6.02 formulas and composition - san diego miramar...

1 05.2015 Formulas and Composition

6.02 Formulas and Composition

Using the Mole in Analysis Calculation

Dr. Fred Omega Garces Chemistry 152 Miramar College


One mole substance equivalent to: Atomic Weight (in grams) - atoms Molecular Weight (in grams) - molecules Formula Weight (in grams) - ionic Compounds

Atoms Compound mole mass C6H12O6 C6H12O6 180 g molecule 1 mole (1 mol)

C 6-C atoms 6 mol C 6(12.0) = 72g C H 12-H atoms 12 mol C 12(1.0) = 12g H O 6-O atoms 6 mol O 6(16.0) = 96g O 180 g total From Chemical Formula we obtain the conversion factors: 1 molecule C6H12O6 ! 6C atoms ≅ 12H atoms ≅ 6 O atoms

or "#1 mole C6H12O6 ! 6 C moles ≅ 12 H moles ≅ 6 O atoms

Counting by Mass: A Review


Percent Composition by Mass

What is meant by % ? Consider the % of pennies in 20¢ that contains 3 nickels and 5 pennies

a) % pennies in terms of coins. b) % pennies in terms of $ c) % pennies in terms of mass (1¢ = 2.5 g, 5¢ = 5.0 g) % Composition ! [part/Whole] •100

3"nickels"+""5"pennies"="8"coins

%"pennies"by"coins"=" 5"pennies8"coins

3100"="62.5%

b) % pennies in terms of $:

5"pennies"="5"¢,""Total"$"amount"="20"¢

%"pennies"by"$""=" 5"¢20"¢"

"8"100"="25%

c) % pennies in terms mass: 1"pennies"="2.5"g",""1"nickel"="5.0"g

"mass"of"pennies""="5"pennies ⋅ 2.5"g"1"penny"

"="12.5g

""""""mass"of"nickel""="3"nickel⋅ 5.0"g"1"nickel"

"="15.0g

"%"mass"of"pennies""="

""""""""""=""5"pennies ⋅ 12.5"g"pennies"(12.5"g"+"15.0"g)"total"

⋅100"="45.5"%

a) % pennies in terms of coins


Percent Composition by Mass

% composition of elements by mass. Similarly for a compound the % composition (by mass) of each element can be determine by part/whole analysis.

C6H12O6 - glucose Consider 100 g Sample What is % C, %H and %O ? i) Atomic basis:

% C atom = [6/24]•100 = 25%; % H atom = [12/24]•100 = 50%; % O atom = [6/24]•100 = 25% ii) Mass basis:

Consider 1 mol of C6H12O6 : 6 C, 12 H, 6 O

C: 6 (12.0g/mol) = 72 g C / 1 mol

H: 12 (1.0g/mol) = 12 g H /mol

O: 6 (16.0g/mol) = 96 g O /mol

Total mass = 180 g Total /mol

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%C =72g/mol

180 g/mol• 100 = 40%

%H =12g/mol

180 g/mol• 100 = 6.7%

%O =96g/mol

180 g/mol• 100 = 53%


Formula Composition from % Composition

Suppose given % composition , What is the chemical formula? Actually can only calculate Empirical (simplest) formula. What is the Chemical formula for with 40% C; 6.7%H and 53% O

-Assume 100 g mass. 100 g CxHyOz g 40 g C; 6.7g H; 53g O

Note: 1.0 g of H not H2 !!

Ratio: C3.3 H6.7 O 3.33 Reduce to simplest whole # ratio divide by 3.33 $ C1 H2 O1

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mol C = 40 g C mol12.0 g

= 3.3 mol C

mol H = 6.7 g C mol1.0 g

= 6.7 mol H

mol O = 53 g C mol16.0 g

= 3.3 mol O

Note the best formula that can be obtain is the simplest (Empirical) formula without other information such as the Molar mass or the total number of atoms in the compound.


Chemical Formula from % Composition and Molar Mass

Suppose given % composition and molar mass, what is the chemical formula?

What is the chemical formula of the compound with an empirical formula of CH2O if the molar mass for this compound was determined to be 180.0 g/mol ?

n"=" MolarMassEmpirical"Mass"

"=" 180.0"g/mol30.01"g/mol"

"="6

Correct"formula:C1(x6) "H2(x6) "O1(x6) "="C6H12O6

MolarMass'='n' ⋅ 'Empirical'Mass

Empirical'Mass'='Weight'of'atoms'in'the'empirical'formulan'='multiplier'of'subscript'to'give'correct'chemical'formula.

Empirical'Mass'(C1H2O1)'='12.01'g'(C)'+'2.0'g'(H)'+'16.0'g'(O)'='30.01'g


Empirical Formula: Determination

Calculate the Empirical formula for the following sample: 12.1 % C; 16.2% O; 71,7% Cl Assume 100 g sample

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mol C = 12.1 g C • mol12.0 g

=1.008 mol C

mol O = 16.2 g O • mol16 g

=1.0125 mol H

mol Cl = 71.7 g Cl • mol35.5 g

= 2.02 mol Cl

CC l

C lOCOCl 2

MWt = 98.91 g/mol


Empirical Formula: Combustion Analysis

In determining the empirical formula for an unknown sample combustion analysis is used.

Basic idea: CxHyOz + O2 → CO2 + H2O

C in CO2 is from Cx

H in H2O is from Hy

O - will be determine by difference

Strategy:

Wt sample → wt CxHyOz

Wt. CO2 → wt. C

Wt H2O → wt. H

Wt O = [wt sample] - [wt C] - [wt H]

Wt. C → mol

Wt. H → mol H

Wt. O → mol O

C mole C H mole H O mole O


Empirical Formula: Combustion Analysis Example1 B&L 3.53b: Menthol, the substance we can smell in mentholated cough drops, is composed C, H and O. A 0.1005-g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula for menthol? If the compound has a molar mass of 156 g/mol, what is its molecular formula ?

mass C = 0.2829 g CO2 • 12.0 g/mol C44.0 g/mol CO2

= 7.72 • 10−2 g C

mass H = 0.1159 g H2O• 2(1.0 g/mol) H18.0 g/mol H2O

= 1.29 •10−2 g H

Wt. Sample 0.1005 g

Wt. C 7.72•10-2 g

Wt. H 1.29•10-2 g

Wt. O 1.05•10-2 g

C : 7.72 • 10-2g • mol C12.0 g C = 6.43 •10−3 mol C

H : 1.29 •10-2g • mol H1.0 g H = 1.29 •10−2 mol H

O : 1.05• 10-2 • mol O16 g O = 6.56 •10−4 mol O

C : 6.43• 10−3 mol C 6.56 •10−4 = 9.80 mol C

H : 1.29 •10−2 mol H6.56 •10−4 = 19.67 mol H

O : 6.56 • 10−4 mol O 6.56 •10−4

= 1.0 mol O

1. Mass CO2 and H2O to Mass C and H 2. Mass of O by difference

3. Mass of C, H and O to moles 4. Simplify moles ratio of C, H and O

C9.8H19.67O1 round-off! → ! ! ! C1 0H2 0O1 (Emp Formula)

Emp.Wt • n = Mol. Wt.156.0 • n = 156.0

n = 1C1 0H2 0O1

Emp to Molc' Formula! → ! ! ! ! ! C1 0H2 0O1

5. Empirical Formula to Molecular Formula


Empirical Formula: Combustion Analysis Example2 Consider the following: Combustion of 0.157 g CxHyNwOz , yields 0.213 g CO2 and 0.0310 g H2O. A separate analysis of 0.103 g sample yields 0.0230 g NH3. What is the molecular formula of this substance if it contains 12-O atoms in its chemical formula.

mass C = 0.213 g CO2 • 12.0 g/mol C44.0 g/mol CO2

= 5.81• 10−2 g C

mass H = 0.0310 g H2O• 2(1.0 g/mol) H18.0 g/mol H2O

= 3.44 •10−3g H

mass N = 0.0230 g NH3 • 1(14.0 g/mol) N17.0 g/mol NH3

= 1.894 •10−2 g N

% N in sample = 1.894 • 10−2 N0.103 g sample •100 = 18.39% N

Amt N in 0.157 g sample = 0.157 g •18.39% N = 2.887• 10-2 g N

Wt. Sample 0.157 g

Wt. C 5.81•10-2 g

Wt. H 3.44•10-3 g

Wt. N 2.887•10-2 g

Wt. O 6.668•10-2 g

C : 5.81• 10-2g • mol C12.0 g C = 4.84 • 10−3 mol C

H : 3.44 •10-3g • mol H1.0 g H = 3.44 • 10−3 mol H

N : 2.887 •10-2 • mol N14 g N = 2.06 •10−3 mol N

O : 6.668• 10-2 • mol O16 g O = 4.17 •10−3 mol O

C : 4.84 •10−3 mol C 2.06 •10−3

= 2.35mol C

H : 3.44 •10−3 2.06 •10−3

= 1.66mol H

N : 2.06 •10−3 2.06 •10−3 = 1mol N

O : 4.17• 10−3 2.06 • 10−3

= 2mol O

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C2.35H1.66N1O2 x 3" → " C7H5N3O6


Empirical Formula: In Class

Consider the following: Combustion of 0.157 g CxHyNwOz , yields 0.213 g CO2 and 0.0310 g H2O. A separate analysis

of 0.103 g sample yields 0.0230 g NH3. What is the molecular formula of this substance if it contains 12-O atoms in

its chemical formula.


Summary: Chem Formula of CxHyOz Determine the mass of C in CO2 . Determine the mass of H in H2O . Determine the mass of O by difference Mass CxHyOz – MC – MH = MO. Convert mass of C, H, and O to moles (subscript - x, y, and w). Take mole ratio of Cx, Hy, and Oz and find simplest integer ratio. Simplest ratio (or whole number ratio) is the empirical formula. Divide Molc’ Wt . by the Emp Wt. to get multiplier n for Mol. Formula.

CxHyRw

g CO2

g H2O

mol CO2

= mol C

mol H2O = 2 mol H

Burn in O2

1 mol CO2 44.01 g

1 mol H2O 18.02 g

Empirical Formula

1 mol H2O 18.02 g

g O2

Mass by O2 difference

Mol O (z)

12 g C 1 mol C

g C

1 g H 1 mol H

g H Mol H (y)

Mol C (x)

Molecular Formula

n


Summary

C6H12O6 Glucose

Moles and Chemical Formulas

6.02 formulas and composition - san diego miramar...

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