CHEM 1415 1st Edition Lecture 1Unit: Calculating Molarity Outline of Current Lecture I. Formulas II. Calculating Molaritya. Example 1b. Example 2III. Hydrocarbonsa. Example 3Current LectureI. Formulasmolarity=moles of soluteliter of solutionUnits=(molL)=(M )II. Calculating Molaritya. Example1:25 g of NaOH500 mL of solution, what is the molarity?NaOHMolecule Number of molecules times atomic numberNa 1 * 23 23O 1 * 16 16H 1 * 1 140 g/molThese 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)(1 mol40 g)=0.625 mLCoverting mililiters¿liters−(500 mL)(1 L1000 mL)=0.5 Lmolarity=0.625 mol.5 L=1.25 mol / L=1.25 Mb. Example 2:How many moles of HCL are in 56.5 mL solution with a 3.0 M ?molarity=moles of soluteliter of solutionMoles of solution=(molarity) (Litersof solution)¿(3.0molL)(0.0565 L)¿0.1695 molIII. Hydrocarbons:- Simplest hydrocarbon – CH4 methane, because there is only one carbono Colorless odorless gas- As number of carbon molecules increases the boiling point of the compound increaseso Compounds move from gas to liquid to solidExample of balancing complete combustion of hydrocarbon ethanol:C2H5OH +O2→ C O2+H2OC2H5OH +3 O2→ 2C O2+2 H2O- Oxygen combines to make carbon dioxide in complete combustion instead of carbon monoxidea. Example 3:100 g of C8H18 octaneFor a complete combustion, how many grams of O2 are needed?C8H18+O2→ C O2+ H2OC8H18+(25 /2)O2→ 8C O2+9 H2OC8H18Molecule Number of molecules times atomic numberC 8 * 12 96H 18 * 1 18114 g/molmolesof octane∈100 g−(100 g)(1 mol114 g)=0.88 molO2Molecule Number of molecules times atomic numberO 2 * 16 3232 g/molmolesof oxygen ∈100 g−(0.88 moles of octane)(25 mol2 mol)=11 mol of O2grams of oxygen needed−(11 mol)(32 gmol)=352 g of
View Full Document
Unlocking...