Calculating Percent Composition and Mass of Compounds using Molar Mass, Lecture notes of Chemistry

Download Calculating Percent Composition and Mass of Compounds using Molar Mass and more Lecture notes Chemistry in PDF only on Docsity! Chemistry 11-Unit 5-The Mole Concept Tutorial 5-1 HELP Chemistry 11-Unit 5-The Mole Concept - Tutorial 5-1 HELP PAGE 1 TUTORIAL 5-1 HELP ANSWER TO QUESTION 1 ON TUTORIAL 5-1: Question 1. Find the percent composition by mass of sodium phosphate, Na3PO4. Step 1: Find the molar mass of Na3PO4 (The mass of one mole of Na3PO4 molecules) Molar Mass = 3(23.0) + 31.0 + 4(16.0) = 164 g/mol . Step 2: Find the total mass of all the sodium atoms in one mole of the compound. To do this, multiply the atomic mass of sodium by the subscript of sodium in the formula.(Na3...) Mass of sodium = 23.0 g/mol x 3 mol = 69.0 g of sodium Step 3: Divide the mass of sodium by the molar mass of Na3PO4 and multiply by 100% to get percent mass. Percent mass of sodium = 69.0 g x 100% = 42.1 % Na 164 g Step 4: Find the total mass of all the phosphorus atoms in one mole of the compound. To do this, multiply the atomic mass of phosphorus by the subscript of phosphorus in the formula.(...P..) (no subscript, so call it “1”) Mass of phosphorus = 31.0 g/mol x 1 mol = 31.0 g of phosphorus Step 5: Divide the mass of phosphorus by the molar mass of Na3PO4 and multiply by 100% to get percent mass. Percent mass of phosphorus = 31.0g x 100% = 18.9 % P 164 g Step 6: Find the total mass of all the oxygen atoms in one mole of the compound. To do this, multiply the atomic mass of oxygen by the subscript of oxygen in the formula.(...O4) Mass of oxygen = 16.0 g/mol x 4 mol = 64.0 g of oxygen Step 7: Divide the mass of oxygen by the molar mass of Na3PO4 and multiply by 100% to get percent mass. Percent mass of oxygen = 64.0 g x 100% = 39.0 % O 164 g Now, we can summarize the percent composition by mass of Na3PO4. So the percent composition by mass of Na3PO4 is: 42.1 % Na, 18.9% P and 39.0 % O Chemistry 11-Unit 5-The Mole Concept Tutorial 5-1 HELP Chemistry 11-Unit 5-The Mole Concept - Tutorial 5-1 HELP PAGE 2 ANSWER TO QUESTION 2 ON TUTORIAL 5-1 Question 2 Find the percent composition by mass of the compound ammonium phosphate, (NH4)3PO4 . Step 1: Find the molar mass of (NH4)3PO4 (The mass of one mole of (NH4)3PO4 molecules) Molar Mass = 3(14.0) + 12(1.0) + 31.0 + 4(16.0) = 149 g/mol . Step 2: Find the total mass of all the nitrogen atoms in one mole of the compound. To do this, multiply the atomic mass of nitrogen by the number of nitrogen atoms in the formula. (NH4)3... = 3 x 1 = 3 Mass of nitrogen = 14.0 g/mol x 3 mol = 42.0 g of nitrogen Step 3: Divide the mass of nitrogen by the molar mass of (NH4)3PO4 and multiply by 100% to get percent mass. Percent mass of nitrogen = 42.0 g x 100% = 28.2 % N 149g Step 4: Find the total mass of all the hydrogen atoms in one mole of the compound. To do this, multiply the atomic mass of hydrogen by the number of hydrogen atoms in the formula. (NH4)3... 3 x 4 = 12 “H” atoms. Mass of hydrogen = 1.0 g/mol x 12 mol = 12.0 g of hydrogen Step 5: Divide the mass of hydrogen by the molar mass of (NH4)3PO4 and multiply by 100% to get percent mass. Percent mass of hydrogen = 12.0g x 100% = 8.1% H 149 g Step 6: Find the total mass of all the phosphorus atoms in one mole of the compound. To do this, multiply the atomic mass of phosphorus by the subscript of phosphorus in the formula.(...P...) 1 atom of “P”. Mass of phosphorus = 31.0 g/mol x 1 mol = 31.0 g of phosphorus Step 7: Divide the mass of phosphorus by the molar mass of (NH4)3PO4 and multiply by 100% to get percent mass. Percent mass of phosphorus = 31.0 g x 100% = 20.8 % P 149g Chemistry 11-Unit 5-The Mole Concept Tutorial 5-1 HELP Chemistry 11-Unit 5-The Mole Concept - Tutorial 5-1 HELP PAGE 5 Empirical Molecular Formula CH2O C3H6O3 Mass 30.0 90.0 A good thing to do now is to figure out the molecular mass using your molecular formula (C3H6O3) and make sure it is the same as the molar mass given (90.0 g/mol): C3 H6 O3 3(12.0) + 6(1.0) +3(16.0) = 90 g/mol is the molar mass. So C3H6O3 must be the correct molecular formula for this compound. ANSWERS TO THE SELF-TEST ON TUTORIAL 5-1 1. Determine the percent composition of the compound calcium nitrate ( Ca(NO3)2 ) (That is, find the % calcium, the % nitrogen and the % oxygen in this compound.) Solution: First, find the molar mass of Ca(NO3)2. In this formula there is 1 Ca, 2 “N”s and 6 “O”s The molar mass is : 40.1 + 2(14.0) + 6(16.0) = 164.1 g/mol The % Ca = 40.1 x 100% = 24.4 % 164.1 The % N = 2(14.0) x 100% = 17.1 % 164.1 The % O = 6(16.0) x 100% = 58.5 % 164.1 So, the percent composition of calcium nitrate is 24.4% “Ca”, 17.1% “N” and 58.5% “O”. x 3 x 3 Chemistry 11-Unit 5-The Mole Concept Tutorial 5-1 HELP Chemistry 11-Unit 5-The Mole Concept - Tutorial 5-1 HELP PAGE 6 2. Find the mass of carbon contained in 336.16 grams of CO2. Solution: There is 1 mole of “C” atoms in a mole of CO2 The mass of 1 mole of C atoms is 12.0 grams (atomic mass of C) The mass of one mole of CO2 (molar mass) is 12.0 + 2(16.0) = 44.0 grams. So we have a conversion factor: 12.0 g C 44.0 g CO2 So we can now go: 336.16 g CO2 x 12.0 g C = 91.68 grams of C = 91.7 grams of C 44.0 g CO2 So there are 91.68 grams of C in 336.16 grams of CO2. ************************************************************* 3. Find the mass of oxygen contained in 860.0 grams of magnesium nitrate ( Mg(NO3)2 ) Solution: The molar mass of Mg(NO3)2 is 24.3 + 2(14.0) + 6(16.0) = 148.3 g/mol . The mass of “O” in one mole of the compound is 6(16.0) = 96.0 g The conversion factor we can use is 96.0 g “O” 148.3 g “Mg(NO3)2” We can now find the answer: 860.0 g Mg(NO3)2 x 96.0 g “O” = 556.7 g “O” = 557 g “O” 148.3 g Mg(NO3)2 Chemistry 11-Unit 5-The Mole Concept Tutorial 5-1 HELP Chemistry 11-Unit 5-The Mole Concept - Tutorial 5-1 HELP PAGE 7 4. A compound used in photography is called potassium persulphate. A 0.8162 gram sample of the compound was analyzed and found to contain 0.2361 grams of potassium, 0.1936 grams of sulphur and the rest was oxygen. a) Find the mass of oxygen in the sample. Solution: The mass of oxygen = [Total mass of sample] - [(mass of K) + (mass of S)] = [ 0.8162 g] - [ ( 0.2361) + ( 0.1936) ] = [ 0.8162 g] - [ 0.4297 ] = 0.3865 grams of oxygen b) Determine the empirical formula for this compound. Element Mass Atomic Mass Moles Moles Smallest moles Simplest Whole # Ratio potassium 0.2361 g 39.1 g/mol 0.00604 mol 0.00604 = 1.00 0.00603 1 sulphur 0.1936 g 32.1 g/mol 0.00603 mol 0.00603 = 1.00 0.00603 1 oxygen 0.3865 g 16.0 g/mol 0.0242 mol 0.0242 = 4.01 0.00603 4 So the empirical formula is KSO4 c) The molar mass of this compound is 270.4 g/mol. Determine the molecular formula. Solution: The mass of the empirical formula (KSO4 ) is 39.1 + 32.1 + 4(16.0) = 135.2 g/mol Dividing the molar mass by the mass of the empirical formula: 270.4 = 2 135.2 We get the whole number “2”.