Determining empirical formula f from % composition iti When you have Wh h only l % composition iti andd do d nott have h a molecular l l mass, you cannot determine a molecular formula. Instead, what you find is called an empirical formula. An empirical formula differs from a molecular formula in that it tells you the relative ratio of each element to each other element, but it does not tell you the actual total amount of any particular element. Determining empirical formula f from % composition iti For instance, carbohydrates (sugars) have an empirical formula of CH2O. However, different carbohydrates may have the molecular formulas like CH2O, C2H4O2, C6H12O6, or even C12H24O12. While each compound has a different molecular formula, with a different number of C C’ss, H’s H s, and O O’ss, they all have the same ratio of C:H:O so they have the same empirical formula, CH2O. Determining empirical formula f from % composition iti Typical empirical formula problem: A compound contain 85.63% carbon and 14.37% hydrogen; What is the empirical formula of this compound? The usual way to attack this kind of problem is to start by assuming you have 100g of the compound compound. This is pretty arbitrary, arbitrary but it helps because it is a nice round, easy to multiply number. What you do then is to multiply the 100g x the % composition to find out how many g of each element would be in a sample of this size. (Next page) A compound p contain 85.63% carbon and 14.37% hydrogen; y g ; What is the empirical formula of this compound? Step 1: Assume a 100 g sample, find g of each element via % composition. 100g x (85.63%C/100%C) = 85.63g C 100g x (14 (14.37%H/100%) 37%H/100%) = 14.37g 14 37g H Notice how these answers are simply the % compositions, that is why we use the 100g sample so you don’t have to do any work at this step A compound p contain 85.63% carbon and 14.37% hydrogen; y g ; What is the empirical formula of this compound? Step 2: Convert the grams of each element to moles of each element using the atomic mass conversion factor. (atomic mass = 1 mole) C 85 63g C x (1 mole/12.01g) 85.63g mole/12 01g) = 7.130 7 130 mole C H 14.37g H x (1 mole/1.008g) = 14.26 mole H A compound p contain 85.63% carbon and 14.37% hydrogen; y g ; What is the empirical formula of this compound? Step 3: Find the element with the smallest number of moles. Now divide the total number of moles of each element by this smallest number to get the mole ratio. Smallest # of moles = 7.13 mole of C Di i i Divisions: C: 7.130 mole C/ 7.130 = 1C H: 14.26 mole H/7.13 = 2H Empirical formula = C1H2 or simply CH2 Determining empirical formula f from % composition iti Occasionallyy when yyou gget to the final mole ratios,, yyou have some funny fractions that take a little more work to clear up. Try this example: What is the empirical formula of a compound that is 43.64% P and 56.36% O? What is the empirical p formula of a compound p that is 43.64% P and 56.36% O? Step p 1 Find ggrams of element in a 100g g sample: p P 100g x (43.64/100) = 43.64g P O 100g x (56 (56.36%/100%) 36%/100%) = 56.36g 56 36g O What is the empirical p formula of a compound p that is 43.64% P and 56.36% O? Step p 2 Convert from ggrams to moles: P 43.64g P x (1 mole/30.97g) = 1.41 mole P O 56 56.36g 36g O x (1 mole/16 mole/16.00g) 00g) = 33.52 52 mole O What is the empirical p formula of a compound p that is 43.64% P and 56.36% O? Step p 3 Find the element with the smallest number of moles: P 1.41 mole P / 1.41 moles = 1 O 33.52 52 mole O / 1.41 1 41 mole = 2.5! 2 5! You can only have integer numbers in an empirical formula, so something is off. When this happens you should be able to multiply the fraction by a whole number to get another whole number. In g for additional step) p) this case 2.5 x 2 = 5. ((Next ppage What is the empirical p formula of a compound p that is 43.64% P and 56.36% O? Step p 4 ((Only y if needed)) Multiply p y ALL mole numbers by y the same whole number that gets rid of any fractional moles: Number needed to get rid of fractional mole : 2 Moles P = 1 mole P x 2 = 2 Moles O = 2.5 mole O x 2 = 5 Empirical formula P2O5 Determining empirical formula f from % composition iti Here is one pproblem for yyou to try: y A compound contains 15.62% H, 53.17% C and 31.01% N. What is the empirical formula of the compound? Determining empirical formula f from % composition iti A compound p contains 15.62% % H,, 53.17% % C and 31.01% % N. What is the empirical formula of the compound? Step 1: 15.62 g H 53.17 g C 31.01 g N Determining empirical formula f from % composition iti A compound p contains 15.62% % H,, 53.17% % C and 31.01% % N. What is the empirical formula of the compound? Step 2: 15.62 g H x (1/1.008) = 15.50 mole H 53.17 g C x (1/12.01) = 4.43 mole C 31.01 g N x (1/14.01) = 2.21 mole N Determining empirical formula f from % composition iti A compound p contains 15.62% % H,, 53.17% % C and 31.01% % N. What is the empirical formula of the compound? Step 3: Smallest mole = 2.21 N 15.50 mole H/ 2.21 = 7 H 4.43 mole C / 2.21 = 2 C 2.21 mole N / 2.21 = 1 N Empirical formula = C2NH7