Presentation on theme: "% Composition, Empirical and Molecular Formulas"— Presentation transcript:
1% Composition, Empirical and Molecular Formulas Chemical QuantitiesThe Mole,% Composition, Empirical and Molecular Formulas
2How you measure how much? You can measure mass, or volume, or you can count pieces.We measure mass in grams.We measure volume in liters.We count pieces in MOLES.
3Moles Defined as the number of carbon atoms in exactly 12 grams of carbon-12.1 mole is 6.02 x particles.Treat it like a very large dozen6.02 x is called Avagadro’s number.
4Representative particles The smallest pieces of a substance.For a molecular compound it is a molecule.For an ionic compound it is a formula unit.For an element it is an atom.
5Types of questions 3 12 3 2 5 How many oxygen atoms in the following? CaCO3Al2(SO4)3How many ions in the following?CaCl2NaOH312325
6Types of questions using the equality; 1 mole = 6.02 x 1023 How many molecules of CO2 are the in 4.56 moles of CO2 ?4.56 mole x x1023 mc =moleHow many moles of water is 5.87 x molecules?5.87 x mc x mole =x1023 mc2.75x1024mcmole
7Types of questions using the equality; 1 mole = 6.02 x 1023 How many atoms of carbon are there in 1.23 moles of C6H12O6 ?1.23 moles x x1023 mc x 6 atoms =mole mcHow many moles is 7.78 x 1024 formula units of MgCl2?7.78x1024 FU x 1 mole = molex1024 FU4.44x1024 atoms
8Measuring Moles Remember relative atomic mass? The amu was one twelfth the mass of a carbon 12 atom.Since the mole is the number of atoms in 12 grams of carbon-12,the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.
9Gram Atomic Mass The mass of 1 mole of an element in grams. 12.01 grams of carbon has the same number of pieces as grams of hydrogen and grams of iron.We can right this as g C = 1 moleWe can count things by weighing them.
10Examples How much would 2.34 moles of carbon weigh? 2.34 moles C x 12 g CmoleHow many moles of magnesium in g of Mg?24.31 g Mg x mole =g Mg= g1.013 mole
11How many atoms of lithium in 1.00 g of Li? 1.00 g Li x 1 mole x 6.02x1023 atomsg Li moleHow much would 3.45 x 1022 atoms of U weigh?3.45x1022 atoms U x 1 mole x g Ux1023atoms 1 mole8.60x1022 atoms13.6 g
12What about compounds?in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atomsTo find the mass of one mole of a compounddetermine the moles of the elements they haveFind out how much they would weighadd them up
13What about compounds? What is the mass of one mole of CH4? 1 mole of C = 12 g4 mole of H x 1 g = 4g1 mole CH4 = = 16gThe Gram Molecular mass of CH4 is 16.05gThe mass of one mole of a molecular compound.
14Gram Formula Mass The mass of one mole of an ionic compound. Calculated the same way.What is the GFM of Fe2O3?2 moles of Fe x 56 g = 112 g3 moles of O x 16 g = 48 gThe GFM = 112 g + 48 g = 160g
15Molar Mass The generic term for the mass of one mole. The same as gram molecular mass, gram formula mass, and gram atomic mass.
16ExamplesCalculate the molar mass of the following and indicate what type it is.Na2S2 (23) + 32 =N2O42(14) + 4(16) =C= 78g1mole Na2S = 78gGram Formula Mass= 92g1 mole N2O4 = 92gGram Molecular Mass12g1 mole C = 12 gGram Atomic Mass
22For example How many moles is 5.69 g of NaOH? need to change grams to moles
23For example How many moles is 5.69 g of NaOH? need to change grams to molesfor NaOH
24For example How many moles is 5.69 g of NaOH? need to change grams to molesfor NaOH1mole Na = 23g 1 mol O = g 1 mole of H = 1 g
25For example How many moles is 5.69 g of NaOH? need to change grams to molesfor NaOH1mole Na = 23g 1 mol O = 16 g mole of H = 1 g1 mole NaOH = 40 g
26For example How many moles is 5.69 g of NaOH? need to change grams to molesfor NaOH1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g1 mole NaOH = 40 g
27For example How many moles is 5.69 g of NaOH? need to change grams to molesfor NaOH1mole Na = 23g 1 mol O = 16 g mole of H = 1 g1 mole NaOH = 40 g
28Examples How many moles is 4.56 g of CO2? 4.56g CO2 x 1 mole 1 44gCO2 How many grams is 9.87 moles of H2O?9.87 moles x 18g H2Omole= moles= 178g
29Examples How many molecules in 6.8 g of CH4? 6.8g CH4 x 1 mole x 6.02x1023 mcg CH mole49 molecules of C6H12O6 weighs how much?49 mc x 1 mole x g C6H12O6 =x1023 mc mole8820 g____= 1.5x10-20 g6.02x1023= 2.56x1023 mc
31Gases Many of the chemicals we deal with are gases. They are difficult to weigh.Need to know how many moles of gas we have.Two things effect the volume of a gasTemperature and pressureScientists compare gases at Standard Temperature and Pressure
32Standard Temperature and Pressure 0ºC and 1 atm pressureabbreviated STPAt STP 1 mole of gas occupies 22.4 LCalled the molar volumeAvogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.
33Examples What is the volume of 4.59 mole of CO2 gas at STP? 4.59mole x 22.4LmoleHow many moles is L of O2 at STP?5.67L x 1 moleLWhat is the volume of 8.80g of CH4 gas at STP?8.80g CH4 x 1mole x 22.4Lg CH mole= = 103L= .523moles= = 12.3L
34We have learned how to change moles to grams moles to atoms moles to formula unitsmoles to moleculesmoles to litersmolecules to atomsformula units to atomsformula units to ions
39Mass Volume Moles 6.02 x 1023 Representative Particles 22.4 L Periodic Table22.4 LMoles6.02 x 1023RepresentativeParticles
40Mass Volume Moles 6.02 x 1023 Representative Particles Atoms 22.4 L Periodic TableMoles6.02 x 1023RepresentativeParticlesAtoms
41Mass Volume Moles 6.02 x 1023 Representative Particles Ions Atoms Periodic TableMoles6.02 x 1023RepresentativeParticlesIonsAtoms
42Percent Composition Like all percents Part x 100 % whole Find the mass of each component,divide by the total mass.
43ExampleCalculate the percent composition of each element in a compound that is 29.0 g of Ag with 4.30 g of S.Ag gS g33.3g/33.3= x 100 = 87.09%/33.3= x 100 = 12.91%
44Getting % from the formula If we know the formula, assume you have 1 mole.Then you know the pieces and the whole.
45Examples Calculate the percent composition of C2H4? C 2(12g)=24 H 4(1g) = +428g/28= x 100 = 85.71%/28= x 100 = 14.29%
46Example Calculate the percent composition of Aluminum carbonate. Al2(CO3)3Al 2(27g)= 54C 3(12g)= 36O 9(16)= 144234g/234= x 100 = 23.08%/234= x 100 = 15.38%/234= x 100 = 61.54%
47• Step 2: Multiply the elements %, by the mass of the compound given. You can also calculate the mass of an element in a given amount of a compound using % composition.• Step 1: calculate the % comp. only of the element you want to find the mass of.• Step 2: Multiply the elements %, by the mass of the compound given.Example: Calculate the mass of sulfur in 3.54g of H2S.MM of H2S = H 2 (1) =S 1(32) = +3234g H2S% S = 32/34 x 100 = 94.1% S94.1% x 3.54g = 3.33g S
48Calculate the mass of nitrogen in 25g of (NH4)2CO3. H 8(1g) =C 1(12g) =O 8(16g) = +128176g (NH4)2CO3%N = 28/176 x 100 = 15.91%15.91% x 25g = 4.0g N
49Calculate the mass of magnesium in 97.4g of Mg(OH)2. Mg 1 (24g) = 24O 2(16g) = 32H 2 ( 1g) = +258g Mg(OH)2%Mg = 24/58 x 100 = 41.38%41.38% x 97.4g = 40.3g Mg
50From percentage to formula Empirical FormulaFrom percentage to formula
51The Empirical FormulaThe lowest whole number ratio of elements in a compound.The molecular formula the actual ration of elements in a compound.The two can be the same.CH2 empirical formulaC2H4 molecular formulaC3H6 molecular formulaH2O both
52Calculating Empirical Just find the lowest whole number ratioC6H12O6CH4NIt is not just the ratio of atoms, it is also the ratio of moles of atoms.In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.In one molecule of CO2 there is 1 atom of C and 2 atoms of O.
53Calculating Empirical Means we can get ratio from percent composition.Assume you have a 100 g.The percentages become grams.Can turn grams to moles.Find lowest whole number ratio by dividing by the smallest moles.
54ExampleCalculate the empirical formula of a compound composed of % C, % H, and %N.Assume 100 g so38.67 g C x 1mol C = mole C g C16.22 g H x 1mol H = mole H g H45.11 g N x 1mol N = mole N g N
55Example The ratio is 3.223 mol C = 1 mol C 3.222 mol N 1 mol N The ratio is mol H = 5 mol H mol N mol NC1H5N1
56A compound is 43.64 % P and 56.36 % O. What is the empirical formula? 43.64 g P x 1mol P = mole P g P56.36 g O x 1mol O = mole O g OThe ratio is mol O = 2.5 mol O mol P mol PCan not have 2.5 atoms! DoubleP2O5
57Caffeine is 49. 48% C, 5. 15% H, 28. 87% N and 16. 49% O Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?49.48 g C x 1mol C = mole C g C5.15 g H x 1mol H = 5.15 mole H g H28.87 g N x 1mol N = mole N g N16.49 g O x 1mol O = mole O g O
58The ratio is 4.123 mol C = 4 mol C 1.031 mol O 1 mol O The ratio is 5.15 mol H = 5 mol H mol O mol OThe ratio is mol N = 2 mol N mol O mol OC4H5N2O1
59Empirical to molecular Since the empirical formula is the lowest ratio the actual molecule would weigh more.By a whole number multiple.Divide the actual molar mass by the the mass of one mole of the empirical formula.Caffeine has a molar mass of 194 g. what is its molecular formula?C4H5N2O1 = 97g194/97 = 2Molecular Formula = C8H10N4O2
60Example 71.65 g Cl x 1mol Cl = 2.047 mole Cl 35 g Cl A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be g. What is its molecular formula?71.65 g Cl x 1mol Cl = mole Cl g Cl24.27 g C x 1mol C = mole C g C4.07 g H x 1mol H = 4.07 mole H g H
61The ratio is 2.047 mol Cl = 1 mol Cl 2.023 mol C 1 mol C The ratio is 4.07 mol H = 4 mol H mol C mol CEmpirical Formula = CClH4= 51g98.96/51 = 2Molecular Formula = C2Cl2H8