Chemical Calculations: Empirical Formulas & Mass Relationships in Equations - Prof. Thomas, Study notes of Chemistry

Download Chemical Calculations: Empirical Formulas & Mass Relationships in Equations - Prof. Thomas and more Study notes Chemistry in PDF only on Docsity! Chapter 5 Chemical Calculations Recall that Dalton’s atomic theory was developed to account for quantitative relationships in chemical compounds. As we shall see, the same logic is easily extended to chemical reactions, but only with a well defined and comprehensive symbolism to encode the information. P the element phosphorous an atom of phosphorous 31.0 amu phosphorous An event in which chemical bonds are formed, broken, or rearranged is described as a chemical reaction or chemical change and are represented by chemical equations: H2 + O2 H2O Reactant(s) Product(s) This equation is read: hydrogen (H2) reacts with oxygen (O2) to form water: Chemical Symbols for Chemical Reactions The Law of Conservation of Matter states that matter may be neither created nor destroyed The total mass of the reactants is always the same as the mass of the products. Observed Fact: The total number of each type of atom must be the same before and after the reaction. Theoretical Explanation: H2 + O2 H2O Reactant Atom Count Product Atom Count H 2 2 O 2 1 The number of oxygen atoms is not balanced. We understand this equation to show that 1 molecule of methane (CH4) reacts with 2 molecules of oxygen (O2) to produce 1 molecule of carbon dioxide (CO2) and two molecules of water (H2O). CH4 + 2O2  CO2 + 2H2O Mass Relationships in Chemical Equations CH4 + 2O2  CO2 + 2H2O But it also means that 16 amu of methane reacts with 64 amu of oxygen to form 44 amu of carbon dioxide and 36 amu water. This mass focused interpretation of the chemical equation implies a constant mass ratio between any pair of the reactants and/or products. 16 amu 64 amu 44 amu 36 amu Just as for chemical formulae, in which the ratio of contained element masses (gravimetric ratios) is constant (C3H8 shown below)… 36.0 amu C 8.0 amu H E:E ratio element:element E:C ratios element:compound 36.0 amu C 44.0 amu propane 8.0 amu H 44.0 amu propane CH4 + 2O2  CO2 + 2H2O 16 amu 64 amu 44 amu 36 amu How much oxygen is needed to combust a kilogram of methane? 16.0 amu CH4 64.0 amu O2 16.0 amu CH4 44.0 amu CO2 16.0 amu CH4 36.0 amu H2O 64.0 amu O2 44.0 amu CO2 64.0 amu O2 36.0 amu H2O 44.0 amu CO2 36.0 amu H2O CH4 + 2O2  CO2 + 2H2O 16 amu 64 amu 44 amu 36 amu How much oxygen is needed to combust a kilogram of methane? 16.0 amu CH4 64.0 amu O2 1.0 kg CH4 x kg O2 = Theoretical mass relationship derived from the Chemical equation for the combustion of methane. Observed, predicted, measured, etc. mass relationship. CH4 + 2O2  CO2 + 2H2O 16 amu 64 amu 44 amu 36 amu How much oxygen is needed to combust a kilogram of methane? (16.0 amu CH4)*(x kg O2) (64.0 amu O2)* (1.0 kg CH4) = 16.0 amu CH4 64.0 amu O2 1.0 kg CH4 x kg O2 = Crossmultiply: CH4 + 2O2  CO2 + 2H2O 16 amu 64 amu 44 amu 36 amu How about the amount of carbon dioxide released in the combustion of 600 kg of methane, roughly what is required to heat the average house? 16.0 amu CH4 64.0 amu O2 16.0 amu CH4 44.0 amu CO2 16.0 amu CH4 36.0 amu H2O 64.0 amu O2 44.0 amu CO2 64.0 amu O2 36.0 amu H2O 44.0 amu CO2 36.0 amu H2O Analyzing Compounds: Empirical Formula We have seen how constant composition in compounds was used to produce a table of relative masses. With an accurate table of relative atomic masses, the constant composition of a compound may be used to calculate its empirical formula. The empirical formula gives the combining atom ratio for a compound. 2 g1 g Counting Atoms: The Mole Concept Imagine two types of marbles: blue ones with a mass of 1.0 g and green ones with a mass of 2.0 g. For one of each, we say that the mass ratio is 1:2. mass ratio 1 : 2 marble ratio 1 : 1 16.0 amu Rather than marbles, consider atoms of the elements sulfur and oxygen. oxygen 32.1 amu sulfur If we wish to define a standard number, or pile, of atoms, a convenient choice is to say that the atomic mass, expressed as grams represents the atom counting standard. oxygen sulfur We may now say that these two samples contain the same number of atoms. This number (6.02×1023) is called a mole. Besides offering a convenient way to “count” atoms, it is the conversion factor between grams and amu: 16.0 g 32.1 g 6.02× 1023 amu 1.0 g atomic mass of element in grams 1.0 mole atoms of the element= Because the number of atoms is linearly related to the mass, calculations using the mole concept are subject to the linear analysis previously outlined. mass of element in grams moles of atoms of the element complete ratio: the defining ratio for 1 mole of the element incomplete ratio Quantitative analysis of a compound of carbon and chlorine shows 89.9% chlorine. What is the empirical formula for the compound? For a 100 g sample, this means 89.9 g chlorine and 10.1 g carbon. First, we convert these masses to mole quantities. 89.9 g Cl x moles Cl 35.5 g Cl 1.0 mole Cl= x = 2.53 moles Cl 10.1g C y moles C 12.0 g C 1.0 mole C= y = 0.842 moles C Quantitative analysis of a compound of carbon and chlorine shows 89.9% chlorine. What is the empirical formula for the compound? C0.842Cl2.53 This ratio can be normalized by dividing through by the smallest mole quantity: x = 2.53 moles Cl y = 0.842 moles C C0.842Cl2.53 0.842 0.842 CCl3