Calculating Molecular Mass and Percent Composition of Compounds - Prof. Charles Campbell, Study notes of Chemistry

Download Calculating Molecular Mass and Percent Composition of Compounds - Prof. Charles Campbell and more Study notes Chemistry in PDF only on Docsity! LOST CLICKERS? YOU MAY BE ABLE TO FIND THEN IN BAGLEY 271. ALSO, PLEASE TURN THEM IN TO ME OR BAGLEY 271. Calculate the Molecular Mass of Glucose: C6H12O6 • Carbon • Hydrogen • Oxygen Calculating the Molecular Mass of a Compound Problem: Using the data in the periodic table, calculate the molecular mass of the following compounds: a) Tetraphosphorous decoxide b) Ammonium sulfate Plan: We first write the formula, then multiply the number of atoms (or ions) of each element by its atomic mass, and find the sum. Solution: a) The formula is P4O10. Molecular mass = (4 x atomic mass of P ) +(10 x atomic mass of O ) = ( 4 x 30.97 amu) + ( 10 x 16.00 amu) = 283.88 = 283.9 amu b) The formula is (NH4)2SO4 Molecular mass = ( 2 x atomic mass of N ) + ( 8 x atomic mass of H) + ( 1 x atomic mass of S ) + ( 4 x atomic mass of O) = ( 2 x 14.01 amu) + ( 8 x 1.008 amu) + ( 1 x 32.07 amu) + ( 4 x 16.00 amu) = 132.154 amu = 132.15 amu Calculating the Moles and Number of Formula Units in a given Mass of Cpd. Problem: Sodium phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = Converting mass to moles: # Formula units = DEMO: P4O10 makes phosphoric acid Calculating the Moles and Number of Formula Units in a given Mass of Cpd. Problem: Sodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.23545 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4= 1.46 x 1023 formula units Mass Fraction and Mass % Mass of Red Balls = Mass Fraction Red = Mass % Red = Mass Fraction Purple = Similarly, mass fraction yellow = Check: Mass Fraction and Mass % Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g Mass Fraction Red = 9.0 g / 16.0 g total = 0.56 Mass % Red = 0.56 x 100% = 56% red Mass Fraction Purple = 2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25% Similarly, mass fraction yellow = 3x1.0/16.0 = 0.19 Check: 56% + 25% + 19% = 100% Calculate M and % composition of NH4NO3. • 2 mol N x • 4 mol H x • 3 mol O x Molar mass = M = %N = x 100% =g N g cpd %H = x 100% =g H g cpd %O = x 100% = g O g cpd Check: 100% total ? Mass Percent Composition of Na2SO4 Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen Elemental masses (g) 2 x Na = 2 x 22.99 = 45.98 1 x S = 1 x 32.07 = 32.07 4 x O = 4 x 16.00 = 64.00 142.05 Percent of each Element % Na = Mass Na / Total mass x 100% % Na = (45.98 / 142.05) x 100% =32.37% % S = Mass S / Total mass x 100% % S = (32.07 / 142.05) x 100% = 22.58% % O = Mass O / Total mass x 100% % O = (64.00 / 142.05) x 100% = 45.05%Check % Na + % S + % O = 100% 32.37% + 22.58% + 45.05% = 100.00% Calculate the Percent Composition of Sulfuric Acid H2SO4 Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol %H = x 100% = 2.06% H2(1.008g H) 98.09g %S = x 100% = 32.69% S1(32.07g S) 98.09g %O = x 100% = 65.25% O4(16.00g O) 98.09 g Check = 100.00% Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Mass % of X Mass fraction of X Mass (g) of X in one mole of compound Multiply by M (g / mol of X) Divide by mass (g) of one mole of compound Multiply by 100 % Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = Mass % of O = x 100% = (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = mol H x M of H mass of 1 mol sucrose mol O x M of O mass of 1 mol sucrose Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose x = 10.25 g C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose Calculating the Mass of an Element in a Compound: Ammonium Nitrate Ammonium Nitrate = NH4NO3 How much Nitrogen is in 455 kg of Ammonium Nitrate? The Formula Mass of Cpd is: 4 x H = 4 x 1.008 = 4.032 g 2 x N = 2 X 14.01 = 28.02 g 3 x O = 3 x 16.00 = 48.00 g 80.052 g Therefore mass fraction N: 28.02 g Nitrogen 80.052 g Cpd = Mass N in sample = Some Examples of Compounds with the same Elemental Ratio’s Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6 Empirical and Molecular Formulas Name Molecular Empirical water H2O hydrogen H2O2 DEMO: Genie in a bottle peroxide ethane C2H6 sulfur S8 acetic acid CH3COOH Empirical and Molecular Formulas Name Molecular Empirical water H2O H2O hydrogen H2O2 HO peroxide ethane C2H6 CH3 sulfur S8 S acetic acid CH3COOH COH2 Determining Empirical Formulas from Measured Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = 0.2469 mol Na Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr Moles of O = 7.902 g O x = 0.4939 mol O 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Converting to integer subscripts (dividing all by smallest subscript): Rounding off to whole numbers: Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na0.2469 Cr0.1235 O0.4939 Converting to integer subscripts (dividing all by smallest subscript): Na1.99 Cr1.00 O4.02 Rounding off to whole numbers: Na2CrO4 Sodium Chromate