Percentage Composition and Empirical Formula OBJECTIVES: Describe how to calculate the percent by mass of an element in a compound. Why do we care? allows us to convert from one to the other,,,but most helpful in the lab…I can mass grams and use the PT to help me determine moles and I can use moles to compare, cannot compare masses, only moles…thus we can calculate formula relationships from experimental data Use example of a crucible.

Percentage Composition and Empirical Formula OBJECTIVES: Interpret an empirical formula.

Percentage Composition and Empirical Formula OBJECTIVES: Distinguish between empirical and molecular formulas.

Percentage Composition If given the masses of each constituent element, divide the mass of each by the total mass Multiply by 100 to get a percentage

Example Calculate the percentage composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S. 29.0 g Ag X 100 = 87.1 % Ag 33.3 g total Total = 100 % 4.30 g S X 100 = 12.9 % S 33.3 g total

Getting it from the formula If we know the formula, assume you have 1 mole You know the mass of the elements and you can calculate the mass of the whole compound (these values come from the periodic table).

Examples Calculate the percent composition of C2H4 Find the molar mass of compound Divide the molar mass of each element by the molar mass of the compound (x 100) 85.6% C, 14.4 % H

Try One! How about aluminum carbonate? Al2 (CO3)3 Remember to consider that you have 3 molecules of CO3 ( so 3 atoms of carbon and 9 of oxygen) Molar mass of the compound: 233.87 g mol-1 (could be 233.99 if used different PT table) 23.02% Al, 15.41% C, and 61.57% O

Empirical Formula AKA simplest formula Indicates the elements present in the compound Indicates the simplest whole number ratio of these elements C6H12O6 is the molecular formula for glucose ( the “true” formula) The empirical formula is CH2O (all divided by the highest common factor of 6)

Distinction All ionic compound formulas are already in empirical formula “mode” Covalent compound formulas may or may not be in empirical formula “mode” Examples: H2O is the empirical and molecular formula for water C6H6 is the molecular formula for benzene, but CH is the empirical formula

Calculating Empirical We can get a ratio from the percent composition. Assume you have a 100 g sample and the percentages become grams (75.1% = 75.1 grams) Convert grams to moles. Find lowest whole number ratio by dividing each number of moles by the smallest value.

Calculate the Empirical formula If given grams: convert all grams to moles Divide each by the smallest number of moles You will now have the ratio and therefore the coefficients of the elements If this doesn't give you whole numbers, then multiply by the smallest value that will give you whole numbers.

Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g sample 38.67 g C x 1mol C = 3.2198 mole C 12.01 g C 16.22 g H x 1mol H = 16.0594 mole H 1.01 g H 45.11 g N x 1mol N = 3.2198 mole N 14.01 g N Now divide each value by the smallest value

Example Continued The ratio is 3.2198mol C = 1 mol C 3.2198mol N 1 mol N The ratio is 16.22 mol H = 5 mol H 3.2198 mol N 1 mol N The coefficients are C = 1, H = 5, N = 1 so C1H5N1 which is = CH5N

Practice A compound is 43.64% P and 56.36% O. What is the empirical formula? P2O5 Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? C4H5N2O

Empirical 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 empirical formula mass – you get a whole number to increase each coefficient in the empirical formula Caffeine has a molar mass of 194 g. what is its molecular formula? = C8H10N4O2

Example A compound made of C and H contains 92.24% by mass of C. Its Mr = 78.1 Determine its molecular formula 92.24% = 92.24 g C = 7.6803 mol C 100% - 92.24% = 7.76 % = 7.76 g H = 7.6832 mol H Empirical formula is CH Relative empirical mass is 13.02

Continued Mr = 78.1 and the relative empirical mass = 13.02 Divide 78.1 by 13.02 = 6 Multiply each subscript by 6 Molecular formula is C6H6

Experimental Methods All the carbon is converted to carbon dioxide One way to find the percentage composition of organic compounds is to burn a known mass of the compound with excess O2 All the carbon is converted to carbon dioxide All the hydrogen is converted to water Mass of oxygen can be found by subtracting the sum of these two from the initial mass

Example 2.80 g of an organic compound containing only C and H forms 8.80 g of CO2 and 3.60 g of H2O when it undergoes complete combustion. Determine the empirical formula. Amount of CO2 = amount of C Amount of H2O = ½ amount of H Convert each original amount to moles

Example Continued 8.80 g of CO2 = 0.200 mol CO2 = mol C 3.60 g of H2O = 0.200 mol H2O, so the amount of H is 2 x 0.200 = 0.400 mol H Divide by the smallest value and you get 1 for C and 2 for H The empirical formula is CH2

Another Example Vitamin C contains C, H, and O only. On combustion of 1.00g of Vitamin C, 1.50 g CO2 and 0.408 g H2O are produced. What is the empirical formula for vitamin C? 1.50 g CO2 = 0.03408 mol CO2 = 0.03408 mol C 0.408 g H2O = 0.02264 mol H2O 2 x 0.02264 = 0.0453 mol H

Continued Need to find how many grams of C and H are present to find how many grams of O are present 0.03408 mol C = 0.409 g C 0.04528 mol H = 0.0457 g H 1.00 g – (0.409 + 0.0457) = 0.54 g O Now change g of O to mol of O = 0.0346 Divide by the smallest value (0.03408)

Continued You will get coefficients of 1, 1.3, and 1 Multiply each by 3 to get 3, 4, 3 Empirical formula for vitamin C is C3H4O3