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The S-word Stoichiometry

The S-word Stoichiometry. Chapter 3. Micro World atoms & molecules. Macro World grams. Atomic mass is the mass of an atom in atomic mass units (amu). By definition: 1 atom 12 C “weighs” 12 amu. On this scale 1 H = 1.008 amu 16 O = 16.00 amu. 3.1. (7.42% x 6.015) + (92.58% x 7.016).

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The S-word Stoichiometry

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  1. The S-wordStoichiometry Chapter 3

  2. Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu 3.1

  3. (7.42% x 6.015) + (92.58% x 7.016) 100 Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium “weighted average of all naturally occurring isotopes of an element: = 6.941 amu 3.1

  4. Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.0221367 x 1023 Avogadro’s number (NA) 3.2

  5. eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 3.2

  6. One Mole of: S C Hg Cu Fe 3.2

  7. 1 g = 6.022 x 1023 amu 1 amu = 1.66 x 10-24 g Or MM = molar mass in g/mol M NA= Avogadro’s number 3.2

  8. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 3.2

  9. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 3.3

  10. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? 3.3

  11. Heavy Heavy Light Light KE = 1/2 x m x v2 v = (2 x KE/m)1/2 F = q x v x B 3.4

  12. n x molar mass of element x 100% molar mass of compound C2H6O Percent composition of an element in a compound = n is the number of moles of the element in 1 mole of the compound 3.5

  13. Types of Formulas • Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula • Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

  14. To obtain an Empirical Formula 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. Divide each by the smallest number of moles to obtain the simplest whole number ratio. • If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers *Be careful! Do not round off numbers prematurely

  15. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance.

  16. Calculation of the Molecular Formula A compound has an empirical formula of NO2. The colorless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

  17. Empirical Formula from % Composition A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance?

  18. Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O 3.6

  19. Example – Determining a Molecular Formula Many homes in rural America are heated by propane gas, a compound that contains only carbon & hydrogen. Complete combustion of a sample of propane produced 2.641 g of carbon dioxide and 1.442g of water. Find the empirical formula of propane.

  20. Mass Changes in Chemical Reactions • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units 3.8

  21. Other units • Molarity • Moles solute / L solution • Gases • 22.4 L = 1 mole of ANY GAS at STP

  22. 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? 3.8

  23. Example – Stoichiometry Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach: Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an antacid: Which is more effective per gram?

  24. 6 green used up 6 red left over Limiting Reagents 3.9

  25. Method 1 • Pick A Product • Try ALL the reactants • The lowest answer will be the correct answer • The reactant that gives the lowest answer will be the limiting reactant

  26. Method 2 • Convert one of the reactants to the other REACTANT • See if there is enough reactant “A” to use up the other reactants • If there is less than the GIVEN amount, it is the limiting reactant • Then, you can find the desired species

  27. LimitingReactant Limiting Reactant • 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3

  28. 2Al + Fe2O3 Al2O3 + 2Fe Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. 3.9

  29. 3.9

  30. Finding Excess Practice • 10.0g of aluminum reacts with 35.0 grams of chlorine gas 2 Al + 3 Cl2 2 AlCl3 • We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are produced. How much aluminum is left unreacted?

  31. % Yield = x 100 Actual Yield Theoretical Yield Percent Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. 3.10

  32. Example – Calculating % Yield Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combining gaseous carbon monoxide & hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H2(g). Calculate the theoretical yield of methanol. If 3.57x104 g of methanol is actually produced, what is the percent yield of methanol?

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