stoichiometry calculations based on chemical reactions

StoichiometryStoichiometry

Calculations based on Calculations based on chemical reactionschemical reactions

StoichiometryStoichiometry

Stoichiometry is a Greek word that Stoichiometry is a Greek word that means using chemical reactions to means using chemical reactions to calculate the amount of reactants calculate the amount of reactants needed and the amount of products needed and the amount of products formed.formed.

Amounts are typically calculated in Amounts are typically calculated in grams (or kg), but there are other grams (or kg), but there are other ways to specify the quantities of ways to specify the quantities of matter involved in a reaction.matter involved in a reaction.

Balancing Chemical Balancing Chemical Equations- ProblemEquations- Problem

Sodium metal reacts with water to Sodium metal reacts with water to produce aqueous sodium hydroxide produce aqueous sodium hydroxide and hydrogen.and hydrogen.



Na(s) + HNa(s) + H22O(O(l l ) )



Na(s) + HNa(s) + H22O(O(l l ) ) NaOH(aq) NaOH(aq) + H+ H22(g)(g)


Sodium metal reacts with water to produce Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen.aqueous sodium hydroxide and hydrogen.

Na(s) + HNa(s) + H22O(O(l l ) ) NaOH(aq) + H NaOH(aq) + H22(g)(g)

The equation is not yet balanced. Hydrogens The equation is not yet balanced. Hydrogens come in twos on the left, and three come in twos on the left, and three hydrogens are on the right side of the hydrogens are on the right side of the equation.equation.



Na(s) + HNa(s) + H22O(O(l l ) ) NaOH(aq) NaOH(aq) + H+ H22(g)(g)

Try a “2” in front of the water.Try a “2” in front of the water.


Sodium metal reacts with water to Sodium metal reacts with water to produce aqueous sodium hydroxide and produce aqueous sodium hydroxide and hydrogen.hydrogen.

Na(s) + 2 HNa(s) + 2 H22O(O(l l ) ) NaOH(aq) + NaOH(aq) + HH22(g)(g)

We now have two O atoms on the left, so We now have two O atoms on the left, so we need to put a 2 before NaOH.we need to put a 2 before NaOH.



Na(s) + 2 HNa(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + 2NaOH(aq) + HH22(g)(g)

The two sodium atoms on the right require The two sodium atoms on the right require that we put a 2 in front of Na on the left.that we put a 2 in front of Na on the left.


Sodium metal reacts with water to produce Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen.aqueous sodium hydroxide and hydrogen.

2 Na(s) + 2 H2 Na(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + 2NaOH(aq) + HH22(g)(g)

The two sodium atoms on the right require The two sodium atoms on the right require that we put a 2 in front of Na on the left. that we put a 2 in front of Na on the left. The equation is now balanced.The equation is now balanced.




Left SideLeft Side Right SideRight SideNa- 2Na- 2 Na- 2Na- 2H- 4H- 4 H- 4H- 4O- 2O- 2 O- 2O- 2

Chemical EquationsChemical Equations


The balanced chemical equation The balanced chemical equation can be interpreted in a variety of ways. can be interpreted in a variety of ways.

It could say that 2 atoms of sodium It could say that 2 atoms of sodium react with 2 molecules of water to react with 2 molecules of water to produce 2 molecules of sodium produce 2 molecules of sodium hydroxide and a molecule of hydrogen.hydroxide and a molecule of hydrogen.



The balanced chemical equation The balanced chemical equation can be interpreted in a variety of ways. can be interpreted in a variety of ways.

It could say that 200 atoms of It could say that 200 atoms of sodium react with 200 molecules of sodium react with 200 molecules of water to produce 200 molecules of water to produce 200 molecules of sodium hydroxide and 100 molecules of sodium hydroxide and 100 molecules of hydrogen.hydrogen.


2 Na(s) + 2 H2 Na(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + H2NaOH(aq) + H22(g)(g)The balanced chemical equation can be The balanced chemical equation can be interpreted in a variety of ways. interpreted in a variety of ways.

It is usually interpreted as 2 moles of It is usually interpreted as 2 moles of sodium will react with 2 moles of water to sodium will react with 2 moles of water to produce 2 moles of sodium hydroxide and produce 2 moles of sodium hydroxide and 1 mole of hydrogen.1 mole of hydrogen.

The balanced equation tells us nothing The balanced equation tells us nothing about the about the massesmasses of reactants or of reactants or products.products.

Stoichiometry ProblemStoichiometry Problem


How many grams of sodium are How many grams of sodium are needed to produce 50.0g of hydrogen?needed to produce 50.0g of hydrogen?

Since the balanced chemical equation Since the balanced chemical equation gives us information about gives us information about molesmoles of of reactants and products, we cannot reactants and products, we cannot answer the question until we convert answer the question until we convert the mass of hydrogen into moles.the mass of hydrogen into moles.



How many grams of sodium are How many grams of sodium are needed to produce 50.0g of hydrogen?needed to produce 50.0g of hydrogen?


? grams? grams 50.0 g 50.0 g



? grams? grams 50.0 g50.0 g

Although the question doesn’t state it, Although the question doesn’t state it, you can assume enough water is present you can assume enough water is present for complete reaction.for complete reaction.

We can map out the problem:We can map out the problem:

g Hg H22 moles H moles H22 moles Na moles Na grams Na grams Na



? grams? grams 50.0 g 50.0 g We can map out the problem:We can map out the problem:


We use the molar mass of HWe use the molar mass of H22 to go from to go from grams of Hgrams of H22 to moles of H to moles of H22..


2 Na(s) + 2 H2 Na(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + H2NaOH(aq) + H22(g)(g)



molar mass of Hmolar mass of H22





molar mass of Hmolar mass of H22

We use the coefficients from the balanced We use the coefficients from the balanced equation to go from moles of Hequation to go from moles of H22 to moles of to moles of Na.Na.



? grams? grams 50.0 g50.0 g We can map out the problem:We can map out the problem:

g Hg H22 moles H moles H22 moles Na moles Na grams grams NaNa

molar mass of Hmolar mass of H2 2 coefficientscoefficients





molar mass of Hmolar mass of H2 2 coefficientscoefficients We use the molar mass of Na to go from We use the molar mass of Na to go from

moles of Na to grams of Na.moles of Na to grams of Na.




g Hg H22 moles H moles H22 moles Na moles Na grams grams NaNa

molar mass of Hmolar mass of H2 2 coefficients molar mass coefficients molar mass of Na of Na



? grams? grams 50.0 g50.0 g


molar mass of Hmolar mass of H2 2 coefficients molar mass coefficients molar mass of Na of Na

(50.0 g H(50.0 g H22) ) (1 mol H(1 mol H22)) (2 moles Na)(2 moles Na) ( 22.99 g Na)( 22.99 g Na) (2.02 g H(2.02 g H22) (1 mol H) (1 mol H22) (1 mol Na)) (1 mol Na)



(50.0 g H(50.0 g H22) ) (1 mol H(1 mol H22)) (2 moles Na)(2 moles Na) ( 22.99 g ( 22.99 g

Na)Na) = = (2.02 g H(2.02 g H22) (1 mol H) (1 mol H22) (1 mol Na)) (1 mol Na)

= 1,138 grams Na = 1.14 x 10= 1,138 grams Na = 1.14 x 103 3 g Na = g Na = 1.14 kg Na1.14 kg Na

Limiting Reagent Limiting Reagent ProblemsProblems

Sometimes you are given quantities of Sometimes you are given quantities of more than onemore than one reactant, and asked to reactant, and asked to calculate the amount of product formed. calculate the amount of product formed. The quantities of reactants might be The quantities of reactants might be such that both react completely, or one such that both react completely, or one might react completely, and the other(s) might react completely, and the other(s) might be in excess. These are called might be in excess. These are called limiting reagentlimiting reagent problems, since the problems, since the quantity of one of the reacts will limit the quantity of one of the reacts will limit the amount of product that can be formed.amount of product that can be formed.

Limiting Reagent - Limiting Reagent - ProblemProblem

Aluminum and bromine react to form Aluminum and bromine react to form aluminum bromide. If 500. g of aluminum bromide. If 500. g of bromine are reacted with 50.0 g of bromine are reacted with 50.0 g of aluminum, what is the theoretical aluminum, what is the theoretical yield of aluminum bromide?yield of aluminum bromide?1. First write the formulas for reactants 1. First write the formulas for reactants

and products.and products.

Al + BrAl + Br22 AlBr AlBr33


Aluminum and bromine react to form Aluminum and bromine react to form aluminum bromide. If 500. g of aluminum bromide. If 500. g of bromine are reacted with 50.0 g of bromine are reacted with 50.0 g of aluminum, what is the theoretical aluminum, what is the theoretical yield of aluminum bromide?yield of aluminum bromide?2. Now balance the equation by adding 2. Now balance the equation by adding

coefficients.coefficients.

2 Al +3 Br2 Al +3 Br22 2 AlBr2 AlBr33


Aluminum and bromine react to form Aluminum and bromine react to form aluminum bromide. If 500. g of aluminum bromide. If 500. g of bromine are reacted with 50.0 g of bromine are reacted with 50.0 g of aluminum, what is the theoretical aluminum, what is the theoretical yield of aluminum bromide?yield of aluminum bromide?


The The theoretical yieldtheoretical yield is the maximum is the maximum amount of product that can be formed, amount of product that can be formed, given the amount of reactants. It is given the amount of reactants. It is usually expressed in grams.usually expressed in grams.



Given: 50.0gGiven: 50.0g 500.g500.g ? ? gramsgrams

There are several ways to solve this problem. There are several ways to solve this problem. One method is to solve the problem twice. One method is to solve the problem twice. Once, assuming that all of the aluminum Once, assuming that all of the aluminum reacts, the other assuming that all of the reacts, the other assuming that all of the bromine reacts. The correct answer is bromine reacts. The correct answer is whichever assumption provides the whichever assumption provides the smallest smallest amountamount of product. of product.



Given: 50.0gGiven: 50.0g 500.g500.g ? ? gramsgrams

The problem can be mapped:The problem can be mapped:

Al: grams Al Al: grams Al moles Al moles Al moles AlBr moles AlBr33 g AlBrg AlBr33

molar mass Al coefficients molar molar mass Al coefficients molar mass mass AlBrAlBr33

Limiting Reagent - Limiting Reagent - ProblemProblem2 Al +3 Br2 Al +3 Br22 2 AlBr2 AlBr33

Given:Given: 50.0g50.0g 500.g ? grams 500.g ? grams

The problem can be mapped:The problem can be mapped:

Al: grams Al Al: grams Al moles Al moles Al moles AlBr moles AlBr33 g AlBr g AlBr33

molar mass Al coefficients molar mass molar mass Al coefficients molar mass AlBrAlBr33

(50.0 g Al) (50.0 g Al) ( 1 mol Al )( 1 mol Al ) ( 2 mol AlBr( 2 mol AlBr33)) (266.7 g (266.7 g AlBrAlBr33))

(26.98 g Al) ( 2 mol Al) (mol (26.98 g Al) ( 2 mol Al) (mol AlBrAlBr33) )



Given:Given: 50.0g 500.g50.0g 500.g ? ? gramsgrams

Al: Al: (50.0 g Al) (50.0 g Al) ( 1 mol Al )( 1 mol Al ) ( 2 mol AlBr( 2 mol AlBr33)) (266.7 g AlBr(266.7 g AlBr33))

(26.98 g Al)(26.98 g Al) ( 2 mol ( 2 mol Al) (mol AlBrAl) (mol AlBr33) )

= 494. g AlBr= 494. g AlBr33 (if all of the Al reacts) (if all of the Al reacts)



Given:Given: 50.0g 500.g50.0g 500.g ? grams ? grams

The calculation is repeated for BrThe calculation is repeated for Br22..

g Brg Br22 moles Br moles Br22 moles AlBr moles AlBr33 g g AlBrAlBr33

molar mass Brmolar mass Br22 coefficients molar coefficients molar mass AlBrmass AlBr33



Given:Given: 50.0g 500.g50.0g 500.g ? grams ? grams

The calculation is repeated for BrThe calculation is repeated for Br22..

g Brg Br22 moles Br moles Br22 moles AlBr moles AlBr33 g AlBr g AlBr33

molar mass Brmolar mass Br22 coefficients molar mass AlBr coefficients molar mass AlBr33

(500. g Br(500. g Br22) ) (1 mol Br(1 mol Br22) ) (2 moles AlBr(2 moles AlBr33)) (266.7 g (266.7 g AlBrAlBr33)) (159.8 g Br(159.8 g Br22) (3 moles Br) (3 moles Br22) (1 mol ) (1 mol AlBrAlBr33))

= 556. grams of AlBr= 556. grams of AlBr33



Given: 50.0gGiven: 50.0g 500.g500.g ? grams ? grams Summary:Summary:

We have enough Al to produce 494. g We have enough Al to produce 494. g AlBrAlBr33

We have enough BrWe have enough Br22 to produce 556. grams of to produce 556. grams of AlBrAlBr33

The theoretical yield is 494. grams of AlBrThe theoretical yield is 494. grams of AlBr33..



Given: 50.0g 500.g 494. gGiven: 50.0g 500.g 494. gSummary:Summary:

All of the Al reacts, so Al is limiting.All of the Al reacts, so Al is limiting. Bromine is in excess.Bromine is in excess. Additional questions:Additional questions:

1. How much bromine is left over?1. How much bromine is left over?2. If 418 grams of AlBr2. If 418 grams of AlBr33 is obtained, is obtained,

what is the % yield?what is the % yield?



Given: Given: 50.0g 500.g 494. g 50.0g 500.g 494. g

1. How much bromine is left over?1. How much bromine is left over?

Since all 50.0 g of the Al reacts, the Since all 50.0 g of the Al reacts, the product must contain 494.g -50.g = product must contain 494.g -50.g = 444. g of bromine. Therefore, 500.g- 444. g of bromine. Therefore, 500.g- 444.g = 56. g of Br444.g = 56. g of Br22 are left over. are left over.



Given: Given: 50.0g 500.g 494. g 50.0g 500.g 494. g

2.2. If 418 grams of AlBrIf 418 grams of AlBr33 is obtained, what is is obtained, what is the % yield?the % yield?The The percent yieldpercent yield is is (actual yield) (actual yield) (100%) (100%)

(theoretical yield)(theoretical yield)

% yield = % yield = (418 g)(418 g) (100%) = 84.6 % (100%) = 84.6 % (494g)(494g)

Concentration of Concentration of SolutionsSolutions

There are many ways to express There are many ways to express the concentration of a solution. The the concentration of a solution. The one most commonly used for one most commonly used for solution stoichiometry is solution stoichiometry is molaritymolarity (M).(M).

Molarity = M =Molarity = M =

moles of soluteliter of solution


Solutions are often used to Solutions are often used to determine how much of a certain determine how much of a certain substance is present in a sample. As substance is present in a sample. As a result, it is important to know how a result, it is important to know how to perform calculations using to perform calculations using solution stoichiometry.solution stoichiometry.


40.0 grams of sodium hydroxide are 40.0 grams of sodium hydroxide are dissolved in enough water to make dissolved in enough water to make 250. mL of solution. Calculate the 250. mL of solution. Calculate the molarity of the solution.molarity of the solution.

grams NaOHgrams NaOH moles NaOH moles NaOH M M NaOHNaOH


40.0 grams of sodium hydroxide are 40.0 grams of sodium hydroxide are dissolved in enough water to make dissolved in enough water to make 250. mL of solution. Calculate the 250. mL of solution. Calculate the molarity of the solution.molarity of the solution.

grams NaOHgrams NaOH moles NaOH moles NaOH M M NaOHNaOH

molar mass of NaOH

M=moles/volume


What volume of the previous What volume of the previous solution contains 0.125 moles of solution contains 0.125 moles of NaOH?NaOH?


How would you make 100. mL of a How would you make 100. mL of a 0.075M solution of NaCl?0.075M solution of NaCl?

molarity

= moles solute

_________________

(liter of solution)


How would you make 100. mL of a How would you make 100. mL of a 0.075M solution of NaCl?0.075M solution of NaCl?

moles solute = M(V)moles solute = M(V)

moles NaCl = (0.075mol/L) (.100L)moles NaCl = (0.075mol/L) (.100L)

moles NaCl = 0.0075 molmoles NaCl = 0.0075 mol

mass NaCl = 0.0075 mol (58.5 g/mol)mass NaCl = 0.0075 mol (58.5 g/mol)

mass NaCl = 0.439 g = 0.44 g NaClmass NaCl = 0.439 g = 0.44 g NaCl

The Preparation of a The Preparation of a Standard Aqueous SolutionStandard Aqueous Solution

DilutionDilution

Most solutions are purchased in Most solutions are purchased in highly concentrated form. They often highly concentrated form. They often need to be diluted before they can be need to be diluted before they can be used.used.

When a solution is diluted, the When a solution is diluted, the moles of solute remain unchanged. moles of solute remain unchanged. Dilution involves the addition of more Dilution involves the addition of more solvent (water), while the amount of solvent (water), while the amount of solute stays the same.solute stays the same.

DilutionDilution

Dilution problems can be easily Dilution problems can be easily solved using the following solved using the following relationships.relationships.

mols concentrated solute = mols mols concentrated solute = mols diluted solutediluted solute

MMconcconc(V(Vconcconc) = M) = Mdildil(V(Vdildil))

DilutionDilution How many milliliters of 6.00M HHow many milliliters of 6.00M H22SOSO44

are needed to make 30.0 mL of 1.25M are needed to make 30.0 mL of 1.25M HH22SOSO44??

Mcon

c

(Vconc

)= Mdil(Vdil)

________

DilutionDilution How many milliliters of 6.00M HHow many milliliters of 6.00M H22SOSO44 are are

needed to make 30.0 mL of 1.25M needed to make 30.0 mL of 1.25M HH22SOSO44??

VVconcconc= = 1.25M(30.0 mL)1.25M(30.0 mL)(6.00M)(6.00M)

Mcon

c

(Vconc

)= Mdil(Vdil)

________

DilutionDilution

V = V = (0.0375 mol) (0.0375 mol)

V = 0.00625 L = 6.25 mLV = 0.00625 L = 6.25 mL

The solution requires 6.25 mL of The solution requires 6.25 mL of 6.00M H6.00M H22SOSO44 diluted to a final diluted to a final volume of 30.0 mL.volume of 30.0 mL.

(6.00 mol/L)

Solution StoichiometrySolution Stoichiometry

How much 0.100M AgNOHow much 0.100M AgNO33 is required is required to precipitate all of the chloride from to precipitate all of the chloride from 25.00 mL of 0.173M CaCl25.00 mL of 0.173M CaCl22??

1. Write a balanced chemical 1. Write a balanced chemical equation.equation.

2AgNO2AgNO33(aq) + CaClCaCl22((aqaq) ) Ca(NOCa(NO33))22((aqaq) + ) + 2 AgCl(2 AgCl(ss))


How much 0.100M AgNOHow much 0.100M AgNO33 is required to is required to precipitate all of the chloride from precipitate all of the chloride from 25.00 mL of 0.173M CaCl25.00 mL of 0.173M CaCl22??

2. Map out the problem.2. Map out the problem.2AgNO2AgNO33(aq) + CaClCaCl22((aqaq) ) Ca(NOCa(NO33))22((aqaq) + 2 ) + 2

AgCl(AgCl(ss))

VVCaCl2CaCl2MMCaCl2CaCl2=moles CaCl=moles CaCl22 moles AgNO moles AgNO33 VVAgNO3AgNO3

coefficients

V=(mol)/(M)


3. Solve the problem.3. Solve the problem.

2AgNO2AgNO33(aq) + CaClCaCl22((aqaq) ) Ca(NOCa(NO33))22((aqaq) + 2 ) + 2 AgCl(AgCl(ss))

VVCaCl2CaCl2MMCaCl2CaCl2=moles CaCl=moles CaCl22 moles AgNO moles AgNO33 V VAgNO3AgNO3

(25.00mL CaCl(25.00mL CaCl22) (0.173 mol/L CaCl) (0.173 mol/L CaCl22) = 4.325 mmol CaCl) = 4.325 mmol CaCl22

(4.325 mmol CaCl(4.325 mmol CaCl22)(2 mmol AgNO)(2 mmol AgNO33/1 mmol CaCl/1 mmol CaCl22) =) =

8.65 mmol AgNO8.65 mmol AgNO33

VVAgNO3AgNO3 = = 8.65mmol 8.65mmol AgNOAgNO33 = 86.5 mL AgNO= 86.5 mL AgNO33

0.100mmol AgNO0.100mmol AgNO33/mL/mL

coefficients

V=(mol)/(M)

TitrationsTitrations

In a titration, a specific volume of In a titration, a specific volume of a solution of unknown concentration a solution of unknown concentration is reacted with a solution of known is reacted with a solution of known concentration. The volume needed concentration. The volume needed for complete reaction is determined for complete reaction is determined using an using an indicator indicator and a buret to and a buret to measure the precise volume needed.measure the precise volume needed.

Titrations may be used for many Titrations may be used for many types of reactions, and they are often types of reactions, and they are often used in acid-base neutralization.used in acid-base neutralization.

Acid-Base TitrationsAcid-Base Titrations

In this titration, In this titration, NaOH is in the buret, NaOH is in the buret, and HCl and the and HCl and the indicator are in the indicator are in the flask. The NaOH flask. The NaOH solution is added until solution is added until the first permanent the first permanent pale pink color is pale pink color is formed in the flask. formed in the flask.

Titration StoichiometryTitration Stoichiometry

10.00 mL of a sulfuric acid solution 10.00 mL of a sulfuric acid solution requires 17.83 mL of a 0.100M NaOH requires 17.83 mL of a 0.100M NaOH solution for complete neutralization. solution for complete neutralization. Determine the concentration of the Determine the concentration of the sulfuric acid.sulfuric acid.

1. Write the balanced chemical reaction.1. Write the balanced chemical reaction.

HH22SOSO44((aqaq) +2 NaOH() +2 NaOH(aqaq)) Na Na22SOSO44(aq) (aq) +2H+2H22O(O(ll))


10.00 mL of a sulfuric acid solution 10.00 mL of a sulfuric acid solution requires 17.83 mL of a 0.100M NaOH requires 17.83 mL of a 0.100M NaOH solution for complete neutralization. solution for complete neutralization. Determine the concentration of the sulfuric Determine the concentration of the sulfuric acid.acid.

2. Map out the solution.2. Map out the solution.

HH22SOSO44((aqaq) + 2 NaOH() + 2 NaOH(aqaq) ) Na Na22SOSO44(aq) + (aq) + 2H2H22O(O(ll))

VVNaOHNaOHMMNaOHNaOH=moles NaOH=moles NaOH moles H moles H22SOSO44 MMH2SO4H2SO4

coefficients

M=mol/liter

Titration StoichiometryTitration Stoichiometry3. Solve the problem.3. Solve the problem.HH22SOSO44((aqaq) + 2 NaOH() + 2 NaOH(aqaq) ) Na Na22SOSO44(aq) + 2H(aq) + 2H22O(O(ll))10.00 mL10.00 mL 17.83mL; 0.100M 17.83mL; 0.100M

VVNaOHNaOHMMNaOHNaOH=moles NaOH=moles NaOH moles H moles H22SOSO44 M MH2SO4H2SO4

(17.83 mL)(0.100 mol NaOH/L) = 1.783 mmol NaOH(17.83 mL)(0.100 mol NaOH/L) = 1.783 mmol NaOH

(1.783 mmol NaOH) ((1.783 mmol NaOH) (1mmol H1mmol H22SOSO44) = 0.892 mmol) = 0.892 mmol (2 mmol NaOH)(2 mmol NaOH) H H22SOSO44

coefficients

M=mol/liter


3. Solve the problem.3. Solve the problem.HH22SOSO44((aqaq) + 2 NaOH() + 2 NaOH(aqaq) ) Na Na22SOSO44(aq) + (aq) +

2H2H22O(O(ll))

10.00 mL; 0.892mmol10.00 mL; 0.892mmol 1.78 mmol1.78 mmol

VVNaOHNaOHMMNaOHNaOH=moles NaOH=moles NaOH moles H moles H22SOSO44 M MH2SO4H2SO4

(0.892 mmol H(0.892 mmol H22SOSO44)/10.00 mL = 0.0892 mol/L)/10.00 mL = 0.0892 mol/L

= 0.0892 M= 0.0892 M

coefficients

M=mol/liter

Chemical CompositionChemical Composition

Chemical composition can be Chemical composition can be expressed in several ways, including expressed in several ways, including percentages by mass, or chemical percentages by mass, or chemical formulas. formulas.

For example, water contains For example, water contains 11.2% hydrogen and 88.8% oxygen 11.2% hydrogen and 88.8% oxygen by mass. This information must be by mass. This information must be consistent with the chemical formula consistent with the chemical formula for water, Hfor water, H22O.O.


For example, water contains 11.2% For example, water contains 11.2% hydrogen and 88.8% oxygen by mass. This hydrogen and 88.8% oxygen by mass. This information must be consistent with the information must be consistent with the chemical formula for water, Hchemical formula for water, H22O.O.

2 H atoms = 2(1.008 amu) = 2.016 amu2 H atoms = 2(1.008 amu) = 2.016 amu1 O atom =1( 16.00 amu) =1 O atom =1( 16.00 amu) =16.00 amu16.00 amumolecular mass of water = 18.02 amumolecular mass of water = 18.02 amu

% H = (2.016/18.02) x 100% = 11.19%H% H = (2.016/18.02) x 100% = 11.19%H% O = (16.00/18.02) x 100% = 88.79%O% O = (16.00/18.02) x 100% = 88.79%O


The early scientists analyzed new The early scientists analyzed new chemical compounds to determine chemical compounds to determine their composition and chemical their composition and chemical formulas. Modern analytical formulas. Modern analytical laboratories still provide this service.laboratories still provide this service.


Usually, the compound is Usually, the compound is combusted in the presence of oxygen. combusted in the presence of oxygen. Any carbon in the compound is Any carbon in the compound is collected as carbon dioxide (COcollected as carbon dioxide (CO22), and ), and any hydrogen is collected as water any hydrogen is collected as water (H(H22O).O).


Similar techniques exist to analyze Similar techniques exist to analyze for other elements.for other elements.

The formula obtained for the The formula obtained for the compound is the simplest whole number compound is the simplest whole number ratio of the elements in the compound, or ratio of the elements in the compound, or the the empirical formulaempirical formula. It may differ from . It may differ from the actual formula. For example, the actual formula. For example, hydrogen peroxide is Hhydrogen peroxide is H22OO22, but chemical , but chemical analysis will provide an empirical analysis will provide an empirical formula of HO.formula of HO.

Determining Empirical Determining Empirical FormulasFormulasIf given % compositionIf given % composition::

1. Assume a quantity of 100 grams of 1. Assume a quantity of 100 grams of the compound.the compound.

2. Determine the number of moles of 2. Determine the number of moles of each element in the compound by each element in the compound by dividing the grams of each element dividing the grams of each element by the appropriate atomic mass.by the appropriate atomic mass.

3. To simplify the formula into small 3. To simplify the formula into small whole numbers, divide the moles of whole numbers, divide the moles of each element by the smallest number each element by the smallest number of moles.of moles.

Determining Empirical Determining Empirical FormulasFormulas

4. If necessary, multiply each 4. If necessary, multiply each number of moles by a factor that number of moles by a factor that produces whole number produces whole number subscripts.subscripts.

5. If you know the approximate 5. If you know the approximate molar mass of the compound, molar mass of the compound, determine the molecular formula.determine the molecular formula.

% Composition Problem% Composition Problem

An oxide of titanium contains 59.9% An oxide of titanium contains 59.9% titanium. Determine the empirical titanium. Determine the empirical formula of the compound.formula of the compound.


If given combustion dataIf given combustion data::

The ultimate goal is to get the The ultimate goal is to get the simplest whole number ratio of the simplest whole number ratio of the elements in the compound. Usually elements in the compound. Usually the compound contains carbon, the compound contains carbon, hydrogen and perhaps oxygen or hydrogen and perhaps oxygen or nitrogen.nitrogen.

1. Use the information about CO1. Use the information about CO22 to to determine the moles and mass of determine the moles and mass of carbon in the compound.carbon in the compound.


If given combustion dataIf given combustion data::

2. Use the information about H2. Use the information about H22O to O to determine the moles and mass of determine the moles and mass of

hydrogen hydrogen in the compound.in the compound.

3. The mass and moles of oxygen (or a 3. The mass and moles of oxygen (or a third element) can be obtained by third element) can be obtained by difference.difference.

4. Once moles of each element is 4. Once moles of each element is obtained, find the relative number of obtained, find the relative number of moles and empirical formula (as with moles and empirical formula (as with % composition).% composition).

Formulas Formulas from from Combustion Combustion DataData

This method assumes the compound contains only C and H.

Empirical Formula using Empirical Formula using Combustion Data- ProblemCombustion Data- Problem

A compound, which contains C, H A compound, which contains C, H and O, is analyzed by combustion. If and O, is analyzed by combustion. If 10.68 mg of the compound produces 10.68 mg of the compound produces 16.01 mg of carbon dioxide and 4.37 16.01 mg of carbon dioxide and 4.37 mg of water, determine the mg of water, determine the empirical formula of the compound.empirical formula of the compound.

If the compound has a molar mass If the compound has a molar mass of 176.1 g/mol, determine the of 176.1 g/mol, determine the molecular formula of the compound.molecular formula of the compound.

stoichiometry calculations based on chemical reactions

Documents

aqueous sodium hydroxide

molecules of sodium

sodium atoms

atoms of sodium

moles of sodium hydroxide

chemical equations2

balanced chemical equation

grams of sodium