Chemistry Notes Essay Example For Students

Chemistry Notes Essay LECTURE 6 NOTES, CHM 101, SEC. 01SOLUTION CONCENTRATIONSTHE ACTUAL WEIGHING OF REACTANTS OFTEN PROVES IMPRACTICAL OR INCONVENIENT. THIS IS ESPECIALLY TRUE IF REACTANTS ARE GASES, LIQUIDS, OR VERY REACTIVE. HOW CAN WE DELIVER KNOWN WEIGHTS OF REACTANTS WITHOUT WEIGHING THEM FIRST?CONSIDER: HNO3 + NaOH = NaNO3 + H2ONEITHER NITRIC ACID, NOR SODIUM HYDROXIDE CAN BE WEIGHED EASILY. WHY?HOWEVER, SOLUTIONS OF KNOWN AMOUNT OF SOLUTE PER UNIT VOLUME CAN BE DELIVERED IN KNOWN VOLUME TO GIVE PRECISELY-KNOWN AMOUNTS OF REACTANTS. WHY?BECAUSE: (MOLES/LITER) X LITERS = MOLESIF YOU KNOW THE NUMBER OF MOLES CONTAINED IN ONE LITER, YOU CAN CALCULATE THE MOLES CONTAINED IN ANY MEASURED VOLUME OF THAT SOLUTION. TYPICALLY, A FLASK CALLED A VOLUMETRIC FLASK IS USED TO PREPARE A SOLUTION OF KNOWN CONCENTRATION,(MOLES SOLUTE/LITERS SOLUTION). SOLUTE IS WEIGHED AND ADDED TO THE FLASK. THEN THE SOLVENT IS ADDED TO MAKE UP A KNOWN FIXED VOLUME OF SOLUTION. THE AMOUNT OF SOLUTE PER UNIT VOLUME SOLUTION(TYPICALLY MOLES/LITER) IS CALLED THE CONCENTRATION OF THE SOLUTION. IT HAS TO BE DEFINED QUANTITATIVELY. THIS CAN BE DONE IN SEVERAL WAYS. WE WILL USE THE DEFINITION MOST COMMONLY USED IN ANALYSIS, MOLARITY. SOLUTION MOLARITY = MOLES SOLUTE/LITERS SOLUTIONTHE MOLARITY IS GIVEN THE SYMBOL, M. IF A SOLUTION IS 1.000 M, 1.000 LITERS OF SOLUTION CONTAIN EXACTLY 1.000 MOLES OF SOLUTE. THE USEFULNESS OF THE CONCEPT REVOLVES AROUND THE FACT THATMOLES = M x LITERS(moles/L)x LYOU MUST KNOW HOW TO PREPARE DILUTE SOLUTIONS FROM CONCENTRATED ONES. BASICALLY, MOLES = M x VLSOLUTION PREPARATIONIN MOST LABORATORIES REAGENTS ARE PURCHASED IN CONCENTRATED FORM DUE TO SHIPPING AND PACKAGING COSTS . THEY ARE RARELY USED AS DELIVERED. RATHER THEY ARE DILUTED FOR USE AS NEEDED. IF YOU EVER WORK IN A LABORATORY YOU MUST KNOW HOW TO PREPARE DILUTE SOLUTIONS FROM MORE CONCENTRATED ONES. ACS-CERTIFIED REAGENT-GRADE HCl IS SOLD IN BOTTLES CONTAINING 12.1 M SOLUTION. HOW WOULD YOU GO ABOUT PREPARING 1 LITER OF 0.100 M HCl?USE THE RELATION:MOLES CON. HCl = MOLES DILUTE HClMCON x VCON = MDIL x VDIL12.1 x VCON = 0.100 x 1.00VCON = 0.100/12.1 = 0.00826 LITERS0.00826 L(1000 mL/Liter) = 8.26 mL8.26 mL OF CONCENTRATED HCl WHEN ADDED TOMAKE ONE LITER OF SOLUTION RESULTS IN A SOLUTION WHICH IS 0.100 M. IT IS A LITTLE EASIER TO MAKE UP SOLUTIONS OF PRECISE MOLARITY VALUES IF THE SOLUTE IS A SOLID AND CAN BE WEIGHED. SILVER NITRATE, AgNO3, IS SUCH A REAGENT. HOW MANY GRAMS OF SILVER NITRATE MUST BE ADDED TO A 100.00 mL FLASK TO PRODUCE A 0.200 M SOLUTION. Ag NO3(S) = Ag+(aq) + NO3-(aq) 0.200M 0.200MSTRATEGY: FIND THE MOLES AgNO3 NEEDED FOR THE SOLUTION. THEN FIND WHAT THOSE MOLES WEIGH. MOLES AgNO3 = MxV = 0.200 x 0.10000 = 0.0200MASS AgNO3 = MOLES x(GRAMS/MOLE)0.0200 x 169.8731 = 3.3975 gWITH WHAT PRECISION DO WE NEED TO WEIGH OUT THE SILVER NITRATE?CHAPTER 3 PROBLEMS:25. Calculate the mass, in grams, of 1.12 mol CaH2. HOW DO WE FIND THE MASS IF WE KNOW THE NUMBER OF MOLES AND THE MOLECULAR FOLMULA?FROM THE MOLECULAR FORMULA WE CALCULATE THE AVERAGE MASS OF ONE MOLE, THE MOLAR MASS IN GRAMS. ONCE WE HAVE THE MOLAR MASS WE MULTIPLY THE MOLAR MASS BY THE NUMBER OF MOLES, 1.12. MASS = (GRAMS/MOLE) X MOLESWE MUST KNOW THE MOLAR MASS TO MAKE THE CALCULATION. Molecular mass = mass atoms in the molecule. =mass Ca + 2x mass H = 40.077 + 2(1.00794) =42.093u/molecule or = 42.093 g./moleTHAT IS THE MOLAR MASS OF CALCIUM HYDRIDE IS 42.093 GRAMS/MOLE. TO CALCULATE THE MASS IN 1.12 MOLES:MASS 1.12 MOLES = = (42.093 GRAMS/MOLE)(1.12 MOLES)= 47.1 g. NOTE THE 3 SIG. FIGS. 27. Calculate the number of moles corresponding to 98.6 g of nitric acid, HNO3. RECOGNIZE: MOLES = GRAMS/(GRAMS/MOLE)AGAIN WE NEED TO KNOW THE MOLAR MASS. THIS TIME WE MUST CALCULATE THE MOLAR MASS OF NITRIC ACID. .u93bdd1fcef2ef13de635f5e0839270b5 , .u93bdd1fcef2ef13de635f5e0839270b5 .postImageUrl , .u93bdd1fcef2ef13de635f5e0839270b5 .centered-text-area { min-height: 80px; position: relative; } .u93bdd1fcef2ef13de635f5e0839270b5 , .u93bdd1fcef2ef13de635f5e0839270b5:hover , .u93bdd1fcef2ef13de635f5e0839270b5:visited , .u93bdd1fcef2ef13de635f5e0839270b5:active { border:0!important; } .u93bdd1fcef2ef13de635f5e0839270b5 .clearfix:after { content: ""; display: table; clear: both; } .u93bdd1fcef2ef13de635f5e0839270b5 { display: block; transition: background-color 250ms; webkit-transition: background-color 250ms; width: 100%; opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #95A5A6; } .u93bdd1fcef2ef13de635f5e0839270b5:active , .u93bdd1fcef2ef13de635f5e0839270b5:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .u93bdd1fcef2ef13de635f5e0839270b5 .centered-text-area { width: 100%; position: relative ; } .u93bdd1fcef2ef13de635f5e0839270b5 .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .u93bdd1fcef2ef13de635f5e0839270b5 .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .u93bdd1fcef2ef13de635f5e0839270b5 .ctaButton { background-color: #7F8C8D!important; color: #2980B9; border: none; border-radius: 3px; box-shadow: none; font-size: 14px; font-weight: bold; line-height: 26px; moz-border-radius: 3px; text-align: center; text-decoration: none; text-shadow: none; width: 80px; min-height: 80px; background: url(https://artscolumbia.org/wp-content/plugins/intelly-related-posts/assets/images/simple-arrow.png)no-repeat; position: absolute; right: 0; top: 0; } .u93bdd1fcef2ef13de635f5e0839270b5:hover .ctaButton { background-color: #34495E!important; } .u93bdd1fcef2ef13de635f5e0839270b5 .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .u93bdd1fcef2ef13de635f5e0839270b5 .u93bdd1fcef2ef13de635f5e0839270b5-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .u93bdd1fcef2ef13de635f5e0839270b5:after { content: ""; display: block; clear: both; } READ: Service and Helping Others and Three Reasons People are Hesitant to Help EssayMOLAR MASS = mmass N + mmass H + 3x mmass O14.0067 + 1.00794 + 3(15.9994) == 63.0128 g/moleMOLES = 98.6g / (63.0128 g/mole) = 1.56 moles29. Calculate the number of molecules in 4.68 mol H2O?YOU HAVE TO RECOGNIZE THAT ONE MOLE CONTAINS 6.02 x 1023 MOLECULES. THUS,THE NUMBER OF MOLECULES = = MOLES X (MOLECULES/MOLE) =MOLECULES = 4.68 moles(6.02 x 1023 molecules/mole) = 28.2 x 1023 = 2.82 x 1024 molecules. HOW MANY TOTAL ATOMS ARE IN THE ABOVE MOLECULES?35. What is the mass per cent oxygen in the compound having the formula, HOOCCH2CH(CH3)COOH?YOU MUST RECOGNIZE THAT:MASS % O = MASS O per mole/molar massx100THUS, WE NEED TO KNOW THE MASS OF OXYGEN IN ONE MOLE OF MATERIAL AND THE MOLAR MASS OF THE MATERIAL. THIS CAN BE DONE GIVEN THE ATOMIC MASSES AND THE FORMULA. MASS O per mole = 4 x(15.9994) g. = 63.9976 g. Molar mass = 5x(atomic mass C) + 4x(atomic mass O) + 8x(atomic mass H) = 5 x 12.011 + 4 x 15.9994 + 8 x 1.00794 = 132.116 g/mole%O = (mass O)/(mass/mole) x 100 = =(63.9976/132.116)x100 = 48.4405 %39. THE EMPIRICAL FORMULA OF PARA-DICHLOROBENZENE IS C3H2Cl. IF THE MOLECULAR MASS IS 147 u, WHAT IS THE FORMULA?RECOGNIZE THAT THE MOLECULAR MASS IS AN INTEGRAL MULTIPLE OF THE EMPIRICAL FORMULA MASS. THE EMPIRICAL FORMULA MASS = 3x mass C + 2 x mass H + mass Cl =3(12.011) + 2(1.00794) + 35.453 = 73.502 uTWO TIMES THE EMPIRICAL FORMULA MASS IS EXACTLY THE MOLECULAR MASS. THIS MEANS THAT THE MOLECULAR FORMULA ISC3H2Cl2 OR C6H4Cl2. 43. Resorcinol, is composed of 65.44% C, 5.49 % H and 29.06 % O. Its molecular mass is 110 u. Determine the molecular formula. RECOGNIZE THAT YOU HAVE BEEN GIVEN THE INFORMATION NEEDED TO CALCULATE THE EMPIRICAL FORMULA. THE EMPIRICAL FORMULA MASS MULTIPLIED BY A SMALL WHOLE NUMBER WILL GIVE THE MOLECULAR MASS. GIVEN PERCENTAGE COMPOSITION DATA, PROCEED ON THE BASIS OF A 100.000 GRAM SAMPLE. THAT IS, ASSUME YOU HAVE A 100 g SAMPLE AND THAT IT CONTAINS 65.44g C, 5.49 g. H AND 29.06 g. O. FIND THE NUMBER OF MOLES OF EACH ELEMENT IN THE SAMPLE. MOLES C = 65.44 g/ (12.011 g/mole) = 5.448MOLES H = 5.49 g./ (1.00794 g/mole) = 5.45MOLES O = 29.06 g./ (15.9994 g./mole) = 1.816MOLES C/MOLES O = 5.448/1.816 = 3.000MOLES H/MOLES O = 5.45/ 1.816 = 3.00THE EMPIRICAL FORMULA = C3H3OTHE FORMULA MASS = 312.011 + 31.00794 + 15.9994 = 55.056 uSINCE THE MOLECULAR MASS IS 110 u, THECORRECT MOLECULAR FORMULA WILL BE C3H3O2 = C6H6O2 , OR THE EMPIRICAL FORMULA MULTIPLIED BY TWO. 55. Balance the following:Cl2O5 + H2O HClO3LATER IN THE COURSE YOU WILL BE GIVEN BETTER DEVICES FOR BALANCING EQUATIONS. AT THIS POINT TRY TO BALANCE THE ELEMENTS OTHER THAN OXYGEN AND HYDROGEN FIRST AND THEN BALANCE THE OXYGEN OR HYDROGEN WITH WHAT EVER MOLECULES YOU HAVE AVAILABLE. START WITH ClCl2O5 2 HClO3 NOTE BY INSPECTION THAT ADDING ONE WATER TO THE LEFT SIDE WILL PRODUCE BALANCE. Cl2O5 + H2O = 2 HClO3ON BOTH SIDES YOU HAVE 2 Cl, 2 H AND 6 O. BALANCE Al + O2 Al2O3BALANCE Al FIRST:2Al Al2O3 NOTE YOU NEED AN EVEN NUMBER OF OXYGENS. 4 Al 2 Al2O3 SEE 3 O2 WILL BALANCE O. 4 Al + 3 O2 = 2 Al2O3 BALANCED. 61. 2 C8H18 + 25 O2 = 16 CO2 + 18 H2Ojjj. How many moles of carbon dioxide are produced when 1.8 x 104 moles of octane are burned?THE STOICHIOMETRIC COEFFICIENTS TELL US:MOLES CO2/ MOLES OCTANE = 16/2 = 8/1 = 8THAT IS, MOLES CARBON DIOXIDE = 8 X MOLES OCTANE. MOLES CARBON DIOXIDE = 8 x 1.8 x 104 = 14.4 x 104 = 1.4 x 105. 67. Kerosene is a mixture of hydrocarbons used in heating and as a jet fuel. Assume that kerosene can be represented by C14H30 and that it has a density of 0.763 g/mL. How many grams of carbon dioxide are produced by the combustion of 3.785 L of kerosene?Strategy: 1. Partially balance the equation for combustion. 2. Note that you have been given the density and the volume of the kerosene. From d = m/v you can calculate the mass of the kerosene. From the mass you can calculate moles after calculating the molar mass of kerosene. .u10424c53ecd8c22ec94861157236f931 , .u10424c53ecd8c22ec94861157236f931 .postImageUrl , .u10424c53ecd8c22ec94861157236f931 .centered-text-area { min-height: 80px; position: relative; } .u10424c53ecd8c22ec94861157236f931 , .u10424c53ecd8c22ec94861157236f931:hover , .u10424c53ecd8c22ec94861157236f931:visited , .u10424c53ecd8c22ec94861157236f931:active { border:0!important; } .u10424c53ecd8c22ec94861157236f931 .clearfix:after { content: ""; display: table; clear: both; } .u10424c53ecd8c22ec94861157236f931 { display: block; transition: background-color 250ms; webkit-transition: background-color 250ms; width: 100%; opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #95A5A6; } .u10424c53ecd8c22ec94861157236f931:active , .u10424c53ecd8c22ec94861157236f931:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .u10424c53ecd8c22ec94861157236f931 .centered-text-area { width: 100%; position: relative ; } .u10424c53ecd8c22ec94861157236f931 .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .u10424c53ecd8c22ec94861157236f931 .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .u10424c53ecd8c22ec94861157236f931 .ctaButton { background-color: #7F8C8D!important; color: #2980B9; border: none; border-radius: 3px; box-shadow: none; font-size: 14px; font-weight: bold; line-height: 26px; moz-border-radius: 3px; text-align: center; text-decoration: none; text-shadow: none; width: 80px; min-height: 80px; background: url(https://artscolumbia.org/wp-content/plugins/intelly-related-posts/assets/images/simple-arrow.png)no-repeat; position: absolute; right: 0; top: 0; } .u10424c53ecd8c22ec94861157236f931:hover .ctaButton { background-color: #34495E!important; } .u10424c53ecd8c22ec94861157236f931 .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .u10424c53ecd8c22ec94861157236f931 .u10424c53ecd8c22ec94861157236f931-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .u10424c53ecd8c22ec94861157236f931:after { content: ""; display: block; clear: both; } READ: Essay on Different Methods Of Quantitative Research Essay3. From the balanced equation you determine the ratio of moles carbon dioxide to kerosene. 4. From moles carbon dioxide you calculate grams carbon dioxide. C14H30 + x O2 = y H2O + 14 CO2 Note moles carbon dioxide = 14 x moles kerosene. Mass kerosene = d x v = (.763 g/mL)(3785mL) = 2.89103 g. MOLAR MASS KEROSINE = 14x 12.011 + 30 x 1.00794 = 198.392 g/moleMOLES KEROSENE = 2.89 x 103/198.392 = 14.6MOLES CARBON DIOXIDE = 14.00 X 14.6 = 204GRAMS CARBON DIOXIDE = 204moles X 44.01g/mole = 8978 g = 8.98 KG. 73. Lithium hydroxide absorbs carbon dioxide to form lithium carbonate and water:2 LiOH + CO2 = Li2CO3 + H2OIf a reaction vessel contains 0.150 mol LiOH and 0.080 mol CO2, which compound is the limiting reagent? How many moles of the carbonate can be produced?0.150 mol of LiOH will react with 0.075 mol of carbon dioxide. (Moles carbon dioxide)/(moles LiOH) = . Thus, the LiOH is the limiting reagent and all the carbon dioxide is not used up. The number of moles of the carbonate will be the same as the number of moles carbon dioxide used, 0.075. 83.a Calculate the molarity of a solution prepared from 6.00 mol HCl in 2.50 L solution. M = moles solute/liters solution = 6.00 moles/2.50 L = 2.4 moles/liter87. How many mL of a 0.215 M solution are required to contain 0.0867 mol NaBr?Note M x VL = moles = 0.0867 = 0.215 x VLVL = 0.0867/.215 = 0.403 liters = 403 mLScience Essays