Slides Shown Friday Oct. 10 2008

An Acid: Any substance that An Acid: Any substance that increases the [Hincreases the [H++] in water] in water

(g) Gas (l) Liquid (s) Solid

(aq) Aqueous (water) solution

HCl(g): hydrogen chloride gas

HCl(aq) hydrochloric acid

Naming Acids

1) Binary acids solutions form when certain gaseous compounds dissolve in water. For example, when gaseous hydrogen chloride (HCl) dissolves in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid hydrochloric acid

2) Oxoacid names are similar to those of the oxoanions, except for two suffix changes: Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite” suffix becomes an “-ous” suffix in the acid. The oxoanion prefixes “hypo-” and “per-” are retained. Thus, BrO4

-

is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and

HIO2 is iodous acid.

Naming acids

Names of Acids that do Contain Oxygen

Acid Name

HF hydrofluoric acid

HCl hydrochloric acid

HBr hydrobromic acid

HI hydroiodic acid

HCN hydrocyanic acid

H2S hydrosulfuric acid

Names of some Oxygen-Containing Acids

Acid Name

HNO3 nitric acid

HNO2 nitrous acid

H2SO4 sulfuric acid

H2SO3 sulfurous acid

H3PO4 phosphoric acid

HC2H3O2 acetic acid

Naming of the Oxoacids of Chlorine

Acid Anion Name

HClO4 perchlorate perchloric acid

HClO3 chlorate chloric acid

HClO2 chlorite chlorous acid

HClO hypochlorite hypochlorous acid

Mass spectrometer

Relative intensities of the signals recorded when natural neon is injected into a mass spectrometer.

Mass spectrum of natural copper

Oxygen, Isotopes

168O 8 Protons 8 Neutrons

99.759% 15.99491462 amu

• 178O 8 Protons 9 Neutrons

0.037% 16.9997341 amu

• 188O 8 Protons 10 Neutrons

0.204 % 17.999160 amu

Calculating the “Average” Atomic Mass (aka the Atomic Weight) of an Element

24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu

24.309268 amu

With Significant Digits = 24.3 amu

Problem: Calculate the average atomic mass of Magnesium. Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).

Calculate the Average Atomic Mass of Zirconium, Element #40

Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr.

Isotope (% abd.) Mass (amu) (%) Fractional Mass

90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu96Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu

91.2215 amu

With Significant Digits = 91.22 amu

Problem: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol.

Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0

Solution: X(78.918336) + Y(80.91629) = 79.904

X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904

78.918336 - 78.918336 Y + 80.91629 Y = 79.904

1.997954 Y = 0.985664 or Y = 0.4933

X = 1.00 - Y = 1.00 - 0.4933 = 0.5067

%X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br

During a perplexing dream one evening you come across 200 atoms of Einsteinium. What would be the total mass of this substance in grams?

SOLUTION:

200 Atoms of Es X 252 AMU/Atom = 5.04 x 104 AMU

5.04 x 104 AMU x (1g / 6.022 x 1023 AMU) =

8.37 x 10-20 g of Einsteinium

Atomic Definitions II: AMU, Dalton, 12C Std.

Atomic mass Unit (AMU) = 1/12 the mass of a carbon-12 atom on this scale Hydrogen has a mass of 1.008 AMU.

Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12C has a mass of 12.00 daltons.

Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12C the chosen standard.

Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

MOLE

• The Mole is based upon the following definition:• The amount of substance that contains as many

elementary parts (atoms, molecules, or other?) as there are atoms in exactly

• 12 grams of carbon-12.

• 1 Mole = 6.022045 x 1023 particles

One-mole samples of copper, sulfur, mercury, and carbon

One moleof commonSubstances

CaCO3

100.09 gOxygen 32.00 gCopper 63.55 gWater 18.02 g

MolesMoles

Molecules

Avogadro’s Number

Molecular Formula

Atoms

6.022 x 1023

Mole - Mass Relationships of Elements

Element Atom/Molecule Mass Mole Mass Number of Atoms

1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms

1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms

1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms

1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms

1 molecule of O2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule

1 molecule of S8 = 256.56 amu 1 mole of S8 = 256.56 g = 6.022 x 1023 molecules

Calculate the Molecular Mass of Glucose: C6H12O6

• Carbon 6 x 12.01 g/mol = 72.06 g

• Hydrogen 12 x 1.008 g/mol = 12.096 g

• Oxygen 6 x 16.00 g/mol = 96.00 g

180.16 g

Molecular Mass - Molar Mass ( M )

The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu

Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = 18.02 g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

How many carbon atoms are present in a 2.0 g tablet of Sildenafil citrate (C28H38N6O11S) ?

SOLUTION:

MW of Sildenafil citrate = 28 X 12 amu (C) +

38 X 1 amu (H) +

6 X 14 amu (N) +

11 X 16 amu (O) +

1 X 32 amu (S) =

666 AMU

2.0 g (C28H38N6O11S) X 1 mol/666g =

3.0 X 10-3 mol (C28H38N6O11S)

3.0 X 10-3 mol (C28H38N6O11S) X

6.022 X 1023 molecules / 1 mol (C28H38N6O11S) =

1.8 X 1021 molecules of C28H38N6O11S

1.8 X 1021 molecules of C28H38N6O11S X

28 atoms of C / 1 molecules of C28H38N6O11S =

5.1 X 1022 Carbon Atoms

Calculating the Number of Moles and Atoms in a Given Mass of Element

Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal?Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number!Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol

NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 1020 atoms of Tungsten

1 mol W183.9 g W

6.022 x 1023 atoms 1 mole of W

Calculating the Moles and Number of Formula Units in a given Mass of Cpd.

Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample?Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients.Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol

Converting mass to moles:Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4

= 0.23545 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4= 1.46 x 1023 formula units

Flow Chart of Mass Percentage Calculation

Moles of X in one mole of Compound

Mass % of X

Mass fraction of X

Mass (g) of X in onemole of compound

M (g / mol) of X

Divide by mass (g) of one mole of compound

Multiply by 100

Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I

Problem: Sucrose (C12H22O11) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose?

(a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd

= 0.421046 To find mass % of C = 0.421046 x 100% = 42.105%

Mass Fraction of C = =

Calculating Mass Percents and Masses of Elements in a Sample of Compound - II

(a) continued Mass % of H = x 100% = x 100%

= 6.479% H

Mass % of O = x 100% = x 100%

= 51.417% O

(b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose)

Mass (g) of C = 24.35 g sucrose X = 10.25 g C

mol H x M of H 22 x 1.008 g Hmass of 1 mol sucrose 342.30 g

mol O x M of O 11 x 16.00 g Omass of 1 mol sucrose 342.30 g

0.421046 g C 1 g sucrose

Mol wt and % composition of NH4NO3

• 2 mol N x 14.01 g/mol = 28.02 g N• 4 mol H x 1.008 g/mol = 4.032 g H• 3 mol O x 15.999 g/mol = 48.00 g O

80.05 g/mol

%N = x 100% = 35.00% N28.02g N2

80.05g

%H = x 100% = 5.037% H 4.032g H2

80.05g

%O = x 100% = 59.96% O48.00g O2

80.05g99.997%

Calculate the Percent Composition of Sulfuric Acid

H2SO4

Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol

%H = x 100% = 2.06% H2(1.008g H2) 98.09g

%S = x 100% = 32.69% S1(32.07g S) 98.09g

%O = x 100% = 65.25% O4(16.00g O) 98.09 g

Check = 100.00%