Cpt 4 chemical stoichiometry
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Cpt 4. CHEMICAL STOICHIOMETRY. Definition : Study of proportions of * atoms inside substances ; * substances in solutions or chemical reactions Objectives: Calculate: * the composition & formulas of chemical compounds * the concentrations of solutions

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Cpt 4 chemical stoichiometry l.jpg
Cpt 4. CHEMICAL STOICHIOMETRY

  • Definition:Study of proportions of

    * atoms inside substances;

    * substances in solutions or chemical reactions

  • Objectives:

  • Calculate:

    * the composition & formulas of chemical compounds

    * the concentrations of solutions

    * the quantitative outcome of chemical reactions


4 1 mole and molar mass l.jpg
4.1. Mole and Molar mass

  • Thought teaser

    Weight Species Number

    1 g H 6.022E23 atoms

    16 g O , , , ,

    18 g H2O , , molecules

    23 g Na(+) , , cations

    35.5 g Cl(-) , , anions

    What trend is uncovered by the information

    from the table? See RQ2-19


Rq2 19 l.jpg
RQ2-19

What trend is uncovered by the information from the table on the previous slide?

  • Any amount of substance with a density equal to its atomic or molecular density contains the same number of components: 6.022E23

  • Any amount of substance with a weight equal to its atomic or molecular weight contains the same number of components: 6.022E23

  • Substances with the same weight contain the same number of components: 6.022E23


The mole l.jpg
The Mole

  • mole = 6.022E23 = Avogadro's number

    of components (atoms in an element,

    molecules in a compound) of a substance.

  • Number of components of a substance in a weight in grams equal to the atomic or the molecular weight of the substance.

  • Molar mass (MW) : mass of a mole : same magnitude but different units as the atomic or molecular mass.

  • unit of molar mass: g/mol

  • Mass - Molar mass relation:

  • # of moles = (mass/molar mass)


Rq2 19b previous material review l.jpg
RQ2-19B: Previous Material Review

  • On the chemical standpoint a mole is a: a, rodentthat lives in underground burrows; b, measure of amount of substance; c, measure of weight of substance

  • The number of moles of a substance shows how much of a substance there is by showing the: a, weight of substance; b, volume of substance; c, the number of components (atoms or molecules) of the substance.


Word of wisdom learning steps l.jpg
WORD OF WISDOMLearning Steps

Resolve, Dare to make the first step in the learning process: REVIEW

Reviewing leads to UNDERSTANDING

Don’t understand? Ask quickly for help to keep the momentum


Word of wisdom before the gain l.jpg
WORD OF WISDOMBefore the Gain

Unconfortable when studying? Expect the unconfort.

Welcome the unconfort. It is beneficial to your learning

You’re changing your thoughts

From UNFOCUSED TO FOCUSED

Say in your mind: “I’m happy to do it”. You feel a smile forming in your mind


Word of wisdom l.jpg
WORD OF WISDOM

Not enoughtime?

MAKE the time!

Sacrifice something.

Your better future: worth the sacrifice

Tolerate Hardship on the way to Prosperity


Word of wisdom hard work l.jpg
WORD OF WISDOMHard Work

Work is too hard?

Keep trying. It’ll geteasier and easier

Ask for help

When the going gets tough, the tough gets going


Concept illustration l.jpg
Concept Illustration

A 60 kg person has in average 36 kg of

water. How many moles of water does

that represent?

#mol = wt / MW; MW(H2O) = 18.02 g/mol

#mol = ?


Illustration 2 l.jpg
Illustration 2

  • Assume you have 0.123 mol of C14H18O4, benzoyl peroxide. What mass does that correspond to?

  • Wt = #mol x MW

    MW (C14H18O4) = (14*12.01 + 18 + 4*16) g/mol = ?

  • Extra exercise: #81, 82, pg 80


Rq2 20 l.jpg
RQ2-20

  • The adult human body has about 1014 cells. Does the body of an adult human have more or less than a mole of DNA?

    a. More, because the average adult has about 2000 moles of water and DNA is larger than water.

    b. Less, because the human body can have no more than 1014 DNA molecules, which is less than a mole

    c. Less, because the average adult has about 2000 moles of water and DNA is smaller than water.


4 2 percent composition l.jpg
4.2. Percent Composition

  • Restriction: wt/wt

  • Definition:% by weight =

    * (wt of atom / wt of molecule) x 100

    * (wt of component / wt of whole item)x100

  • Law of Definite Composition: “the proportion of each element in a sample of pure compound is constant.”


Illustration l.jpg
Illustration

  • Calculate the weight % of iron in Fe2O3. What mass of iron is in 25.0 g of Fe2O3?

  • Extra Exercise: Find the % of C, H and O in glucose: C6H12O6.

  • Extra exercise: #69,pg 126


Illustration continued l.jpg
Illustration (Continued)

  • Information provided

  • Formula: Fe2O3

  • Information requested: % of Fe in Fe2O3.

  • Way to answer: use MW’s

  • % of Fe = 100 x (2 x MW(Fe) / MW(Fe2O3)) = ?

  • Extra example: #65, pg 126


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Illustration (Continued 2)

  • Information provided b:

  • Weight of Fe2O3: 25. 0 g

  • Information requested: mass of Fe in the Fe2O3 sample

  • Way to answer:

  • Wt(Fe) = % x Wt(Fe2O3) / 100 = ?

  • Extra exercise: # 77, pg 127


Rq2 20b previous material review l.jpg
RQ2-20B: Previous Material Review

  • The number of moles of a substance indicates the: a, weight; b, number of components; c, density of the substance

  • The percentage by weight of salt in a water solution indicates for example the number of grams of salt in: a, 100g; b, 100 mL; c,10 g of solution


Rq2 20b2 previous material review l.jpg
RQ2-20B2: Previous Material Review

  • If you are given the #mol of a substance for which you know the molar weight, you can find the weight of the substance by: a, multiplying; b, dividing: c, adding #mol and MW.

  • If you are given the % composition of solute in a solution for which you know the weight, you can find the weight of solute by using this expression:

    a, wt(solute) = % / wt(solution) / 100;

    b, wt(solute) = % x wt(solution) x 100;

    c, wt(solute) = % x wt(solution) / 100


Slide19 l.jpg

SUCCESS

UNDERSTAND

REMEMBER

REVIEW

PRACTICE

WANT

WISH


Word of wisdom hard work20 l.jpg
WORD OF WISDOMHard Work

Work is too hard?

Keep trying. It’ll geteasier and easier

Ask for help

When the going gets tough, the tough gets going


Word of wisdom before the gain21 l.jpg
WORD OF WISDOMBefore the Gain

Unconfortable when studying? Expect the unconfort.

Welcome the unconfort. It is beneficial to your learning

You’re changing your thoughts

From UNFOCUSED TO FOCUSED

Say in your mind: “I’m happy to do it”. You feel a smile forming in your mind


Word of wisdom22 l.jpg
WORD OF WISDOM

Not enoughtime?

MAKE the time!

Sacrifice something.

Your better future: worth the sacrifice

Tolerate Hardship on the way to Prosperity


4 3 derivation of formulas l.jpg
4.3. Derivation of Formulas

  • Types of formulas:

  • Empirical formula: shows the simplest ratios of atoms (groups of atoms) in a molecule

  • Molecular Formula: shows the actual numbers of atoms in a molecule

    a. Derivation of Empirical Formulas

  • General procedure:

    Weights of components Moles of components Raw Formula Intermediate formula Empirical formula



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Detailed Procedure

  • Raw formula: obtained using (calculated) # of moles for subscripts

  • Intermediate formula: obtained dividing all subscripts by the smallest subscript

  • Empirical formula: obtained after operation to round all fractional subscripts into the smallest whole numbers.


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Special Fractional Subscripts

  • Unrounded Subscripts that end with .33, .66 .25, .5 and .75

  • Procedure:

  • Multiply the fractional subscript in the intermediate formula by the smallest number needed to produce the smallest whole number (or a number close to it).


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Illustration

  • Information provided:

    % composition of mandelic acid (an alpha hydroxy acid): C(63.15%), H(5.30%), and O(31.55%)

  • Information requested: EF of mandelic acid.

  • How do you proceed to solve the problem? See RQ2-20b


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RQ2-20b

  • How do you proceed to solve the problem in the previous question using the KNU method. Justify your answer.

  • a. Find RF from IF, itself from EF, itself found using the #mol of components of mandelic acid.

  • b. Find EF from IF, itself from RF, itself found using the #mol of components of mandelic acid.

  • c. Find IF from EF, itself from RF, itself found using the #mol of components of mandelic acid.


Rq2 20c previous material review l.jpg
RQ2-20C: Previous Material Review

  • The empirical formula of a compound is determined first by using as subscripts the: a, weight; b, volume; c, number of moles of the atoms that make up the compound

  • If one of the subscripts in the intermediate formula ends with .67, the subscript should be: a, rounded to the next whole number; b, multiplied by 3; c, divided by 3 to get the empirical formula


Illustration solution l.jpg
Illustration Solution

  • Empirical Formula <- Intermediate Formula <- Raw Formula: CxHyOz

    x, y, z = #mol of C, H, and O

  • #mol = wt / MW

  • Wt = unknown. Use % values

    * Wt of C = ?

    MW = 12.0 g/mol

    * Wt of H = ?

    MW = 1.01 g/ mol

    * Wt of O = ?

    MW = 16.0 g/mol


Illustration solution continued l.jpg
Illustration Solution (Continued)

  • Plug info of Wt and MW in the expression of #mol

  • Plug info of #mol in the expression of raw formula

  • Use the raw formula to find the intermediate formula

  • Use the intermediate formula to find the empirical formula

  • Extra exercise: #81, pg 127


Molecular formulas l.jpg
Molecular Formulas

  • Empirical Formula (EF)-> Empirical weight (EW)

  • Molecular Formula (MF)-> Molecular weight (MW)

  • MW/EW = Molecular/Empirical Ratio (MER)

  • MF/EF = MER -> MF = EF x MER


Molecular formulas example l.jpg
Molecular Formulas (Example)

  • Information provided:

    * MW of cacodyl = 210 g/mol

    * % C: 22.88 %

    * % H: 5.76 %

    * As: 71.36 %

  • Information requested : Empirical and molecular formulas of cacodyl


Molecular formulas example continued l.jpg
Molecular Formulas (Example Continued)

  • EF Intermediate formula <- Raw formula: CxHyAsz

  • x, y, and z = #mol of C, H, and As

    #mol = wt / MW;

    Find wt using %’s.

    What do you use the EF information for?

    See RQ2-20C


Ef and mf of cacodyl continued l.jpg
EF and MF of Cacodyl (Continued)

  • EF -> EW = ?

  • MW / EW = MER = ?

  • MF = EF x MER = ?;

  • Extra exercise: #117, pg 128


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Molecular Formulas (Example 2)

  • Info provided:

    * wt of S: 1.256 g

    * wt of SFx: 5.722 g

  • Info requested:

    Value of x in SFx.


Molecular formulas example 2 solution l.jpg
Molecular Formulas (Example 2 Solution)

  • EF <- Raw formula: SmFn;

    n, m = #mol of S and F

    #mol = wt / MW;

    #mol of S = ?

    Find wt of F using wt of SFx – wt of S

    #mol of F =?

  • Extra example: #121, pg 128


Ef mf example 3 using results of reactions 120 pg 128 l.jpg
EF/MF Example 3: Using Results of reactions#120, pg 128

  • Information provided

    * Weight of estrone (made of C, H, and O): 1.893 g

    * Weight of CO2 from combustion of estrone: 5.545 g

    * Weight of H2O from combustion of estrone : 1.388 g

    * MW of the estrone : 270.36 g/mol

  • Information requested: MF of estrone


Ef mf example 3 solution l.jpg
EF/MF Example 3 (Solution)

  • MF = EF x MER

  • EF <- IF <- Raw formula, CxHyOz.

    x, y, and z = #mol of C, H and O.

  • #mol of C = #mol of CO2.

  • #mol of H = 2 x #mol of H2O

  • #mol of O = wt/MW

  • Wt of O = total wt (sample) – wt of C and H

  • Continued on next slide


Ef mf example 3 solution continued l.jpg
EF/MF Example 3 (Solution, Continued)

  • Plug resulting weight into the expessions of #mol of O

  • Use #mol of C, H, and O to get the RF.

  • Use the RF to get the IF

  • Use the IF to get the EF

  • Use the EF to calculate the EW

  • Use the EW and the MW to get the MER

  • Use the MER and the EF to get the MF

  • Extra exercise: #119, pg 128.


Rq2 21 l.jpg
RQ2-21

  • What type of information about moles does the molecular formula provide?

  • a. The actual number of moles ofcomponent atoms per mole of compound

  • b. The simplest ratio of moles ofcomponent atoms per mole of compound

  • c.It does not give any information about moles.Only the actual numbers ofcomponent atoms per molecule of compound


4 4 solutions l.jpg
4.4. Solutions

  • Solution: homogeneous mixture of two or more substances.

  • Solvent: most abundant component.

  • Solute: least abundant

  • Concentration: amount of solute in a given amount of solvent

  • Molarity (M) = Measure of the concentration of a solution

    = moles of solute / volume (liters) of solution. M = #mol/V

  • Units: mol/L

  • Commercial unit : molar (M)


Molarity illustration l.jpg
Molarity (Illustration)

Info provided:

  • Wt of K2Cr2O7: 2.335 g

  • Volume of final solution: 500 mL = 0.500 L

    Info requested:

  • Molar concentration of K2Cr2O7

  • Molar concentrations of K(+) and Cr2O7(2-)

  • Extra ex: #53, pg 178


Dilution concentration of solutions l.jpg
Dilution/Concentration of Solutions

  • Dilution -> decreased final molarity

  • Concentration -> increased final molarity.

  • Total #mol of solute = constant = same before and after dilution/concentration

    Initial #mol(solute) = Final #mol (solute)

    Mi x Vi = Mf x Vf

    I = initial; f = final


Dilution concentration of solutions illustration l.jpg
Dilution/Concentration of Solutions (Illustration)

Info provided:

  • Volume of original CuSO4 solution: 4.00 mL

  • Molarity of original CuSO4 solution: 0.0250 M

  • Final volume of CuSO4 solution: 10.0 mL

    Info requested: final molarity of CuSO4 solution.

    Extra ex: #55, 178


Rq2 21b previous material review l.jpg
RQ2-21B: Previous Material Review

  • If you divide the #mol of solute by the volume (in L’s) of solution, you are determining the: a; density; b, percent composition; c, molarity of the solution

  • If you want to determine the #mol of solute in a solution, all you have to do is: a, multiply; b, divide; c, add the molarity by the volume of the solution


Rq2 22 l.jpg
RQ2-22

  • Compare the stability of % composition by weight and molarityin function of temperature

  • a. % by weight is stableandmolarity is not. Unlike volumes, weights changewith temperature

  • b. molarity is stableand% by weightis not. Unlike weights, volumes do not changewith temperature

  • c. % by weight is stableandmolarity is not. Unlike volumes, weights do not changewith temperature


4 5 calculations based on equations and reactions l.jpg
4.5. Calculations based on Equations and Reactions

  • Study case: C3H8 + 5O2 -> 3CO2 + 4H2O

  • Info fromEquations -> Identity & theoretical proportionsbetweenreactants & products of a reaction.

    Proportions -> mole ratios in equations

  • Info from Actual reactions: amounts of reactants actually used / products actually formed

    Weights -> # of moles -> mole ratios in actual reactions.

  • General Principle:

    mole ratios from actual reaction= mole ratios from equation


A target 1 amounts of reactants or products l.jpg
a. Target # 1: amounts of reactants or products

  • Condition: Reactants proportions are as shown by the mole ratios from equations

  • General Principle:

    mole ratios from equation = mole ratios from actual reaction

  • Mol ratio = #mol of target substance / #mol of known substance

  • #mol of target substance = mol ratio of (Target/Known) x #mol of known substance

  • How do you determine the weight of target substance? See RQ2-23


Rq2 23 l.jpg
RQ2-23

  • Use the expression of #mol of target substance to determine the weight of target substance

  • a. wt of target substance=mol ratio of (Target/Known)x #mol of knownx MW of target substance

  • b. wt of target substance=mol ratio of (Known/Target)x #mol of knownx MW of known substance

  • c. wt of target substance=mol ratio of (Target/Known)x #mol of known/ MW of target substance


Finding amounts of reactants or products illustration l.jpg
Finding amounts of reactants or products (Illustration)

Info provided:

  • CH4 + 2O2 -> CO2 + 2H2O

  • Wt of CH4: 25.5 g

    Info requested: wt of O2 needed to burn the CH4 completely

    What relation do you use to solve the problem? See RQ2-24


Rq2 24 l.jpg
RQ2-24

  • What relation do you use to solve the problem in the previous question? Which parameter is known and which one should be determined next?

  • a. Wt(O2) = #mol (O2) / MW. MW is known and #mol(O2) should be determined next.

  • b. Wt(O2) = MW / #mol (O2). #mol(O2)is known and MW should be determined next.

  • c. Wt(O2) = #mol (O2) x MW. MW is known and #mol(O2) should be determined next.


Illustration solution53 l.jpg
Illustration (Solution)

  • Wt(O2) = #mol (O2) x MW

  • #mol(O2) = mole ratio of (O2/CH4) x #mol(CH4)

  • #mol(CH4) = ?

  • Use #mol(O2) to find wt(O2).

  • Extra exercise: #33, pg 177


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Finding amounts of reactants or products (Illustration 2)

Info provided:

  • Volume of HNO3 solution: 50.0 mL

  • Molarity of the HNO3 solution: 0.125 M

  • Reaction equation:

    Na2CO3 + 2HNO3 -> 2NaNO3 + CO2 + H2O

    Info requested: wt of Na2CO3 needed for the complete reaction.


Illustration 2 solution l.jpg
Illustration 2 (Solution)

  • Wt(Na2CO3) = #mol x MW (Na2CO3)

  • #mol (Na2CO3) = mole ratio of (Na2CO3)/ HNO3) x #mol(HNO3)

  • What parameter is not known in the previous expression? How do you find it?

  • See RQ2-25


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RQ2-25

  • What parameter is not known in the previous expression #mol (Na2CO3) = mole ratio of (Na2CO3)/ HNO3) x #mol(HNO3) ? How do you find it?

    a. Unknown: #mol(HNO3), found using the relation #mol = wt/MW

    b.Unknown: #mol(HNO3), found using the relation #mol = M x V

    c.Unknown: mole ratio, found using the relation wt = #mol x MW


Illustration 2 solution continued l.jpg
Illustration 2 (Solution, Continued)

  • #mol(HNO3) = M x V = ?

  • Use #mol(Na2CO3) to find wt(Na2CO3).

  • Extra exercise: # 59, pg 179


Rq2 25b previous material review l.jpg
RQ2-25B: Previous Material Review

  • If you know the #mol of reagent A in the reaction A + 3B -> AB3, you can find the #mol of B using the: a, weight; b, mole; c, volume ratio of B to A.

  • In order to find the weight of B in the previous reaction, the mole ratio of B to A is multiplied by the #mol of A and the molar weight of: a, B; b, AB3; c, A


B target 2 limiting reagent lr l.jpg
b. Target 2: Limiting Reagent (LR)

  • LR = reactant that is used in less than enough amount to complete the reaction with the other reactant.

  • LR limits the reaction. It stops when LR is used up.

  • General procedure

    * Find #mol of reactants

    * Find the simplest reactant/reactantmole ratio

    from actual reaction in decimal form

    * Compare reactant/reactant mole ratios from

    actual reaction and from equation (also in decimal form) the : insufficient reactant = limiting reagent


Illustration60 l.jpg
Illustration

Info provided

  • Reaction equation: 2Al + 3Cl2 -> 2AlCl3

  • Wt of Al: 2.70 g

  • Wt of Cl2: 4.05 g

    Info requested: limiting reactant = ?


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Way to the Solution

  • LR = insufficient reactant through comparison of reactant/reactant mole ratios from equation and actual reaction.

  • Al/Cl2 mole ratio from equation: 2/3

  • Al/Cl2 mole ratio from actual reaction = ?

  • #mol (Al) = ?

  • #mol (Cl2) = ?


Way to solution continued l.jpg
Way to Solution (Continued)

  • Mole ratio of Al/Cl2 from actual reaction = ?

  • Put the ratio in same form as 2/3 = 0.67 from the equation. Divide numerator and denominator by same number that produces 2 at the numerator (for example).

  • Compare the two mole ratios. The insufficient reactant = LR = ?

  • Extra ex: #37, pg 177


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Limiting Reagent and Co-reactant leftover

  • What amount of unreacted other reactant is left?


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c. Target # 3: Reaction Yield

  • Yield = Quantitative result of an actual reaction

  • *1. Theoretical yield (TY)

  • TY = weight of product expected from a reaction that runs completely and perfectly

  • TY = #mol x MW of target product(TP).

  • #mol(TP) = Mol ratio of (TP / LR) x #Mol of LR


Reaction yield continued l.jpg
Reaction Yield (Continued)

  • *2. Actual yield (AY)

    AY = Weight of material actually collected from the reaction

    *3. (%) Yield

    (%) Yield= (AY / TY) x 100


Illustration66 l.jpg
Illustration

  • Under certain conditions the formation of ammonia from nitrogen and hydrogen has a 7.82% yield. Under these conditions, how many grams of NH3 will be produced from the reaction 25.0 g N2 with 2.00 g H2?

  • The reaction occurs according to the following equation:

    N2(g) + 3 H2(g) -> 2 NH3(g)


Way to the solution67 l.jpg
Way to the Solution

Info provided:

* Reaction: N2 + 3H2 -> 2NH3

* Wt of N2: 25.0 g

* Wt of H2: 2.0 g

* % yield of the reaction: 7.82%

Info requested

* Wt of NH3 produced (AY)


Way to solution 2 l.jpg
Way to Solution (2)

  • Expression of requested wt of NH3 is AY = %Y x TY / 100

  • Expression of TY = mole ratio of (NH3/LR) x #mol of LR x MW of NH3

  • Limiting reagent = insufficient reactant between N2 and H2.

  • How do you determine the limiting reagent?

  • See RQ2-26


Rq2 26 l.jpg
RQ2-26

  • How do you determine the limiting reagent?

    a. Compare mole ratios from actual reaction to those from equation

    b. Compare moles from actual reaction to those from equation

    c. Compare weight ratios from actual reaction to those from equation


Way to solution continued70 l.jpg
Way to Solution (Continued)

  • Compare mole ratios from actual reaction to those from equation

  • Mole ratio from equation = ?

  • Mole ratio from actual reaction = #mol of H2 used / #mol of N2 used

  • #mol of N2 used = ?

  • #mol of H2 used = ?


Way to solution 4 l.jpg
Way to Solution (4)

  • TY = (mol ratio of NH3/H2) x #mol of H2 x MW of NH3 = (2/3 x 0.990 mol) x 17.0 g/mol = 11.22 g

  • Requested wt of NH3 is AY = (% x TY)/ 100 = 7.82 x 11.22 g / 100 = 0.877 g

  • Extra Exercise: # 45, pg 179


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Calculations on Reactions (Additional Illustration)

  • A mass of 2.052 g of a metal carbonate, MCO3, is heated to give the metal oxide and 0.4576 g CO2.

  • MCO3(s) -> MO(s) + CO2(g)

  • What is the identity of the metal?


Empirical formula additional illustration l.jpg
Empirical Formula (Additional Illustration)

  • A 4.236 g sample of a hydrocarbon is combusted to give 3.810 g of H2O and 13.96 g of CO2. What is the empirical formula of the compound?


Calculations on reactions additional illustration 2 l.jpg
Calculations on Reactions (Additional Illustration 2)

  • A 1 g of chewable vitamin C requires 27.85 mL of 0.102 M Br2 solution for titration to the equivalence point. What is the mass of vitamin C in the tablet?


Calculations on reactions additional illustration 2b l.jpg
Calculations on Reactions (Additional Illustration 2b)

Info provided

* Reaction of Vit C

* Wt of Vit C Tablet

* Volume of Br2 solution

* Molarity of , , , ,

  • Info requested: wt of Vit C in tablet


Empirical formulas example 2 l.jpg
Empirical Formulas (Example 2)

  • A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound?


Empirical formula problem2 analysis l.jpg
Empirical Formula Problem2 (Analysis)

  • Information provided

    * Weight of the compound: 35.0 mg

    *Weight of CO2: 33.5 mg

  • Weight of H2O: 41.1 mg

    Information requested

    Empirical formula of the compound


Empirical formula problem 2 solution l.jpg
Empirical Formula Problem 2 (Solution)

  • Determination of EF requires raw formula: CxHyNz. x, y, z = #mol of C, H, and N.

  • #mol of C = same as #mol of CO2

  • #mol of H = 2 x #mol of H2O

  • #mol of N = wt/MW. Wt of N = Total wt – wt of C and N.