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# Empirical, molecular formulas & % Composition

## on Mar 22, 2013

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## Empirical, molecular formulas & % CompositionPresentation Transcript

• PERCENT COMPOSITION, EMPIRICAL &MOLECULAR FORMULAS 1
• Percent CompositionoNew food labels are required to describe the ingredients using percents of the daily reccom- mended allowance • These numbers tell what part of the total # of calories can be ob- tained from a product • AKA percent composition 2
• Percent Compositiono To get the information found on food labels the chemists had to know what fraction of the whole was each component • Component/total and then multiply by 100 • There are a couple of procedures used to calculate percent compositions 3
• Calculating PC given formula What percentage of Hydrogen and Oxygen is in Water (H2O)?Assume you have 1 mole ofwater, and calculate its molarmass(2•1.008g) + (1•15.994g) = 18.01g 4
• Calculating PC given formulaoThere are 2 mols of H atoms for every 1 mol of Water moleculesoHow much do 2 mols of H atoms weigh? H: (2•1.008g)= 2.016g Ho Percent of H in Water? 2.016g H X 100%= 11.2% 18.01 g H2O 5
• Calculating PC given formulaoThere is 1 mol of O atoms for every 1 mol of Water moleculesoHow much does 1 mol of O atoms weigh? O: (1•15.994g)= 15.994g Oo Percent of O in Water? 15.994 O X 100%= 88.8% 18.01 g H2O 6
• Percent Compositiono Another method of calculating the percent composition is by experimental analysis. • the overall mass of the sample is measured. • then the sample is decomp- osed or separated into its component elements 7
• Percent Compositiono The masses of the component elements are then determined and the percent composition is calculated as before • by dividing the mass of each element by the total mass of the sample • then multiplying by 100 8
• Calculating PC given sampleFind the percent composition of a compnd that contains 1.94g of carbon, 0.48g of Hydrogen, and 2.58g ofSulfur in a 5.0g sample of the compnd. 9
• Calculating PC given sampleo Calculate the percents for each element much like you would calculate the percents for anything. C: 1.94g/5.0g X 100% = 38.8% H: 0.48g/5.0g X 100% = 9.6% S: 2.58g/5.0g X 100% = 51.6% 10
• Empirical Formulaso Once the percent compositions are determined then they can be used to calculate a simple chem formula for the compnd • key is to convert the percents by mass into amounts in moles • Then, compare the moles using ratios to determine coefficients 11
• Calculating Empirical Formulas What is the empirical formula of a compound that is 80%C and 20%H by massoSince we have been given per- cents rather than masses we need to make an assumption. • Let’s suppose we have a total sample that weighs 100 g. 12
• Calculating Empirical Formulaso This allows us to say that if we had a 100 grams of sample, • 80 g is Carbon • 20 g is Hydrogeno Now that we have a set of masses we need to convert them to moles • Divide by the molar masses from the Periodic Table 13
• Calculating Empirical Formulas 1 mole C80g C = 6.7mol C 12 g C 1 mole H20g H = 20 mol H 1gH• Now calculate the simplest ratio of each by dividing both values by the smallest value 14
• Calculating Empirical Formulas Divide each mole value by the smaller of the two values: C: 6.7/6.7=1 H: 20/6.7 = 2.98  3Ratio is 1 C’s for every 3 H’s;so the formula is = CH3 15
• Calculating Empirical Formulas Determine the empirical formula of a compound containing 25.9g of N and 74.1g of O. Notice we have masses this time not percents, we can convert masses directly to moles 16
• Calculating Empirical Formulas 1 mol N25.9g N = 1.85 mol N 14 g N 1.85 mol 1 mol O74.1g O = 4.63 mol O 16 g O 1.85 mol 17
• Calculating Empirical Formulas Is the final answer N1O2.5? Of course not! We need a whole number ratio…Each part of the ratio is multiplied by a number that converts the fraction to a whole number N2(1)O2(2.5)= N2O5 18
• Molecular Formulaso The empirical formula indicates the simplest ratio of the atoms in the compnd • However, it does not tell you the actual numbers of atoms in each molecule of the compnd • For instance, glucose has the molecular formula of C6H12O6 • Empirical form would be CH2O 19
• Molecular Formulaso The empirical formula of CH2O, could be several compnds. • C2H4O2 or C3H6O3 or C100H200O100o It’s more important to know the exact numbers of atoms involved The numbers of atoms define the properties of the compnd 20
• Molecular Formulaso The molecular formula is always a whole-number multiple of the emp. formulao In order to calculate the molecular formula you must have 2 pieces of information • Empirical formula • Molar mass of the unknown compound (must be given) 21
• Calculating Molecular Formulas Find the molecular formula of a compound that contains 56.36 g of O and 54.6 g of P. If the molar mass of the compound is 189.5 g/mol.1) Find the Empirical Formula2) Find the MM of the Emp. Form.3) Find the ratio of the 2 molar masses (Mol MM/Emp MM) 22
• 1)Find the Empirical Formula56.36g O 1 mol O = 3.5 mol O 16 g O 1.8 mol 1 mol P54.6g P = 1.8 mol P 31g P 1.8 mol Empirical formula: P1O2 23
• 2) Find the MM of the Emp Form.MM of PO2: (1•31g P) + (2•16g O) = 63g/mol3)Find the ratio of the 2 molar masses (mol MM/emp MM) GIVEN 189.5 g/mol = 3.00 CALCULATED 63 g/mol 24
• Calculating Molecular Formulaso So the Molecular formula is 3 times heavier than the Empirical formula • Therefore, the molecular formula has 3 times more atoms than the emp. formula P3(1)O2(3)= P3O6 25