Working with Unknowns Determining Formula

Working with Unknowns – Determining Formula

If we are given an unknown and want to know the formula a particular process must be followed. The first step is using the info we know and finding the % composition in terms of mass. The next step is determining the empirical formula by converting the mass to moles. We can then use the empirical formula and the actual mass of the sample to determine the formula of the unknown compound. This is remarkably useful. A combustion analyzer is one way to determine the % composition of a substance. By burning the hydrocarbon in question, we can carefully measure the amount of each substance in the compound. From this the empirical formula can be found. The molecular formula is far more useful. Recall that this is the way that covalent compounds are often reported. C2H2 (acetylene) is very different than C6H6 (benzene) even though they both have the same empirical formula, or lowest ratio, CH. Examples: CO2 has empirical formula CO2 and molecular formula CO2 N2O4 has the empirical formula NO2 PCB’s (C6H4Cl2)2 has the empirical formula C3H2Cl

Calculating empirical and molecular formulas

What is the empirical formula of a compound with 39.99% C , 6.727% H and 53.28% O?

1) Assume 100g of the substance. Thus, 39.99g would be C, 6.727g H & 53.28g O. g 2) Convert each mass to moles: nC = 39.99 g / (12.01 /mol) = 3.330 mol C g nH = 6.727 g / (1.01 /mol) = 6.66 mol H g nO = 53.28 g / (16.00 /mol) = 3.330 mol O 3) divide all 3 by the smallest number of moles C = 1.000, H = 2.00, O = 1.000 4) write the empirical formula CH2O

To find the molecular formula, simply divide the molar mass of the known sample by the empirical mass.

From our example above, the sample was known to have a molecular mass of 180.18g/mol . What is the molecular formula of this compound? To continue the question from where we left off…. empirical mass = (1) (12.01 g/mol) + 2 (1.01 g/mol) + 1 (16.00 g/mol) = 30.03 g/mol molecular mass / empirical mass = 180.18 g/mol 30.03 g/mol = 6.00 Thus the molecular formula is C6H12O6

1. The molecular formulas of some substances are as follows. Write their empirical formulas. (a) Acetylene, C2H2 (used in oxyacetylene torches) (b) Octane, C8H18 (a component of gasoline) (c) Ammonium nitrate, NH4NO3 (a component of fertilizer)

2. The composition of nicotine is 74.0% C, 8.7% H, and 17.3% N. The molecular mass of nicotine is 162. What is its molecular formula?

3. Potassium persulphate (Anthion) is used in photography to remove the last tracers of hypo from photographic papers and plates. A 0.8162 gram sample was found to contain 0.2361 grams of potassium and 0.1936 grams of sulphur; the rest was oxygen. The formula weight of this compound was measured to be 271. What are the empirical and molecular formulas of potassium persulphate? (Arrange the atomic symbols in the formulas in the order KSO.)

4. Realgar (re-Al-gar) is a deep red pigment used in painting. A 0.6817 grams sample was found to contain 0.4774 grams of arsenic; the remainder was sulphur. The formula weight of realgar was found to be 428. What are the empirical and molecular formulas of this pigment? (Arrange the symbols in the order AsS.)

5. One compound of mercury with a formula weight of 519 contains 77.26% Hg, 9.25% C, and 1.17% H, and the remainder is oxygen. Calculate its empirical and molecular formulas, arranging the atomic symbols in the order Hg, C, H, and O.

Answers: 1a) C1H1 or CH (b) C4H9 (c) N2H4O3

2. By assuming that you have a 100 g sample, there is 74g C, 8.7 g H and 17.3 g N. nC = 74 g / 12.01 g/ mol = 6.184 mol nH = 8.7 g / 1.01 g/mol = 8.614 mol nN = 17.3 g / 14.01 g/mol = 1.236 mol Dividing each by 1.236 mol the ratio becomes C5H7N Its empirical mass would be 5(12.01 g/mol) + 7(1.01 g/mol) + 14.01 g/mol = 81.13g/mol The ratio of molecular mass to empirical mass is 162/81 or 2 Therefore the molecular formula for nicotine would be C10H14N2

3. nK = 0.2361 g / 39.10 g/mol = 0.00604 mol nS = 0.1936 g / 32.06 g/mol = 0.00604 mol nO = 0.3865 g / 16.00 g/mol = 0.02415 mol Dividing each by 0.00604 mol the ratio becomes KSO4 (empirical mass of 135.16 g/mol) The ratio of molecular mass to empirical mass is 271/135 =2 Therefore the molecular formula for Anthion would be K2S2O8.

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