The Mole Concept Avagadro’s Number = 6.022 x 1023 Objectives: To state the value of Avagadro’s number: 6.02 x1023. To state the mass of Avagadro’s number of atoms for any element by referring to the periodic table. To relate the moles of a substance to the number of particles.

Counting Atoms Chemistry is a quantitative science - we need a "counting unit." The MOLE 1 mole is the amount of substance that contains as many particles (atoms or molecules) as there are in 12.0 g of C-12.

The Mole is Developed Carbon Atoms Hydrogen Atoms Mass Ratio Number Mass (amu) Number Mass (amu) Mass carbon / Mass hydrogen 12 1 12 amu = 12 1 amu 1 24 [2 x 12] 2 [2 x 1] 24 amu = 12 2 amu 1 120 [10 x 12] 10 [10 x 1] 120 amu = 12 10 amu 1 600 [50 x 12] 50 [50 x 1] 600 amu = 12 50 amu 1 (6.02 x 1023) x (12) Avogadro’s number (6.02 x 1023) x (1) (6.02 x 1023) x (12) = 12 (6.02 x 1023) x (1) 1 Avogadro’s number

Amedeo Avogadro (1776 – 1856) Particles in a Mole A more precise value of Avagadro’s number = 6.022045 E23. How could you determine the number of sugar crystals in a bowl of sugar? Is it possible to get a reasonably close estimate? ? quadrillions thousands trillions billions millions 1 mole = 602213673600000000000000 or 6.022 x 1023 There is Avogadro's number of particles in a mole of any substance.

Careers in Chemistry - Philosopher Q: How much is a mole? A: A mole is a quantity used by chemists to count atoms and molecules. A mole of something is equal to 6.02 x 1023 “somethings.” 1 mole = 602 200 000 000 000 000 000 000 Q: Can you give me an example to put that number in perspective? A: A computer that can count 10,000,000 atoms per second would take 2,000,000,000 years to count 1 mole of a substance.

Counting to 1 Mole Is that right? A computer counting 10 million atoms every second would need to count for 2 billion years to count just a single mole. Lets look at the mathematics. Therefore 1 year has 31,536,000 seconds or 3.1536 x 107 sec. A computer counting 10,000,000 atoms every second could count 3.153 x 1014 atoms every year. Finally, 6.02 x 1023 atoms divided by 3.1536 x 1014 atoms every year equals 1,908,929,477 years or approximately 2 billion years!

How Big is a Mole? One mole of marbles would cover the entire Earth (oceans included) for a depth of three miles. One mole of $100 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times. 6.02 x 1023 softballs = volume of Earth 6.02 x 1023 Olympic shot puts = mass of Earth 6.02 x 1023 atoms H laid side by side would encircle Earth ~1,000,000 times It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

Avogadro’s Number A MOLE of any substance contains as many elementary units (atoms and molecules) as the number of atoms in 12 g of the isotope of carbon-12. This number is called AVOGADRO’s number NA = 6.02 x 1023 particles/mol The mass of one mole of a substance is called MOLAR MASS symbolized by MM Units of MM are g/mol Examples H2 hydrogen 2.02 g/mol He helium 4.0 g/mol N2 nitrogen 28.0 g/mol O2 oxygen 32.0 g/mol CO2 carbon dioxide 44.0 g/mol

1 Mole of Particles

Molar Mass Objectives: To calculate the molar mass of a substance given its chemical formula.

Molecular Weight and Molar Mass Molecular weight is the sum of atomic weights of all atoms in the molecule. example: NaCl has a molecular weight of 58.5 a.m.u. this is composed of a single molecule of NaCl Molar mass = molecular weight in grams. example: NaCl has a molar mass of 58.5 grams this is composed of a 6.02 x1023 molecules of NaCl

The Molar Mass and Number of Particles in One-Mole Quantities Substance Molar Mass Number of Particles in One Mole Carbon (C) 12.0 g 6.02 x 1023 C atoms Sodium (Na) 23.0 g 6.02 x 1023 Na atoms Iron (Fe) 55.9 g 6.02 x 1023 Fe atoms NaF (preventative 42.0 g 6.02 x 1023 NaF formula units for dental cavities) CaCO3 (antacid) 100.1 g 6.02 x 1023 CaCO3 formula units C6H12O6 (glucose) 180.0 g 6.02 x 1023 glucose molecules C8H10N4O2 (caffeine) 194.0 g 6.02 x 1023 caffeine molecules

Chemical Equations

Molar Mass and Mole Calculations n M

Molar Mass Molar mass is the mass in grams of one mole of particles (atoms, ions, molecules, formula units). Equal to the numerical value of the average atomic mass (get from periodic table) 1 mole of C atoms = 12.0 g 1 mole of Mg atoms = 24.3 g 1 mole of Cu atoms = 63.5 g So, Molar Mass is the atomic mass expressed in units grams/mole (g/mol)

The number is the same, just change the unit. Molar Mass Chlorine has an atomic mass of 35.453 amu. So what is its molar mass? 35.453 grams per mole The number is the same, just change the unit.

Molar Mass How do we find the molar mass of a compound? Find the molar mass of each element in the compound Determine how many of each element are in the compound. Multiply the number of each element by its molar mass. Add these mumbers.

Molar Mass Aluminum Bromide AlBr3 In one mole of Aluminum Bromide, there is 1 mole of aluminum atoms and 3 moles of bromine atoms.

Molar Mass To find the molar mass of AlBr3, we do the following: 1 mol Al x 26.9815 g/mol Al = 26.9815 g Al 3 mol Br x 79.904 g/mol Br = 239.712 g Br 26.9815 g Al + 239.712 g Br = 266.6935 g AlBr3 per mole or 266.6935 g/mol AlBr3

Mole Calculations Where, n is the # of moles (mol) m is the mass (g) n = m / M M = m/n m = n x M m n M Where, n is the # of moles (mol) m is the mass (g) M is the molar mass (g/mol)

Relationship Between Moles, Mass and Particles

Mole Calculations What is the mass of 3.0 moles of chromium (III) iodide? CrI3

Mole and # of Particles 6.02x1023 molecules in a mole

A mole contains 6.02 x 1023 particles. How many particles are in 2.0 moles of particles? 2 mol x 6.02 x 1023 particles/mol = 12.04 x 1023 particles = 1.2 x 1024 particles When changing moles to number of particles simply Multiply the number of moles of particles by 6.02 x 1023 particles/mol

A mole contains 6.02 x 1023 particles. How many particles are in 4.0 moles of particles? 4 mol x 6.02 x 1023 particles/mol = 24.08 x 1023 particles = 2.4 x 1024 particles How many particles are in 0.35 moles of particles? 0.35 mol x 6.02 x 1023 particles/mol = 2.1 x 1023 particles

A mole contains 6.02 x 1023 particles. How many moles of particles are in 1.02 x 1025 particles = 1.02 x 1025 particles / 6.02 x 1023 particles/mole = 16.9 moles To change # of particles to moles simply Divide the # of particles by 6.02 x 1023 particles/mole

A mole contains 6.02 x 1023 particles. How many moles of particles are in 4.74 x 1024 particles = 4.74 x 1024 particles / 6.02 x 1023 particles/mole = 7.873 moles = 7.87mol 3.24 x 1022 particles = 3.24 x 1022 particles / 6.02 x 1023 particles/mole = 0.05382 moles = 5.38x10-2 moles

A mole contains 6.02 x 1023 particles. How many particles are in 45.6 moles of particles? 44.6 moles x 6.02 x 1023 particles/mol = 24.08 x 1023 particles = 2.68 x 1025 particles How many moles are in 4.5 x 1021 particles? 4.5 x 1021 particles / 6.02 x 1023 particles/mol = 0.0075 mol

How many molecules are there in 4.78 g of chlorine dioxide? ClO2 a) Find the number of moles of ClO2 first n = m / M = 4.78 g / 67 g/mol =0.0713 mol b) Now convert # of mol into particles = 0.0713 mol x 6.02 x 1023 particles/mol =4.3 x 1022 molecules of ClO2

How many molecules are there in 2.3 g of aluminum bromide? AlBr3 a) Find the number of moles of AlBr3 n = m / M = 2.3 g / 267 g/mol = 0.0086 mol of AlBr3 b) Now convert # of mol into particles = 0.0086 mol x 6.02 x 1023 particles/mol = 5.2 x 1021 molecules of AlBr3

What is the mass of 8.2 x 1021 molecules of C2H5? a) First find the # of moles. 8.2 x 1021 molecules / 6.02 x 1023 molecules/mol = 0.01362 mol b) Convert from moles to mass by multiplying the number of moles by the molar mass of C2H5. MM = (12 x 2 + 5 x 1 ) = 29 g/mol = (0.01362mol)x 29 g/mol = 0.395 g

What is the mass of 8.2 x 1021 molecules of C2H5? a) First find the # of moles. 8.2 x 1021 molecules / 6.02 x 1023 molecules/mol b) Convert from moles to mass by multiplying the number of moles by the molar mass of C2H5. MM = (12 x 2 + 5 x 1 ) = 29 g/mol = (8.2 x 1021 molecules / 6.02 x 1023) x 29 g/mol = 0.395 g

What is the mass of 3.7 x 1024 molecules of C4H10? First find the # of moles, then find the mass. 3.7 x 1024 molecules / 6.02 x 1023 molecules/mol (= 6.15mol) b) Convert from moles to mass by multiplying the number of moles by the molar mass of C4H10. MM = (12 x 4 + 10 x 1 ) = 58 g/mol 1 step: (3.7 x 1024 molecules / 6.02 x 1023) x 58 g/mol = 356 g = 3.6x102g

Converting Particles, Moles and Mass 1. Find the mass of each of the following: a) 7.8 x 1026 molecules of N2O4 b) 5.6 moles of C2H5OH c) 1.56 x 1021 molecules of C3H6 d) 7.4 moles of ClS3 e) 3.6 x 1025 molecules of CO2 2. Find the number of particles in each of the following: a) 75 g of NF5 b) 0.0034 moles of C3H7OH c) 2.8 g of C5H10 d) 2.5 moles of FBr3 e) 435 g of CO2