CFISD ChemistryCFISD ChemistryEOC PrepEOC Prep

1 Dozen = 12 of something1 Dozen = 12 of something

1 ream = 500 sheets of paper1 ream = 500 sheets of paper

1 yard1 cubic yard (y1 cubic yard (y33) = 27 ft) = 27 ft33 of stuff of stuff

1 Stone= 14 pounds (140 lbs =10 stones)1 Stone= 14 pounds (140 lbs =10 stones)

•1 Liter = 1000 mL1 Liter = 1000 mL

•1 gram = 1000 mg1 gram = 1000 mg

•1 m = 100 cm1 m = 100 cm

Metric Prefixes

Tera 10 12 T

Giga 10 9 G

Mega 10 6 M

Kilo 10 3 k

Hecto 10 2 h

Deca 10 1 da

Deci 10 -1 d

Centi 10 -2 c

Milli 10 -3 m

Micro 10 -6

Nano 10 -9 n

Pico 10 -12 p

Litersgramsmolesmetersseconds

1 mole = 6.02 X 10 1 mole = 6.02 X 10 2323 particles particles

MoleculesMoleculesAtomsAtomsFormula Formula

UnitsUnits IonsIons

6.02 x 1023 particles

Smallest particle of an element that maintains Smallest particle of an element that maintains the identity (#protons) of that element.the identity (#protons) of that element.

C CC1212

66

Carbon-12Carbon-12

1s22s22p2 or [He]2s22p2CC

particles that contain one or more covalently bonded particles that contain one or more covalently bonded (nonmetal) atoms.(nonmetal) atoms.

O=C=OCO2 moleculeHydrogen

molecule (diatomic)

HH

• the lowest ratio of cations (metals) to anions (nonmetals) in a the lowest ratio of cations (metals) to anions (nonmetals) in a neutral compound.neutral compound.

NaCl

BaClBaCl22 1 Ba 1 Ba+2+2 + 2 Cl + 2 Cl-1-1

• formed when an atom gains or loses an electron(s)formed when an atom gains or loses an electron(s) anion (-) cation (+) polyatomic ion ( + or -)anion (-) cation (+) polyatomic ion ( + or -)•combine to form ionic compounds (metals and nonmetals)combine to form ionic compounds (metals and nonmetals)

BaBa33(PO(PO44))22 33 Ba Ba+2+2 + + 22 PO PO44-3-3

Formula UnitsFormula Units

Substance Name atom Molecule Formula Unit

ion

C6H12O6

Al2(SO4)3

Zn

FeCl3

NO3-1

Br2

NO2

• Name each substance and identify the types of particle present in the sample by placing an X in the corresponding box.

• Pause the presentation to give yourself time to work prior to moving on.

Substance Name atom Molecule Formula Unit

ion

C6H12O6 Sugar X

Al2(SO4)3 Aluminum sulfateX

Zn zinc X

FeCl3 Iron (III) chloride X

NO3-1 nitrate x

Br2 Bromine X

NO2 Nitrogen dioxide X

We know…We know…•particles are too small to be counted particles are too small to be counted by hand….by hand….•we can measure mass and we can measure mass and volume…….volume…….•the relationships between moles and the relationships between moles and mass an moles and volume.mass an moles and volume.

6.02 x 1023 particles

Molar mass

Equivalent amounts of SiliconEquivalent amounts of Silicon28.1 g Si = 1 mole Si = 6.02 x 10 28.1 g Si = 1 mole Si = 6.02 x 10 2323 atoms Si atoms Si 6.02 x 1023 particles

Molar Mass

2 H 2 H 2(2(1.0g1.0g)= 2.0 g)= 2.0 g

*Since there are 2 hydrogen atoms in each molecule, we could *Since there are 2 hydrogen atoms in each molecule, we could determine the number of hydrogen atoms by.. determine the number of hydrogen atoms by.. 2(6.02 x 102(6.02 x 102323) = 1.20 x 10) = 1.20 x 102424 hydrogen atoms hydrogen atoms

2.0g H2 = 1 mole H2 = 6.02 x 1023 H2 molecules 2.0g H2 = 1 mole H2 = 6.02 x 1023 H2 molecules

1 Mg1 Mg 1(24.3g) = 24.3 g 1(24.3g) = 24.3 g2 02 02(16.0g)= 32.0 g2(16.0g)= 32.0 g2 H2 H2(1.0g) = 2(1.0g) = 2.0 g 2.0 g Mg(OH)Mg(OH)22 = 58.3 g = 58.3 g

**Since there are 2 hydroxide ions in each formula unit, we Since there are 2 hydroxide ions in each formula unit, we could determine the number of hydroxide ions by.. could determine the number of hydroxide ions by.. 2(6.02 x 102(6.02 x 102323) = 1.20 x 10) = 1.20 x 102424 hydroxide ions hydroxide ions..

58.3g Mg (0H)58.3g Mg (0H)22 = 1 mole Mg(0H) = 1 mole Mg(0H)22 = 6.02 x 10 = 6.02 x 102323Mg(0H)Mg(0H)22 formula formula unitsunits

Substance Molar Mass

C6H12O6

Al2(SO4)3

Zn

FeCl3

KNO3

Br2

NO2

• Determine the molar mass of each substance to the 10th place. • Pause the presentation to give yourself time to work prior to moving

on.

Substance Molar Mass

C6H12O6 180 g/mole

Al2(SO4)3 342 g/mole

Zn 65.4 g/ mole

FeCl3 162.2 g/mole

KNO3 101.1 g/mole

Br2 159.8 g/mole

NO2 46.0 g/mole

Identify equivalent relationships for that Identify equivalent relationships for that substancesubstanceWrite ratios using number given in the problem Write ratios using number given in the problem and equivalentsand equivalentsSolve: unit on answer, sig. fig or roundSolve: unit on answer, sig. fig or round

Given Given = = X (?)X (?)Equiv EquivEquiv Equiv

Determine the number ofDetermine the number of moles moles in in 48 g48 g of of hydrogen gas. Express your answer to hydrogen gas. Express your answer to 22 significant significant figures.figures.

48g H48g H22 X moles H X moles H22

2.0g H2 = 1 mole H2 = 6.02 x 1023 H2 molecules 2.0g H2 = 1 mole H2 = 6.02 x 1023 H2 molecules

Given Given 48 g H48 g H22 = = X moles HX moles H22

Equivalents 2.0 g HEquivalents 2.0 g H22 1 mole H 1 mole H22

1 mole x 1 mole x 48 g 48 g HH22= = X moles HX moles H2 2 x 1 molex 1 mole 2.0 g H2.0 g H22 1 mole H 1 mole H22

Answer: X = 24 moles of HAnswer: X = 24 moles of H22

Mole

6.02 x 1023

Molar Mass

6.02 x 1023 particles

Determine the number of hydrogen Determine the number of hydrogen moleculesmolecules in in 48 g48 g of hydrogen gas. Express your answer to of hydrogen gas. Express your answer to 33 significant figures.significant figures.

48g H48g H22 X molecules H X molecules H22

2.0g H2 = 1 mole H2 = 6.02 x 1023 H2 molecules 2.0g H2 = 1 mole H2 = 6.02 x 1023 H2 molecules

48 g H48 g H22 = = X molecules HX molecules H22

2.0 g H2.0 g H22 6.02 x 10 6.02 x 102323 molecules H molecules H22

6.02 x 106.02 x 102323molecules Hmolecules H22 x x 48 g 48 g = = X molecules X molecules x 6.02 x 10 x 6.02 x 102323 molecules molecules 2.0 g 6.02 x 102.0 g 6.02 x 102323 molecules H molecules H22

Answer: X = 1.44 x 10Answer: X = 1.44 x 102525 molecules of H molecules of H22

Determine the Determine the massmass of of 3.6 x 10 3.6 x 10 2424 formula units formula units of of Mg(OH)Mg(OH)22. Express your answer to 2 significant figures.. Express your answer to 2 significant figures.

X g Mg(OH)X g Mg(OH)22 3.6 x 10 3.6 x 102424 fm fm

Given Given 3.6 x 103.6 x 102424 fm fm = = X gX g

Equivalents 6.02 x 10Equivalents 6.02 x 102323 fm 58.3 g fm 58.3 g

calcualtor : X = 348.63 g of Mg(OH)calcualtor : X = 348.63 g of Mg(OH)22

Correct Answer x = 350 g of Mg(OH)Correct Answer x = 350 g of Mg(OH)22

58.3g Mg (0H)58.3g Mg (0H)22 = 1 mole Mg(0H) = 1 mole Mg(0H)22 = 6.02 x 10 = 6.02 x 102323Mg(0H)Mg(0H)22 formula formula unitsunits

58.3 g x 58.3 g x 3.6 x 103.6 x 102424 fm fm = = X gX g x 58.3 g x 58.3 g

6.02 x 106.02 x 102323 fm 58.3 g fm 58.3 g

22.4 L

Molar Mass22.4 L

Mole

6.02 x 1023 particles

= = =

2.0 g H2.0 g H22 = 22.4 L H = 22.4 L H22 = 1 mole H = 1 mole H22 = 6.02 x 10 = 6.02 x 102323 HH22moleculesmolecules

Determine the Determine the volume in litersvolume in liters of of 1.0 g1.0 g of H of H22 gas at gas at STP. Express your answer to 3 significant figures.STP. Express your answer to 3 significant figures.

1.0 g H1.0 g H22 X Liters H X Liters H22

Given Given 1.0 g H1.0 g H22 = = X L HX L H22

Equivalents 2.0 g HEquivalents 2.0 g H22 22.4 L H 22.4 L H22

Answer: X = 11.2 Liters of HAnswer: X = 11.2 Liters of H22

2.0 g H2.0 g H22 = 22.4 L H = 22.4 L H22 = 1 mole H = 1 mole H22 = 6.02 x 10 = 6.02 x 102323 H H22 molecules molecules

22.4 L H22.4 L H22x x 1.0 g 1.0 g HH22= = X LX L H H22 x 22.4 L x 22.4 L

2.0 g H2.0 g H22 22.4 L H 22.4 L H22

MV=mol

Liters of Solution x Molarity = Liters of Solution x Molarity = moles of solutemoles of solute x Liters of solutions x Liters of solutions Liter of SolutionLiter of Solution

• Liters of Solution = VolumeLiters of Solution = Volume• Volume x Molarity = moles of soluteVolume x Molarity = moles of solute• VM = mol of solute VM = mol of solute

Advanced Level QuestionAdvanced Level Question

Determine the Determine the ## moleculesmolecules of of CC66HH1212OO66 present in present in 2.0 2.0 L L ofof 0.1 M 0.1 M CC66HH1212OO66 solution to 2 sig figs. solution to 2 sig figs. VM =Mole VM =Mole (2.0)(0.1)= 0.2molesC(2.0)(0.1)= 0.2molesC66HH1212OO66 X molecules C X molecules C66HH1212OO66

6.02 x 106.02 x 102323 molec x molec x 0.2 moles C0.2 moles C66HH1212OO66 = = X molec CX molec C66HH1212O O x 6.02 x 10 x 6.02 x 102323 molec molec 1 mole C1 mole C66HH1212OO6 6 6.02 x 106.02 x 102323molec Cmolec C66HH1212OO66

Answer: X = (1.204) or 1.2 x 10Answer: X = (1.204) or 1.2 x 102323 molecules molecules

180g C180g C66HH1212OO66= 1 mole C= 1 mole C66HH1212OO66 = 6.02 x 10 = 6.02 x 1023 23 CC66HH1212OO6 6 moleculesmolecules

Given Given 0.2 mole 0.2 mole CC66HH1212OO66= = X molecules X molecules CC66HH1212OO66

Equivalents Equivalents 1.0 mole C1.0 mole C66HH1212OO66 6.02 x 10 6.02 x 10 2323 Molecules C Molecules C66HH1212OO66

Molar Mass22.4 L

Mole

6.02 x 1023 particles

= = =

2.0 g H2.0 g H22 = 22.4 L H = 22.4 L H22 = 1 mole H = 1 mole H22 = 6.02 x 10 = 6.02 x 102323 HH22moleculesmolecules

VM=mol

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