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General Chemistry 101 CHEM. د. عبد العزيز بن علي الغامدي 1434-1435 هـ. References . Chemistry the General Science 11E T. L. Brown, H. E. LeMay, B. E. Bursten and C. J. Murphy. الكيمياء العامة. د.أحمد العويس، د.سليمان الخويطر، د.عبدالعزيز الواصل، د.عبدالعزيز السحيباني. . - PowerPoint PPT Presentation
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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 146أ2وaalghamdia@ksu.edu.saاأليميل •
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 = massVolume
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.
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Products appear on the right side of the equation.
Anatomy of a Chemical 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.
Anatomy of a Chemical Equation
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 amu30.0 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.
Examples1- 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.
Calculating Empirical Formulas
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.
Calculating Empirical Formulas
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
Calculating Empirical Formulas
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?
moles of solutevolume of solution in liters
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 VV3.0 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 =
moles of solutevolume of solution in liters
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
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