CONSERVATION OF MASS During a chemical reaction, matter is neither created nor destroyed. - i. e. the number of atoms of each element remains constant

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1 1 CHEMICAL REACTINS Example: Hydrogen + xygen Water H + H Note there is not enough hydrogen to react with oxygen - It is necessary to balance equation. reactants products + H + H (balanced equation) CNSERVATIN F MASS During a chemical reaction, matter is neither created nor destroyed. - i. e. the number of atoms of each element remains constant BALANCING CHEMICAL EQUATINS - ensures # of reactant atoms = # of product atoms CH4 + C + H Cu + S8 CuS Fe3 + C Fe + C Sc3 + H Sc(H)3 CH3NH + C + H + N - Sometimes it is more convenient to balance groups of atoms (polyatomic ions) than individual atoms. AgN3 + CaCl Ca(N3) + AgCl - Balance N3 - ions rather than N and atoms. Ba(Cl4) + NaS4 BaS4 + NaCl4 - Balance Cl4 - and S4 - ions.

2 FRMULA AND MLECULAR MASS Formula Mass (Weight) sum of atomic masses in chemical formula Molecular Mass (Weight) same as formula mass for molecular compounds - technically not defined for ionic compounds Clarification of definitions - often the words mass and weight are used interchangeably even though they are technically not the same thing. - often formula mass and molecular mass are used interchangeably even though they are technically not the same things Calculating Formula Mass Example: C3H6 (cyclopropane) 3 (1.011 amu) + 6 ( amu) = amu Example: Ca(H) (calcium hydroxide) amu + ( amu) + ( amu) = amu THE MLE ***A mole is 6.0 x 10 3 items.*** 1 mole = 6.0 x 10 3 items Analogy 1 dozen = 1 items 18 eggs = 1.5 dz. - to convert from eggs from dozen, we need to multiply by conversion factor Example: How many eggs in 3. dozen? Analogy # 1 gross = 144 items Example: How many gross is 68 pencils?

3 3 1 mole = 6.0 x 10 3 items x 10 3 is called Avogadro s number and is abbreviated NA. 6.0 x 10 3 molecules = 1 molemolecule x 10 3 molecules = molemolecule Example: How many moles of atoms is 7.43 x 10 1 atoms? Example: How many moles of ions is.5 x 10 5 ions? Example: How many molecules are in mol of molecules? MLAR MASS Definition of Molar Mass Mass of 1 mole 1 C is g BY DEFINITIN Recall 1 atom of 1 C is amu *This is not a coincidence* Definition of amu and a mole are made to ensure coincidence 1 amu = 1.66 x 10-7 kg = 1/NA grams 1 molc-1 = g molc-1 = g gC molC molc g mol C 1 ATMIC MLAR MASS 1 C: 1 atom = 1 amu 1 mol = 1 g M( 1 C) = 1 g/mol 4 Mg: 1 atom = 4 amu 1 mol = 4 g M( 4 Mg) = 4 g/mol 56 Fe: 1 atom = 56 amu 1 mol = 56 g M( 56 Fe) = 56 g/mol C 1

4 AVERAGE ATMIC MLAR MASS - Elements often have two or more naturally occurring isotopes - Average atomic molar mass is the average of all atomic molar masses of the naturally occurring isotopes according to each isotope s relative abundance - Except for units, identical to average atomic mass - Note: mass of isotopes not integers because of nuclear forces Example: Neon nuclide abundance mass(g/mol) 0 Ne 90.48% Ne 0.7% Ne 9.5% M(Ne) = (19.99 g/mol) (0.994 g/mol) (1.991 g/mol) = 0.18 g/mol - Note: example is the same as example for average atomic mass Example: How many moles of atoms are in 96.3 grams of carbon? Example: How much mass does moles of uranium have? FRMULA MLAR MASS - add molar masses for all atoms within a chemical formula Example: ethene, CH4 M(CH4) = x g/mol + 4 x g/mol = g/mol Ethene (ethylene) is used to ripen fresh fruit. It is also used to make polyethylene, which is used to make milk jugs. Example: How many moles are in 538 g of Ba(N3) First calculate formula weight g/mol x g/mol 6 x g/mol g/mol 538g 538g 06. mol g Barium nitrate is used to color fireworks green.

5 5 SCHEME: Converting mass to moles to number Mass (g) M Molar mass Moles (mol) NA Avogadro s number Note: Molar mass and Avogadro s number are conversion factors Number (atoms or molecules) Example: How many atoms are in 0.1 g of He? Example: How many F atoms are in g of CaF? EMPIRICAL FRMULA FRM MASS PERCENT ANALYSIS Given: Percent Mass Composition Find: Empirical Formula Strategy: 1) Assume 100 g of matter. ) Multiply 100 g by mass percent to find amount of each element. 3) Convert mass of each element to moles using molar mass. 4) Find whole number ratios by dividing each number of moles by lowest number of moles.

6 6 Example: Find the empirical formula for a compound with the following mass percentages: % Cl % C.6 % H Cl: g x = g C: g x = g H: g x 0.06 =.6 g 3 H: C: Cl: 4 Empirical Formula is CH3Cl3

7 7 CMBUSTIN ANALYSIS A mass of hydrocarbon is burnt with oxygen to produce C, H and N. From the mass of C, H and N produced and mass of oxygen used, find the empirical formula for the compound. Ultimately, we want to find molar ratios that we ll use to find the empirical formula. 1. Convert mass of C to moles of carbon atoms. - calculation yields moles of carbon in hydrocarbon sample. C (g) + KH (s) KHC3 (s). Convert mass of H to moles of hydrogen atoms. - because there are two moles of hydrogen to one mole of water, multiply result by to yield moles of hydrogen in hydrocarbon sample. H (g) + CaCl (s) CaCl H (s) 3. Convert mass of N to moles of nitrogen atoms. - because there are two moles of nitrogen atoms in one mole of nitrogen molecules, multiply result by to yield moles of nitrogen in hydrocarbon sample. 4. Calculate moles of oxygen in products. - moles of oxygen = moles of C + moles of H 5. Calculate moles of oxygen in hydrocarbon - moles of hydrocarbon oxygen = moles of oxygen atoms in products moles of oxygen atoms used 6. Divide molar amounts by smallest of the values to find molar ratios in terms of whole numbers.

8 Example: Find the empirical formula for a sample of hydrocarbon that produces 7.61 g of C, 4.15 g of H and uses 7.39 g of during combustion. 8 Moles of carbon Moles of hydrogen C C 7.61g 0.173molC 44.0 gc C H mol 18.0 g H 4.15g 0.46 H H H Moles of oxygen in products C mol H mol products 7.61g 4.15g 44.0g 18.0g Moles of oxygen in hydrocarbon C C H H mol mol mol mol hydrocarbon products mol mol 7.39g 3.0g mol 0.46 mol 0.115mol mol Molar ratios 0.173molC 3mol mol mol 0.46H 4 mol mol C H Empirical formula is C3H8

9 THERETICAL STICHIMETRY - coefficients of balanced equations relate moles of reactants to moles of products Example: N (g) + 3 H (g) NH3 (g) mole of N is stoichiometrically equivalent to moles of NH3. - in other words, for every 1 mole of N reacted, moles of NH3 are produced. - N mol NH 3 - equivalence is only true for specific chemical reaction - equivalence can be considered a conversion factor N mol NH 3 N mol NH or mol NH N other equivalences are - N 3molH - 3mol H mol NH 3 The Haber process is essential in the production of fertilizer, which in turn, is essential for sustenance of Earth s 7,000,000,000 people. Example: a) What are all of the stoichiometric equivalences for the reaction, CH (g) + 5 (g) 4 C (g) + H (g)? molch 5mol C H molc CH H 5mol 4molC 5mol mol H mol C H b) How many moles of carbon dioxide are formed when 5 moles of acetylene (CH) is combusted? c) How many moles of oxygen are needed to fully burn 9.8 moles of acetylene (CH)? Acetylene is a welder s fuel.

10 PRACTICAL STICHIMETRY - can t measure moles directly in the real world. - must measure amount of substance with grams. ***- cannot compare substances stoichiometrically by mass, must convert to moles.*** 10 SCHEME: M Molar mass Mass of reactant (g) Mass of product (g) M Molar mass Moles of reactant (mol) Balanced equation Moles of product (mol) Example: For the reaction, NH3 (g) + HCl (g) NH4Cl (s), a) how much NH3 is needed to react with 9.3 g of HCl? 1) First convert grams of reactant to moles of reactant ) Compare moles of one reactant to other reactant. 3) Convert moles of other reactant to grams. The reaction of ammonia with hydrogen chloride gas is used to create a smokescreen. b) How much ammonium chloride is produced when 9.3 g of HCl is fully reacted? - Note with dimensional analysis, we can do problems all on one line.

11 Example: For the reaction 4BaC3 (s) + Y(C3)3 (s) + 6 CuC3 (s) YBaCu37 (s) + 13 C (g) + 3 (g) a) calculate how many grams of CuC3 is needed to fully react with g of BaC3, 11 b) calculate how many grams of YBaCu37 is formed from g of BaC3 fully reacting. Yttrium barium copper oxide (YBC) is a superconducting ceramic. It is superconducting below a temperature of 95 K.

12 1 Example: Epsom salts are used in foot baths to soften skin and relieve itching. Epsom salt is a hydrated crystal of magnesium sulfate. An intense heat source such as a Bunsen burner flame can drive off the water from the crystal leaving the anhydrous salt. Given the data below, write the correct chemical formula for Epsom salt. Mass of beaker Mass of beaker and Epsom salt Mass of beaker and dried salt g g g Find the mass of the Epsom salt g g = g Find the mass of the dried salt g g = 5.48 g Find the mass of water that has left g 5.48 g = g Find the moles of water Find the moles of MgS4 Find the number of water molecules per MgS4 unit The correct chemical formula is MgS4 7 H

13 13 LIMITING REAGENTS - ften starting materials are not available in proper stoichiometric proportions. - Given unbalanced amounts of reactants, we would like to know how much product could be produced. Analogy: Bicycle Factory The equation to make a bicycle is wheels + 1 frame + 1 handlebar 1 bicycle If the parts inventory is as follows: 40 wheels 150 frames 135 handlebars, we ask ourselves - What reactant limits production? - How much product can be produced? Limiting reactant: wheels Production: 10 bicycles ***In limiting reagent problems, we need to compare moles to moles*** - need to convert all masses to moles - To find limiting reactant, calculate number of moles of product formed from each number of moles of reactant - Limiting reactant will yield lowest number of moles produced. (The lowest number of moles produced is the actual number of moles of product produced.) Example: For the reaction S (g) + (g) + H (l) HS4 (aq), if 5.6 mol of S, 4.8 mol of and 6.0 mol of H are reacted together, a) how many moles of HS4 are produced? HS4 For the S: 56. mols 56. mol molhs4 For the : 48. mol 96. mol S H S 4 HS4 H S4 For the H: 6.0 molh 6.0 molh S4 H S is limiting reactant and therefore 5.6 moles of HS4 is produced.

14 b) how much is remaining after reaction is complete? - to answer the question, we need to know how much was used. - then remaining is the amount of reacted subtracted from the starting amount of. used: 56. mols 8. mol mol S remaining = starting used = 4.8 mol.8 mol =.0 mol Sulfur dioxide is a pollutant from burning coal that is a contributor to acid rain. Sulfur dioxide is removed from air with calcium oxide.s (g) + Ca (s) CaS 3 (s) 14 Example: For the reaction Zn (s) + CuCl (aq) ZnCl (aq) + Cu (s) a) How much copper metal is produced from the reaction of.00 g of Zn and.00 g of CuCl? Thus CuCl is the limiting reactant and the amount of copper produced is b) How much reactant was left over? Since CuCl is the limiting reactant, all of it was consumed in the reaction. Thus some Zn is left over. Moles of Zn used Zn molcu mol Total original moles moles used = moles left over Cu Zn

15 15 REACTIN YIELDS - an actual chemical process is rarely perfect - actual yield is less than theoretical perfect yield - we have been calculating theoretical yields - we often want to compare actual yield to theoretical yield Example: For the reaction Cr3 (s) + Al (s) Cr (s) + Al3 (s) 18.7 g of Chromium (III) oxide reacts to form 10.8 g of chromium metal. What the percent yield of this process? % yield Theoretical yield of chromium metal is actual yield 100% theoretical yield Thus the percent yield is

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