Chemistry 3.2 Determining the Formula of an Unknown Compound the Molecular Nature of 3.3 Writing and Balancing Chemical Equations Matter and Change Fifth Edition

Unit IV - Lecture 11 Mole - Mass Relationships in Chemical Systems

3.1 The Mole Chemistry 3.2 Determining the Formula of an Unknown Compound The Molecular Nature of 3.3 Writing and Balancing Chemical Equations Matter and Change Fifth Edition

Martin S. Silberberg

The Mole Figure 3.1 Counting objects of fixed relative mass.

mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.

This amount is 6.022x1023. The number is called Avogadro’s number and is abbreviated as N.

23 One mole (1 mol) contains 6.022x10 entities (to four 12 red marbles @ 7g each = 84g significant figures) 12 yellow marbles @ 4g each = 48g 55.85 g Fe = 6.022 x 1023 atoms Fe 32.07 g S = 6.022 x 1023 atoms S

Table 3.1 Summary of Mass Terminology Figure 3.2 Oxygen 32.00 g Term Definition Unit One mole of common Isotopic mass Mass of an isotope of an amu substances. element Atomic mass Average of the masses of the amu (also called naturally occurring isotopes of an atomic weight) element weighted according to their abundance Molecular Sum of the atomic masses of the amu (or formula) mass atoms (or ions) in a molecule (or Water (also called formula unit) 18.02 g molecular weight) CaSO 2H O 4* 2 Molar mass (M) Mass of 1 mole of chemical entities g/mol 172.19 g (also called gram- (atoms, ions, molecules, formula units) Copper molecular weight) 63.55 g Table 3.2 Information Contained in the Chemical Formula of Interconverting Moles, Mass, and Number of Chemical Entities Glucose C6H12O6 ( M = 180.16 g/mol)

Carbon (C) Hydrogen (H) Oxygen (O) no. of grams Mass (g) = no. of moles x g Atoms/molecule 6 atoms 12 atoms 6 atoms 1 mol of compound Moles of atoms/ 6 mol of 12 mol of 6 mol of 1 mol mole of compound atoms atoms atoms No. of moles = mass (g) x M no. of grams Atoms/mole of 6(6.022 x 1023) 12(6.022 x 1023) 6(6.022 x 1023) compound atoms atoms atoms 6.022x1023 entities No. of entities = no. of moles x Mass/molecule 6(12.01 amu) 12(1.008 amu) 6(16.00 amu) 1 mol of compound =72.06 amu =12.10 amu =96.00 amu 1 mol Mass/mole of 72.06 g 12.10 g 96.00 g No. of moles = no. of entities x compound 6.022x1023 entities

Figure 3.3 Sample Problem Calculating the Mass in a Given Number of Summary of the mass-mole- 3.1 Moles of an Element number relationships for elements. PROBLEM Silver (Ag) is used in jewelry and tableware but no : longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag? PLAN: To convert mol of Ag to g, use the # g Ag/mol Ag, the molar mass M.

amount (mol) of Ag multiply by M of Ag (107.9 g/ mol) mass (g) of Ag

SOLUTION 107.9 g : 0.0342 mol Ag Ag = 3.69 g x mol Ag Ag

Sample Problem Calculating Number of Atoms in a Given 3.2 Mass of an Element Summary of the mass-mole- Figure 3.4 number relationships for PROBLEM: Iron (Fe), the main component of steel, is the most important compounds. metal in industrial society. How many Fe atoms are in 95.8 g of Fe? PLAN: To convert g of Fe to atoms, first find the # mols of Fe and then convert mols to atoms. mass (g) of Fe divide by M of Fe (55.85 g/mol) amount (mol) of Fe multiply by 6.022x1023 atoms/mol

atoms of Fe

mol SOLUTION 95.8 g Fe Fe = 1.72 mol : x 55.85 g Fe Fe 6.022x1023 atoms = 1.04 x 1024 atoms 1.72mol Fe Fe x mol Fe Fe Sample Problem Calculating the Moles and Number of Formula Mass Percent from the Chemical Formula 3.3 Units in a Given Mass of a Compound PROBLEM: How many formula units are in 41.6 g of ammonium carbonate? Mass % of element X = PLAN:Write the formula for the compound. Calculate M. Convert the given mass to mols and then mols to formula units. mass (g) of (NH4)2CO3 divide by M atoms of X in formula x atomic mass of X (amu) SOLUTION: The formula is (NH ) CO x 100 4 2 3. molecular (or formula) mass of compound (amu) amount (mol) of multiply by 6.022 x 1023 formula(NH 4units/)2CO3 mol number of (NH4)2CO3 formula Mass % of element X = M = (2 x 14.01 g/mol N) + (8 x 1.008 g/mol H) units + (12.01 g/mol C) + (3 x 16.00 g/mol= 96.09 O) g/mol moles of X in formula x molar mass of X (g/mol) x 100 mol 6.022x1023 formula units mass (g) of 1 mol of compound 41.6 g (NH4)2CO3 x = (NH4)2CO3 (NH ) CO x 96.09 g 4 2 mol3 (NH4)2CO3 (NH4)2CO3 2.61x1023 formula units

(NH4)2CO3

Sample Problem 3.4 Calculating the Mass Percents and Masses Sample Problem Calculating the Mass Percents and Masses of Elements in a Sample of Compound 3.4 of Elements in a Sample of Compound

PROBLEM: Glucose (C6H12O6) generates chemical potential energy. SOLUTION (a) Per mole glucose there are: (a) What is the mass percent of each element in : 6 moles of C, 12 moles H, and 6 moles O glucose? (b) How many grams of carbon are in 16.55 g of 12.01 g C 1.008 g 6 mol C x 12 mol H PLAN: Findglucose? the total mass of each element and of the glucose molecule. = 72.06 g H = 12.096 g mol C x mol H Divide the mass of each element by the mass of the molecule and C H multiply by 100. 16.00 g 6 mol O x = 96.00 g O M = 180.16 g/mol amount (mol) of element X in 1 mol Omol O compound multiply by M (g/mol) of (b) 72.06 g C X mass percent of C = 0.3999x 100 = 39.99 mass mass (g) of X in 1 mol of compound 180.16 g = %C glucose divide by mass (g) of 1 mol of 12.096 g compound = x 100 = 6.714 mass mass fraction of X mass percent of H 180.16H g 0.06714 %H = glucose multiply by 100 96.00 g O mass percent of O = 0.5329x 100 = 53.29 mass 180.16 g mass % X in = %O glucose compound

Sample Problem 3.5 Determining the Empirical Formula from Masses of Elements Empirical and Molecular Formulas PROBLEM:Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the Empirical Formula - empirical formula and name of the compound? PLAN: Find the relative number of moles of each element. Divide by the The simplest formula for a compound that agrees with the lowest mol amount to find the relative mol ratios (empirical elemental analysis and gives rise to the smallest set of formula). mass (g) of each SOLUTION: whole numbers of atoms. mol Na element 2.82 g Na = 0.123 mol divide by M (g/ 22.99 g mol) Na amount (mol) of each Na mol Cl use # of moles elementas subscripts Molecular Formula - 4.35 g Cl = 0.123 mol Cl 35.45 g Cl preliminary formula mol O The formula of the compound as it exists; it may be a 7.83 g O = 0.489 mol O change to integer multiple of the empirical formula. 16.00 g O subscripts 0.123 0.489 empirical formula Na and Cl = = 1 and O = = 3.98 0.123 0.123

Na1 Cl1 O3.98 or NaClO4 NaClO4 is sodium perchlorate. Sample Problem Determining a Molecular Formula from Sample Problem Determining a Molecular Formula from 3.6 Elemental Analysis and Molar Mass 3.6 Elemental Analysis and Molar Mass PROBLEM: Elemental analysis of lactic acid (M = 90.08 g/mol) shows it contains 40.0 mass % C, 6.71 mass % H, and 53.3 mass % SOLUTION: Assuming there are 100. g of lactic acid, the constituents O.(a) Determine the empirical formula of lactic are (b)acid. Determine the molecular formula. 40.0 g C mol C 6.71 g H mol H 53.3 g mol O 12.01 g 1.008 g O 16.00 g PLAN: assume 100 g lactic acid and find the mass of each C H O divide each mass byelement mol mass = 3.33 mol C = 6.66 mol H = 3.33 mol O (M) amount (mol) of each element C3.33 H6.66 O3.33 use # mols as 3.33 3.33 3.33 CH2O empirical subscripts formula preliminary formula molar mass of 90.08 g convert to integer 3 C H O is the subscripts lactate 3 6 3 empirical formula mass of CH2O 30.03 g molecular formula divide mol mass by mass of empirical formula to get a molecular formulamultiplier

Sample Problem Determining a Molecular Formula from Figure 3.5 Combustion apparatus for determining formulas 3.7 Combustion Analysis of organic compounds. PROBLEM: When a 1.000 g sample of vitamin C (M = 176.12 g/mol) is placed in a combustion chamber and burned, the following data are obtained:

mass of CO2 absorber after combustion = 85.35 g

mass of CO2 absorber before combustion = 83.85 g

mass of H2O absorber after combustion = 37.96 g

PLAN: difference mass of (after H2O - absorber before) = before mass ofcombustion oxidized = 37.55 g What is the molecularelement formula of vitamin C?

m m C H + (n+ ) O = n CO(g) + H O(g) find the mass of each element in its combustion n m 2 2 2 2 product preliminary empirical molecular find the mols formula formula formula

Sample Problem 3.7 Determining a Molecular Formula from Table 3.3 Some Compounds with Empirical Formula Combustion Analysis CH O 2 (Composition by Mass: 40.0% C, 6.71% H, SOLUTION: CO2 85.35 g - 83.85 g = 1.50 H2O 37.96 g - 37.55 g = 0.41 53.3% O) g g Whole-Number M 12.01 g C 0.409 g Molecular = 0.0341 mol Multiple (g/mol) Use or 1.50 g CO2 = 0.409 g Name Formula 12.01C g Function 44.01 g CO2 C C C 1 30.03 disinfectant; biological preservative 0.046 g H formaldehyd CH2O 2.016 g H = 0.0456 mol e 0.41 g H2O = 0.046 g acetic acid C H O 2 60.05 acetate polymers; vinegar (5% soln) 1.008 g H H 2 4 2 18.02 g H2O H lactic acid C3H6O3 3 90.09 sour milk; forms in exercising muscle 0.545 g O 4 1.000 g - (0.409 + 0.046) g = 0.545 = 0.0341 mol erythrose C4H8O4 120.10 part of sugar metabolism g O 16.00 g O O 5 ribose C5H10O5 150.13 component of nucleic acids and B2 0.0341 0.0341 0.0456 6 180.16 major energy source of the cell C = 1; H = 1.34; O = 1 glucose C6H12O6 0.0341 0.0341 0.0341

C1H1.34O1 = C3H4.01O3 C3H4O3 Insert structures from bottom table 3.3

176.12 g/ CH O = 2.000 C H O 2 C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6 mol 6 8 6 88.06 mol g Table 3.4 Two Pairs of Constitutional Isomers

C4H10 C2H6O Property Butane 2- Ethanol Dimethyl Ether Methylpropane M(g/mol) 58.12 58.12 46.07 46.07

Boiling -0.50 -11.060C 78.50C -250C Point C Density at 200C 0.579 g/ 0.549 g/ 0.789 g/ 0.00195 g/ mL mL mL mL (gas) (gas) (liquid) (gas) Structural formulas

Space-filling models