general chemistry 101 chem د. عبد العزيز بن علي الغامدي 1434-1435 هـ

General Chemistry

101 CHEM

الغامدي. علي بن العزيز عبد دهـ 1434-1435

References

• Chemistry the General Science 11E

T. L. Brown, H. E. LeMay, B. E. Bursten and C. J. Murphy.

العامة .الكيمياء . سليمان. د العويس، أحمد د

. الواصل، عبدالعزيز د الخويطر،السحيباني. عبدالعزيز .د

األول • الفصلي بعد: االختبار يحدد .لم

• : الثاني الفصلي بعد االختبار يحدد .لم

قريبا 134أ2المكتب • سيكون 146أ2و

[email protected]األيميل •

MATTER

We define matter as anything that has mass and takes up space.

Matter

• Atoms are the building blocks of matter.• Each element is made of the same kind of atom.• A compound is made of two or more different kinds of

elements.

States of Matter

Units of Measurement

• SI Units

• Système International d’Unités• A different base unit is used for each quantity.

Metric SystemPrefixes convert the base units into units that are appropriate for the item being measured.

Volume

• The most commonly used metric units for volume are the liter (L) and the milliliter (mL).– A liter is a cube 1 dm long

on each side.– A milliliter is a cube 1 cm

long on each side.

Density

Density is a physical property of a substance.

d = mV

Density =mass

Volume

Chemical Equations

Chemical equations are concise representations of chemical reactions.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Reactants appear on the left side of the equation.


CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Products appear on the right side of the equation.


CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

The states of the reactants and products are written in parentheses to the right of each compound.


CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Coefficients are inserted to balance the equation.

Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule.

Subscripts and Coefficients Give Different Information

• Coefficients tell the number of molecules.

Examples of balancing chemical equation

HCl + Zn ZnCl2 + H2

C3H8 + O2 CO2 + H2O

Zn + HNO3 Zn(NO3)2 + H2 2

5 3 4

2

Reaction Types

Combination &

decomposition reactions

Combustion in Air

Combination Reactions

• In this type of reaction two or more substances react to form one product.

• Examples:– 2 Mg (s) + O2 (g) 2 MgO (s)

– N2 (g) + 3 H2 (g) 2 NH3 (g)

– C3H6 (g) + Br2 (l) C3H6Br2 (l)

Decomposition Reactions

• In a decomposition one substance breaks down into two or more substances.

• Examples:– CaCO3 (s) CaO (s) + CO2 (g)

– 2 KClO3 (s) 2 KCl (s) + O2 (g)

– 2 NaN3 (s) 2 Na (s) + 3 N2 (g)

Combustion Reactions

• These are generally rapid reactions that produce a flame.

• Most often involve hydrocarbons reacting with oxygen in the air.

• Examples:– CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

– C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)

Formula Weights(FW)• A formula weight is the sum of the atomic

weights for the atoms in a chemical formula.

• So, the formula weight of calcium chloride, CaCl2, would be

Ca: 1(40.1 amu*) + Cl: 2(35.5 amu)

111.1 amu

• Formula weights are generally reported for ionic compounds.

*atomic mass unit

Molecular Weight (MW)• A molecular weight is the sum of the

atomic weights of the atoms in a molecule.

• For the molecule ethane, C2H6, the molecular weight would be

C: 2(12.0 amu)+ H: 6(1.0 amu)

30.0 amu

Percent Composition

One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

% element =(number of atoms)(atomic weight)

(FW of the compound)x 100

Percent Composition

So the percentage of carbon in ethane is…

%C =(2)(12.0 amu)

(30.0 amu)

24.0 amu

30.0 amu= x 100

= 80.0%

Avogadro’s Number

• 6.02 x 1023

• 1 mole of 12C has a mass of 12 g.• 1 mole of H2O has a mass of 18g.

Molar Mass

• By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).– The molar mass of an element is the mass

number for the element that we find on the periodic table.

– The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.

• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

Examples

1- Calculate how many atoms there are in 0.200 moles of copper. The number of atoms in one mole of Cu is equal to the Avogadro number = 6.02 x 1023 . Number of atoms in 0.200 moles of Cu = (0.200 mol) x (6.02x1023 mol-1 ) = 1.20 x 1023 .

2- Calculate how molecules of H2O there are in 12.10 moles of water. Number of water molecules = (12.10 mol)x(6.02 x 1023) = 7.287 x 1024

3- Calculate the number of moles of glucose (C6H12O6) in 5.380 g of C6H12O6.

Moles of C6H12O6 = = 0.02989 mol.

4- Calculate the mass, in grams, of 0.433 mol of Ca(NO3)2.

Mass = o.433 mol x 164.1 g/mol = 71.1 g.

5.380 g

180.0 gmol-1

5- How many glucose molecules are in 5.23 g of C6H12O6? How many oxygen atoms are in this sample?

Molecules of C6H12O6 = x (6.02x1023)

= 1.75 x 1022 molecules

Atoms O = 1.75 x 1022 x 6 = 1.05 x 1023

5.23 g180.0 gmol-1

Finding Empirical Formulas

Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition.


The compound para-aminobenzoic acid is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%).

Find the empirical formula of PABA.


Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g

Calculating Empirical FormulasCalculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol


These are the subscripts for the empirical formula:

C7H7NO2

Combustion Analysis

• Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.– C is determined from the mass of CO2 produced.

– H is determined from the mass of H2O produced.– O is determined by difference after the C and H have

been determined.

Determining Empirical Formula by Combustion Analysis

*Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol.

Grams of C = x 1 x 12 gmol-1 = 0.153 g 0.561 g CO2 44 gmol-1

Grams of H = x 2 x 1 gmol-1 = 0.0343 g 0.306 g H2O 18 gmol-1

Grams of O = mass of sample – (mass of C + mass of H) = 0.255 g – ( 0.153 g + 0.0343 g) = 0.068 g

Moles of C = = 0.0128 mol 0.153 g C 12 gmol-1

Moles of H = = 0.0343 mol

Moles of O = = 0.0043 mol

0.0343 g H 1 gmol-1

0.068 g O 16 gmol-1

Calculate the mole ratio by dividing by the smallest number of moles:

C:H:O2.98:7.91:1.00

C3H8O

Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.

Stoichiometric Calculations

Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

Stoichiometric CalculationsC6H12O6 + 6 O2 6 CO2 + 6 H2O

Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6…use the coefficients to find the moles of H2O…and then turn the moles of water to grams.

Limiting Reactants

• The reactant that is completely consumed in a reaction is called the limiting reactant (limiting reagent). – In other words, it’s the reactant that run out first (in this

case, the H2).

• In example below, O2 would be the excess reactant (excess reagent).

Limiting Reactants

Theoretical Yield

• The theoretical yield is the maximum amount of product that can be made.

• The amount of product actually obtained in a reaction is called the actual yield.

Theoretical Yield

• The actual yield is almost always less than the theoretical yield.

Why?• Part of the reactants may not react.• Side reaction.• Difficult recovery.

Percent Yield

• The percent yield of a reaction relates to the actual yield to the theoretical (calculated) yield.

Actual YieldTheoretical Yield

Percent Yield = x 100

Examples

Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)

If we start with 150 g of Fe2O3 as the limiting reactant, and found actual yield of Fe was 87.9 g, what is the percent yield?

The percent yield = x 100Actual Yield

Theoretical Yield

Theoretical yield = 105 g.

The percent yield = x 100 = 83.7 %

Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)

150 g Fe2O3

150 g 159 g mol-1

0.943 mol Fe2O32 mol Fe1 mol Fe2O3

X 1.887 mol Fe

1.887 mol x 55.85 g mol-1

105 g Fe

87.9 g105 g

Solutions

• Solutions are defined as homogeneous mixtures of two or more pure substances.

• The solvent is present in greatest abundance.

• All other substances are solutes.

Molarity

• Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.

• Molarity is one way to measure the concentration of a solution.

moles of solutevolume of solution in liters

Molarity (M) =

Mixing a Solution

• To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.

• The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

Examples

Calculate the molarity of a solution made by dissolving 0.750 g of sodium sulfate (Na2SO4) in enough water to form 850 mL of solution.

Moles of Na2SO4 = = 0.0053 mol

Molarity = = 0.0062 M

0.750 g 142 g mol-1

0.0053 mol0.850 L

How many moles of KMnO4 are present in 250 mL of a 0.0475 M solution?

M =

Moles of KMnO4 = 0.0475 x 0.25 L

= 0.012 mol

molL

molL

How many milliliters of 11.6 M HCl solution are needed to obtain 0.250 mol of HCl?


Molarity (M) =

volume in liters = 0.022 L = 22 mL.

11.6 M = 0.250 mol volume in liters

What are the molar concentrations of each of the ions present in a 0.025 M aqueous solution of Ca(NO3)2?

Ca2+ = 0.025 MNO3

- = 0.025 x 2 = 0.05 M

Dilution• One can also dilute a more concentrated solution

by– Using a pipet to deliver a volume of the solution to a

new volumetric flask, and– Adding solvent to the line on the neck of the new flask.

DilutionThe molarity of the new solution can be determined from the equation

Mc Vc = Md Vd

where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions.

Dilution

Mc Vc = Md Vd

Moles solute before dilution = moles solute after dilution

Examples

How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4 ?

Mc Vc = Md VV

3.0 M x Vc = 0.10 M x 450 mL

Vc = 15 mL

Ways of Expressing Concentrations of Solutions

mass of A in solutiontotal mass of solution 100Mass % of A =

• Mass Percentage, ppm, and ppb Mass Percentage

Example: 36% HCl by mass contains 36 g of HCl for each 100 g of solution (64 g H2O)

Parts per Million (ppm)

ppm =

Parts per Billion (ppb)

mass of A in solutiontotal mass of solution

106

ppb = mass of A in solutiontotal mass of solution

109

Examples

Calculate the mass percentage of Na2SO4 in a solution containing 10.6 g Na2SO4 in 483 g water.

= 2.15 %

Mass % of Na2SO4 =10.6 g

(483 + 10.6) g 100

An ore contains 2.86 g of silver per ton of ore. What is the concentration of silver in ppm?

= 2.86 ppm

ppm = 2.86 g106 g

106

• Mole Fraction, Molarity, and Molality

Mole Fraction (X)

Molarity (M)

moles of Atotal moles in solutionXA =


Molarity (M) =

Molality (m)

moles of solute Kilograms of solventm =

Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.

Since volume is temperature-dependent, molarity can change with temperature.

ExamplesAn aqueous solution of hydrochloric acid contains 36 % HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.

Moles HCl = = 0.99 mol

Moles H2O = = 3.6 mol

36 g 36.5 g mol-1

64 g 18 g mol-1

XHCl = =

moles HCl moles H2O + moles HCl

0.99 mol HCl 0.064 kg H2O

0.99 3.6 + 0.99

= 0.22

Molality of HCl = = 15.5 m

general chemistry 101 chem د. عبد العزيز بن علي الغامدي 1434-1435 هـ

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