Mass Relationships in Chemical Reactions Chapter 3 Mass Relationships in Chemical Reactions

3.1 Atomic Mass Mass depends on number of protons, neutrons, and electrons Atomic Mass Unit (amu): a mass exactly equal to 1/12th the mass of one carbon-12 atom Atomic masses on the PTE are average atomic masses Average atomic mass = (% abundance)(mass) + (% abudance)(mass)…. For an example, see pg. 77 Example 3.1

3.2 Avogadro’s Number and the Molar Mass of an Element Mole: the amount of a substance that contains as many elementary entities as there are atoms in exactly 12 grams of carbon-12 1 mole = 6.0221415 x 1023 entities The mass of 1 mole of an element is equal to its mass in amu (on the periodic table) We therefore can convert between mass of an element, moles of an element, and number of atoms of an element. Look at examples 3.2, 3.3, 3.4

3.3 Molecular Mass Molecular Mass: the sum of the atomic masses in amu in the molecule See example 3.5 The molar mass of a compound is numerically equal to it’s molecular mass

3.5 Percent Composition of Compounds Percent Composition by Mass: percent by mass of each element in a compound % Composition = (Mass of element/Mass of Compound) x 100 See example 3.8

3.6 Experimental Determination of Empirical Formulas Empirical formulas can be determined from percent composition data Example: A sample of a compound contains 30.46% nitrogen and 69.54% oxygen. The molar mass of the compound is between 90 g and 95 g. Determine the empirical and molecular formulas.

3.7 Chemical Reactions and Chemical Equations Chemical Reaction: A process in which a substance or substances is changed into one or more new substances Chemical Equations: uses chemical symbols to show what happens during a chemical reaction 2H2(g) + O2(g)  2H2O(l) Reactions must be balanced Reactants are on the left side, products are on the right side Do we need to review balancing equations?

3.8 Amounts of Reactants and Products How do we determine how much product will be formed from a specific amount of starting materials? Or How much starting material must be used to obtain a specific amount of product? Stoichiometery: the quantitative study of reactants and products in a chemical reaction

3.8 Amounts of Reactants and Products Example: The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C6H12O6) to carbon dioxide (CO2) and water (H2O). If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of CO2 produced?

3.9 Limiting Reagents Limiting Reagent: the reactant used up first in the reaction Excess Reagent: the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent

3.9 Limiting Reagents Example: Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide. In one process, 637.2 g of NH3 are treated with 1142 g of CO2. (a.) Which of the reactants is the limiting reagent? (b.) Calculate the mass of (NH2)2CO formed. (c.) How much excess reagent, in grams, is left at the end of the reaction? 2NH3(g) + CO2(g)  (NH2)2CO(aq) + H2O(l)

3.10 Reaction Yield The amount of limiting reagent present at the start of a reaction determines the theoretical yield. Theoretical yield: the amount of product that would result if all the limiting reagent reacted (maximum obtainable amount). Actual yield: the amount of product actually obtained from a reaction Percent yield: the proportion of the actual yield to the theoretical yield Percent yield = (actual yield)/(theoretical yield) x 100

3.10 Reaction Yield Example: Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 950 C and 1150 C. In a certain industrial operation 3.54 x 107 g of TiCl4 are reacted with 1.13 x 107 g of Mg. (a.) Calculate the theoretical yield of Ti in grams. (b.) Calculate the percent yield if 7.91 x 106 g of Ti are actually obtained. TiCl4(g) + 2Mg(l)  Ti(s) + 2MgCl2(l)