General Chemistry 101CHEM 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.
الاختبار الفصلي الأول: لم يحدد بعد. • الاختبار الفصلي الثاني: لم يحدد بعد. • المكتب 2أ134 و سيكون قريبا 2أ146 • الأيميل email@example.com
Matter We define matter as anything that has mass and takes up space.
Matter • Atoms are the building blocks of matter. • Eachelementis made of the same kind of atom. • Acompoundis made of two or more different kinds of elements.
Units of Measurement • SI Units • Système International d’Unités • A different base unit is used for each quantity.
Metric System Prefixes 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.
mass Volume m V d = Density = Density Density is a physical property of a substance.
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 • 2 • C3H8 + O2 CO2 + H2O • 5 3 4 • Zn + HNO3 Zn(NO3)2 + H2 • 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
(number of atoms)(atomic weight) x 100 % element = (FW of the compound) 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:
(2)(12.0 amu) %C = (30.0 amu) x 100 24.0 amu = 30.0 amu Percent Composition So the percentage of carbon in ethaneis… = 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
5.380 g 180.0 gmol-1 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.
5.23 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
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.
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 mol 12.01 g 1 mol 1.01 g 1 mol 14.01 g 1 mol 16.00 g Calculating Empirical Formulas
C: = 7.005 7 H: = 6.984 7 N: = 1.000 O: = 2.001 2 5.09 mol 0.7288 mol 0.7288 mol 0.7288 mol 5.105 mol 0.7288 mol 1.458 mol 0.7288 mol Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles:
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 CO2produced. • 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. 0.561 g CO2 44 gmol-1 Grams of C = x 1 x 12 gmol-1 = 0.153 g 0.306 g H2O 18 gmol-1 Grams of H = x 2 x 1 gmol-1 = 0.0343 g Grams of O = mass of sample – (mass of C + mass of H) = 0.255 g – ( 0.153 g + 0.0343 g) = 0.068 g
0.153 g C 12 gmol-1 Moles of C = = 0.0128 mol 0.0343 g H 1 gmol-1 Moles of H = = 0.0343 mol 0.068 g O 16 gmol-1 Moles of O = = 0.0043 mol Calculate the mole ratio by dividing by the smallest number of moles: C:H:O 2.98:7.91:1.00 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 Calculations C6H12O6 + 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).
Limiting Reactants • In example below, O2 would be the excess reactant (excess reagent).
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.
Actual Yield Theoretical Yield Percent Yield = x 100 Percent Yield • The percent yield of a reaction relates to the actual yield to the theoretical (calculated) yield.