Chemistry Bond

Chemistry Review

Chemical Quantities

The mole

• amount of someting measured by: -counting, by mass, by volume -definir cantidad basica de materia -siempre tenemos el mismo mole

• chemists also need a convenient method for counting accurately the # of atoms, mol, or formula units

• SI unit measure amt of substance

• # of rep part (carbon atoms) is usually 12g of pure carbon 12

• one mole= 6.02x10 23 rep part(avogadros #) -1Kgof Fe-->1mol Fe

• term rep part refers to species present in substance: atoms, mol, formula units, electrons or ions

Mole/Particle Rel.

1 mol= 6.02x10 23 part particle to mol mol to particle 1 mol/6.02x10 23 6.02x10 23/ 1 mol

Ex. 1 Calculate # of rep part:

a. 2.50 mol of iron-->1 mol

1mol Fe-->6.02x10 23 at Fe

2.50 [6.02x10 23 at Fe/ 1 mol Fe] 1.51 x 10 24 at Fe

Ex. 2 Find # of moles

a. 1.51x 10 24 atom of Ca--># of mol

1mol of Ca--> 6.02 x 10 23 at Ca

1.51x10 24 [1mol Ca/6.02x10 23 at Ca] 2.51 mol Ca

Mol/ Mass Rel.

• Molar mass: mass in grams of one mole of any substance -molar mass of any element is numerically equal to its atomic mass and has the unit of g/mol

Mol to Mass Mass to Mol

# of mol=(grams/mol) mass=(1mol/grams)

Ex. 3 Dtermine the mass in grams of 3.50 mol of copper

Molar mass= 63.5g/mol

3.50 mol Cu[63.5/1mol Cu] 222g Cu

Ex.4 Determine the # of moles 237 g of copper

237g Cu[1mol/63.5 g Cu] 3.73 mol Cu

• because a formula tells you what atoms or ions are present in a compound tou are able to determine the molar mass of the compound

• molar mass of comp = the + of molar mass of the # of atoms in formula

Ex. 5 calculate molar mass of following comp

a. KBr

1K=39.1 1Br=80.0 molar mass KBr 119.1g/mol

Ex. 6 Dtermine the mass in g of 3.75mol of potassium chlorate

K(ClO3) 1K=39.1g.mol 1Cl=35.5 3O=48.0

Molar mass 122.6 g/mol

3.75mol KClO3[ 122.6g/1mol] 460g KClO3

Ex. 7 Determine the # of mole in 574.5g of iron (III) oxide

Fe2O3 2Fe=11.6 3O=48.0

Molar mass 160g/mol

574.5mol Fe2O3[1mol Fe2O3/ 160g] 3.59 mol Fe2O3

Mass/ Particle Rel.

• to convert a given mass to # of part Given mass-->moles-->part

• to find mass of # of part # of part-->moles-->mass

Ex.8 Determine mass in grams of 1.22x1022 atoms of silver

1.22x10 22[ 1mol/6.02 x 10 23Ag] 0.02mol Ag

0.020 mol Ag[ 108gAg/1mol] 2.16g Ag

Mole/Volume Rel

• the particles maijng up diff gases varies greatly in size

• according to the kinetic theory the particles in a gas are usually far enough apart that size has a neglible influence on volume occupied by a fixed # of particles -size doesnt matter for gases- particle behavior is independent , individual atom or molecules and do not influence, force of attraction is neglibe -lo que determina el vol del gas es la distancia de partiulas

• 1000 paticles of K occupies the same volume as 1000 He -does not depend on size but on motion

Avogadros Hypotesis

• equal volume og fases at same temperature and pressure contain equal # of particles

• misma presion y temp- mismo # de particulas

• molar volume for a gas is the volume that one mole occupies at 0.00C and 101.3KPa(1.00 atom)

• these conditions of temp and pressure are known as standard temp and pressure (STP)

• avogadro showed that one mole of any gas will occupy a volume of 22.4Lat STP -todo gas bajo presion y temperatura & mismo # de particulas si definimos STP vamos a tener 1 mol de cualquier gas -ej 1 mol de Br at 0.00 C y 101.3K no existe porque es un liquido y las particulas interactuan

Ex 10 Calculate the volume thta 0.881 mol of gas at STP will occupy

1 mol= 22.4L 0.881[22.4L/1mol] 20.0L at STP

Ex Calculate the volume that 2.0 Kg of methane gas (CH4) will occupy at STP

2.0x10 3 gCH4[1mol/16.04g CH4]= 125mol CH4

125mol CH4[22.4L/1mol]= 2.80x10 3 L

Percent Composition and Chemical Formulas

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