SOLVING EMPIRICAL FORMULA PROBLEMS

Transcription

1 SOLVING EMPIRICAL FORMULA PROBLEMS Why do we want to use Empirical Formulas? 1)Substances that do not consist of discrete units, such as in a crystal (ionic solid) of NaCl---we dont want to write Na456Cl910 or something like this. Instead we want to write it in simplest form, its Empirical Formula- NaCl. 2) Best understood by a classic Empirical formula problem: Lets supposed you work in a lab and someone brings you a white powder and tells you to figure out what it is, called a qualitative analysis. The analysis tells you what elements are in the compound but not the percentages of each compound. e.g. C H O The next step is to determine what percentage of each element are present. This is called the percent composition and we find that: 40%C 67%H 53.3%O What is the compounds Formula?? The formula does not tell us the ratio of the elements by mass but rather it tells us the ratio of the elements by moles. H2O tells us there are two moles of H for every one mole of O How to convert these molar masses into a chemical formula??? One way is this: imagine you have a sample of the element that is 100g CxHyOz at 100g <100g is more convenient because we are dealing with percentages> Hence, from our percent composition analysis we get: 40g C 6.7g H 53.3g O

2 So, in a 100g sample we have the ratio of the 3 masses but we want the ratios of the moles of the 3 elements, then we better turn these masses into moles. 40g C 6.7g H 53.3g O 40g C 12g/mol! =! 3.33 mol C 6.7g H 1g/mol! =! 6.7 mol H 53.3g O 16g/mol! =! 3.33 mol O Therefore the Formula is C3.33H6.7O3.33 Hence, the final ratio can be found the following way: 1) make the smallest subscript 1 2) then divide all the numbers by this smallest number As such, we get: C3.33H6.7O3.33!! = CH2O is the Impirical Formula is this the actual molecular formula?? Can t tell at this the point. We need to now do another experiment to find Molecular Formula:

3 Finding Molecular Formulas For this, we need to know molar mass: CH2O(Empirical Formula) = (1.0) = 30.0g/mol {This is the Empirical Mass} (Impirical Formula-is NOT molecular formula) From lab work, we find that the actual Molar Mass = 180g/mol You may have noticed that 30g is six times smaller than 180g 180/30 = 6 So the actual molecular formula is 6 times the empirical formula OR 6(CH2O) = C6H12O6 SUMMARY: TO GET THE MOLECULAR FORMULA of an unknown substance YOU MUST DO 2 EXPERIMENTS: 1) FIND THE EMPIRICAL FORMULA 2) FIND EMPIRICAL MASS 3) FIND THE MOLAR MASS 4) COMPARE EMPIRICAL MASS AND MOLAR MASS 1) FIND THE EMPIRICAL FORMULA(use a 100g sample and use atomic weights to determine percent composition for each atom) 2) FIND EMPIRICAL MASS (use atomic weights as per percent composition) 3) FIND THE MOLAR MASS(from lab work<examples given at the end of these notes>)

4 4) COMPARE EMPIRICAL MASS AND MOLAR MASS Explanaition of how Molar Mass is found through lab work: You dont have to know the following but if you were wondering what lab work can be done to determine the molar mass of an unknown substance here are some techniques, beginning with a sample problem/solution: problem: A g sample of an unknown substance was dissolved in 20.0 ml of cyclohexane. The density of cyclohexane is 0.779g/mL. The freezing-point depression was 2.5ºC. Calculate the molar mass of the unknown substance. (Kf cyclohexane = 20.0ºC/m) solution: Mass cyclohexane = 20.0 ml x g/ml = g=> Kg delta T = Kf x m 2.5 = 20.0 C/m x m m = molality solution = = moles solute / mass solution = moles solute / Kg moles solute = molar mass = g / mol =125.1 g/mol Other techniques: 1)Use colligative properties. Dissolve a precisely weighted amount of the solid in a known mass of some solvent (say ethanol, or propanone) and measure the solution s boiling point elevation or its freezing point depression, relative to the pure solvent. This temperature difference is related to the solution s molality, and as you know how much solvent you used, you can discover how many moles of the organic solid acid were present in the solution. As you also know the mass of solid you added, making (mass of solid / number of moles) yields its molar mass. 2)Take a weighed amount of it, and add more than enough sodium hydroxide, of known concentration, to dissolve it. Then titrate the excess alkali with HCl of known concentration.

5 3)Use mass spec.. it will give you the relative masses of the different atoms in your molecule. To identify the actual molecule itself, use NMR and IR to determine structure relationships. Remember, you will not be able to understand the stereochemistry from this, only the basic structure.) Lets now try some practice problems...

6 Found by reading from the periodic table the atomic mass of Sodium (Na) 43.4% Na % C 45.28% O Assume a 100g sample: (if I do this, then all the percentages convert nicely into grams directly: hence, 43.4% becomes 43.4g) 43.4g Na 23g/mol = 1.89mol Na 11.32g C 12g/mol = 0.94mol C 45.28g O 16g/mol = 2.83mol O Na1.89C0.94O2.83 -now divide each subscript by the smallest number (.94) and get = Na2CO3 (Sodium Carbonate)

7 First we need to find the Empirical Formula: 62.07% C 10.34% H 27.59% O 62.07g C 12.0g/mol = 5.17 mol C 10.34g H 1.0g/mol = mol H 27.59g O 16.0g/mol = 1.72 mol O C5.17H10.34O1.72 = 1.72 C3H6O (Empirical Formula) = 58.0g/mol (Empirical Mass) but the problems states molar mass of the compound is 116g/mol therefore: 116.0(actual molar mass) 58.0 (empirical mass) = 2 therefore the actual molar mass is twice the empirical mass The answer is thus: 2(C3H6O) = C6H12O2

8 Assume we have a 100g sample: 26.58g K 39.1g/mol = mol K 35.35g Cr 52.0g/mol = mol Cr 38.07g O 16.0 g/mol = 2.38 mol O K0.680Cr0.680O2.38 = KCrO This is a formula that has a half (3.5). But you canʼt have a half so the way to get rid of the half is double all the subscripts, thus giving you an answer of: = K2Cr2O7 (Empirical Formula)