole Round to the hundredths place!!!!!!

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Atoms/Molecules and Grams what number particles of Cu are available in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X 10 23 iotas Cu 63.5 g Cu 1 mol Cu = 3.4 X 10 23 particles Cu

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Learning Check! What number of molecules of K are available in 78.4 g of K? 78.4 g K 1 mol K 6.02 X 10 23 iotas K 39.1 g K 1 mol K = 1.20 X 10 24 particles K

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Empirical and Molecular Formulas 11.4

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Percent Composition Mass of component x 100 = percent by Mass of compound Example: Determine the percent piece of NaHCO 3 . Focus the mass of every component first! %mass component = mass of component in 1 mol compound x 100 molar mass of compound Percent Na = 22.99 g Na x 100 = 27.37% Na 84.01 g NaHCO 3 You need to finish this progression for every component in the compound!!!!

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Determine the Percent Composition from the Formula C 2 H 5 OH Divide the mass of every component by the molar mass of the compound and increase by 100%

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Types of Formulas Empirical Formula The recipe of an aggravate that communicates the littlest entire number proportion of the iotas present. Sub-atomic Formula The equation that expresses the genuine number of every sort of iota found in one particle of the compound.

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Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Convert the rates to grams by expecting you have 100 g of the compound Step can be skipped if given masses

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Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Convert the grams to moles

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Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Divide each by the littlest number of moles

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Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen If any of the proportions is not an entire number, reproduce every one of the proportions by a variable to make it an entire number If proportion is ?.5 then duplicate by 2; if ?.33 or ?.67 then increase by 3; if ?.25 or ?.75 then duplicate by 4 Multiply every one of the Ratios by 3 Because C is 1.3

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Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen Use the proportions as the subscripts in the observational equation C 4 H 6 O 3

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Molecular Formulas The atomic recipe is a various of the exact equation To focus the sub-atomic recipe you have to know the experimental equation and the molar mass of the compound

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Determine the Molecular Formula of Benzopyrene on the off chance that it has a molar mass of 252 g and an exact recipe of C 5 H 3 Divide the given molar mass of the compound by the molar mass of the exact recipe Round to the closest entire number

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Determine the Molecular Formula of Benzopyrene in the event that it has a molar mass of 252 g and an exact equation of C 5 H 3 Multiply the experimental recipe by the ascertained component to give the sub-atomic equation (C 5 H 3 ) 4 = C 20 H 12

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