Stoichiometry (part II)

= 6.022 x 1023 units of that thing (Avogadro’s number) Review Slide: 1 mole of anything = 6.022 x 1023 units of that thing (Avogadro’s number) = molar mass of that thing Thus, the molar mass of an element is numerically equal to its atomic mass (reflected on the periodic table) molar mass = atomic mass

Percent Composition of Compounds Review Slide: Percent Composition of Compounds Mass percent of an element: Calculate the mass % of Fe in iron(III) oxide, (Fe2O3):

Calculate the percent composition by mass of this material Review Problem: In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen (77K) was discovered. The approximate formula of this substance is YBa2Cu3O7. Calculate the percent composition by mass of this material

What is the empirical formula? What is the molecular formula? e.g. #1 The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? What is the molecular formula? 5

Formulas Empirical formula - Simplest whole-number ratio > Base the calculation on 100g of compound. > Determine the number of moles of each element present in 100g of compound using the atomic masses of the elements present. > Divide each value of the number of moles by the smallest of the values. * If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula. * If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers

Actual formula of the compound = (empirical formula)n where: Formulas Molecular formula Actual formula of the compound = (empirical formula)n where: n => integer = molar mass empirical formula mass

Chemical Stoichiometry Review Slide: Chemical Stoichiometry Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Stoichiometric Calculations Chemical equations can be used to relate the masses of reacting chemicals.

Notes on Balancing Chemical Equations Review Slide: Notes on Balancing Chemical Equations The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts must not be changed to balance an equation. Coefficients can be fractions, although they are usually given as lowest integer multiples. A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.

Fe2S2 + Cl2 => FeCl3 + S2Cl2 e.g. #2 Balance the following chemical equation: Fe2S2 + Cl2 => FeCl3 + S2Cl2

e.g. #3 Balance the following chemical equation: C6H6 + H2O2 => CO2 + H2O

Review Slide: Limiting Reactants Limiting reactant – the reactant that is consumed first and therefore limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.

Review Slide: Reaction Yield An important indicator of the efficiency of a particular laboratory or industrial reaction. % yield = (actual yield / theoretical yield) x 100%

e.g. #4 (Stoichiometry Review) The reaction between gasoline (octane) and oxygen that occurs inside automobile engines. 2C8H18 + 25O2 => 18H2O + 16CO2 a) If 25g of C8H18 is reacted with 25g of O2, Which is the limiting reactant in terms moles of H2O ? b) What mass of product will be produced? c) What would be the % yield of the reaction if 9.65 g of H2O are actually produced at the end?

Molarity (M) - concentration of solutions - expressed as molar concentrations in M (mol/L) M = number of moles volume

e.g. #5 Fe2O3 + 6H2C2O4 => 2Fe(C2O4)3 + 3H2O + 6H Oxalic Acid (H2C2O4) is used for the removal of rust (Fe2O3). The equation is shown as: Fe2O3 + 6H2C2O4 => 2Fe(C2O4)3 + 3H2O + 6H a) Calculate the number of grams of rust that can be removed with 1 L of 0.10M solution of oxalic acid. b) Which is the limiting reactant if Fe(C2O4)3 is our concerned product and 0.10 mole Fe2O3 reacted with 0.10 mole H2C2O4? c) What mass of product will be produced? d) What would be the % yield of the reaction if only 2.50g of Fe(C2O4)3 are actually isolated at the end?3

e.g. #6a The reaction between gasoline (octane) and oxygen that occurs inside automobile engines. 2C8H18 + 25O2 => 18H2O + 16CO2 If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

e.g. #6b The reaction between gasoline (octane) and oxygen that occurs inside automobile engines. 2C8H18 + 25O2 => 18H2O + 16CO2 How many grams of water would be produced if 20.0 liters of oxygen were burned at a temperature of -10.0°C and a pressure of 1.3 atm?

Molar Volume of an Ideal Gas Additional Information: Molar Volume of an Ideal Gas For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is 22.42 L. STP = standard temperature and pressure 0°C and 1 atm Therefore, the molar volume is 22.41 L at STP. V = nRT = P (1 mol) 0.08206 L atm (273.15K) = 22.41L mol K 1 atm

Molar Mass of a Gas MM = Molar Mass = DRT P D = density of gas Additional Information: Molar Mass of a Gas MM = Molar Mass = DRT P D = density of gas T = temperature in Kelvin P = pressure of gas R = universal gas constant