lecture 3 stoichiometry

7/27/2019 Lecture 3 Stoichiometry

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Expressing mass in chemistry Formula weight Molecular weight Molar mass

Concept of the moleEmpirical formula (simplest formula)Atomic mass %

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Formula weight (FW) sum of masses of all atoms in the chemical formula

e.g. NaCl FW = 23.0 amu + 35.5 amu = 58.5 amu

Molecular weight (MW) sum of masses of atoms in the molecular formula

e.g. H 2O MW = 21.0 amu + 16.0 amu = 18.0 amu

Molecular compounds have FW and MWNon-molecular compounds have only FW

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Number of atoms in 12 g of 12 C 1 mole = 6.02210 23 Avogadros number (NA) 6.022x10 23 mol -1 (note the units) Molar mass (M) mass of 1 mol of molec. formula e.g. M HCl = 36.5 g mol -1

NOTE: Molar mass (in g/mol) is numericallyequal to the FW (in amu)

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Empirical formula : Relative numbers of atoms in a substance can be obtained from elemental analysis dataElemental analysis data relative masses of elements in compound can convert to relative numbers of atoms using the

atomic masses

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A compound contains 75.9% carbon, 17.7%nitrogen, and 6.4% hydrogen by mass.What is the empirical formula?

Step 1 : Assume 100 g of sample.100 g contains 75.9 g of C 17.7 g of N 6.4 g of H.

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The empirical formula is C 5H5N Note:

5.01 is not exactly 5

Remember:It is a measurementMeasurements always have errors

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A compound contains C (63.8%), H (6.4%) andN (29.8%). What is the empirical formula?

In 100 g of sample there are:63.8 g C (1 mol/12.011 g) = 5.31 moles C6.4 g H (1 mol/1.0079 g) = 6.3 moles H29.8 g N (1 mol/14.007 g) = 2.13 moles N

The mole ratio is therefore:C : H : N = 2.49 : 3.0 : 1

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Mass %AnalysisData

Work out molesof atoms in100g of sample

Divide by lowestnumber of molesto get mole ratio

Ismole ratio

simple, whole

numbers?

Multiply by smallestfactor that will result

in whole-number ratio No

Report result(empiricalformula)

Yes

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Need one more thing: Molar mass of cmpd

e.g. MW of compound is 136.15 g/mol.Empirical formula is C 4H4O. What is

molecular formula?FW = (4x12.01)+(4x1.01)+16.00

= 68.08 g mol -1

About half the molar massTherefore , the molecular formula is C 8H8O2

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Quantitative aspects of reactionsRelative moles of species in the reactionKey: balanced chemical equation

From the equation can derive mole ratios Like conversion factors in dimensional analysis Convert reagent consumed into product formed

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Used to find empirical formulae of moleculescontaining only C, H & O C converted to CO 2 H is converted to H 2O

CO2 and H 2O trapped and weighed mass % of C and H are found from stoichiometry 1 C atom per CO 2 trapped; 2 H atoms per H 2ORemaining mass attributed to O

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Stoichiometric equn for combustion of acompound of C, H and O is:

CXHYOZ + { X+Y

/ 4-Z

/ 2} O2 X CO2 +Y

/ 2 H2O

moles CO 2 (RHS) = moles C atom (LHS)moles H 2O (RHS) = 1 / 2 moles H atom (LHS)

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1.663 g of a C-H-O compound is combusted,producing 3.48 g of CO 2 and 0.712 g of H 2O.What is the empirical formula of the compound?

1 mole of C atom produces 1 mole of CO 2 ,therefore,

Amt. C =

22

2COg48.3

= 0.0791 mol C2

2

COg01.44COmol1

2COmol1Cmol1


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1 mole of H 2O produced implies 2 moles of Hin the compound

Amt H =

H and C are in a 1:1 ratio in the original cmpdCombined mass of H and C is: (0.0791 mol)(12.01 g/mol) C +

(0.0790 mol)(1.008 g/mol) H = 1.03 g

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OHg.71202

= 0.0790 mol HOHg8.021

OHmol1

2

2

OHmol1

Hmol2

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The remainder of the compound is O

Mass of O = Mass of Cmpd - Mass of C+H

= 1.663 g - 1.030 g= 0.633 g

Moles of O = (0.633 g)/(16.00 g mol -1 )= 0.0396 mol

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The mole ratio, C:H:O is0.0791 : 0.0790 : 0.0396

Or, dividing by 0.0396,

2.00 : 1.99 : 1

Therefore, the empirical formula is C2H2O

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Students sometimes use stoichiometric ratioswith mass data, without converting to moles

e.g. Find mass O 2 reqd to form 16.7 g Fe 2O3 from Fe 3O4?

Fe 3O4(s ) + O 2(g ) Fe 2O3(s )

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2 31/2

Balance First

Amateur error: Do not do this in a test! mass O 2 = 16.7 g Fe 2O3

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2

OFe3O5.0

= 2.78 g O2


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Stoichiometric ratios are MOLAR ratiosConvert moles of one substance into moles of another substance

moles O 2 = 16.7g Fe 2O3

29

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2OFemol3 Omol5.032

32OFeg59.71 OFemol1

Convert to

moles Fe 2O3

Convert to

moles O 2

= 1.74 10 -2 mol O 2

mass O 2 = 1.74 10 -2 mol O 22

2Omol1Og32.00

= 0.557 g O 2


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In chemistry, the fundamental unit of amount of a substance is the mole Stoichiometry shows quantitativerelationship between reactants andproducts in a reactionStoichiometric info is summarised in thebalanced chemical equationData in mass units usually has to beconverted to moles before it is of ANY useThe conversion factor between mass andmoles is the molar mass (units: g mol -1 )

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What mass of H 2SO4 is produced from the followingreaction if 21.0 g of SO 2 , 5.00 g of O 2 and excessH2O are available, and the reaction goes tocompletion?

(MSO2 = 64.06; M H2SO4 = 98.08; M O2 = 32.00)

Rxn: 2 SO 2 (g ) + O 2 (g ) + 2 H 2O ( l ) 2 H 2SO4 (aq )

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Determine whether O 2 or SO 2 is the limitingreagent from the stoichiometry.

(0.328 mol SO 2) (1mol O 2)/(2mol SO 2)

= 0.164 mol O 2Need 0.164 mol O 2 to react with available SO 2 Have only 0.156 mol O 2 Therefore, O 2 is limiting reagent

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How much H 2SO4 if all the O 2 is consumed?(0.156 mol O 2)(2 mol H 2SO4)/(1 mol O 2)

= 0.312 mol H 2SO4Convert to mass:

(0.312 mol H 2SO4)(98.07 g/mol) = 30.6 g H 2SO4This is the Theoretical Yield of H 2SO4

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Convert masses to molesDivide moles of each reagent by itsstoichiometric constant

Lowest ratio is limiting reagent.e.g. In the preceding example, SO2 : (0.328 mol)/(2 mol) = 0.164

O2 : (0.156 mol)/(1 mol) = 0.156(limiting reagent)

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Theoretical yield : Mass of product formed if a reaction goes to completion (limitingreagent completely consumed)Percent yield : Ratio of actual mass of productformed, divided by the theoretical yield,expressed as a percentage.N.B. To find the theoretical yield, you mustfirst determine the limiting reagent.

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4.21g of cyclohexene (C 6H10 ) reacts inpresence of 10.0 g Br 2 to form 12.1 g of 1,2-dibromocyclohexane (C 6H10 Br2) as:

C6H

10(l ) + Br

2(l ) C

6H

10Br

2(l )

Find the theoretical and percent yields:McHx = 82.14 g/molMBr2= 159.8 g/mol

McHxBr2 = 241.9 g/mol

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Step 2: Find the limiting reagentC6H10 (both stoichiometric constants are 1)

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Step 3: Find theoretical yield of C 6H10 Br2 amount: (0.0512 mol C 6H10 )(1mol C 6H10 Br2 /1mol

C6H10 )= 0.0512 mol C 6H10 Br2

mass: (0.0512 mol)(241.9 g/mol) = 12.4 g

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Stoichiometry is the quantitative relationshipbetween the reactants and productsThink in MOLES! molar mass converts mass to moles and vice versa

Limiting reagent determines theoretical yield Know how to determine the limiting reagent quickly

lecture 3 stoichiometry

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