Alternative transcript:

not exactly one mole MOLES TO MOLECULES •conversion factors for Avogrado's number (6.02 x 1023) I molt 6.02x10 mc's OR I make 6.02 x10 23 mc's *mas molecules example How many carbon atoms are in 2.7 moles? 2.7 males/6.02 x to ¹ arom= 1.6 x 1024 atoms I mol 107.251x1025 molecule of CO₂ represents how many more?! 7.251x10²5 mcs I mal 120.4 moll X 6,02 10²3 C • How many moles are contained in 5x10²1 arom of gold' 5x10m Imal = (0.008 moi 6.02 X10 11.55 represents how many molecule? 1.55 mar x6 02x10 ²9.33 x 1023 mcs 1 DA FIVE STAR FIVE STAR DI MOLE BRIDGE grams (mass) by molar mass * by molar mass X by 6.02 x 1023 molcs 2 WAYS = 1.0 x10 25 molecules example @How many molecules in 46.0g of Agz Te? Find mol 107.87 (2) +127.6 = 343.34 g/mol soluc 460g Imol 16.02x1025 mcs= 343.34 g/mol Imol by 6.02x10² How many grams in 1.0 x10²⁰ mc of 1₂ 126.90 (2) = 253.80 g/mol Imol 6.02 X10¹ -aroms particles molecules -rons -furmula vnit 8.07x1022 mics x 11 mol 253.80g = 4100g Imol How many grams in 381.3 moles Of Cu(NO3)2? 63.55 + 14.01 (2) + 16.00 (6)= 187.57 g/mol 71520 g 381.3 moles 187.57g I mul 9/8/22 MASS PERCENT - obtain To of element in a Find percent weight of Al in Alz (104); 6 STEP 1: Find mass of each element. If not given use molar mass Al:2 (26.98)-53.96 5:3 (32.07) = 96.21 012 (16.00) - 192 STEP 2: Divide mass of compound *100=7 Cu = 1 (63.55) = 63.55 N = 2 (14.01) = 28.02 06 (16.00) element Total mass Which has a higher percentage of copper ( (OH) 2 63.55 187.57 compound 96.00 187.57 *100= 33.88% 342.11g/mol each element by total mass of 174.27 (U=1 (63.55) = 63.55 N: 2 (1600) = 32.00 H-2 (1.01) = 2.02 97.57 65.876 63.55 91.57 What is the moss percent of each atom in compound K₂ JO 4 K₂ = 2 (39.10) 78.20 X100 - 44.87% S = 32.07 K= 174 21 0-4 (16-00) S = x 100 = 32.01 174-27 0 = 64 ? (v (NO₂)2 or 174.27 x 100 = 18.407 *100 = 36.72% O divide # of g with malar mass H = 00343 Imal of element 1.01 divide answer w/ smallest # @get ine mass (120) and divide with total moss of EF 3 MF formula Isopropyl alcohol fr compored of 0.1539 of C₂ 0.0343 g of H, and 0.0689 of O. Find EF. If the molecular mass i 120.0g/mol, find mf C = 0.1539| /mol. 12.01g element (C+|++0) multiply Product with of to get mif 040069 Imal | 16.009 001279 0.004 -0.0340 0.0043 0.0043 0.0043 C3 Hg OEF (3= 12.01 (3) + Hg = 1:01 (8) + 0 = 1600 = 60.11 = 120.09 - 199 22 ((31180) (2) = (( & HR O₂) Oz 60-11 120 60.11 1=MF EMPIRICAL - smallest ratio of atoms in a molecule •H₂O2 → HO - (4 H8 - CH₂ •C6H12O6 (02- → -(SHIO CH2O cannot be rounded further - CH2 CH 2 molecular Formula (MF) STEP 4 (EF) formula (EF) example a somble of carbon weighing 43.29 burns in oxygen to form a compound of mass of 159.0 g. Find EF (STEP 11: Find mass of cach element C= 43.2 0-159-43.2=115.89 STEP 2 change q to mols for each crement 43.2g Imol 12-01 g = 3.60mol STEP 3: divide both valves by the smaller mol number 0:7.238 mal 2.01 = 2 3.60 mal C: 3.60 mol = 1 3.60 mol use whole numbers to write EF (round) COz) 4.209/14019 115.8g/ Imal | 16.00 g R A compound is known to contain only nitrogen 3 oxygen. Find EF if sample contains 4.20 g of nitrogen and 14.40 of oxygen N = Im ol 0.300 mol 0.300 mol = 7.238 mal 0 = NO3 9/12/22 round to whole # if clase (0.9/0-1) ૧.૧૦ ૧( loº =0.0000mar 0.700mal ⇓ m O al S M N a L n e (3) Find EF of 48.8 % Cd, 20.8% C, 2.6% Ht 27.8% 0 cd = 48.89 | 1001 Pb (= = 62.89 Pb Cl 20.09|11:01) H = 2.69|1 mul 1.019 0=27.89 (mal 116.00g 62-8 112.419 Pb ₂ C1 z Imor 207.209 Imol 35.45 0.434 mol = 1 0.434 mol = 6.73 mol 104159 520.75 : 12.11 g <1 (EF) 0.434 mol 2.6 mol 0.434 mal 2 1.74 mol 0.434 mul 3 =6036 3.99 4 = 4.01 = 4 subscripts by this inicger _Q.4545 0.303 6 If molecular mass is 1041.5g? what is of divicle molecular mass = 0.303 mol = 1 x 2 = 2 0.303 mal t =15x2=3 by empircral mars Pl CI 2 (207.20) + 3(34.45) = 520.75 g/mol ca+H0+ not close enough to round so multiply Pb ₂C13 muttiply to Pbz Cl 3 x2 ý multiply Pby Cis FIVE FIVE STAR. S BAL BALANCING EQUATIONS • Chemical reactions (4) Fe +(3)02 →>Fe₂O3 Creactants → products) law of conservation of mass → atoms are conserved example STEP Count all atoms on each side + ac12 → Naci i Na ONal STEP 2: balance by adding coefficents. (Don't change subscripts of formula 2 Na + Cl2 → ZNOCI wall lick STEP 3: Check and reduce if Necessary H₂ + 02 BAD 11011 1242 CIO + → H₂O 02 → 2H₂OX (3)5r+ N2 → Sr 3 N2 ||| SR ||| ||N|| 1 (2) K² + 1₂ => FORD WIN Mg(NO3)2 (2) kl 1 (¹)P₂0₂ + (6) C/ ₂ → (4)PC(5 +(1D02 1 1 HUPI 1 JUNG HATT ICT WAT LAT 11 Na₂ 504 + NaNO3+ Mg 104 Il Nall 15041 IMG! AN0₂11 |_ No SO4 + Mg(NO₂)2 → 2 NaNO₂ + Mgsoy. and H₂O. Write and balance → (7) CO₂ 1 C ₂ H ₁4 + (2) Oz (11 11 UHK G example A hydrocarbon, C₂H₁4, burns in Oz to produce CO₂ equation + (7) H 20 = 2C₁HM 210₂ →14 (02+ 14 H/₂0 I 110 TH TUT THT HAT CLASSIFYING REATION types of reactions Synthesis reaction -two or more substances combine to form one compound 5+0₂= 5O₂ → only one product Decomposition Reaction -one substance break down to form two or more products 2H₂O → 2H2 + O₂ • Single Replacement Reaction -one element is replaced by another in a reaction - must be element + a compound • Cu + AgNO3 → CUNO 3 + Ag element which places Double Replacement Reaction elements switch groups • Na₂S + Ca(NO3)2 beginning crement switch group Combustion 1 reacant → 2NaNO3 + cas -buins with O2 and giver off large amounts of energy and light 2 reactant •CH4 + 20₂ →→ CO² + 2H+₂0 "key indicator for coefficents represent marccuics in formula - 0110 moto Cumbustion reaction must be on reft of reaction. Cannot be combined with another element STOICHIOMETRY • involves measuring / calculating the amount of element/ Compounds involved in chemical reaction Quantity of given. unknown →> moles of given → males of quantity of unknown example (STEPS) Find grams of Na and Cl₂ needed to produce 737 g Of Naci. NO + C1₂-NaCI STEPT Balance equation: 2 Na + C1₂-2NaCl STEP 2 draw lines Cone for mol, one for grams) STEP 3 12.6 ₂6.30 12.6 mole g (STEP Get to top of line by changing to mores STEP S advance on mole line by using mole ratios Croefficent of unknown on top) STEP 6 Change moles to grams. x 22.99 12.6 6.30 47 12.6 mol C 441 709 737 9 L 298 12.6 mol Na 22.999 Imal = 39.10 +35.45 +16 (3) 737 298 th 2 KC/03 → 2KCI @ 0.963 118 7379 Imol 158.44 ↑ example How many grams of O2 are produced when we have 1189 of KC/03? 6.30 mol 10.904 = 447 Imol + 30² 1.44 46.19 12.6 mol Naci (1 2 =6.30 mol C 2 NOCI 12.6 mol Naci/ 2 Na 12 NOCI = 12.6 mol 9. * round to same SF 9/27/22 €12-6 1189 Imol = 0.963 mal 122.55g 189|1 0.963 mol kclos/302 2KC10₂ 31.44 mol 0₂ 32.009 = 46.19 Ima =1.44 mul Ог • Actual held - amount under "normal" conditions Theoretical/calculated yield amound under "perfect' lab conditions • Percent vield = actual yield (g) Theoretical yreld (g) X 100 • Percent error = 100-percent yield OR actual-Theo Theoretical X/00 example What is the percent yield if you produce only 40.3 grams of KClO3? *100 = 87.270 Ax100 = 40.3 46.2 FIVE STAR FIVE STAR. 1 ✔ LIMITING • limiting reactant/reagent (run out of first) Excers reactant/reagent (left over after reaction done) Usually, in chemical reactions, one reactant is used up before the other and the reaction stops 1- Rarely are reactions in perfect ratros example 2M9 Y + 02 / limiting reactant excess reactant (LR), is vied left over up first REACTANT example If 79.1g of Zinc reacts with 96.79 of HCl to produce / ZnCl2 and H2. *65.39 2mg0. find limiting Reactant How much 2nCl2 should by produced 'If you produce 159.4g what is the % yreld? Zn + 2HCl ZnCl₂ + H₂ 1.21 mol -2.65 79.1 96.7 1.21 LR always determines how much product will be made Į - 165 e 10 write + balance equation 19.19 Imal 1.21 mol = 1.21 96.79 Imal 65.39 I 36.46 find moles of each reatant divide the molt by their coefficient 14 smallest indicatos LR use LR to find mols of product (mole ratio) 16 multiply mois by molar mass to get grams % yield = 159.4 x 100 = 96.676 165 136.29 : 2.65 mol = 1.33 2 1.21 mol 2n/1 ZnCl₂ 1 2n 1.21 136.29 Imal I I 10/3/22 3 LR 1 = 165g IF 23 9 of B₂03 reacts with 1.5 grams of H₂0 to produce/ B (OH)s, what is the limiting reactant? How much B(OH3) is produced? How much excess reactant is left over • Use LR ✔ 13203 + 3H₂O -> 2B(OH)3 0.42 23 1895 1.5 B₂03 239 | 1mol = 0.33m01:-(0.33) H₂O 7.5g | imol = 0 42 mol (0.14) 169.629 3 18.02g - 1 0.42 mol H₂6/2 (OH) = 0.28 mol (61.849 = 179 3 H₂0 Imal 0.42 mol H₂O18₂03 0-14 mol 169.62g = 9.79 (vied up) 3/20 limal subtract og mass w/ used up = 23-9.7 = 13g left over