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MOLE CONCEPTATOMIC MASS

* By international agreement, an atom of the carbon isotope (carbo-12) that has six protons and sixneutrons has a mass of exactly 12 atomic mass units (amu).

1 amu - a mass exactly equal to one-twelfth the mass of one carbon-12 atom

MOLAR MASS OF AN ELEMENT AND AVOGADRO'S NUMBER

Mole(mol) - the amount of substance that contains as many elementary entities (atoms, molecules, orother particles) as there are atoms in exactly 12 grams of the carbo-12 isotope: 1 mole = 6.022045x1023particles. This number (6.022x1 023

) is called Avogadro's number, in honor of the Italian scientistAmedeo Avogadro.

Molar Mass - the mass (in grams or kilograms) of 1 mole of units (such as atoms or molecules) of asubstance. Is numerically equal to the atomic mass of an element.

11 g = 6.022x1 023 amu

Molecular mass - is the sum of he atomic masses (in amu) in the molecule.

no. of moles = mass/ molar mass

no. of molecules (or atoms) = no. of moles x Avogadro's no.

Illustrative Examples:

1. Zinc (Zn) is a silvery metal that is used to form brass (with copper) and plate iron to prevent corrosion.(a) How many grams of zinc are there in 0.356 mole of zinc? (b) How many moles of zinc are in 668 g ofZn? Ans. (a) 23.3 g (b) 10.2 mol

2. Silver (Ag) is a precious metal used mainly in jewelry. What is the mass (in grams) of one Ag atom?Ans.1.792x10·22 g

3. Sulfur is a nonmetallic element. Its presence in coal gives rise to the acid rain phenomenon. How manyatoms are in 16.3 g of S? Ans.3.06x1023atoms

4. Calculate the molecular masses of: (a) sulfur dioxide (802) (b) calcium nitrate Ca(N03h.Ans. (a) 64.07 amu

5. Methane (CH4) is the principal component of natural gas. How many moles of CH4 are present in 6.07g of CH4? Ans. 0.378 mol

6. How many hydrogen atoms are present in 25.6 grams of urea [(NH2hCO], which is used as fertilizer, inanimal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g.

Ans. 1.03x1 024 atoms

PERCENT COMPOSITION OF COMPOUNDS

percent composition by mass - is the percent by mass of each element the compound contains.molar mass of element

%element in a given cmpd = x 100molar mass of cmpd

Illustrative Example:Phosphoric acid (H3P04) is used in detergents, fertilizers, and toothpastes. It is also the

ingredient responsible for the tangy flavor of carbonated beverages. Calculate the percent composition bymass of H, P, and 0 in this compound.Solution:

4x16.00 = 64.001x30.97 = 30.973x1.008 = 3.02

97.99 amu

%H=3(1.008~

97.99~x 100 = 3.086%

O/OP= 1(30.97~

97.99 :at1'f[(x 100 = 31.61%

4(16.~mu)

%0= x 100 = 65.31%97.99a~

EMPIRICAL AND MOLECULAR FORMULAS

Empirical formula - the simplest formula of a given compound, which gives the smallest whole numberratio of atoms that make up the compound.

Steps:1. If not given directly, find the relative number of grams of each element present. That is, if the

composition is given as a percentage, consider a total of 100 grams and determine from that whatthe relative number of grams of each element would be.

2. Convert each of these quantities in grams to moles of atoms, using the atomic weight of eachelement.

3. Convert the ratio of numbers obtained to the smallest possible whole numbers, through dividingby the smallest value of number of moles, and then multiplied to whatever number is necessary toobtain whole numbers.

Illustrative Example:Ascorbic acid (vitamin C) cures scurvy and help prevent the common cold. It is composed of

40.92%C, 4.58%H, and 54.50%0 by mass. Determine its empirical formula.

Solution:Step 1: The given should be in nO.of grams; but in this problem elements are expressed in percentage.

To get the grams values, have a basis of 100 g of ascorbic acid.Basis: 100 g ascorbic acid

me = 40.92g mH = 4.58g mo = 54.50g

Step 2: me 40.92 r(ne = ---

mWc 12'0J.-9"/mol

mH 4.58'nH= ---

mWH 1.00~/mol

mo 54.50 ~ne = ---

mWo 16.0~/mol

Step 3:

= 3.407 mol

= 4.54 mol

= 3.406 mol--7 the smallest value

3.407 4.54 3.406C: ---=1 H: = 1.33 0: = 1

3.406 3.406 3.406

This gives us CH1330 as the formula of ascorbic acid. Next, we need to convert 1.33, thesubscript for H, into an integer. This can be done by trial and error procedure:

1.33 x 2= 2.66 1.33 x 3 = 3.99",4Because 1.33 x 3 gives as an integer(4), we can multiply all the subscripts by 3 and obtain C3H403 as theempirical formula for ascorbic acid.

Molecular formula - the true formula which shows the actual number of atoms of each element presentin one molecule of the compound. To calculate molecular formula from the empirical formula, themolecular mass must be known.

Illustrative example:A compound is found to contain 27.3 g of carbon, 4.55 g of ydrogen, and 36.4 g of oxygen. The

molecular mass is found to be 90.0. Determine the molecular formula of the compound.

Solution:Step 1: The given were already expressed in grams. Proceed to Step 2.Step 2: me 27.3 )Y

ne = = 2.27 mol --7 the smallest value

mWc 12.0~mol

mH 4.55 g' = 4.51 molnH= ---mWH 1.00~/mol

mo 36.4g..-/= 2.28 molnc = ---

mWo 16.0~/mol

Step 3:4.51 2.282.27

C: = 1 H: = 1.99",,2 0: = 1.00

2.27 2.27 2.27

This gives us CHzO as the empirical formula of the compound.

Step 4: From the empirical formula, the empirical molar mass would be:CH20\ I' 1 x 16.00 amu = 16.00amu

2 x 1.008 amu = 2.02amu1 x 12.01 amu = 12.01amu

30.03amu

Step 5: Divide the molecular mass (molar mass) by the empirical molar mass in determining themolecular formula.

Molecular massEmpirical molar mass

= 90.0 = 330.03

Multiplying the subscripts of each element in the given compound by the ratio of themolecular mass to the empirical molar mass, which is 3:The molecular formula would be C3H603.

CHEMICAL RACTIONS AND CHEMICAL EQUATIONS

Chemical reaction - is a chemical changeChemical equation - uses chemical symbols to show what happens during a chemical reaction.Reactants - are the starting materials in a chemical reaction.Products - are the substances formed as a result of a chemical

Reaction.

AMOUNTS OF REACTANTS AND PRODUCTSStoichiometry - shows the mass relationships among reactants and

products in a chemical reaction.

The mole method consists of the following steps:1. Write correct formulas for all reactants and products, and balance the resulting equation.2. Convert the quantities of some or all given or known substances (usually reactants) into moles.3. Use the coefficients in the balanced equation to calculate the number of moles of the sought or

unknown quantities (usually products) in the problem.4. Using the calculated numbers of moles and the molar masses, convert the unknown quantities to

whatever units are required (typically grams). .

Calculating the Amount of ProductIllustrative Example:

An alkali metals react with water to produce hydrogen gas and the corresponding alkali metalhydroxide. A typical reaction is that between lithium and water:

2Li(s) + 2H20(I)---7 2LiOH(aq) + Hz(g)

(a) How many moles of Hz will be formed by the complete reaction of 6.23 moles of lithium withwater? Ans. 3.12 mol

(b) How many grams of Hz will be formed by the complete reaction of 8.57 g of lithium withwater? Ans. 11.70 g

Practice Exercise:The reaction between nitric oxide(NO) and oxygen to form nitrogen dioxide(N02) is a key step in

photochemical smog formation:2NO(g) + 02(g) --7 2N02(g)

(a) How many moles of N02 are formed by the complete reaction of 0.254 mole of 02?(b) How many grams of N02 are formed by the complete reaction of 1.44 g of NO?

LIMITING REAGENTS AND YIELDS OF REACTIONS

Stoichiometric amounts - the proportions indicated by the balanced equation.Limiting reagent - the reactant used up first in a reaction, on which the maximum amount of productformed depends on how much of this reactant was originally present.Excess reagent - the reactant present in quantities greater than those needed to react with the quantityof the limiting reagent.Theoretical yield - the amount a product that would result if all the limiting reagent reacted.Actual yield - the quantity of product that actually results fro the reactions.Percent yield - the proportions of the actual yield to the theoretical yield.

Actual yield%yield = x 100

theoretical yield

Illustrative Example:Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, and jet

engines. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 950°Cand 1150°C:

TiCI4(g) + 2Mg(l) ---7 Ti(s) + 2MgCI2(1)

In a certain operation, 3.54x107 g of TiCI4 are reacted with 1.13x1 07 g of Mg.(a) Determine the limiting reagent.(b) Determine the number of moles of the excess reagent.(c) Calculate the theoretical yield of Ti in grams.(d) Calculate the percent yield if 7.91x1 06g of Ti are actually obtained.

Solution:

(a) First, we calculate the number of moles of the reactants (TiCI4 and Mg).1 mol TiCI4

Moles of TiCI4 = 3.54x107~X

= 1.87x105 mol TiCI4

Moles ofMg = 1.13x10~ x

= 4.65x1 05 mol Mg

1 mol Mg

24.31 iM9

From the balanced equation we see that 1 mol TiCI4 "" 2mol Mg; therefore, the number of moles ofmagnesium needed to react with 1.87x105 moles of TiCI4 is

"_____ 2 mol Mg

1.87x10:.morriC14 x ~ = 3.74x105 mol Mg1~

Since 4.65x1 05 moles of Mg are present, more than is needed to react with the amount of TiCI4 we have,Mg must be the excess reagent and TiCI4 the limiting reagent.

(b) no. of moles of excess reagent = 4.65x105 mol Mg - 3.74x105 mol Mg= 0.91 x105 mol Mg

(c) Since the limiting reagent is TiCI4, the theoretical amount of Ti should depend with the amount of TiCI4.And since 1 mol TiCI4 ::::1 mol Ti, the theoretical yield of Ti is: ./

1 mol T 4 1 rnolTi 47.88g Titheo yield = 3.54x107

~ X X ----- X

~ 1j;rli= 8.93x1 06 9 Ti

(d) To find the percent yield, we writeActual yield

%yield = x 100theoretical yield

%yield =7.91x106 g

8.93x106 gx 100 = 88.6%

GasesNature and Properties of Gases1. Gases have no definite shape and volume.2. Gases diffuse very readily, and any two gases will mix completely when combined.3. Gases can be compressed very readily.4. Gases have much lower densities than liquids or solids.

Kinetic Molecular Theory of Gases1. A gas is composed of molecules that are separated from one another by distances far greater thantheir own dimensions. The molecules can be considered to be "points"; that is, they posses mass buthave negligible volume.2. Gas molecules are in constant motion in random directions' and frequently collide with one another.Collisions among molecules are perfectly elastic. Although energy may be transferred from one moleculeto another as a result of a collision, the total energy of all the molecules in a system remains the same.3. Gas molecules exert neither attractive nor repulsive forces on one another.4. The average kinetic energy of the molecules is proportional to the temperature of he gas in Kelvin. Anytwo gases at the same temperature will have the same average kinetic energy. The average kineticenergy of a molecule is given by

KE = 1/2 m\?Where m is the mass of the molecule and v is its speed.

Some substances Found as Gases at 1 atmH2 Xe HF NH3 HCNN2 He HCI NOO2 Ne HBr N0203 Ar HI N20F2 Kr CO S02CI2 Rn CO2 H2S

Pressure of a GasPressure - force applied per unit area.Manometer - use to measure the pressure of gases other than atmosphere.Barometer - use to measure atmospheric pressure.5.1. unit of pressure: Pascal (Pa)

Conversion factors:1 Pa = 1 N/m2

1 atm = 760 mmHg1 atm = 760 torr1 atm = 101,325 Pa1 atm = 1.01325x102kPa

The Gas Laws1. Boyle's Law: The Pressure-Volume Relationship

- studied by Robert Boyle- states that the volume of a fixed amount of gas maintained at constant temperature is inverselyproportional to the gas pressure.