Chemistry (AHS 111) Faculty of Applied Health Sciences Technology PDF

‫بسم هللا الرحمن الرحيم‬ Faculty of Applied Health Sciences Technology CHEMISTRY (AHS 111) (1st level, 1st semester) (3) ‫ حجاج حسن‬/‫دكتور‬  properties of various forms of electromagnetic ra...

‫بسم هللا الرحمن الرحيم‬ Faculty of Applied Health Sciences Technology CHEMISTRY (AHS 111) (1st level, 1st semester) (3) ‫ حجاج حسن‬/‫دكتور‬  properties of various forms of electromagnetic radiation Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted energy. Waves A wave is a periodic oscillation that transmits energy through space. The distance between corresponding points on adjacent waves is the wavelength (). Important Properties of Waves (a) Wavelength (λ in meters), frequency (ν, in Hz), and amplitude. (b) The wave with the shortest wavelength has the highest frequency). If two waves have the same frequency and speed, the one with the greater amplitude has the higher energy. wavelength (λ, ) The distance between two corresponding points in a wave—for example, between two troughs are described by a unit of distance. The frequency (ν, nu) of a wave is the number of oscillations that pass a particular point per second (1/s = s−1), which in is called the hertz The amplitude, or vertical height, of a wave is defined as half the peak-to-trough height The distance traveled by a wave per unit time is its speed v, which is typically measured in meters per second (m/s). Electromagnetic Radiation All electromagnetic radiation travels at the same velocity: the speed of light (c), 3.00  108 m/s. Therefore, c =  The Nature of Matter Matter, like electromagnetic radiation, also exhibits both wave-like properties and particle-like properties. the energy of a molecule is “quantized.” Molecules can store energy in a variety of ways. They rotate in space, their bonds vibrate like springs, their electrons can occupy a number of possible molecular orbitals, and so on. For atoms and molecules a line spectrum of discrete wavelengths is observed.  Stoichiometry: Calculations with Chemical Formulas and Equations  Chemical Equations ◼ Chemical reactions occur when bonds (between the electrons of atoms) are broken or formed ◼ All chemical equations have reactants and products. ◼ We express a chemical equation as follows: Reactants → Products C + O2 → CO2 ◼ This reads “carbon plus oxygen react to yield carbon dioxide” Equations show the formulae of the reactants and the products. show the relationship between the numbers of each substance involved; this is known as the STOICHIOMETRY can show in which state the substances exist. ▪ Chemical reactions involve ▪ changes in the chemical composition of matter ▪ the making of new materials with new properties ▪ energy changes: ▪ Bond breaking absorbs Energy (endothermic process) ▪ Bond making releases Energy (exothermic process) Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al (s) + 3 O2 (g) ---> 2 Al2O3 (s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds. Balancing a Chemical Equation ▪ A chemical equation is balanced when the ions or atoms found on the reactant side of the equation equals that found on the product side. ◼ Because of the principle of the conservation of matter (matter can not be created or destroyed) an equation must be balanced. ◼Energy changes are written in (endo-/ exothermic reactions) ◼ Law of Conservation of Energy MUST ALSO BE FOLLOWED! Chemical Equations Chemical equations are Concise representations of chemical reactions  Anatomy of a Chemical Equation 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 Products appear on the right side of the equation. 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 Coefficients tell the number of molecules Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule Coefficients tell the number of molecules Sample Exercise Interpreting and Balancing Chemical Equations Practice In the following diagram, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH3 molecules should be shown in the right (products) box?  BALANCING EQUATIONS – WORKED EXAMPLE Step 1 sodium + water ——> sodium hydroxide + hydrogen Step 2 Na(s) + H2O(l) ——> NaOH(aq) + H2(g) Step 3 Count up the atoms : Left... 1 x Na, 2 x H, 1 x O Right... 1 x Na, 3 x H, 1 x O. The equation doesn’t balance; an extra H is needed on the LHS. However the formula must not change. One can only get extra H’s by having two waters; multiply H2O by two. Step 4 Na(s) + 2 H2O(l) ——> NaOH(aq) + H2(g) This doesn’t solve the problem as we now have too many O’s (2) and H’s (4) on the LHS; multiplying the NaOH by two will solve this problem. Na(s) + 2 H2O(l) ——> 2 NaOH(aq) + H2(g) However, it creates yet another problem because it has introduced an extra Na on the RHS; multiply the Na on the LHS by two. 2 Na(s) + 2 H2O(l) ——> 2 NaOH(aq) + H2(g) Step 5 Check the equation; it balances. As you can see it can take time but with a little effort a balanced equation can be achieved.  Reaction Types 1- Combination Reactions In combination reactions two or more substances react to form one product. Examples: N2 (g) + 3 H2 (g) ⎯⎯→ 2 NH3 (g) C3H6 (g) + Br2 (l) ⎯⎯→ C3H6Br2 (l) 2 Mg (s) + O2 (g) ⎯⎯→ 2 MgO (s)  2- Decomposition Reactions In a decomposition reaction 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) 3- Combustion Reactions Combustion reactions are generally rapid reactions that produce a flame, most often involve hydrocarbons reacting with oxygen in the air. Examples: CH4(g) + 2O2(g) ⎯⎯→ CO2(g) + 2H2O(g) 2CH3OH(l) + 3O2(g) ⎯⎯→ 2CO2(g) + 4H2O(g) Write the balanced equation for the reaction that occurs when ethanol, C2H5OH(l), burns in air. Formula Weight (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 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 Atomic mass unit and the mole Avogadro’s Number 6.02 x 1023 1 mole of 12C has a mass of 12 g At STP, one mole of any gas = 22.4 L In Terms of Volume 2H2 + O2 → 2H2O (2 x 22.4 L H2) + (1 x 22.4 L O2) → (2 x 22.4 L H2O) Sample Exercise Estimating Numbers of Atoms Without using a calculator, arrange these samples in order of increasing number of O atoms: 1 mol H2O 1 mol CO2 3  1023 molecules O3. Sample Exercise Converting Moles to Number of Atoms How many oxygen atoms are in: (a) 0.25 mol Ca(NO3)2 (b) 1.50 mol of sodium carbonate? 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 Mole calculations Suppose we have 100.0 grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? 100.0 g Fe moles Fe = 55.8 g/mol = 1.79 moles of Fe Conversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass? mass M g = ( 5.75 moles)  (24.3 g/mol) = 140 grams of M g 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 Calculations Mass, in grams, of 1.50 x 10-2 mol CdS? Mass, in grams, of 1.50 x 1021 molecules of aspirin, C9H8O4? 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: (number of atoms)(atomic weight) % element = x 100 (FW of the compound) So the percentage of carbon in ethane (C2H6) is… (2)(12.0 amu) %C = (30.0 amu) 24.0 amu = x 100 30.0 amu = 80.0% Finding Empirical Formulas Combustion Analysis gives % composition 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 Percent Composition What is the % composition of C6H12O6? 12.011g C 72.066 6 moles C -------------- = 72.066 g C x100%=40.0% C 1 mole C 180.155 1.0079g H 12.0948 12 moles H -------------- = 12.0948 g H x100%=6.7% H 1 mole H 180.155 15.999g O 6 moles O -------------- = 95.994g O 95.994 x100%=53.3% O 1 mole O 180.155 180.155 % composition is 40.0% C, 6.7% H, and 53.3% O Presentation of Lecture Outlines, 3–29 The compound para-aminobenzoic acid (PABA) composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products 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 Calculations 4 KO2 + 2 CO2 → K2CO3 + 3O2 How many moles of O2 are produced when 0.400 mol of KO2 reacts? How many grams of KO2 are needed to form 7.50 g of O2? How many grams of CO2 are used when 7.50 g of O2 are produced? Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount In other words, it’s the reactant you’ll run out of first (in this case, the H2, while the O2 would be the excess reagent. 2 H2 + O2 → 2 H2O  Theoretical Yield The theoretical yield is the amount of product that can be made It’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, the amount one actually produces and measures Percent Yield The percent yield is a comparison of the amount actually obtained to the amount it was possible to make Actual Yield Percent Yield = x 100 Theoretical Yield Theoretical yield = 30 X 157/78 = 60 g Actual yield = 56.7 g