Presentation on theme: "THE MOLE CONCEPT AND ITS APPLICATIONS"— Presentation transcript:
1 THE MOLE CONCEPT AND ITS APPLICATIONS CHEMICAL QUANTITIESTHE MOLE CONCEPT AND ITS APPLICATIONS
2 Chemical Quantities Objectives Students will be able to Understand and knowthe definitions of the mole and Avogadro’s Numberrelationship between moles and massrelationship between moles and particlesCalculate the formula mass of a compound given the periodic tableCalculate the mass percent composition of a compound given the formula, formula mass, and periodic table.
3 Drill & Objectives How many eggs are there in a dozen? If I bought 3 dozen eggs, how many eggs do I have?What are some ways that you can measure things? (such as the amount of a substance)ObjectivesStudents will be able toUnderstand and knowthe definitions of the mole and Avogadro’s Numberrelationship between moles and massrelationship between moles and particlesCalculate the formula mass of a compound given the periodic tableCalculate the mass percent composition of a compound given the formula, formula mass, and periodic table.
4 Chemical QuantitiesHow many eggs are there in a dozen?12!
5 Chemical QuantitiesHow many roses are there in a dozen roses?12!
6 Chemical QuantitiesDoes it matter what the _____________ is?
7 Chemical Quantities Does it matter what the _____________ is? NO! A Dozen is a dozen! No matter if it’s flowers, eggs, bagels, ect.It represents a number of _______
8 Chemical Quantities In Chemistry the same concept is valid, Let’s talk about “The Mole” and how it relates to Chemistry…
11 The Mole? “The Mole” or “A Mole” is an amount of a substance There are 6.02 x1023 atoms (representative particles of a substance) in 1 Mole (of that substance).This number (6.02 x1023) is called Avogadro’s numberNamed after Amedeo Avogadro di Quarenga (1776 – 1856), an Italian scientist.
12 The Mole! Let’s put this into perspective 6.023 x 1023!! How much is 1 mol of a substance?6.023 x 1023!!
13 The Mole!IF you had 6.02 X 1023 Watermelon Seeds…
14 The Mole! IF you had 6.02 X 1023 Watermelon Seeds… it would be found inside a watermelon slightly larger than the moon!
15 The Mole!If you had 6.02 X 1023 Grains of Sand…
16 The Mole! IF you had 6.02 X 1023 Grains of Sand… it would be more than ALL the sand on Miami Beach.
17 The Mole?A mole is the amount of substance or representative particles of any substance.Therefore, as we have just seen1 mol of N2 there are 6.02x1023 MOLECULES of N21 mol of C12H22O11 has how many molecules?6.02x1023 can be used for “atoms” if they are the same elementORIt can be used for “molecules” for many elements that are together, via compounds.
18 Chemical QuantitiesHow many moles of Magnesium is 3.01x1022 atoms of magnesium?Step 1 - What do we know?We know that there are 3.01x1022 atoms of MgStep 2 - What do we want know?We want to know the number of moles of MgStep 3 – What do we know that can be used for these conversions?We know that there are 6.02x1023 atoms in 1 molStep 4 – Solve the problem
19 Chemical Quantities = 5.00x10-2 mol Mg How many moles of Magnesium is 3.01x1022 atoms of magnesium?1 mol Mg3.01x1022 atoms MgX6.02x1023 atoms of Mg= 5.00x10-2 mol Mg
20 Chemical Quantities Individual Problems How many moles are 1.20x1025 atoms of phosphorus?How many atoms are in mol of Zinc?How many molecules are there in 4 mol of glucose, C6H12O6?How many molecules are there in 0.44 mol N2O5
21 Chemical QuantitiesNow there is a difference between asking how many atoms are there in the entire compound and how many atoms of one of the compound there are.We said that there are the same number or atoms in a compound or a diatomic molecule. However, there is a difference between these questions…How many atoms are there in aluminum fluoride?VS.How many fluoride ions are there in aluminum fluoride?
22 Chemical Quantities So let’s answer the question. How many fluoride ions are there in 1.46 mol of aluminum fluoride?If we follow the four steps that we went through prior
23 NO! Chemical Quantities How many fluoride ions are there in 1.46 mol of aluminum fluoride?Step 1 - What do we know?We know that there are 1.46 mol of aluminum fluorideStep 2 - What do we want know?We want to know the number of F-Step 3 – What do we know that can be used for these conversions?NO!We know that there are 6.02x1023 atoms in 1 molBUT is that enough information??
24 AlF3 Chemical Quantities How many fluoride ions are there in 1.46 mol of aluminum fluoride?We want to know the number of F-But how can we find out the number of fluoride ions we have from what is given?We can get the information we need by writing out the chemical formula of the compound.AlF3
25 Chemical QuantitiesHow many fluoride ions are there in 1.46 mol of aluminum fluoride?6.02x1023 Formula units of AlF31.46 mol AlF3XX1 mol AlF33 F- ions= x 1023 F- ions1 formula unit AlF3
26 Individual WorkHow many ammonium ions are in mol ammonium phosphate, (NH4)3PO4?
27 DRILL What is Avogadro's number? How many particles are there in 1 mol of a substance?How many atoms are there in 1.45 mol of Na?
29 Chemical QuantitiesChemists have defined the gram atomic mass as the number of grams of an element that is numerically equal to the atomic mass in amuThe Gram atomic mass is the mass of one mole of atoms of a monatomic elementThis can also be described as MOLAR MASS – in place of gram formula mass to refer to the mass of a mole of any element or compound.I will be using the term Molar Mass more frequently
31 Chemical Quantities Therefore, there are 47.88 grams/mol of Ti This number tells you how many grams of this element there are in 1 mol of the elementTherefore,there are grams/mol of TiOr there are grams of Ti IN 1 mol of Ti
32 The molar mass in grams/mol Chemical QuantitiesBCNOF10.811gggg18.998gThe molar mass in grams/mol
33 Chemical QuantitiesYou can calculate the total molecular weight of a molecule by adding up the molar masses of each element.The molar mass of the molecule SO4 is equal toS = g/mol x 1 S = g/molO = g/mol x 4 O = g/molSO4 = g/molI will specify the amount of significant figures needed in your answer
34 Individual Work What is the molar mass of the following compounds? REMEMBER YOUR UNITS!PCl3Sodium CarbonateC8H18Aluminum Sulfate(NH4)2CO3
35 Chemical Quantities Why is this important? Let’s do a calculation… This is important because it will allow us to covert from moles to grams and grams to moles, allowing for quantitative experimentation.Let’s do a calculation…How many grams are in 7.20 mol of dinitrogen trioxide?
36 Chemical QuantitiesHow many grams are in 7.20 mol of dinitrogen trioxide?Step 1 – add up the total molar mass of the compoundIf it is in word form, write the chemical formulaStep 2 – Set up the proper conversion factorsStep 3 – Solve
37 Chemical QuantitiesHow many grams are in 7.20 mol of dinitrogen trioxide?Step 1 – add up the total molar mass of the compoundIf it is in word form, write the chemical formulaN2O3N = 14.0 g/mol x 2 N = 28.0 g/molO = 16.0 g/mol x 3 O = 48.0 g/molN2O3 = 76.0 g/mol
38 Chemical QuantitiesHow many grams are in 7.20 mol of dinitrogen trioxide?Step 2 – Set up the proper conversion factorsStep 3 – Solve
39 Individual Work Find the mass of the following 3.32 mol K5.08 mol Ca(NO3)24.52x10-3 mol K2CO3Find the number moles of the followingg Li2HPO4847g (NH4)2CO3195g calcium nitrate
40 Exit TicketDoes g of Mn have the same number of moles as that of g of Cd? Explain why or why not.
41 DRILL What is Avogadro’s number? How many moles are there in 245 kg of CH2COOH?Does g of Mn have the same number of moles as that of g of Cd? Explain why or why not.
42 Group Work Find the mass (g) of the following: 10.0 mol Cr2.20x10-3 mol Sn0.720 mol Be2.40 mol N24.52x10-3 mol C20H42mol Potassium CarbonateFind the number of moles of the following:72.0 g Ar3.70x10-1 g B333 g Tin (II) Fluoride7.21x10-2 g He27.4 g TiO2
43 The Volume of a Mole of Gas Chemical QuantitiesThe Volume of a Mole of Gas
44 Chemical QuantitiesHow would we measure the amount of moles in a gas?
45 Chemical QuantitiesThe volume of a gas is usually measured at STP (Standard Temperature and Pressure)STP conditionsTemperature0°CPressure1 atmosphere (atm)
46 Chemical QuantitiesAt STP conditions 1 mol of ANY gas occupies a volume of 22.4 Liters (L) 22.4 Liters of a gas / 1 mol of the gas 22.4 L of a gas contains 6.02 x 1023 representative particles of that gas This is known as the molar volume of a gas
47 Chemical Quantities What were the units of the molar volume of a gas? LitersWhat kind of unit is liters?VOLUMETherefore, 1 mol of any gas occupies the same volume not mass
48 Chemical Quantities He 22.4 L N2 6.02 x 1023 molecules of N2 28g of N2 Ne22.4 L CO26.02 x 1023 molecules of CO244g of CO2CO222.4 L He6.02 x 1023 molecules of He4g
49 Individual WorkWhat is the volume at STP of these gases? 5.40 mol O x 10-2 mol CO2 Assuming STP conditions, how many moles are there in these volumes 74.6 L SO2 5.78x10-2 N2
50 Chemical QuantitiesBecause the molar volume of a gas is a VOLUME, we can relate density (mass over volume) of a gas to determine the mass of the gas. If we have a gas that has a density of g/L we can multiply the density with the molar volume of a gas (22.4 L) to calculate the mass of the gas x 22.4 L = 44.0g
51 Group WorkThe densities of gases A, B, & C are 1.25 g/L, 2.86 g/L, and g/L, respectively. Calculate the molar mass of each of these substances and compare them to the molar mass of ammonia, sulfur dioxide, chlorine, nitrogen, and methane Identify the gasses A, B, & C
52 Chemical Quantities Atoms (6.02 x 10 23) 1mol Molar Mass Solids and Liquids1 mol22.4 LGases
53 Homework problems Copy them down! Calculate the volume of each of these gases at STP9.6 mol He4.8 mol N2How many moles is each of the following, assuming STP56.o L N2O0.224 L O2Find each of the following quantitiesThe mass of 18.0L of CH4 (STP)The volume in liter, of 835 g of SO3 (STP)
54 Drill What are STANDARD conditions? How many liters are there in 1 mol of gas?Does it matter which gas it is? Why or why not?
56 Chemical Quantities- % Composition Calculating Percent CompositionPercent CompositionThe percent by mass of each element in a compoundThey must add up to 100%Ex. K2CrO440.3% K26.8% Cr32.9% O100% Total
57 Chemical Quantities - % Composition Percent Mass of an element in a compound is the number of grams of the element divided by the grams of the compound, multiplied by 100% % mass = x 100%Grams of elementGrams of compound
58 Chemical Quantities- % Composition Let’s look at our example againK2CrO4STEP 1Calculate the Total Molecular Mass of K2CrO4– Go ahead a calculate that nowK = 39 g x 2K = 78 gCr = 52 gO = 16 g x 4O = 64gK2CrO4 = 194gHave them calculate it out
59 Chemical Quantities- % Composition Let’s look at our example againK2CrO4STEP 2Using the individual molar mass of each element, divide each element with the total molar massGo ahead and calculate that now with 3 Significant figuresK = 39 g x 2K = 78 g K ÷194g = .402Cr = 52 g ÷194g = 0.268O = 16 g x 4O = 64g ÷194g = 0.330
60 Chemical Quantities- % Composition Let’s look at our example againK2CrO4STEP 3Multiply the number calculated from Step 2 and multiply by 100%Go ahead and calculate that nowK = 39 g x 2K = 78 g K ÷194g = .402 x 100% = 40.2% KCr = 52 g ÷194g = x 100% = 26.8% CrO = 16 g x 4O = 64g ÷194g = x 100% = 33% O
61 Chemical Quantities- % Composition Individual WorkCalculate the mass of carbon in 82g in C3H8Calculate the percent composition of each of these compounds.Propane, C3H8Sodium bisulfate, NaHSO4Calcium acetate, Ca(C2H3O2)2Hydrogen cyanide, HCN
62 ClosureTake the remainder of this period and write down in complete sentences what you learned today, approximately 2-3 paragraphs.
63 DRILL What is percent composition? What is the formula for percent composition?
64 DRILLCalculate the mass of oxygen in 142g of Sodium bisulfate, NaHSO4
66 Chemical Quantities- Empirical Formulas Empirical Formula Gives the lowest whole number ratio of the elements in a compound. This ratio is not necessarily the molecular formula! Calculating the empirical formula is going to require the use of almost everything you’ve learned so far. Each element in the compound would require different amounts depending on the mass and % composition
67 Chemical Quantities- Empirical Formulas Steps to solve for an Empirical FormulaWrite down the base formula AxBYConvert whatever you have into moles using dimensional analysisDivide each element by the lowest number of moles in the compoundRound if necessary to the nearest 10ths place and/or multiply by a number to get a whole number ratioWrite down the Empirical Formula
68 Chemical Quantities- Empirical Formulas What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen?STEP 1 - Write down the base formula AxBY…NxOy
69 Chemical Quantities- Empirical Formulas What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 2 - Convert whatever you have into moles using dimensional analysisBecause this is a % composition, we know that everything has to add up to 100%; therefore, we can assume that if we had 100g total of the compound, there would be 25.9g of N and 74.1g of O
70 Chemical Quantities- Empirical Formulas What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen? STEP 2 - Convert whatever you have into moles using dimensional analysis1 mol N25.9g of N x= 1.85 mol N14.0 g N1 mol O74.1g of O x= 4.63 mol O16.0 g O
71 Chemical Quantities- Empirical Formulas What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen?STEP 3 - Divide each element by the lowest number of moles in the compound1 mol N25.9g of N x = 1.85 mol N74.1g of O x = 4.63 mol O÷ 1.85 mol = 1 mol N14.0 g N1 mol O÷ 1.85 mol = 2.5 mol O16.0 g O
72 Chemical Quantities- Empirical Formulas What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen?STEP 4 - Round if necessary to the nearest 10ths place and/or multiply by a number to get a whole number ratio1 mol N x ____ =2.5 mol O x ____ =22 mol N25 mol O
73 Chemical Quantities- Empirical Formulas What is the empirical formula of a compound that is 25.9% Nitrogen and 74.1% Oxygen?STEP 5 - Write down the Empirical Formula2 mol N 5 mol ON2O5
74 Chemical Quantities- Molecular Formulas The Molecular formulais the actual formula of a molecular compoundWhile the empirical formula give us the base ratio of the compound, the molecular formula will give the actual molecular formula by using the molar mass.x (empirical formula) = molecular formulaRatio of the compounds
75 Chemical Quantities- Molecular Formulas Therefore, if you have the molar mass of the compound and divide it by the molar mass of the empirical formula, you will get the ratio between the two.Multiply the ratio to each element of the compound to get the the molecular formula.
76 Chemical Quantities- Molecular Formulas Ex. If the empirical formula was calculated to be CH4N and the known molecular formula molar mass is = to 60g1st – Take the molar mass of the empirical formula CH4NIn this case it is calculated to be 30g/mol2nd – Divide the known molar mass of 60g/mol and divide it by 30 g/mol 60/30 = 23rd – Take the ratio of the two molar masses and then multiply it by each element of the compound C1x2H4x2N1x2= C2H8N2
77 Chemical Quantities- Molecular Formulas Write the Molecular Formula for the problem belowThe compound methyl butanoate has a percent composition of 58.8% C, 9.8% H, 31.4%O. If its molecular mass is 102 g/mol, what is it’s molecular formula?Write the Empirical Formula for the following79.8% C 20.2% H67.6% Hg 10.8% S 21.6% O27.59% C 1.15% H 16.09%N 55.17%O17.6% Na 39.7%Cr 42.7%O
78 DRILLWhat is the difference between an empirical formula and molecular formula?
79 Chemical Quantities – Review Determine the Empirical Formula with the parametersbelow71.72% Cl, 16.16% O, and 12.12%CWhat is the molecular formula for the compound below?The compound’s empirical formula and molar mass is givenBelowHgCl, g/molDetermine the molecular formula for the compound94.1%O and 5.9% H ; molar mass = 34gFind the empirical formula for each compound from its percent composition72.4% Fe and 27.6% O52.8% Sn, 12.4% Fe, 16.0% C, and 18.8% NA&W pg 13945b, 47b, 48b, 52a, 52c
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