1 PERCENT COMPOSITION, EMPIRICAL & MOLECULAR FORMULAS

2 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 obtained from a product AKA 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 obtained from a product AKA percent composition Percent Composition

3 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 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 Percent Composition

4 What percentage of Hydrogen and Oxygen is in Water (H 2 O)? What percentage of Hydrogen and Oxygen is in Water (H 2 O)? Calculating PC given formula Assume you have 1 mole of water, and calculate its molar mass (21.008g) + (21.008g) + ( g) = ( g) = 18.01g 18.01g

5 oThere are 2 mols of H atoms for every 1 mol of Water molecules H: (21.008g)= 2.016g H oPercent of H in Water? 2.016g H g H 2 O X 100%= 11.2% Calculating PC given formula

6 oThere is 1 mol of O atoms for every 1 mol of Water molecules Calculating PC given formula O: ( g)= g O oPercent of O in Water? O g H 2 O X 100%= 88.8%

7 oAnother method of calculating the percent composition is by experimental analysis. the overall mass of the sample is measured. Then the sample is separated into its component elements The equation is the SAME as before! oAnother method of calculating the percent composition is by experimental analysis. the overall mass of the sample is measured. Then the sample is separated into its component elements The equation is the SAME as before! Percent Composition

8 oThe 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 oThe 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 Percent Composition

9 Find the percent composition of a compound that contains 1.94g of carbon, 0.48g of Hydrogen, and 2.58g of Sulfur in a 5.0g sample of the compound. Calculating PC given sample

10 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% oCalculate the percentage for each element much like you would calculate the percentage for anything. Calculating PC given sample

11 oOnce the percent compositions are determined then they can be used to calculate a simple chem formula for the cmpnd key is to convert the percents by mass into amounts in moles Then, compare the moles using ratios to determine subscripts oOnce the percent compositions are determined then they can be used to calculate a simple chem formula for the cmpnd key is to convert the percents by mass into amounts in moles Then, compare the moles using ratios to determine subscripts Empirical Formulas

12 oSince we have been given per-cents rather than masses we need to make an assumption. Assume we have a total sample that weighs 100 g. oSince we have been given per-cents rather than masses we need to make an assumption. Assume we have a total sample that weighs 100 g. What is the empirical formula of a compound that is 80%C and 20%H by mass Calculating Empirical Formulas

13 Calculating Empirical Formulas oThis allows us to say that if we had a 100 grams of sample, 80 g is Carbon 20 g is Hydrogen oNow that we have a set of masses we need to convert them to moles Divide by the molar masses from the Periodic Table oThis allows us to say that if we had a 100 grams of sample, 80 g is Carbon 20 g is Hydrogen oNow that we have a set of masses we need to convert them to moles Divide by the molar masses from the Periodic Table

14 80g C 1 mole C 12 g C = = 6.7mol C Calculating Empirical Formulas 20g H 1 mole H 1 g H = = 20 mol H Now calculate the simplest ratio of each by dividing both values by the smallest value

15 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 Calculating Empirical Formulas

16 Determine the empirical formula of a compound containing 25.9g of N and 74.1g of O. Calculating Empirical Formulas Notice we have masses this time not percents, we can convert masses directly to moles Notice we have masses this time not percents, we can convert masses directly to moles

g N 1 mol N 14 g N = = 1.85 mol N Calculating Empirical Formulas 74.1g O 1 mol O 16 g O = = 4.63 mol O 1.85 mol

18 Calculating Empirical Formulas Is the final answer N 1 O 2.5 ? Of course not! Is the final answer N 1 O 2.5 ? 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 N 2(1) O 2(2.5) = N 2 O 5

19 oThe 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 C 6 H 12 O 6 Empirical form would be CH 2 O oThe 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 C 6 H 12 O 6 Empirical form would be CH 2 O Molecular Formulas

20 oThe empirical formula of CH 2 O, could be several compnds. C 2 H 4 O 2 or C 3 H 6 O 3 or C 100 H 200 O 100 oIt’s more important to know the exact numbers of atoms involved  The numbers of atoms define the properties of the compnd oThe empirical formula of CH 2 O, could be several compnds. C 2 H 4 O 2 or C 3 H 6 O 3 or C 100 H 200 O 100 oIt’s more important to know the exact numbers of atoms involved  The numbers of atoms define the properties of the compnd Molecular Formulas

21 oThe molecular formula is always a whole-number multiple of the emp. formula oIn order to calculate the molecular formula you must have 2 pieces of information Empirical formula Molar mass of the unknown compound (must be given) oThe molecular formula is always a whole-number multiple of the emp. formula oIn order to calculate the molecular formula you must have 2 pieces of information Empirical formula Molar mass of the unknown compound (must be given) Molecular Formulas

22 Find the molecular formula of a compound that contains g of O and 54.6 g of P. If the molar mass of the compound is 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) 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) Calculating Molecular Formulas

g O 54.6g P 16 g O 1 mol O = 3.5 mol O 31g P 1 mol P = 1.8 mol P 1)Find the Empirical Formula 1.8 mol Empirical formula: P 1 O 2

24 MM of PO 2 : (131g P) g/mol 2)Find the MM of the Emp Form. + (216g O) = 63g/mol 3)Find the ratio of the 2 molar masses (mol MM/emp MM) 63 g/mol = 3.00 GIVEN CALCULATED

25 oSo the Molecular formula is 3 times heavier than the Empirical formula Therefore, the molecular formula has 3 times more atoms than the emp. formula oSo the Molecular formula is 3 times heavier than the Empirical formula Therefore, the molecular formula has 3 times more atoms than the emp. formula Calculating Molecular Formulas P 3(1) O 2(3) = P 3 O 6