2Counting by weighing Situation: You work in a candy shop. You sell jelly beans.Somebody wants to buy 1000 jelly beans.How are you going to count those out?Are you going to do it one by one??
3Counting by weighing Here’s an idea!! Weigh them!! Lets say that all jelly beans are identical.One jelly bean weighs 5 g.So 1000 x 5g = 5000g or 5kgProblem solved!!!
4Counting by weighing Problem: all jelly beans aren’t identical. Suppose we weigh 10 jelly beans and find:Now we can find he average mass of a bean.
5Counting by weighingNow lets say that there is a customer who wants a bag of jelly beans and a bag of mints. He wants the same number of jelly beans and mints.What do you do?
6Counting by weighing You know: Jelly bean avg mass = 5 gMint avg mass = 15 gOne scoop of jelly beans is 500gWhat mass of mints do you need, to give the same number of mints as there are jelly beans in 500g of jelly beans?
7Counting by weighing Lets compare the avg masses. Avg mass of mints/avg mass of jelly bean15g/5g = 3The mints mass is 3 times bigger than the jelly bean.So, 3 x 500g = 1500g of mints.
8Lets look at some ratios: 1mint x 15g = 15g1 bean x 5g = 5g the ratio is 3/110mints x 15g = 150g10beans x 5g = 50g the ratio is 3/1100mint x 15g = 1500g100 bean x 5g = 500g the ratio is 3/1
9How does this apply to Chemisty? Lets say you have a pile of C and want to know how many O2 molecules you need to convert all the C into CO2.Or you want to know how many O atoms are in a cup of H2O
10Atomic Masses How much does a C atom weight? 1.99 x 10-23g that’s really small!!!To help with these complicated numbers we use atomic mass unit (amu).1 amu = 1.66 x 10-24g
11Atomic Masses Remember how not all jelly beans weighed the same. Not all C atoms have the same mass.Remember what an isotope is?Isotopes are atoms of element with different number of neutrons.For the atomic mass we use the average atomic mass of those isotopes.
12Atomic MassesWhere can we find what the avg atomic mass of an element?Look on the periodic table!!
13Atomic Masses So the common isotopes of C are Carbon 12Carbon 13Carbon 14Based on the abundance found in nature C avg atom mass amu1 carbon atom = amu
14Back to the questionLet’s say you have a pile of C and want to know how many O2 molecules you need to convert all the C into CO2.You weighed the pile of C and it’s 3 x 1020 amu3x1020amu X (1carbon atom/12.01amu) = 2.5x1019atoms (see pg176)
15Practice Remember: conversion factors are all equal to one. Convert 10 ft to inConvert 5 years to days and minsConvert 289g to kgConvert 3749mm to m
16Practice Remember: Convert 3x1020amu of C to atoms Convert 75 Al atoms to amuConvert amu of N to # of atoms
17Homework Convert: 213 Hg atoms to amu 53 P atoms to amu 172 O atoms to amu
18Homework Convert: 12 amu of Cl to atoms 74 amu of Na to atoms 123 amu of Li to atoms
20Now to a mole! Is a counting unit The mole (mol) is the SI unit used to measure the amount of a substance.It is based on the number of atoms in 12g of carbonWe can convert number of particles to moles and moles to particles.
21The Mole is 6.02 X 1023 (in scientific notation) Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000This number is named in honor of Amedeo Avogadro( )He studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present
22What is a mole? Answer: 6.02x1023 eggs, paper clips, shoes One mole of anything consist of 6.02x1023 units of that substance.How many eggs are in a mole of eggs?How many paper clips are in a mole of paper clips?How many shoes are in a mole of shoes?Answer: 6.02x1023 eggs, paper clips, shoes
23Avogadro’s Number as Conversion Factor 6.02 x 1023 particles1 moleor1 moleNote that a particle could be an atom, a molecule or anything!!!
24Just How Big is a Mole?Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.
25The MoleThis photograph shows one mole of salt(NaCl), water(H2O), and nitrogen gas(N2).
26We also know…1 mole of an elements atoms = that element’s average atomic mass (expressed in g)
27We also know…Ex g of Hg has 6.02x1023 atoms or 200.6g of Hg has 1 mol of Hg atomsConversion factor: 200.6g Hg/1 mol HgEx g of S has 6.02x1023 atoms or 32.07g of S has 1 mol of S atomsConversion factor: 32.07g S/1 mol S
28Example We have an unknown # of H atoms. We want to know many H atoms are present?We weigh the sample, results = .5g
29Example So .5g H X (1mol H/1.008g H) = .496 mol of H in the sample. Then .496 mol H atoms X (6.022x1023 H atoms/1 mol H atoms) = 2.99x1023 H atoms in the sample(See pg 181)
30Practice Convert 10g of Al into moles of atoms Convert 8g of S into moles of atomsConvert 25g of Ca into moles of atoms
31Practice Convert .371 mol Al into # of atoms Convert .249 mol S into # of atomsConvert .624 mol Ca into # of atoms
32Summary 1 amu = 1.66 x 10-24g 1 ___ atom = _____ amu 1 C atom = amu1 Si atom = amu1 mol of ___ = 6.022x1023 ____ units1 mol F = 6.022x1023 F atom1 mol Mg = 6.022x1023 Mg atomWhat are the conversion factors?
33Summary 1mol of ___ = ____g of _____ 1mol Hg = 200.6g Hg1mol Cl = 35.45g Cl____g of __ = 6.022x1023 atoms of ____200.6g Hg = 6.022x1023 atoms Hg35.45g Cl = 6.022x1023 atoms ClWhat are the conversion factors?
34Classwork How many moles are in 15 grams of lithium? How many atoms? How many grams are in 2.4 moles of sulfur? How many atoms?How many moles are in 22 grams of argon? How many atoms?
35ClassworkHow many grams are in 88.1 moles of magnesium? How many atoms?How many moles are in 2.3 grams of phosphorus? How many atoms?How many grams are in 11.9 moles of chromium? How many atoms?
37Molar Mass Back to the candy shop. This time the customer wants to buy suckers.When you weigh the suckers what are you weighing? How many parts are there?Answer: 3 parts ( wrapper, candy, stick)One sucker with 3 parts.
38Molar Mass Chemical compounds have multiple parts like the sucker. Ex. Methane (natural gas) CH41 CH4 moleculeWith 5 parts (1 C atom and 4 H atoms)
39Quick NoteIn 1 mol of a molecule there are corresponding moles of its parts.Ex. In 1 mol of CH4 there are 1 mol of C and 4 mol of H
40Molar Mass How do we figure out the mass of 1 mol of methane? We break it down in to its parts and add them up. This is called the molar mass.Molar Mass- is the mass of a moleculeTotal mass of all the parts of a molecule
41Example: SO2 What is the molar mass of SO2? Mass of 1 mol of S = 1 x = 32.07gMass of 1 mol of O = 2 x = 32.00gMass of 1 mol of SO2 = 64.07gThe molar mass of SO2 is 64.07g
42Example: CaCO3 What is the molar mass of CaCO3? Mass of 1 mol of Ca = 1 x ____ = ____Mass of 1 mol of C = 1 x _____ = ____Mass of 1 mol of O = 3 x _____ = ____Mass of 1 mol of CaCO3= ____The molar mass of CaCO3 is _____
43Practice What is the molar mass of water, H2O? What is the molar mass of ammonia, NH3?What is the molar mass of propane, C3H8?What is the molar mass and name of CaSO4?What is the molar mass and name of Na2CO3?What is the molar mass and name of Ba(OH)2?
44Practice What is the molar mass of water, H2O? What is the molar mass of ammonia, NH3?What is the molar mass of propane, C3H8?
45Practice What is the molar mass and name of CaSO4? What is the molar mass and name of Na2CO3?What is the molar mass and name of Ba(OH)2?
47Back to the candy shopA customer comes in and wants 3 different kinds of candy: jelly beans, mints and M&M’s.You give him 2 scoops of jelly beans, 6 scoops of mints, and 1 scoop of M&M’s.He now wants to know what percent of his bag has jelly beans by weight?You’re going to use Percent Composition.
48Back to the candy shop 2 jelly beans, 6 mints, 1 M&M’s. Break it down in to part of the bag.Mass of a bean = 5g x 2 = 10gMass of a mint = 15g x 6 = 90gMass of a M&M = 10g x 1 = 10gTotal mass of 1 bag = 110g
49Back to the candy shop 2 jelly beans, 6 mints, 1 M&M’s. So: percent of beans =mass of beans x 100%mass of 1 mixed bag10g x 100% = 9%110gSo: percent of mints =90g x 100% = 81%
50Percent CompositionPercent composition consists of the mass percent of each element in a compound.Mass Percent – the percent of each element present based upon its mass.AKA: It’s how much stuff is there in a percent form.Mass percent =
51For example…Ethanol(C2H5OH) – used to enhance the octane level of gasoline.Question: What is the percent mass of C?First we need to find the molar mass of the molecule.
52Ethanol(C2H5OH) First we need to find the molar mass of the molecule. Mass of 1 mol of C = 2 x = 24.02gMass of 1 mol of H = 6 x = 6.048gMass of 1 mol of O = 1 x = 16.00gMass of 1 mol of C2H5OH= 46.07gThe molar mass of C2H5OH is 46.07g
53Ethanol(C2H5OH) Question: What is the percent mass of C? Next we divide the mass of C in 1 mole of C2H5OH by the mass of 1 mol of C2H5OHSo: Mass percent of C =mass of C in 1 mol C2H5OH x 100%mass of 1 mol C2H5OH24.02g x 100% = %46.07g
54Ethanol(C2H5OH) Question: What is the percent mass of C? That means that ethanol is 52.14% by mass of carbon.That means 52.14% of ethanol’s mass is C.What about H and O?
55Ethanol(C2H5OH) Question: What about H and O? So: Mass percent of H = mass of H in 1 mol C2H5OH x 100%mass of 1 mol C2H5OH6.048g x 100% = %46.07gThe mass percent of H is 13.13%
56Ethanol(C2H5OH) Question: What about H and O? So: Mass percent of O = mass of O in 1 mol C2H5OH x 100%mass of 1 mol C2H5OH16.00g x 100% = 34.73%46.07gThe mass percent of O is 34.73%
57Ethanol(C2H5OH)Double check your answers. All mass percents should add up to 100%52.14% - C13.13% - H34.73% - O100.00%
58PracticeWhat is the mass percent of C, H and O in the compound Carvone, C10H14O?First find the masses of the individual elements.
59PracticeWhat is the mass percent of C, H and O in the compound Carvone, C10H14O?Now use that information to find the percent of each element.C: ______H: ______O: ______Total: ______
60PracticeWhat is the mass percent of Mg, N and O in the compound Mg(NO3)2 ?Remember to first find the molar mass thenMg: ______N: ______O: ______Total: ______
61ClassworkPenicillin F ( C14H20N2SO4) – Compute the mass percent of each element in this compound.C: ______H: ______N: ______S: ______O: ______Total: ______
62Relationships Three important 1 mol _____ = 6.022x1023 units of _____ 1 mol H20 = x1023 molecules of H201 mol _____ = avg mass(g) of _____1 mol H20 = g H20avg mass(g) of ____ = 6.022x1023 units of ___g H20 = x1023 molecules of H20
63Practice Mass of sample Moles of Sample Atoms in Sample 1) .250 mol Al 2)25.4g Fe3)4)5)6)2.13 x 1024 atoms Au7)1.28 mol Ca8)4.28 g9)10)11)12)3.14 x 1023 atoms C
65Empirical FormulasLets say you mix two chemicals together. From that you get a precipitate, a solid product forms.What is the chemical formula for this solid?How do you figure it out?The empirical formula is the simplest whole number ratio of the atoms present in the compound.
66Empirical Formulas Steps to find the Empirical Formula Obtain the mass of each element in gramsDetermine # of moles of each element.Divide the moles by the smallest # of moles found in previous step.
68Empirical FormulasTo figure out the chemical formula we will first need obtain mass. This time they were given to us!!! Using methods we won’t discuss you figure out that in the .2015g sample, there are:.0806g C.01353g H.1074g O.
69Step 20.0806g C x 1 mol C12.01g C= mol Cg H x 1 mol H1.008g H= mol H0.1074g O x 1 mol O16.00g O= mol O
70Step 3 0.00671 mol C 0.00671 = 1 mol C 0.01342 mol H 0.00671 = 2 mol H mol O= 1 mol O
71Empirical Formulas Now look at the moles of each element. 1 mol C2 mol H1 mol OThe ratio of C to H to O is 1:2:1The empirical formula is CH2O!!
72Empirical FormulasThe empirical formula is the simplest whole number ratio of the atoms present in the compound.So the chemical formula could not be C3H6O3 or C6H12O6 or C4H8C4 or more.All of the compounds have the same empirical formula of CH2O.The molecular formula gives the actual number of each type of atom present.
73Lets try one…Lets say that you heat up 4.151g of Al. Because it is so hot it reacts with the O in the air. After the sample cools you find the mass is 7.843g. This means that 3.692g O reacted.What is the empirical formula?Lets do the math…
74Lets try one… 4.151g Al x 1 mol Al 26.98g Al = 0.1539 mol Al 3.692g O x 1 mol O16.00g O= mol O
75Lets try one…Remember we want to use whole numbers only, so we divide our answer by the smallest # of mol.mol Al = mol Al0.1539mol O = mol ODoes that mean that our empirical formula is AlO1.5? No it doesn’t.
76Lets try one… Remember we want to use whole numbers only!! So we then multiply each element by 2.1.000 Al x 2 = = 2 Al atoms1.500 O x 2 = = 3 O atomsSo our empirical formula is Al2O3. For every 2 aluminum atoms there are 3 oxygen atoms.
77Lets try one…If we end up with a decimal we need to multiply all the sub scripts by the same # to get ride of the decimal.Some easy ones to recognize are:½ = 0.5, ¼ = .25, ¾ = .752/3 = .666, 1/3 = .333
78Lets try one…To convert these decimales to whole #s just multiply by their fractions denominator.Multiply by the bottom #.Ex. 0.5 x 2 = 1Ex x 3 = 2Ex x 4 = 7
79Class wide practiceA 1.500g sample of a compound containing only C and H is found to contain 1.198g of C. What is the empirical formula for this compound?Answer: CH3
80Class wide practiceA sample of lead arsenate, an insecticide used against the potato beetle, contains g of Pb, g of H, g of As, and g of O. What is the empirical formula for lead arsenate?Answer: PbHAsO4(see pg 203)
81Empirical FormulasWhen a compound is analyzed to find the relative amounts of elements present, the results are usually given in terms of % by mass of the various elements.Now can you figure out the EF (Empirical Formula) if you are given the percent composition?
82Empirical FormulasRemember that percent composition is how much of that part, per 100 parts, of the totalEx. 15% C means… the compound contains 15g of C per 100g of the compound.That means if we are given the % mass we can still find the EF.
83For example…Cisplatin is used to treat cancerous tumors. It has a composition of 65.02% Pt, 9.34% N, 2.02% H and 23.63% Cl. Calculate the empirical formula for Cisplatin.
84FirstRemember that percent composition is how much of that part, per 100 parts, of the totalRecognize that:65.02% Pt => g Pt2.02% H => 2.02g H9.34% N => 9.34g N23.63% Cl => 23.63g Cl
85SecondNow we are back to the same thing we were doing before. Find the # of moles of each element.65.02g Pt x 1 mol Pt195.1g Pt= mol Pt2.02g H x 1 mol H1.008g H= 2.00 mol HWhere did the 195.1g Pf come from?
86SecondNow we are back to the same thing we were doing before. Find the # of moles of each element.9.34g N x 1 mol N14.01g N= mol N23.63g Cl x 1 mol Cl35.45g Cl= mol ClWhere did the 23.63g Cl come from?
87SecondNow we are back to the same thing we were doing before. Find the # of moles of each element.Summary of results:mol Pt2.00 mol H0.667 mol Nmol Cl
88Third Divide by the smallest # of moles. 0.3333 mol Pt = 1.000 mol Pt 2.00 mol H = mol H0.667 mol N = mol Nmol Cl = mol Cl
89FinialSince all the ratio numbers came out to be whole numbers those are our numbers for our formula.Answer: PtN2H6Cl2Notice the only new thing we did differently was at the beginning. We changed from percentages of an element to grams of an element.Ex % Pt => g Pt
90Class wide practiceThe most common form of nylon is 63.68% C, 12.38% N, 9.80% H, and 14.4% O. What is the empirical formula for nylon?Answer: C6NH11O
92Finding the molecular formula Since we know “percent composition is how much of that part, per 100 parts, of the total”.Ex % Pt => g PtWe can now find the molecular formula if we are given the molar mass.Quick Review:What is the difference between the molecular formula and the empirical formula?What is the molar mass?How do we find the molar mass?
93Lets take a closer look… A white powder is analyzed and has and empirical formula of P2O5. The compound has a molar mass of What is the molecular formula?
94Lets take a closer look… The empirical formula mass is the total mass of the empirical formula. P2O52 mol P: 30.97g x 2 = 61.94g5 mol O: 16.00g x 5 = 80.00g= g of P2O5The mass of 1 mol of P2O5 is g.Where did 30.97g come from?Where did the “x 5” come from?
95Lets take a closer look… The empirical formula mass is the total mass of the empirical formula.The molar mass is the mass of all the atoms, not just the ones in the empirical formula.Lets compare:Molar Mass = g = 2Empirical Formula mass gThe molar mass is 2 times as big as the EF mass
96Lets take a closer look… Lets compare:Molar Mass = g = 2Empirical Formula mass gThe molar mass is 2 times as big as the EF massThat means that we multiply the compound by 2.(P2O5)2 => P4O10The molecular formula is P4O10
97Class wide practiceCaffeine is a compound containing C, H, N and O. The mass percent composition of caffeine is 49.47% C, 5.191% H, 28.86% N, and 16.48% O. The molar mass is about 194 g/mol.1) What is the empirical formula?2) What is the molecular formula?Answer: C4H5N2OAnswer: C8H10N4O2
98How it all relatesFormula Summary for the sugar glucose: