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Chapter 3: Stoichiometry

Chapter 3: Stoichiometry. Atomic Mass. Atoms are so small, it is difficult to discuss how much they weigh in grams. Use atomic mass units. An atomic mass unit (amu) is 1/12 the mass of a carbon-12 atom. This gives us a basis for comparison.

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Chapter 3: Stoichiometry

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  1. Chapter 3:Stoichiometry

  2. Atomic Mass • Atoms are so small, it is difficult to discuss how much they weigh in grams. • Use atomic mass units. • An atomic mass unit (amu) is 1/12 the mass of a carbon-12 atom. • This gives us a basis for comparison. • The decimal numbers on the table are atomic masses in amu.

  3. Atomic Masses are not whole numbers • Atomic mass is based on a weighted average of isotopes. • Can calculate the average atomic mass from the mass of the isotopes and their relative abundance.

  4. Example 3.1 • When a sample of natural copper is vaporized and injected into a mass spectrometer, 69.09% of Cu-63 and 30.91% of Cu-65 are obtained. • The mass of Cu-63 is 62.93amu and the mass of Cu-65 is 64.93amu. • Calculate the average atomic mass of natural copper.

  5. The Mole • The mole is a very large number. • 6.022 x 1023 of anything is a mole • The number of atoms in exactly 12 grams of carbon-12. • Makes the numbers on the table the mass of the average atom.

  6. Molar mass • Mass of 1 mole of a substance. • Sometimes called molecular weight. • The atomic mass of an element expressed in grams is the molar mass. • To determine the molar mass of a compound, add up the molar masses of the elements that make it up.

  7. Find the molar mass of • CH4 • Mg3P2 • Ca(NO3)3 • Al2(Cr2O7)3 • CaSO4 · 2H2O

  8. Percent Composition • Percent of each element that comprise a compound. • Find the total mass of each element, divide by the molar mass, multiply by 100. • Easiest if you use a mole of the compound. • Find the percent composition of CH4 • Al2(Cr2O7)3 • CaSO4 · 2H2O

  9. Working backwards • From percent composition, you can determine the empirical formula. • Empirical Formulais the lowest whole number ratio of atoms in a molecule. • Based on mole ratios. • A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40% O. Find its empirical formula.

  10. Pure O2 in CO2 is absorbed Hydrocarbon Sample Sample is burned completely to form CO2 and H2O H2O is absorbed

  11. Example • From p.91 • When 0.1156g of a compound is reacted with oxygen, 0.1638g of CO2, 0.1676g of H2O are collected. • If the compound is composed of C, H, and N, what is the empirical formula? • If the compound has a molar mass of 31.06g/mol, what is the molecular formula?

  12. Empirical To Molecular Formulas • Empirical is lowest ratio. • Molecular is actual molecule. • Need molar mass to do problems. • Ratio of empirical to molar mass will tell you the molecular formula.

  13. Example • A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its molecular formula?

  14. Chemical Equations • Are sentences. • Describe what happens in a chemical reaction. • Reactants ® Products • Equations should be balanced. • Have the same number of each kind of atoms on both sides because ...

  15. Balancing equations CH4 + O2® CO2 + H2O Reactants Products

  16. Abbreviations (s) - solid (g) - gas (aq) - aqueous heat D - change in Catalyst - speeds up a reaction without being part of a reaction. Is not used up.

  17. Practice • Ca(OH)2 + H3PO4® H2O + Ca3(PO4)2 • Cr + S8® Cr2S3 • KClO3(s) ® Cl2(g) + O2(g) • Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. • Fe2O3(s) + Al(s) ® Fe(s) + Al2O3(s)

  18. Meaning • A balanced equation can be used to describe a reaction in molecules and atoms. • Not grams. • Chemical reactions happen molecules at a time • or dozens of molecules at a time • or moles of molecules.

  19. Stoichiometry • Given an amount (mass, volume or concentration) of either a reactant or product you can determine other quantities. • use conversion factors from • molar mass (g / mole) • balanced equation (coefficient ratios)

  20. Ex 3.16 • Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?

  21. Ex 3.17 • Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach. • NaHCO3(s) + HCl (aq) → NaCl (aq) + H2O (l) + CO2(aq) • Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an antacid. • Mg(OH)2(s) + 2HCl (aq) → 2H2O (l) + MgCl2(aq) • Which is the more effective antacid per gram?

  22. Limiting Reactant (Reagent) • Reactant that determines the amount of product formed. • The one that runs out of first. • Makes the least product. • There are two ways to determine LR

  23. Limiting reagent (1) • To determine the limiting reagent requires that you do two stoichiometry problems. • Figure out how much product each reactant makes. • The one that makes the least is the limiting reagent.

  24. Limiting reagent (2) • To determine the limiting reagent: • Find the moles of each reactant • Divide moles by the coefficient of each reactant • The smaller reference number is the limiting reactant.

  25. Ex 3.18 • Nitrogen gas can be prepared by passing gaseous ammonia over solid copper II oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed?

  26. Excess Reagent • The reactant left when the reaction stops. • The amount of product produced is the yield. • The theoretical yield is the amount produced if the reaction went perfect. (usually calculated) • The actual yield is what is produced in the lab.

  27. Yield • How much is produced in a chemical reaction. • Typically given as a percent. (Percent Yield)

  28. Percent Yield • % yield = Actual x 100% Theoretical • % yield = produced in lab x 100% calculated product

  29. Ex 3.19 • Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. • Suppose 68.5kg CO (g) is reacted with 8.60 kg H2 (g). Calculate the theoretical yield of methanol. If 3.57 x 104 g CH3OH is actually produced, what is the percent yield of methanol?

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