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PERCENT COMPOSITION, EMPIRICAL & MOLECULAR FORMULAS 1 Percent Composition oNew 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 Composition oTo 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 of water, and calculate its molar mass (2•1.008g) + (1•15.994g) = 18.01g 4 Calculating PC given formula oThere are 2 mols of H atoms for every 1 mol of Water molecules oHow much do 2 mols of H atoms weigh? H: (2•1.008g)= 2.016g H oPercent of H in Water? 2.016g H X 100%= 11.2% 18.01 g H2O 5 Calculating PC given formula oThere is 1 mol of O atoms for every 1 mol of Water molecules oHow much does 1 mol of O atoms weigh? O: (1•15.994g)= 15.994g O oPercent of O in Water? 15.994 O X 100%= 88.8% 18.01 g H2O 6 Percent Composition o 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 Composition o 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 sample Find the percent composition of a compnd that contains 1.94g of carbon, 0.48g of Hydrogen, and 2.58g of Sulfur in a 5.0g sample of the compnd. 9 Calculating PC given sample o 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 Formulas o 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 mass oSince 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 Formulas o This allows us to say that if we had a 100 grams of sample, • 80 g is Carbon • 20 g is Hydrogen o 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 C 80g C = 6.7mol C 12 g C 1 mole H 20g 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 3 Ratio is 1 C’s for every 3 H’s; so the formula is = CH 3 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 N 25.9g N = 1.85 mol N 14 g N 1.85 mol 1 mol O 74.1g 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 Formulas o 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 Formulas o The empirical formula of CH2O, could be several compnds. •C2H4O2 or C3H6O3 or C100H200O100 o It’s more important to know the exact numbers of atoms involved The numbers of atoms define the properties of the compnd 20 Molecular Formulas o The molecular formula is always a whole-number multiple of the emp. formula o 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 Formula 2) 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 Formula 56.36g O 1 mol O = 3.5 mol O 16 g O 1.8 mol 1 mol P 54.6g 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/mol 3)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 Formulas o 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

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Formula for Calculating Percentages document sample

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