General Chemistry 101 CHEM

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## General Chemistry 101 CHEM

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**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 aalghamdia@ksu.edu.sa**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.