Chemical Quantities

Chemistry IA Chemistry IA—Chemical Quantities - The Mole: - MOLE—SI unit of an amount in chemistry - the mole is a COUNTING unit (just like a dozen) - 1 dozen = 12 eggs - 1 MOLE = 6.02 x 1023 particles (Avogadro’s number!!) - particles can be atoms, ions, molecules or formula units - use the FACTOR LABEL METHOD to convert between moles and atoms 1 mole 6.02 x 1023 atoms ---------------------------- or -----------------------------6.02 x 1023 atoms 1 mole How many atoms are in 2.5 moles of C?? 𝟐. 𝟓 𝒎𝒐𝒍 𝑪 𝟔. 𝟎𝟐 𝒙 𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑪 ( )( ) = 𝟏. 𝟓 𝒙 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔 𝑪 𝟏 𝟏 𝒎𝒐𝒍 𝑪 - Molar Mass: - molar mass—the mass of 1 mole of a substance (expressed in grams) - the molar mass is EQUAL in number to the atomic mass in amu - it is much more convenient to measure the mass of MOLES of a substance because we can measure in grams!! - 1 atom H = 1.0079 amu - 1 mole H = 6.02 x 1023 atoms H = 1.0079 grams H - 1 atom O = 15.9994 amu - 1 mole O = 6.02 x 1023 atoms O = 15.9994 grams O - FORMULA MASS—the mass of a compound - you calculate the formula mass by the ALGEBRAIC SUM of all of the atoms making up the compound What is the formula mass of Al2(SO4)3 ?? Al → 2 x 27.0g = 54.0 g S → 3 x 32.1g = 96.3 g O → 12 x 16.0g = 192.0 g --------------------------------342.3 g - use the FACTOR LABEL METHOD to convert from moles to mass Chemistry IA How many grams are in 2.7 moles of O2? 𝟐. 𝟕 𝒎𝒐𝒍 𝑶𝟐 𝟑𝟐. 𝟎 𝒈 𝑶𝟐 ( )( ) = 𝟖𝟔. 𝟒 𝒈 𝑶𝟐 → 𝟖𝟔 𝒈 𝑶𝟐 𝟏 𝟏 𝒎𝒐𝒍 𝑶𝟐 How many moles are there in 15.5 grams NaCl? 𝟏𝟓. 𝟓 𝒈 𝑵𝒂𝑪𝒍 𝟏 𝒎𝒐𝒍 𝑵𝒂𝑪𝒍 ( )( ) = 𝟎. 𝟐𝟔𝟓 𝒎𝒐𝒍 𝑵𝒂𝑪𝒍 𝟏 𝟓𝟖. 𝟓 𝒈 𝑵𝒂𝑪𝒍 - the mole is the CENTRAL UNIT of chemistry!!!!!! - all conversions usually involve the mole!!! Molar Mass (grams) MOLES Number of particles (atoms) - when converting from grams to atoms or atoms to grams you MUST convert to moles first!! Chemistry IA How many molecules are in 12.5 grams H2O? 𝟏𝟐. 𝟓 𝒈 𝑯𝟐 𝑶 𝟏 𝒎𝒐𝒍 𝑯𝟐 𝑶 𝟔. 𝟎𝟐 𝒙 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑯𝟐 𝑶 ( )( )( ) 𝟏 𝟏𝟖. 𝟎 𝒈 𝑯𝟐 𝑶 𝟏 𝒎𝒐𝒍 𝑯𝟐 𝑶 = 𝟒. 𝟏𝟖 𝒙 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑯𝟐 𝑶 How many grams do 5.3 x 1030 molecules of CO2 weigh? 𝟓. 𝟑 𝒙 𝟏𝟎𝟑𝟎 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑪𝑶𝟐 𝟏 𝒎𝒐𝒍 𝑪𝑶𝟐 𝟒𝟒. 𝟎 𝒈 𝑪𝑶𝟐 ( )( ) ( ) 𝟏 𝟔. 𝟎𝟐 𝒙 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑪𝑶𝟐 𝟏 𝒎𝒐𝒍 𝑪𝑶𝟐 = 𝟑. 𝟗 𝒙 𝟏𝟎𝟖 𝒈 𝑪𝑶𝟐 - Percent Composition and Chemical Formulas - percent composition—the percent by mass of each element in a compound mass of element % comp = ---------------------------- x 100 mass of compound - can use numbers from given data or use the MOLAR MASSES for a known compound Calculate the percent composition of each element in NaHCO 3 FORMULA MASS = 84.0 g 23.0 𝑔 % 𝑁𝑎 = 𝑥 100 = 27.4 % 𝑁𝑎 84.0 𝑔 %𝐻 = 1.0 𝑔 𝑥 100 = 1.2 % 𝐻 84.0 𝑔 %𝐶 = 12.0 𝑔 𝑥 100 = 14.3 % 𝐶 84.0 𝑔 %𝑂 = 48.0 𝑔 𝑥 100 = 57.1 % 𝑂 84.0 𝑔 Chemistry IA 9.03 g Mg completely combines with 3.48 g N to form a compound. What is the percent composition of this compound?? Mg + N2 → 9.03g 3.48g MgxNy 12.51g 𝟗. 𝟎𝟑 𝒈 % 𝑴𝒈 = 𝒙 𝟏𝟎𝟎 = 𝟕𝟐. 𝟐 % 𝑴𝒈 𝟏𝟐. 𝟓𝟏 𝒈 𝟑. 𝟒𝟖 𝒈 %𝑵= 𝒙 𝟏𝟎𝟎 = 𝟐𝟕. 𝟖 % 𝑵 𝟏𝟐. 𝟓𝟏 𝒈 - Calculating Empirical Formulas: - empirical formula—gives the LOWEST WHOLE NUMBER RATIO of atoms in the compound - used for formulas of IONIC compounds - may or may not be the same as the molecular formula!! - Solving for the empirical formula from % composition data: STEP 1: Assume 100 g STEP 2: Convert % → g STEP 3: Calculate the number of MOLES of each atom STEP 4: Divide each number of MOLES by the smallest number of MOLES STEP 5: Make sure that this mole ratio is a WHOLE NUMBER RATIO (multiply by different factors until it is) STEP 6: Use the whole number ratio as SUBSCRIPTS in the empirical formula Chemistry IA A compound is 67.6% Hg, 10.8% S, and 21.6 % O. What is it’s empirical formula?? Assume 100 g 𝟔𝟕. 𝟔 𝒈 𝑯𝒈 𝟏 𝒎𝒐𝒍 𝑯𝒈 𝟎. 𝟑𝟑𝟕 𝒎𝒐𝒍 𝑯𝒈⁄ )( )= 𝟔𝟕. 𝟔% 𝑯𝒈 → ( 𝟎. 𝟑𝟑𝟕 = 𝟏 𝟏 𝟐𝟎𝟎. 𝟓𝟗 𝒈 𝑯𝒈 𝟏𝟎. 𝟖% 𝑺 → ( 𝟏𝟎. 𝟖 𝒈 𝑺 𝟏 𝒎𝒐𝒍 𝑺 )( ) = 𝟎. 𝟑𝟑𝟕 𝒎𝒐𝒍 𝑺⁄𝟎. 𝟑𝟑𝟕 = 𝟏 𝟏 𝟑𝟐. 𝟎𝟔𝟓 𝒈 𝑺 𝟐𝟏. 𝟔% 𝑶 → ( 𝟐𝟏. 𝟔 𝒈 𝑶 𝟏 𝒎𝒐𝒍 𝑶 )( ) = 𝟏. 𝟑𝟓 𝒎𝒐𝒍 𝑶⁄𝟎. 𝟑𝟑𝟕 = 𝟒 𝟏 𝟏𝟓. 𝟗𝟗𝟗 𝒈 𝑶 So… HgSO4 - Calculating the molecular formula: - molecular formula—shows the actual number of atoms in a molecule - may not be the smallest whole number ratio - Calculating the molecular formula: STEP 1: Calculate the empirical formula STEP 2: Divide the MOLAR MASS / empirical formula mass STEP 3: Multiply the subscripts in the empirical formula by the answer to step 2 STEP 4: Write the molecular formula using these new subscripts Chemistry IA A compound is 58.8 % C, 9.8 % H, and 31.4% O. It has a molar mass of 102 g/mol. What is the molecular formula??? Assume 100 g 𝟓𝟖. 𝟖 𝒈 𝑪 𝟏 𝒎𝒐𝒍 𝑪 )( ) = 𝟒. 𝟖𝟗𝟔 𝒎𝒐𝒍 𝑪⁄𝟏. 𝟗𝟔𝟑 𝟏 𝟏𝟐. 𝟎𝟏𝟏 𝒈 𝑪 = 𝟐. 𝟓 𝒙 𝟐 → 𝟓 𝟓𝟖. 𝟖% 𝑪 → ( 𝟗. 𝟖 𝒈 𝑯 𝟏 𝒎𝒐𝒍 𝑯 )( ) = 𝟗. 𝟕𝟐𝟑 𝒎𝒐𝒍 𝑯⁄𝟏. 𝟗𝟔𝟑 𝟏 𝟏. 𝟎𝟎𝟕𝟗𝟒 𝒈 𝑯 = 𝟓 𝒙 𝟐 → 𝟏𝟎 𝟗. 𝟖% 𝑺 → ( 𝟑𝟏. 𝟒 𝒈 𝑶 𝟏 𝒎𝒐𝒍 𝑶 )( ) = 𝟏. 𝟗𝟔𝟑 𝒎𝒐𝒍 𝑶⁄𝟏. 𝟗𝟔𝟑 𝟏 𝟏𝟓. 𝟗𝟗𝟗𝟒 𝒈 𝑶 = 𝟏𝒙𝟐→𝟐 𝟑𝟏. 𝟒% 𝑶 → ( So… C5H10O2 is the EMPIRICAL 𝑴𝒐𝒍𝒂𝒓 𝑴𝒂𝒔𝒔 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝑴𝒐𝒍𝒂𝒓 𝑴𝒂𝒔𝒔 𝒆𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 = 𝟏𝟎𝟐 𝒈/𝒎𝒐𝒍 𝟏𝟎𝟐 𝒈/𝒎𝒐𝒍 =𝟏 So … {C5H10O2} x 1 → C5H10O2