Chapter 3. Mass Relationships, Stoichiometry and Chemical Formulas. Announcements. Learning Objectives MOLE

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Announcements HOUR EXAM 1 --Critical To Understand Chapter 3 July 18 6-7:30PM --Skip Combustion Analysis & Isomers (p.82-83 in Principles of Chemistry Text) See me if you donʼt understand!

119 (Principles of Chemistry) Chapter 3: 3.7, 3.11, 3.13, 3.15, 3.17, 3.19, 3.26, 3.29, 3.34,3.36, 3.38, 3.42, 3.45, 3.52, 3.56, 3.62, 3.64, 3.66, 3.72, 3.74, 3.82, 3.88, 3.93, 3.95, 3.97, 3.99, 3.

1 Announcements HOUR EXAM 1 --Critical To Understand Chapter 3 July :30PM --Skip Combustion Analysis & Isomers (p in Principles of Chemistry Text) See me if you donʼt understand! Chapter 3 Relationships, Stoichiometry and Chemical s Chapter 3: 6, 10, 12, 14, 16, 18, 20, 23, 26, 28, 30, 32, 33, 38, 40, 41, 43, 45, 49, 51, 65, 66, 69, 71, 73, 75, 83, 85, 89, 93, 95, (Principles of Chemistry) Chapter 3: 3.7, 3.11, 3.13, 3.15, 3.17, 3.19, 3.26, 3.29, 3.34,3.36, 3.38, 3.42, 3.45, 3.52, 3.56, 3.62, 3.64, 3.66, 3.72, 3.74, 3.82, 3.88, 3.93, 3.95, 3.97, 3.99, 3.101,3.114, 3.118, (4th Chemistry Nature of Matter) Learning Objectives 1.Understand relative atomic masses, average isotopic mass. 2.Connect the dots between amu & grams & the periodic table via the mole. 3.Compute a molecular & molar mass of a substance from a formula. 4.Using factor label method to convert between grams<=>moles<=>molecules 5.% To Empirical and Vis-versa 6. Balancing equations and mastering stoichiometry 7. Limiting Reagent, Yields, Solution Stoichiometry The mole links 6.02 x or molecules to a mass in grams useful in the lab. molecular mass of water MOLE molar mass of water KEY RIFF: the periodic table tells us the mass of 1-atom of any element in amu, and also it tells us the mass of 1-mole of any element in grams! Scientists have defined a counting number called the mole and determined its value by experiment. 1 Mole 12 C = 6.02 x C = # of 12 C in exactly 12 grams of 12 C The mole links the mass of 1 atom in amu to the mass of 1 mole in grams.

00 amu 35.45 amu 58.45 amu The Molar mass of a is the same number as the molecular mass with units of grams per mole. 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.

2 Thinking Small and Thinking Big The mass (or molecular weight) is the sum of the atomic masses (in amu) in a single molecule. Thinking Small and Thinking Big The formula mass (or forumula weight) is the sum of the atomic masses (in amu) in a formula unit. SO 2 1S 2O SO amu + 2 x amu amu 1 Na 1 Cl NaCl amu amu amu The Molar mass of a is the same number as the molecular mass with units of grams per mole. 1 molecule SO 2 = amu 1 mole SO 2 = g SO 2 The molar mass of an ionic is numerically equal to the formula mass, in units of grams/mole. Molar mass of NaCl = g/mol 1 mole of NaCl(s) = grams NaCl(s) 6.022!10 23 NaCl units = grams NaCl(s) Let s Summarize Term Vernacular... Must be able to see Info in a chemical formula! Example: C 6 H 12 O 6 (MW = g/mol) Term Definition Unit Isotopic mass of an isotope of an element amu Atomic mass (also called atomic weight) or formula mass (also called molecular weight) Molar mass (M) (gram-molecular weight) Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the (or ions) in a molecule (or formula unit) of 1 mole of chemical entities (as, ions, molecules or formula units) amu amu g/mol Atoms in 1 glucose molecule of 1 glucose molecule Atoms/mole of Moles in 1 mole of /mole of Carbon (C) (12.01 amu) =72.06 amu 6(6.022 x ) 6 moles Hydrogen (H) 12(1.008 amu) =12.10 amu 12(6.022 x ) 12 moles g g Oxygen (O) 6 6(16.00 amu) =96.00 amu 6(6.022 x ) 6 moles g We must learn how to recognize conversion factors within a chemical formula, and how to convert grams to moles to molecules to and vis versa. MASS(g) of What s In A Chemical? Urea, (NH2)2CO, is a nitrogen containing used as a fertilizer around the globe? Calculate the following for 25.6 g of urea: Molar mass (g/mol) MOLES of MOLECULES or FORMULA UNITS Chemical Avogadro s Number MOLES of elements in ATOMS in a a) the molar mass of urea? b) the number of moles of urea in 25.6? b) # of molecules of urea in 25.6 g of urea? c) # hydrogen present in 25.6 g of urea.

Solution To Urea Problem 1. Calculate the molar mass (MM) of urea, (NH2)2CO MM(NH2)2CO = 2 x MN + 4 x MH + MC + MO MM(NH2)2CO = (2 x 14.07g) + 4 (1.007g) + 12.01g + 15.99g MM(NH2)2CO = 60.06 g 2.

3 Solution To Urea Problem 1. Calculate the molar mass (MM) of urea, (NH2)2CO MM(NH2)2CO = 2 x MN + 4 x MH + MC + MO MM(NH2)2CO = (2 x 14.07g) + 4 (1.007g) g g MM(NH2)2CO = g 2. # moles of (NH2)2CO in 25.6 g Mol (NH 2 ) 2 CO = 25.6 g (NH 2 ) 2 CO 1 mol (NH 2 ) 2 CO g (NH 2 ) 2 CO = mol (NH 2 ) 2 CO 3. # molecules of (NH2)2CO in 25.6 g Molec urea = mol urea molecu urea = molecu urea Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. You are given 20.0 g of borax... (a) what is the formula mass of Na2B4O7 (b) how many moles of borax is 20.0 g? (b) how many moles of boron are present in 20.0 g Na2B4O7? (c) how many grams of boron are present in 20.0 g Na2B4O7? (d) how many of B are present in 20.0g? (e) how many of O are present in 20.0g? (e) how many grams of atomic oxygen are present? 4. # H in 25.6 g (NH2)2CO #H = mol urea 4 mol H H = Borax is the common name of a mineral sodium tetraborate, an industrial cleaning adjunct, Na2B4O7. Suppose you are given 20.0 g of borax Solution: (a) The formula weight of Na2B4O7 is (2! 23.0) + (4! 10.8) + (7! 16.0) = b) # mol borax = (20.0 g borax) / (201.2 g mol 1 ) = 0.10 mol of borax, c) # mol B = 20.0 g borax)/(201.2 g mol 1 ) X 4 mol B/ 1 mol borax = 0.40 mol of B. d) # g B = (0.40 mol)! (10.8 g B/ mol B) = 4.3 g B e) # B = 0.40 mol B x 6.02 X B/1 mol B = 2.41 X B f) # g O = 20.0 g borax)/(201.2 g mol 1 ) X 7 mol O/ 1 mol borax x g O/1 mol O = 11.1 g O The plant photosynthetic pigment chlorophyll contains 2.68 percent magnesium by weight. How many of Mg will there be in 1.00 g of chlorophyll? Solution: Each 100 grams of chlorophyll contains 2.68 g of Mg and the atomic mass of Mg = 24.3 g/mol. Number of moles in this weight of Mg: (.0268 g) / (24.2 g mol 1 ) = mol Number = ( mol)! (6.02 x mol 1 ) = 6.64 X Is this answer reasonable? Yes, because we would expect to have huge numbers in any observable quantity of a substance. A chemical formula determines the % mass of each element in a. n x molar mass of element molar mass of x 100% n is the number of moles of an element in 1 mole of the. C 2 H 4 O2 2 x (12.01 g) %C = x 100% = 40.0% 4 x (1.008 g) %H = x 100% = 6.714% 2 x (16.00 g) %O = x 100% = 53.28% 40.0% % % = 100.0% Empirical and s The formula of a as it actually exists according to experimental data. It is a multiple of the empirical formula. Molar CH 2O C 2H 4O 2 C 3H 6O 3 C 4H 8O 4 C 5H 10O Empirical The simplest formula for a that gives rise to the smallest set of whole numbers of. CH 2O

Percent Empirical Name formaldehyde acetic acid lactic acid erythrose ribose glucose CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 C 6 H 12 O 6 Whole-Number Multiple 1 2 3 4 5 6 M (g/mol)

4 Compounds that have the same % mass of its elements have the same empirical formula! Compounds below have the same % by mass: 40.0% C 6.71% H 53.3% O. A laboratory technique called elemental analysis can determine the % by mass of each element in a. With it we can generate the empirical formula of any unknown. Percent Empirical Name formaldehyde acetic acid lactic acid erythrose ribose glucose CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 C 6 H 12 O 6 Whole-Number Multiple M (g/mol) Use or Function disinfectant; biological preservative acetate polymers; vinegar(5% soln) sour milk; forms in exercising muscle part of sugar metabolism component of nucleic acids and B 2 major energy source of the cell %C = a% %H = b% %O = c% % of element Atomic C x H y Oz ratio of moles If we know the % mass of elements in a we can determine the empirical formula. With molar mass we get molecular formula formula GIVEN Molar Determining the Empirical from es of Elements Elemental analysis of a sample of an ionic gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this? Assume 100 g sample Calculate mole ratio Determining the Empirical from es of Elements (Elemental Analysis) Elemental analysis of a sample of an ionic gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this? SOLUTION: 2.82 g Na x mol Na = mol Na g Na 4.35 g Cl x mol Cl = mol Cl g Cl 7.83 g O x mol O g O = mol O Determining a from Elemental Analysis and Molar Dibutyl succinate is an insect repellent used against household ants and roaches. Elemental analysis or analysis indicates that % mass of the composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate.

Step 1: Determine the mass of each element. Assume a 100 gram sample and use the given data we have: C 62.58 g H 9.63 g O 27.79 g Step 2: Convert masses to amounts in moles.

5 Step 1: Determine the mass of each element. Assume a 100 gram sample and use the given data we have: C g H 9.63 g O g Step 2: Convert masses to amounts in moles. Step 3: Write a tentative empirical formula. C 5.21 H 9.55 O 1.74 Step 4: Convert to small whole numbers. Divide by smallest number of moles C 2.99 H 5.49 O Step 5: Convert to a small whole number ratio. Multiply by 2 to get C 5.98 H O 2 The empirical formula is C 6 H 11 O 2 Empirical formula mass = 6(12.01) (11) + 2 (16.00) = 115 u. Step 6: Now using the empirical formula mass and molecular mass together determine the molecular formula. Empirical formula mass is 115 u. formula mass is 230 u. n = mass/empirical mass = 230 amu/115 amu = 2 The molecular formula is C 12 H 22 O 4 Determining a from Elemental Analysis and Molar During physical activity, lactic acid (M = g/mol) forms in muscle tissue and is responsible for muscle soreness at fatigue. Elemental analysis shows that this contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. Understand what is asked: What is the formula CxHyOz Percent Empirical Determining a from Elemental Analysis and Molar 1. Assume there are 100. g of lactic acid then use % mass: 40.0 g C 1 mol C 6.71 g H 1 mol H 53.3 g O 12.01g C g H 1 mol O g O 3.33 mol C 6.66 mol H 3.33 mol O 2. The red numbers are the number of moles in lactic acid. This is what we use in the formula C 3.33 H 6.66 O 3.33 C 3.33 H 6.66 O molar mass of lactate mass of CH 2 O CH 2 O g g empirical formula 3. The molecular formula will be a whole number multiple of the empirical formula determined BY THE MOLAR MASS ratio 3 C 3H 6 O 3 is the molecular formula QUIZ 2 Please place books on floor. 1-Single sheet of Paper With Name Calculator backs off 10 minute quiz STOP When I say STOP 1. Write the formula or the name the following s. (1a) Iron (III) sulfide (1a) KHCO3 2. Boron has only two naturally-occurring isotopes Boron-10 and Boron-11. The mass of Boron-10 is amu and the mass of Boron-11 is amu. Their relative abundances are 19.91% and 80.09%, respectively. 2a) Calculate the atomic mass of the element boron. 2b) How many protons and neutrons does Boron-11 have? 3. 3-laws make up Dalton s atomic theory. Identify the mass law that the following observation demonstrates: 3a) The mass of a enclosed beaker is g both before and after a chemical reaction had occurred. 3b) The ore magnetite is 72.36% iron and hematite is 69.94% iron. 4. Classify the following s as ionic or covalent: 4a) PCl3 4b) calcium carbonate

You should be reading Chapter 3 and practicing nomenclature from Chapter 2 (lots to know).

Announcements You should be reading Chapter 3 and practicing nomenclature from Chapter 2 (lots to know). Chapter 2: Please expect 1 or perhaps 2 to be collected as a Quiz grade on Tuesday. 1,2,4,13,14,16,18,20,24,27,34,38,40,42,45,50,53,54,