The Mole, Stoichiometry, and Solution Chemistry

H Chemistry I Unit 8

Objectives #1-4 The Mole and its Use in Calculations

I. Fundamentals

1 mole of an element or compound contains 6.02 X 10 23 particles and has a mass = to its molar mass “The Triad of the Mole”

1 mole = 6.02 X 10

23

particles = molar mass

*henceforth the number 6.02 X 10 23 will be known as the Avogadro’s number

*examples of mole quantities: 1 mole of iron = 55.8 63.5 18.0 grams 1 mole of copper = grams 1 mole of water = grams

II.Introduction to Mole Problems

(Follow the procedures outlined in Unit 1 for dimensional analysis problems) 1. Calculate the number of atoms in .500 moles of iron.

*Determine known: .500 atoms Fe

*Determine unknown: atoms Fe *Use Triad of the Mole to determine conversion factor: 1 mole = 6.02 X 10 23 atoms *Use conversion factor solve problem: .500 moles Fe X 6.02 X 10 23 atoms/1 mole

Check answer for units, sig. figs,, and reasonableness 3.01 X 10 23 atoms Fe 2. Calculate the number of atoms in .450 moles of zinc.

.450 moles Zn X 6.02 X 10 23 atoms Zn / 1 mole = 2.71 X 10 23 atoms Zn

3. Calculate the number of moles in 2.09 X 10 25 atoms of sulfur.

2.09 X 10 25 atoms S X 1mole S / 6.02 X 10 4. Calculate the number of moles in 3.06 X 10 22 23 atoms S = 34.7 moles S atoms of chlorine.

3.06 X 10 22 6.02 X 10 23 atoms Cl X 1 mole Cl / atoms Cl = .0508 moles Cl

5. Calculate the number of atoms in 35.7 g of silicon.

35.7 g Si X 6.02 X 10 23 atoms Si / 28.1 g = 7.65 X 10 23 atoms Si

II.Molar Mass

1. Calculate the molar mass of H 3 PO 4 : 3 H at 1.00 g = 3.00 g 1 P at 31.0 g = 31.0 g 4 O at 16.0 g = 64.0 g Total = 98.0 g

2. Calculate the molar mass of Al(OH) 3 : 1 Al at 27.0 g = 27.0 g 3 O at 16.0 g = 48.0 g 3 H at 1.0 g = 3.0 g Total = 78.0 g

3. Calculate the molar mass of BaCl 2 H 2 2 .

2H 2 O: 1 Ba at 137.3 g = 137.3 g 2 Cl at 35.5 g = 71.0 g O at 18.0 g = 36.0 g Total = 244.2 g

III.Mole-Mass Problems

1. How many grams are in 7.20 moles of dinitrogen trioxide?

*determine known: 7.20 moles N grams N 2 O 3 2 O 3 *determine unknown:

*determine molar mass of compound if grams are involved: 76.0 g *use “Triad of the Mole” to determine conversion factor: 1 mole = 76.0 g *use conversion factor to solve problem: 7.20 moles N 547.2 g 2 O 3 X 76.0 g N 2 O 3 / 1 mole =

*check answer for units, sig. figs., and reasonableness: 547 g N 2 O 3

2. What is the mass in grams of 4.52 moles of barium chloride?

4.52 moles BaCl 3.50 g (NH 4 ) 3 PO 2 X 208.2 g / 1 mole = 941 g 3. Calculate the number of ammonium ions in 3.50 grams of ammonium phosphate.

4 X 1 mole (NH 4 ) 3 PO 4 / 149. 0 g X 3 NH 4 +1 / 1 mole (NH 4 ) 3 PO 4 6.02 X 10 23 4.24 X 10 22 ions / 1 mole NH NH 4 +1 ions X 4 +1 ions =

4. Calculate the mass of carbon in 7.88 X 10 26 molecules of C 8 H 18 .

7.88 X 10 26 1 mole C 8 H 18 molecules C 8 H 18 / 6.02 X 10 23 X X 8 moles C / 1 mole C 8 H 18 molecules C X 12.0 g C / 1 mole C = 1.26 X 10 5 g C 8 H 18

5. Calculate the number of molecules present in 2.50 moles of water.

2.50 moles H 2 O X 6.02 X 10 23 molecules / 1 mole = 1.51 X 10 24 molecules H 2 O

6. Calculate the mass in grams of 4.50 X 10 25 molecules of C 6 H 10 .

4.50 X 10 25 molecules C 6 H 10 82.0 g / 6.02 X 10 23 X molecules = 6130 g C 6 H 10

7. Calculate the mass of 1 molecule of propane.

1 molecule C 3 H 8 X 44.0 g / 6.02 X 10 23 molecules = 7.31 X 10 -23 g C 3 H 8

Objective #5 Characteristics of Solutions

Characteristic Components Particle Size Tyndall Effect Result Ability to be separated by filtration Homogeneous or heterogeneous Examples in different phases

solute and solvent .01 – 1 nm (atoms, ions, and molecules) negative no homogeneous solid – brass liquid – saltwater gas - air

Objective #5 Characteristics of Solutions

II.The Solution Process

*dissociation and hydration of solutes: solute is split apart by solvent (dissociation) solute particles are surrounded by solvent particles (hydration or solvation) *the rate of solution formation can be increased by: stirring, raising temperature, and powdering

Objective #5 Characteristics of Solutions

*behavior of ionic, polar, and nonpolar solutes in water: water is polar and will dissolve many ionic and polar solutes NaCl → Na +1 HCl C 6 H 12 → H +1 O 6 NaCl (ionic) HCl (polar) I 2 (nonpolar)  C + Cl -1 + Cl 6 H nonelectrolytes vs. electrolytes *”like dissolves like”: materials that have similar bonds will dissolve each other H H 2 2 CCl 12 -1 O 6 O (polar) O (polar) 4 (nonpolar)

Objective #5 Characteristics of

one phase

Solutions

*miscible vs. immiscible liquids: miscible liquids dissolve in each other and form immiscible liquids don’t dissolve in each other and form two phases *unsaturated vs. saturated vs. supersaturated solutions: Unsaturated (less solute dissolved than possible) Saturated (limit of solute dissolving) Supersaturated (beyond limit of solute dissolving)

Objective #5 Characteristics of Solutions

*effect of pressure on solubility: gases will be more soluble in a liquid if the atmospheric pressure is increased *effect of temperature on solubility: for solids in liquids – increasing temp. usually increases solubility for gases in liquids – increasing temp. decreases solubility (interpreting solubility graphs)

Interpreting Solubility Graphs

Example Problems: 1. Which chemical is the most soluble at 20 oC ? _________ 2. Which chemical is the least soluble at 40 oC ? ________ 3. What is the solubility of KCl at 25 oC ?

________ 4. At 90 50 oC oC you dissolved 10 g of KCl in 100 g of water. Is 5. Amass of 80 g of KNO 3 is dissolved in 100 g of water at solution saturated. _________ oC . How many more

Objective #6 Molarity

*formula: Molarity = moles of solute / liter of solution *examples: 1. Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

11.5 g NaOH X 1 mole / 40.0 g = .288 moles M = .288 moles / 1.50 L = .192 M

2. A chemist requires 1.00 L of .200 M potassium dichromate solution. How many grams of solid potassium dichromate will be required?

Moles = M X L = (.200 M) (1.00 L) = .200 moles K 2 Cr 2 O 7 .200 moles K 2 Cr 2 O 7 X 294.2 g / 1 mole = 58.8 g K 2 Cr 2 O 7

.200 moles K 2 Cr 2 O 7 X 294.2 g / 1 mole = 59 g K CaCl 2 2 Cr .15 M Ca +2 2 O 1 Ca to 2 Cl 7 3. What is the molarity of each ion in the following solutions (assuming all are strong electrolytes) .15 M calcium chloride: 2(.15 M) = .30 M Cl -1

.22 M calcium perchlorate Ca(ClO 4 ) 2 1 Ca to 2 ClO 4 .22 M Ca +2 2 (.22 M) = .44 M ClO 4 -1 .18 M sodium chloride .18 M Na and .18 M Cl

4. What is the molarity of an HCl solution M 1 made by diluting 3.50 L of a .200 M solution to a volume of 5.00 L?

moles before dilution = moles after dilution recall moles = MV V 1 = M 2 V 2

M M 1 2 V 1 = M #1 = concentrated #2 = diluted = M 1 V 2 1 V / V = (3.50 L) (.200 M) / 5.00 L = .140 M 2 2

5. An experiment calls for 2.00 L of a .400 M HCl, which must be prepared from 2.00 M HCl. What volume in milliliters of the concentrated acid is needed to be diluted to form the .400 M solution?

M V 1 1 V 1 = M 2 V 2 = M 2 V 2 / M 1 = (.400 M) (2.00 L) / (2.00 M) = .400 L = 400. ml How would one prepare solution?

Objs. #7-10 Acids, Bases, and pH/pOH

I. Strong vs. Weak Acids and Bases *strong acids and bases completely ionize in water HCl (aq) ---  H +1 (aq) + Cl -1 (aq) NaOH (aq) *common examples: HCl, H SO Ca(OH) 4 2 ---  , HNO 3 Na +1 (aq) , HClO 4 + OH -1 (aq) , NaOH, KOH,

*weak acids and bases don’t completely ionize completely in water HCN (aq) + H 2 O (aq) < > CN -1 (aq) + H 3 O +1 (aq) NH 3(aq) + H 2 O (aq) *common examples: HC 2 H 3 O 2 , NH 3 < > NH 4(aq) +1 + OH -1 (aq)

II. Self-Ionization of Water *water has the ability to act as both a Bronsted-Lowery acid and base H 2 O (l) + H 2 O (l) < > H 3 O (aq) +1 + OH (aq) -1 K w = {H 3 O +1 } {OH -1 } where at 25 1.00 X 10 1.00 X 10 1.00 X 10 -14 -14 -7 oC the K w = {H 3 O +1 } {OH -1 } = x 2 = x = 1.00 X 10 -14 so

*these values for the H however III. pH and pOH 3 O expressing the relative +1 concentrations are valid for room temperature conditions; K does vary with temperature *pH and pOH are convenient methods for H 3 and OH O +1 or -1 OH -1 of a solution *pH is defined as the –log{H 3 O +1 } *pOH is defined as the –log{OH -1 } *at 25 oC pH +pOH = 14

*pH and pOH scales compared:

1-6 7 8-14

pH acidic / pOH basic {H +1 } > 1 X10 -7 M Neutral {H +1 }= {OH -1 }=1X10 -7 M pH basic / pOH acidic {H +1 } < 1 X 10 -7 {OH -1 } < 1 X10 -7 M {OH -1 } > 1 X10 -7 M pH < 7.0

pOH > 7.0

pH = pOH = 7.0

pH > 7.0

pOH < 7.0

*examples: Calculate the pH for each of the following solutions: 1. 1.0 M H +1 2. 1.0 X 10 -9 M H +1 3. 2.5 X 10 -4 M H +1

4. Calculate the hydronium ion concentration of a solution having a pH of 4.5.

5. Calculate the hydronium ion concentration of a solution having a pH of 7.52

6. Calculate the H +1 and OH -1 for a 1.0 X 10 -3 M NaOH solution. Calculate the pH and pOH.

IV.Buffers

*buffers are solutions that resist changes in pH when small amounts of acid or base are added *buffer solutions can be produced from a combination of a weak acid and a salt containing a common ion or a combination of a weak base and a salt containing a common ion

*example: adding acid to a mixture of sodium acetate and acetic acid: CH 3 COO -1 + H +1 ---  CH 3 COOH + H 2 O The hydrogen “captured” by the acetate ion adding base to a mixture of sodium acetate and acetic acid: CH 3 COOH + OH The hydroxide ion of the additional acid will be -1 --  CH 3 COO -1 + H ion of the base will be “neutralized” by the acid 2 O

*Examples: Which of the following mixtures would form buffers?

1. HCl and NaCl 2. HNO 2 and NaNO 2 3. HNO 2 and NaCl

Objective #11 Percentage Composition

I. Formula

% element in compound = mass of element in sample of compound / mass of sample of compound

*examples: 1. Calculate the percentage composition potassium phosphate (K 212.4 g 3 K at 39.1 g each = 117.3 g 1 P at 31.0 g each = 31.0 g 4 O at 16.0 g each = 64.0 g 3 PO 4 ) *determine molar mass of compound: 2. Determine total mass of each component:

3. Divide total mass of each component by molar mass of compound: % K = 117.3 g / 212.4 g = .55

% P = 31.0 g / 212.4 g = .15

% O = 64.0 g / 212.4 g = .30

4. Multiply each result by 100 % and round appropriately.

.55 X 100 = 55 % K .15 X 100 = 15 % P .30 X 100 = 30 % O

2. Calculate the percentage composition of sodium carbonate (Na 2 CO 3 ) molar mass = 106.0 g % Na = (46.0 g / 106.0 g) X 100 = 43% % C = (12.0 g / 106.0 g) X 100 = 11% % O = (48.0 g / 106.0 g ) X 100 = 45%

Objective #12 Determining Empirical Formulas

*empirical vs. molecular formulas: H 2 O 2 --› HO H 2 O --› H 2 O *examples: 1. Determine the empirical formula for a compound that when analyzed contained 70.9% potassium and 29.1% sulfur by mass.

*If given percentages, assume 100 gram sample and convert percents to grams: 70.9% K --› 70.9 g K 29.1% S --› 29.1 g S

*convert masses to moles: 70.9 g K X 1 mole K / 39.1 g = 1.81 moles K 29.1 g S X 1 mole S / 32.1 g =.907 moles S *divide all mole values by the smallest mole value in the set: .907 / .907 = 1 S 1.81 / .907 =1.995 K

*round resulting values to the nearest whole number or multiply by factor: 1 S 2 K *use final values as subscripts and write formula with the elements in order of increasing electronegativity; if compound is organic, list carbon first, then hydrogen, and the remainder by electronegativity: K 2 S

2. A compound of iron and oxygen when Fe 2 analyzed showed 70.0% iron and 30.0% oxygen by mass. Determine the empirical formula.

70. 0 g Fe X 1 mole/55.8 g = 1.25 moles Fe 30.0 g O X 1 mole/16.0 g = 1.88 moles O 1.25 / 1.25 = 1 1.88 / 1.25 = 1.5

1 Fe X 2 = 2 Fe 1.5 O X 2 = 3 O O 3

3. Determine the empirical formula for a compound that contains 10.88 g of calcium and 19.08 g of chlorine.

10.88 g Ca X 1 mole / 40.1 g = .271 mole 19.08 g Cl X 1 mole / 35.5 g = .537 mole .271 / .271 = 1 Ca .537 / .271 = 2 Cl CaCl 2

Objective #13 Determining Molecular Formulas

*examples: 1. Analysis of a compound showed it to consist of 80.0% carbon and 20.0% hydrogen by mass. The gram molecular mass is 30.0 g. Determine the molecular formula.

*determine empirical formula if needed:

80.0 g C X 1 mole / 12.0 g = 6.67 moles C 20.0 g H X 1 mole / 1.0 g = 20.0 moles H 6.67 / 6.67 = 1 C 20.0 / 6.67 = 3 H CH 3 *determine empirical formula mass (efm): 15 g

*if efm matches gram molecular mass (gmm), then empirical formula is the same as molecular 15 g ≠ 30 g *if masses don’t match, divide gmm by efm to determine factor: 30 g / 15 g = 2 *multiply subscripts of empirical formula by factor to obtain molecular formula: C 2 H 6

2. If the empirical formula for a C 6 compound is C emp. mass 60.5 g H 3 Cl 3 2 HCl, determine the molecular formula if the gram molecular mass is 181.5 g.

181.5 g / 60.5 g = 3

3. A compound contains 58.5% carbon, 9.8% hydrogen, 31.4% oxygen and the gram molecular mass is 102 g. Determine the molecular formula.

58.5 g C X 1 mole / 12.0 g = 4.88 mole C 9.8 g H X 1 mole / 1.0 g = 9.8 mole H 31.4 g O X 1 mole / 16.0 g = 1.96 mole O 4.88 / 1.96 = 2.5 C 9.8 / 1.96 = 5 H 1.96 / 1.96 = 1 O

2.5 C X 2 = 5 C 5 H X 2 = 10 H 1 O X 2 = 2 O emp. formula = C 5 H 10 O 2 emp. mass = 102 g molecular formula = C 5 H 10 O 2

Objectives #14-15 Introduction to Stochiometry

*Stoichiometry is Quantitatively a method of calculating amounts in a chemical reaction I. Interpreting Chemical Equations *example: 4Al + 3O 2 --› 2Al 2 O 3 *the following information can be determined from this reaction: 1. Number of particles 2. Number of moles 3. Mass

1. Particles: 4Al + 3O atom molecule formula unit 4 atoms Al 3 molecules O 2 2 2 formula units Al --› 2Al 2 O 3 2 O 3

2. Moles 4Al + 3O 2 4 moles Al 3 moles O 2 2 moles Al 2 O 3 --› 2Al 2 O 3

3. Mass: 4Al + 3O 2 108 g Al 96 g O 2 204 g Al 2 O 3 --› 2Al 2 O 3 What is conserved in a chemical reaction?

4Al + 3O 2 --› 2Al 2 O 3 *playing with the mole ratios: 4 moles Al --› ? Al 2 O 3 4 moles Al --› 2 moles Al 2 O 3 8 moles Al --› ? Al 2 O 3 8 moles Al --› 4 mole Al 2 O 3 2 moles Al --› ? Al 2 O 3 2 moles Al --› 1 mole Al 2 O 3

4Al + 3O 2 --› 2Al 2 O 3 .200945 moles Al --› ? Al 2 O 3 .200945 moles Al X 2 moles Al 2 O 3 / 4 mole Al = .100473 mole Al

Objective #16 Stoichiometric Calculations

*examples: 1. If 20.0 g of magnesium react with excess hydrochloric acid, how many grams of magnesium chloride are produced?

*write balanced chemical equation if not provided: Mg + 2HCl --› MgCl 2 + H 2

Mg + 2HCl -  H 2 + MgCl 2 *determine known: 20.0 g Mg *determine unknown: g MgCl 2 *convert given to moles: 20.0 g Mg X 1 mole Mg / 24.3 g *select and utilize appropriate mole-mole to convert to moles of unknown: X 1 mole MgCl 2 / 1 mole Mg

*if unknown is to be measured in moles skip to step 7. If unknown is to be measured in grams, multiply by unknown’s molar mass. If unknown is to be measured in particles, multiply by Avogadro’s number: X 95.3 g MgCl 2 / 1 mole MgCl 2

*perform math: 78.4 g MgCl 2 *check answer for units, sig figs, and reasonableness

2. How many moles of water can be formed from 7.5 moles of ethyne gas (C 2 H 2 ) reacting with excess oxygen gas?

2C 2 H 2 + 5O 2 7.5 moles C 2 H 2 --› 4CO X 2 2 moles H 2 O / 2 moles C 2 H 2 7.5 moles H 2 O + 2H 2 O =

3. If 50.00 g of Rb reacts with excess S 8 how many formula units of Rb 2 S can be formed?

16 Rb + S 50.00 g Rb X 1 mole Rb / 85.5 g Rb X 8 mole Rb 2 8 --› 8Rb 2 S S / 16 mole Rb X 6.02 X 10 23 f. units Rb 2 S / 1 mole Rb 2 S = 1.760 X 10 23 f. units Rb 2 S

Obj. #17 Solution Stoichiometry

*stoichiometry that involves using molarity Example I: HCl + NaHCO 3 --  NaCl + H 2 O + CO 2 How many grams of NaHCO 3 neutralize 18.0 ml of .100 M HCl?

solution are needed to

Example II: H 2 SO 4 + 2NaOH --  Na 2 SO 4 + 2H 2 O How many grams of sodium sulfate is produced from 25.0 ml of .150 M NaOH solution?

Obj. #18-19 Titration

*description of titration:

*end point: the point in a titration when the indicator changes color for example, phenolphthalein is clear in a pH below 8 and is pink in a pH above 8 *equivalence point: the moles of base the point in a titration when the moles of acid just neutralize

*Example I: H 2 SO 4 + 2NaOH --  Na 2 SO 4 + 2H 2 O What is the molarity of a 25.0 ml sample of sulfuric acid that required 37.5 ml of .15 M sodium hydroxide?

*Example II: HCl + NaHCO 3 -  NaCl + H 2 O + CO 2

Objective #20 Limiting Reagents

*General Terms: excess reagent reactant that is leftover limiting reagent reactant that is used up *Examples: 1. How many grams of carbon dioxide can be obtained by the action of 50.0 g of sulfuric acid on 100.0 g of calcium carbonate?

(follow general steps for stoichiometry, just do it twice)

H 2 SO 4 + CaCO 50.0 g H 2 SO 4 1 mole CO 2 44.0 g CO 2 1 mole CO 2 3 --› CaSO 4 X 1 mole H 2 SO 4 / 1 mole H 2 / 1 mole CO SO 44.0 g CO 2 / 1 mole CO 2 2 / 1 mole CaCO 4 3 X X + CO 2 +H 2 O / 98.1 g X = 22.4 g CO

2

----------------------------------------- 100.0 g CaCO 3 X 1 mole CaCO 3 / 100.1 g X = 43.96 g CO

2

(examine two answers produced; the

smaller of the two is the right answer

of product produced) *the reactant quantity that provides the smaller correct answer is called the limiting reagent *the reactant quantity that provides the larger incorrect answer is called the excess reagent

2. How many grams of hydrogen gas can be produced by the reaction of 100. g of phosphoric acid on 25.0 g of aluminum?

2H 3 PO 4 100. g H 3 PO 4 3 mole H 2.0 g H 2 2 + 2Al --› 2AlPO 4 X 1 mole H 3 PO 4 / 2 mole H 3 PO 4 / 1 mole H 2 X + 3H 2 / 98.0 g X = 3.06 g H 2

25.0 g Al X 1 mole Al / 27.0 g Al X 3 mole H 2 / 1 mole H 2 X 2.0 g H 2 / 1 mole= 2.78 g H 2 2.78 g H 2 (smaller value)

Determine how much of the excess reagent Al is left over.

*determine limiting reagent: *use stochiometry to convert from mass of limiting reagent to find mass of excess reagent actually used: 25.0 g Al X 1 mole Al / 27.0 g Al X 2 mole H 3 PO 4 / 2 mole Al X 98.0 g H 3 PO 4 / 1mole = 90.7 g H 3 PO 4 used

*Subtract amount of excess reagent used from original amount of excess reactant given in problem to determine mass of excess reactant left over: 100. g – 90.7 g = 9 g left over

3. How many grams of sodium chloride can be produced from then reaction of 15.0 g of chlorine gas and 15.0 g of sodium bromide in the above reaction? Determine how much of the excess reagent is left over?

Cl 2 + 2NaBr --› Br 2 + 2NaCl

15.0 g Cl 2 X 1 mole Cl 2 / 71.0 g Cl 2 2 mole NaCl / 1 mole Cl 2 X X 2 mole NaCl / 2 mole NaBr X X 58.5 g NaCl / 1 mole NaCl = 24.7 g NaCl ------------------------------------------- 15.0 g NaBr X 1 mole NaBr / 102.9 g NaBr 58.5 g NaCl / 1 mole NaCl = 8.53 g NaCl

15.0 g NaBr X 1 mole NaBr / 102.9 g NaBr X 1 mole Cl 2 / 2 mole NaBr X 71.0 g Cl 2 / 1 mole Cl 2 = 5.17 g Cl 2 (used) ------------------------------------------- 15.0 g – 5.17 g = 9.8 g Cl 2 leftover

Objective #21 Percentage Yield

*General Formula: Percent Yield = (actual yield / theoretical yield) X 100% *examples: 1. When 9.00 g of aluminum react with an excess of phosphoric acid, 30.0 g of aluminum phosphate are produced. What is the percentage yield of this reaction?

*write balanced chemical equation: 2Al + 2 H 3 PO 30.0 g AlPO 4 4 --› 3H 2 + 2AlPO reaction given in the problem: mass of appropriate product: 4 *determine actual yield of a product in *use given reactant quantity and general stochiometry procedures to determine

9.00 g Al X 1 mole Al / 27.0 g Al X 2 mole AlPO 4 / 2 mole Al X 122.0 g / 1 mole = 40.7 g AlPO percentage yield: 4 *divide actual yield from problem by the theoretical yield calculated to find %Y = (30.0 g / 40.7 g) X 100 = 74%

2. Calculate the percentage yield if 6.00 g of aluminum are produced from the decomposition of 25.0 g of aluminum oxide.

2Al 2 O 3 25.0 g Al --› 4Al + 3O 2 2 O 3 X 1 mole Al 4 mole Al / 2 mole Al 2 O 3 2 O X 3 / 102.0 g X 27.0 g Al / 1 mole Al = 13.2 g Al %Y= (6.00 g / 13.2 g) X 100% = 45%