# Chem 110 04 Stoichiometry

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Published on August 16, 2008

Author: cpesison

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Slide 1: Chapter 04 Stoichiometry Slide 2: Atomic Mass: The weighted average of the isotopic masses of the element’s naturally occurring isotopes. Average Atomic Mass Slide 3: carbon-12: 98.89 % natural abundance 12 amu carbon-13: 1.11 % natural abundance 13.0034 amu Why is the atomic mass of the element carbon 12.01 amu? = 12.01 amu mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111) = 11.87 amu + 0.144 amu Average Atomic Mass Slide 4: Example 4.1 Copper, a metal known since ancient times, is used in electrical cables and pennies among other things. The atomic masses of its two stable isotopes, 6329Cu (69.09 percent ) and 6529Cu (30.91 percent), are 62.93 amu and 64.9278 amu. Calculate the average atomic mass of copper. The relative abundances are given in parentheses. 63.55 amu How to get the Average Atomic Mass of an Element Slide 5: Mole Concept Physically counting atoms is impossible. We must be able to relate measured mass to numbers of atoms. buying nails by the pound or kilogram. using atoms by the gram Slide 6: Mole Concept Mole (mol) is the amount of a substance that contains as many elementary entities (atoms, molecules, or partilcles) as there are atoms in exactly 12 g of the carbon-12 isotope, which is equivalent to 6.022 x 1023 (Avogadro’s Number) Molar Mass is the mass of 1 mole of units (such as atoms or molecules) of a substance. (This a more general term for Atomic Mass, Formula Mass, and Molecular Mass) Slide 14: Example 4.2 Zinc (Zn) is a silvery metal that is used in making brass (with copper) and in plating iron to prevent corrosion. How many moles of Zn are there in 23.3 of Zn? 0.356 moles Zn How to convert mass of an element to moles of an element Slide 15: Example 4.3 Sulfur (S) is a nonmetallic element that is present in coal. When coal is burned, sulfuric acid is converted to sulfur dioxide and eventually to sulfuric acid that gives rise to the acid rain phenomena. How many atoms are in 16.3 g of S? 3.06 x 1023 atoms of S How to convert mass of an element to number of atoms of an element Slide 16: Example 4.4 Silver (Ag) is a precious metal used mainly in jewelry. What is the mass (in grams) of one Ag atom? 1.792 x 10-22 g of Ag How to convert number of atoms of an element to mass of an element Slide 17: Molar Mass, Molecular Weight and Formula Weight Atomic Mass – The mass of one atom of an element. Molecular Mass – The mass of one molecule of a compound Molar Mass – a general term (can be atomic or molecular) it just means that it is the mass of one mole of an element or a compound. Slide 18: Example 4.5 Calculate the molecular masses (in amu) of the follwing compounds (a) sulfur dioxide (SO2) (b) caffeine (C8H10N4O2) 64.07 amu 194.20 amu How to get the molecular mass of a compound Slide 19: Example 4.6 Methane (CH4) is the principal component of natural gas. How many moles of CH4 are present in 6.07 g of CH4? 0.378 mol CH4 How to convert mass of a compound to moles of a compound Slide 20: Example 4.7 How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g. 1.03 x 1024 H atoms How to convert mass of a compound to number of atoms of an element Slide 21: Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023 formula units. One mole of any substance is equivalent to its molecular or formula mass. 1 mole = 28.0 g C2H4: 6.022 x 1023 molecules = 28.0 g 1 mole = 36.5 g 6.022 x 1023 molecules = 36.5 g HCl: Molar Mass, Molecular Weight and Formula Weight Slide 22: Avogadro’s Number (NA): One mole of any substance contains 6.022 x 1023 formula units. One mole of any substance is equivalent to its molecular or formula mass. 1 mole = 28.0 g C2H4: 1 mole = 36.5 g HCl: Molar Mass, Molecular Weight and Formula Weight Slide 23: Percentage Composition Percent composition is the percent by mass of each element in a compound. Slide 24: Example 4.8 Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in this compound. %H = 3.086 % % P = 31.61 % % O = 65.31 % How to get the percentage composition of a compound Slide 25: Empirical Formula The empirical formula of a chemical compound is a simple expression of the relative number of each type of atom in it. The word “empirical” literally means “based only on observation and measurement” Slide 26: Example 4.9 Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 percent carbon (C), 4.58 percent hydrogen (H) and 54.50 percent oxygen (O) by mass. Determine its empirical formula. C3H4O3 How to get the empirical formula from percent composition Slide 27: Example 4.10 Chalcopyrite (CuFeS2) is a principal mineral of copper. Calculate the number of kilograms of Cu in 3.71 x 103 kg of chalcopyrite. Mass of CuFeS2 = 1.28 x 103 kg How to get the mass of an element from mass of a compound Slide 28: Molecular Formula The formula calculated from percent composition by mass is always the empirical formula because the subscripts in the formula are always reduced to the smallest whole numbers. To calculate the actual molecular formula, we must know the approximate molar mass of the compound. Slide 29: Example 4.11 A sample of a compound contains 1.52 g of nitrogen (N) and 3.47 g of oxygen (O). The molar mass of this compound is between 90 g and 95 g. Determine the molecular formula and the accurate molar mass of the compound. Molecular formula of the compound is N2O4 and the molar mass is 92.02 g/mol How to get the mass of an element from mass of a compound Slide 30: Mass-Mass / Weight-Weight Calculations Stoichiometry is the quantitative study of reactants and products in a chemical reaction. When moles are used to calculate the amount of product formed in a reaction, this is called the Mole Method approach, which simply means that the stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance. Slide 31: Example 4.12 When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. This layer prevents further reaction between aluminum and oxygen. And it is the reason that aluminum beverages cans do not corrode. Write a balanced equation for the formation of Al2O3. 4Al + 3O2  2Al2O3 How to write and balance chemical equations Slide 32: Example 4.13 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): C6H12O6 + 6CO2  6CO2 + 6H2O If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of CO2 produced? Mass of CO2 = 1.25 x 103 g CO2 How to get the mass of a compound from the mass of another compound in a chemical equation. Slide 33: Example 4.14 All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is that between lithium and water: 2Li + H2O  2LiOH + H2 How many grams of Li are needed to produce 9.89 g of H2? 68.1 g Li How to get the mass of a compound from the mass of another compound in a chemical equation. Slide 34: Limiting Reactants When a chemists carries out a reaction, the reactants are usually not present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation. Some reactant will be left over at the end of the reaction. The reactant used up first in a reaction is called the Limiting Reagent. Excess Reagents are the reactants present in mole quantities greater than necessary to react with the quantity of the limiting reagent. Slide 35: Example 4.15 Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide: 2NH3 + CO2  (NH2)2CO + H2O In one process, 637.2 g of NH3 are treated with 1142 g of CO2. (a) which of the two reactants is the limitng reagent? (b) calculate the mass of (NH2)2CO formed. (c) How much excess reagent (in grams) is left at the end of the reaction? Mass of CO2 remaining = 319 g How to know the limiting reactant, excess reactant, mass of product formed, and the amount of excess reactant left after the reaction. Slide 36: Theoretical and Percentage Yields The amount of limiting reagent present at the start of a reaction determines the theoretical yield of the reaction, that is, the amount of product that would result if all the limiting reagent reacted. The theoretical yield then is the maximum obtainable yield, predicted by the chemical equation The actual yield, the amount actually obtained from a reaction, is almost always less than the theoretical yield. To determine how efficient a given reaction is, chemists often figure the percent yield, which describes the proportion of the actual yield to the theoretical yield. Slide 37: Example 4.16 Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It prepared by the reaction of titanium(IV) chloride with molten magnesium between 950°C and 1150°C: TiCl4 + 2Mg  Ti + 2MgCl2 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 % yield = 88.4 % How to get the theoretical yield from the masses of the reactants

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