Announcements. 1 point for every question attempted; 0.2 extra credit points for every correct answer

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Morgan Stokes
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1 Announcements Print worksheet #3 prior to your Tuesday discussion section Solutions to worksheets #1 and #2 are posted online now A full schedule of readings and suggested problems is posted on the course website LON-CAPA assignments 1 and 2 are due Friday at 9am Clicker points begin TODAY 1 point for every question attempted; 0.2 extra credit points for every correct answer

2 Molecular Formula number and type of atoms present in a molecule Example: glucose C 6 H 12 O 6 Covalent compound composed entirely of non-metals mix C(s), H 2 (g) and O 2 (g) 1 atom of C 2 x g 1 atom of 12 C (6 protons + 6 neutrons) 12 amu But according to the periodic table, C = amu

3 Average Atomic Mass 98.89% of natural carbon is the 12 C isotope 1.11% of natural carbon is the 13 C isotope *Other isotopes exists, but in very small quantities Measurements with a mass spectrometer reveal the mass of 13 C to be amu The average atomic mass of carbon is equal to ( x 12 amu) + ( x amu) = amu, the mass on the periodic table

4 12 C is assigned a mass of exactly 12 amu; all other reported masses are relative to this standard 12 C is also used to define Avogadro s number (N A ): 12 g 12 C = x atoms 12 C N A = mole 12 g 12 C x atoms x atoms mole = 12 g 12 C mol

5 The mass of 1 mole of an element is equal to its atomic mass in grams C=12.01 amu O=16.00 amu H=1.01 amu C=12.01 g/mol O=16.00 g/mol H=1.01 g/mol 1 molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms 1 mole of glucose contains 6 moles of carbon atoms, 12 moles of hydrogen atoms and 6 moles of oxygen atoms

6 To calculate the molecular mass of glucose, add the masses of each type of atom: C 6 H 12 O 6 6 x amu (6 C atoms) + 12 x 1.01 amu (12 H atoms) 6 x amu (6 O atoms) amu (1 glucose molecule) 1 molecule of glucose has a mass of amu 1 mole of glucose molecules has a mass of grams, or a molar mass of g/mol

7 Suppose a sample of glucose contains g C How many grams of oxygen does it contain? Step 1: Convert grams of carbon to moles of carbon 1mol C g C = mol C 12.01g C Step 2: Use the chemical formula to determine moles of oxygen 6 mol O mol C = mol O 6 mol C Step 3: Convert moles of oxygen to grams of oxygen g O mol O = g O 1mol O

8 We ve determined that the glucose (C 6 H 12 O 6 ) sample contains mol C and mol O. How many moles of hydrogen does the sample contain? A mol H B mol H C mol H D mol H Turn on your clickers and press the button corresponding to the correct answer. Look for the green light to verify that your answer has been received. To change your answer, simply press another button on your remote.

9 C 6 H 12 O 6 Percent composition % (by mass) of each element molar mass = g/mol 6 mol C = (6 x g/mol) = = 40.00% C mol C 6 H 12 O 6 (1 x g/mol) 12 mol H mol C 6 H 12 O 6 = (12 x g/mol) (1 x g/mol) = = 6.72% H 6 mol O = (6 x g/mol) = = 53.28% O mol C 6 H 12 O 6 (1 x g/mol)

10 Using mass % to determine formula of unknown Suppose a compound is 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass. What is the formula for the compound? Step 1: Determine the grams of each component. (Assume you have a 100 gram sample) g C 4.58 g H g O Step 2: Calculate the number of moles of each element 1mol C g C = mol C 12.01g C 1mol H 4.58 g H = 4.53 mol H 1.01g H 1mol O g O = mol O g O

11 Using mass % to determine formula of unknown Step 3: Determine empirical formula Step 3a: Divide each of the mole values obtained in mol C Step 2 by the smallest mole value 4.53 mol H = mol C = 1.33 mol H mol O = mol O C 1 H 1.33 O 1 Step 3b: Find the smallest whole-number ratio of atoms in the compound C 1 H 1.33 O 1 C 3 H 4 O 3

12 Empirical vs Molecular Formula empirical formula - smallest whole-number ratio of atoms in the compound molecular formula actual formula of a compound empirical molecular C 3 H 4 O 3 C 3 H 4 O 3 C 6 H 8 O 6 C 9 H 12 O 9 C 12 H 16 O 12 Need the molar mass (molecular weight) of the compound to determine the molecular formula

13 C Combustion reaction CO H (unbalanced equation) x H yoz + O O Suppose 11.5 g of an unknown compound containing carbon, hydrogen and oxygen is burned, producing 22.0 g CO 2 and 13.5 g H 2 O. Determine the empirical formula of the unknown compound. Step 1: Determine the grams of each component 11.5 g unknown 22.0 g CO g H 2 O Step 2: Calculate moles of each element 22.0 g CO 1mol CO g CO 2 = 22 11mmoollCCO0.500 mol C

14 Combustion reaction CO H (unbalanced equation) Cx H yoz + O O Step 2: Calculate moles of each element 2molH1mol H2O 13.5 g H2 O 1mol= 1.50 mol H g H2O H 2 O 12.01g C 6.01gC1.01g H 1.52gH0.500 mol C = and 1.50 mol H = 1mol C 1mol H gC1.52soxygeng unknown = + + masoxygen4.0go = What about oxygen? Both the unknown and the O 2 (g) contain it. How do we determine how much oxygen came from the unknown? ghma

15 C Combustion reaction CO H (unbalanced equation) x H yoz + O O Step 2: Calculate moles of each element Step 3: Determine empirical formula mol mol mol O 4.0 g O = g O C = 2.0 mol C O = 1.0 mol O 0.25 mol O 1.50 mol 0.25 H = 6.0 mol H Empirical formula of unknown: C 2 H 6 O

Unit III: Quantitative Composition of Compounds

Unit III: Quantitative Composition of Compounds A. Atoms and Isotopes B. Atomic Composition of Chemical Compounds C. Formula and Molecular Mass D. Calculations using Moles of Atoms E. Calculations using