UNT CHEM 1415 - Calculating Molarity (3 pages)

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CHEM 1415 1st Edition Lecture 1 Unit Calculating Molarity Outline of Current Lecture I Formulas II Calculating Molarity a Example 1 b Example 2 III Hydrocarbons a Example 3 Current Lecture I Formulas molarity Units II moles of solute liter of solution molL M Calculating Molarity a Example1 25 g of NaOH 500 mL of solution what is the molarity NaOH Molecule Na Number of molecules times atomic number 1 23 23 O 1 16 16 H 1 1 1 40 g mol These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute Coverting grams per mol mililiters 25 g Coverting mililiters liters 500 mL molarity 140molg 0 625 mL 10001 LmL 0 5 L 0 625 mol 1 25 mol L 1 25 M 5 L b Example 2 How many moles of HCL are in 56 5 mL solution with a 3 0 M molarity moles of solute liter of solution Moles of solution molarity Liters of solution 3 0 mol 0 0565 L L 0 1695 mol III Hydrocarbons Simplest hydrocarbon CH 4 methane because there is only one carbon o Colorless odorless gas As number of carbon molecules increases the boiling point of the compound increases o Compounds move from gas to liquid to solid Example of balancing complete combustion of hydrocarbon ethanol C2 H 5 OH O2 C O2 H 2 O C2 H 5 OH 3 O2 2C O2 2 H 2 O Oxygen combines to make carbon dioxide in complete combustion instead of carbon monoxide a Example 3 100 g of C8 H 18 octane For a complete combustion how many grams of O2 are needed C8 H 18 O2 C O2 H 2 O C8 H 18 25 2 O 2 8C O2 9 H 2 O C8 H 18 Molecule C Number of molecules times atomic number 8 12 96 H 18 1 18 114 g mol moles of octane 100 g 100 g O2 Molecule O 1114molg 0 88 mol Number of molecules times atomic number 2 16 32 32 g mol moles of oxygen 100 g 0 88 moles of octane grams of oxygen needed 11 mol mol 252 mol 11 mol of O 32molg 352 g of O 2 2