Chemical Formulas and Composition Stoichiometry PDF

Chapter 3 Chemical Formulas and Composition Stoichiometry Prof. D. Wael Abou-Elmagd  Stoichiometry Stoichiometry describes the quantitative relationships among elements in compounds (composition stoichiometry) and among substances as they undergo chemical c...

Chapter 3 Chemical Formulas and Composition Stoichiometry Prof. D. Wael Abou-Elmagd  Stoichiometry Stoichiometry describes the quantitative relationships among elements in compounds (composition stoichiometry) and among substances as they undergo chemical changes (reaction stoichiometry). The chemical formula for a substance shows its chemical composition. This represents the elements present as well as the ratio in which the atoms of the elements occur. The chemical formula for a single atom is the same as the symbol for the element. Thus, Na can represent a HO 2 single sodium atom. Number of A subscript following the symbol of an element hydrogen atoms indicates the number of atoms in a molecule. For instance, H2O is a molecule containing two hydrogen atoms and one oxygen atom. The Mole  The mole is the SI unit for amount of substance, it is abbreviated mol.  The mole is defined as the amount of substance that contains as many particles (atoms, molecules,..) as there are atoms in exactly 12 g of pure carbon-12 atoms.  The number of particles in one mole is known as Avogadro’s Number (NA)  Avogadro's number = 6.022x1023 = 602200000000000000000000.  = 6.022x1023 12.000 g of carbon carbon atoms (one mole of carbon) This means we now have Avogadro's number of carbon atoms (NA) = 6.022x1023 on the scales pan One mole of hydrogen (H) contains the same number of particles as 12 g of carbon-12 One mole of helium (He) contains the same number of particles as 12 g of carbon-12 One mole of neon (Ne) contains the same number of particles as 12 g of carbon-12 The number of particle in one mole of hydrogen (H) = The number of particle in one mole of helium (He) = The number of particle in one mole of neon (Ne) =.. The number of particle in one mole of any substance = Avogadro's Number=6.022x1023 The Molar Mass (M) The molar mass of a substance is the mass of one mole of that substance. Since one mole contains Avogadro's number of particles, then The Molar mass = the total mass of Avogadro's number of particles. The mass of one mole (Molar Mass) of atoms of a pure element in grams is numerically equal to the atomic weight of that element in atomic mass units. The molar mass units are grams/mole, also written as g/mol or g mol -1.  For instance, if you obtain a pure sample of the metallic element titanium (Ti), whose atomic weight is 47.88 amu, and measure out 47.88 g of it, you will have one mole, or 6.022 x1023 titanium atoms. One mole of atoms of some common elements. Back row (left to right): bromine, aluminium, mercury, copper. Front row (left to right): sulfur, zinc, iron. Each of these samples has a different mass but contains the same number of atoms; a very large number 6.022x1023 - Avogadro's number How many moles of atoms does 136.9 g of iron (Fe) metal contain? answer We are given the total mass of the Fe sample (m) = 136.9 g We look up the atomic mass of Fe = 55.85 amu Then The mass of one mole (molar mass M) of Fe= 55.85 g/mol Then total mass m number of moles (n)   mass of one mole (molar mass ) M m n M 136.9 g n  2.451mol 55.85 g / mol How many atoms are contained in 2.451 mol of iron? answer We are asked about the total number of atoms (N) We are given the number of moles (n) =2.451 mol We know that one mole of atoms of an element contains Avogadro’s number (NA) of atoms, or 6.022 x1023 mol-1. Then total number of atoms ( N )  number of moles ( n)  avogadro' s number ( N A ) N  n NA N  2.451 mol  6.022  10 mol  1.476  10 23 1 24  Calculate the average mass of one iron atom in grams. answer The atomic mass of Fe = 55.85 amu Then The mass of one mole (molar mass M) of Fe= 55.85 g mol-1 This is the mass of Avogadro’s number of atoms Then molar mass ( M )  mass of one atom  Avogadro' s Number ( N ) A 1 molar mass ( M ) 55.85 g mol mass of one atom    9.274  10 g 23 N A 6.022  10 mol 23 1 Thus, the average mass of one Fe atom is only 9.274 x10-23 g, that is, 0.00000000000000000000009274 g  Molecule, Mole and Molar Mass Molar mass of H2O = mass of one mole of H2O = One Total mass of 6.022x1023 molecule of molecules of H2O = H2O Formula weight in grams of H2O = 18 g mol-1 One mole of H2O (6.022x1023 molecules of H2O) How many moles of substance are contained in each of the following samples? (a) 18.3 g of NH3 (b) 5.32 g of ammonium bromide NH4Br (c) 6.6 g of PCl5 (d) 215 g of Sn answer total mass m number of moles (n)   mass of one mole (molar mass ) M m n M m 18.3 g 18.3 g a)n  NH 3    1.07 mol 14.0 g mol  3  1.0 g mol 1 1 1 NH 3 M NH 3 17.0 g mol What is the formula…? When a chemist has discovered a new compound, the first question to answer is, What is the formula? To answer, you begin by analyzing the compound to determine amounts of the elements for a given amount of compound. This is conveniently expressed as percentage composition— that is, as the mass percentages of each element in the compound. But, what do we really mean by a chemical formula? CHEMICAL EQUATIONS Chemical equations are used to describe chemical reactions, and they show (1) The substances that react, called reactants; (2) the substances formed, called products; and (3) the relative amounts of the substances involved. Number of moles N.B. The amounts of substances are in moles What does this equation tell us? it says that for every CH4 molecule that reacts, two molecules of O2 also react, and that one CO2 molecule and two H2O molecules are formed (produced). If we multiply throughout by Avogadro's number So, it turns out that The equation, therefore, tells us that for every CH 4 mole that reacts, two moles of O2 also react, and that one CO2 mole and two H2O moles are formed (produced). We know the mass of 1 mol (molar mass) of each of these substances, Molar mass of CH4 = 12.0 g mol-1 + 4 x 1.0 g mol-1 = 16 g mol-1 Molar mass of O2 = 2 x 16.0 g mol-1 = 32.0 g mol-1 Molar mass of CO2 = 12.0 g mol-1 + 2 x 16.0 g mol-1 = 44.0 g mol-1 Molar mass of H2O = 2 x 1.0 g mol-1 + 16.0 g mol-1 = 18.0 g mol-1 so we can also write The equation then tells us that 16.0 g of CH4 reacts with 64.0 g of O2 to produce 44.0 g of CO2 and 36.0 g of H2O In fact knowing the BALANCED chemical equation allows us to know a great deal of information; we can calculate for example; Number of Moles Formed/Produced Mass of a Reactant Required Mass of a Product Formed The limiting reactant Percent yield How? Number of Moles Formed How many moles of water (H2O) could be produced by the reaction of 3.5 mol of methane (CH4) with excess oxygen (O2) (i.e., more than a sufficient amount of oxygen is present)? MAKE SURE THAT THE EQUATION IS BALANCED; that is the number of atoms of each element that appears in the LHS appears as well in the RHS. If it is balanced then we can read the equation as x 3.5 mol 3.5 moles x 3.5 mol  2 mol x  7.0 mol 1 mol Then 7.0 moles of water will be produced Mass of a Reactant Required What mass of oxygen is required to react completely with 1.20 mol of CH4? answer The balanced equation tells us 1.20 moles x 2 mol 1.20 mol x  2.40 mol 1 mol Then 2.40 moles of oxygen are required. but we are asked about the mass. We therefore need to calculate the mass of 2.40 moles of oxygen. Mass of one mole of oxygen= molar mass of oxygen (O 2) =16.0 g mol-1x 2= 32.0 g mol-1 The required mass of oxygen (O2)=2.40 mol x 32.0 g mol-1=76.8 g Mass of a Product Formed Calculate the mass of CO2, in grams, that can be produced by burning 6.00 mol of CH4 in excess O2. The molar mass of CO2= 12.0 g mol-1 + 2 x 16.0 g mol-1 =44.0 g mol-1 The molar mass of CH4 = 12.0 g mol-1 + 4 x 1.0 g mol-1 = 16.0 g mol-1 The balanced equation tells us One molar mass of CH4 produces ONE molar masses of CO2 6.00 molar mass of CH4 produces X 6 M  M 6 16.0 g  mol  44.0 g mol 1 1 X CH 4 CO2  1  2.64 10 g 2 M CH 4 16.0 g mol THE CONCEPT OF LIMITING REACTANT Say we want to me make a sandwich; we then need two slices of bread and one slice of cheese (The sandwich-making equation) What if we have six slices of bread and two slices of cheese. Even though you have extra bread, you are limited to making two sandwiches by the amount of cheese you have on hand. Cheese is the limiting reactant. So the number of sandwiches is determined by the ratio of cheese slices to those of bread When an equal number of molecules of H2 and O2 are reacted according to the equation 2H2+O2 →2H2O all of the H2 completely reacts, whereas only half of the O2 is consumed. In this case, the H2 is the limiting reactant and the O2 is the excess reactant.  What is the limiting reactant in the reaction of 16.0 g of CH4 with 48.0 g of O2 according to the following equation? We have 16.0 g of CH4 = 1.0 mol of CH4 We have 48.0 g of O2 = 1.5 mol of O2 The balanced equation tells us that one mole of CH4 reacts with two moles of O2. Then the number of moles of CH4 required to react with 1.5 mol of O2= 0.75 mol we have 1.0 mol of CH4 i.e. we have excess of CH4 Then the number of moles of O2 required to react with 1.0 mol of CH4= 2.0 mol we have only 1.5 mol of O2 i.e. O2 is the limiting reactant Then we should base any further calculation on the amount of O 2 CONCENTRATIONS OF SOLUTIONS The reaction between two solid reactants often proceeds very slowly or not at all. Many chemical reactions are therefore more conveniently carried out with the reactants mixed in solution rather than as pure substances. A solution is a homogeneous mixture, at the molecular level, of two or more substances. Simple solutions usually consist of one substance, the solute, dissolved in another substance, the solvent. The solutions used in the laboratory are usually liquids, and the solvent is often water. These are called aqueous solutions. The general term concentration refers to the quantity of solute in a standard quantity of solution. Molar concentration, or molarity (M), is defined as the number of moles of solute dissolved in one litre (cubic decimeter) of solution. number of moles of solute (n) n Molarity   volume of solution in litres (V ) V The unit of molarity is therefore mol/L, mol L-1 or M “read as molar ” total mass number of moles (n)  mass of one mole (molar mass ) E.g. An aqueous solution that is 0.15 M NH3 (read this as “0.15 molar NH3”) contains 0.15 mol NH3 per litre of solution. If you want to prepare a solution that is, for example, 0.200 M CuSO4, you place 0.200 mol CuSO4 in a 1.000 L volumetric flask or place 0.100 mol CuSO4 in a 0.5 L volumetric flask or … A sample of NaNO3 weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark on the neck, dissolving the solid. What is the molarity of the resulting solution? number of moles of solute (n) n Molarity   volume of solution in litres (V ) V To calculate the molarity, we need the number of moles of solute. Therefore, you first convert gramsNaNO3to moles. Molar mass of NaNO3= 23 g mol-1+ 14.0 g mol-1 + 3x16.0 g mol-1 =85 g mol-1 mass 0.38 g n  1  4.47  103 mol Molar mass 85 g mol V  50.0 mL 50.0 103 L number of moles of solute ( n) n 4.47  10 mol 3 Molarity ( M )     volume of solution in litres (V ) V 50.0  10 L 3  0.089 mol / L  0.089 M read this as “0.089 molar” Calculate the mass of barium hydroxide Ba(OH) 2 required to prepare 2.50 L of a 0.06 M solution of barium hydroxide. number of moles of solute ( n) n Molarity   volume of solution in litres (V ) V mass m n  Molar mass M m Molarity  MV m? V  2.50 L Molarity  0.06M  0.06 mol L 1 M  137.3 g mol  (16.0 g mol  1.0 g mol ) 2  171.3 g mol 1 1 1 1 m 0.06 mol L  1 171.3 g mol  2.50 L 1 m  0.06 mol L  171.3 g mol  2.5 L  25.7 g 1 1 DILUTING SOLUTIONS number of moles of solute ( n) n Molarity ( M )   volume of solution in litres (V ) V MV  n MV n 1 1 MV n 2 2 Number of moles before dilution=Number of moles after dilution=n ( MV ) before dilution  ( MV ) after dilution MV M V1 1 2 2 Note, in here, M is the molarity of the solution (not the molar mass of the solute) How many milliliters of 18.0 M H2SO4 are required to prepare 1.00 L of a 0.900M solution of H2SO4? ( MV ) before dilution  ( MV ) after dilution MV M V1 1 2 2 18.0 mol L  V  0.900 mol L  1.00 L 1 1 1 0.900 mol L  1.00 L 1 V  1  0.05 L  0.05  10 mL  50.0 ml 1 3 18.0 mol L Solution Stoichiometry If we plan to carry out a reaction in a solution, we must calculate the amounts of solutions that we need. If we know the molarity of a solution, we can calculate the amount of solute contained in a specified volume of that solution. Calculate the volume in litters of a 0.324 M solution of sulphuric acid, H2SO4, required to react completely with 2.792 grams of Na2CO3 according to the equation The balanced equations tells us n H 2 SO4 n Na2CO3 For the reaction to go to completion n H 2 SO4  n Na2CO3 mass of Na CO 2.792 g n Na2CO3   2  2.63310 mol 3 1 2 Molar mass of Na CO 106.0 g mol 2 3 n Molarity  ; n H 2 SO4  0.324 mol L.V 1 V 0.324 mol L.V  2.63310 mol  V  0.0813 L  81.3 mL 1 2