The MOLECULAR FORMULA tells us the actual number of atoms of each element in a molecule or formula unit of a compound.
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1 EMPIRICAL FORMULA/MOLECULAR FORMULA The EMPIRICAL FORMULA tells us the simplest formula and tells us the simplest whole number ratio of atoms. It also tells us the simplest mole ratio of elements in the compound. The MOLECULAR FORMULA tells us the actual number of atoms of each element in a molecule or formula unit of a compound. While the empirical and molecular formulae of a compound are sometimes the same, sometimes they are different. Examples COMPOUND EMPIRICAL FORMULA (EF) MOLECULAR FORMULA (MF) MOLECULAR FORMULA IS? TIMES LARGER THAN carbon dioxide CO 2 CO 2 MF = 1 x EF phosphorous P 2 S 3 P 2 S 3 MF = 1 x EF trisulfide butane C 2 H 5 C 4 H 10 MF = 2 x EF phosphorous oxide P 2 O 4 P 4 O 8 MF =2 x EF benzene CH C 6 H 6 MF = 6 x EF EF CALCULATING EMPIRICAL FORMULAE:
2 In order to write the empirical formula of a compound we need to determine the simplest mole ratio in which the elements are combined. Reminder: The number of moles, n, of an element may be calculated using the following formula. n= m/m where m = given mass(g) and M is the molar mass (g/mol) Example: # moles in g of magnesium: n = g 24.3 g/mol = mol Note that the units also are correct. g = g x mol = mol g/mol g Example #1: What is the empirical formula of a compound which has been found by quantitative analysis to contain g carbon and g hydrogen? solution: mass #moles (n= m/m) (g) g = mol g/mol C g = mol 12 g/mol H g = mol 1.01 g/mol actual mole ratio whole # ratio = = empirical formula 1 CH 3 3 Note: the actual mole ratio was not exactly a whole number. However 2.93 is close enough to 3 so we can round it off. Example #2 - What is the empirical formula of a compound whose
3 percentage composition is found to be 69.9% iron and 30.1% oxygen? - the key here is to assume that you are working with 100g of the compound m Fe = 69.9% x g Fe = 69.9 g m O = 30.1% x g O = 30.1 g M Fe = 55.8 g/mol M O = g/mol The molar ratio for Fe:O is 1.25:1.88. Dividing by 1.25 to obtain the lowest ratio, we obtain the molar ratio of Fe:O to be 1:1.5. Multiplying this ratio by 2, we obtain a final 2:3 ratio. The empirical formula of the compound is Fe 2 O 3. CALCULATING MOLECULAR FORMULAE
4 This whole number multiple may be determined by comparing the molar mass (M) to the empirical formula mass (EMF). The following formula may be used. MF = EF X M/EMF Suppose a compound with an empirical formula (EF) of CH 2 O has a molar mass of 180 g. The molecular formula (MF) of this compound would be determined as follows: MF = EF X M = CH 2 O X 180 g/mol EMF 30 g/mol (EMF = (1) + 16 = 30 g/mol) = (CH 2 O) x 6 = C 6 H 12 O 6 Example #1 Calculate the empirical formula and the molecular formula of a compound containing g carbon, g hydrogen and g oxygen. The molar mass of the compound is 168 g/mol Solution eleme nt mass (g) # mol n= m/m actual mol ratio whole # ratio C g = mol mol = g/mol mol H g = mol mol = g/mol O g = mol 16.0 g/mol mol mol = mol The EF, empirical formula of this compound is C 3 H 4 O. The mass of this, the EFM is 3(12) + 4(1) + 16 = = 56 g 1 The molecular formula of the compound is C 9 H 12 O 3. Example #2 - A compound was found to contain 10.06% C, 89.10% Cl, and 0.84% H, by mass. If the molar mass of the compound is g/mol,
5 calculate its molecular formula. Again, the key here is to assume that you are working with 100 g of the compound. m C = 10.06% x g C = g m Cl = 89.10% x g Cl = g m H = 0.84% x g H = 0.84 g M C = g/mol M Cl = g/mol M H = 1.01 g/mol The molar ratio for C:Cl:H is 0.84:2.51:0.83. Dividing by 0.83 to obtain the lowest ratio, we obtain the molar ratio of 1:3:1. The empirical formula of the compound is CCl 3 H. The molecular formula of the compound is CCl 3 H. Combustion Problems Empirical/Molecular Formulae
6 When an organic compound undergoes complete combustion, the compound reacts with oxygen, and all the elements in the original compound are oxidized with the exception of oxygen itself. Thus C (in the original compound) CO 2(g) H (in the orginal compound) H 2 O (g) S (in the original compound) SO 2(g) N (in the original compound) NO 2(g) The mass of carbon in the CO 2(g) formed will be the same as the mass of C in the original compound (Law of Conservation of Mass), and it can be calculated as follows mass of C in CO 2(g) = (mass of CO 2(g) formed) x 12.0 g/44.0 g The mass of hydrogen in the H 2 O (g) formed will be the same as the mass of H in the original compound, and it can be calculated as follows mass of H in H 2 O (g) = (mass of H 2 O (g) formed) x 2.01 g/10.0 g Sulphur and nitrogen are calculated in a similar way It is not possible, however, to calculate the mass of oxygen present in the original compound in this manner, since the compound has taken in additional oxygen from the air. Instead, if the compound contains oxygen, its mass must be calculated by subtracting the sum of the masses of all the other elements from the mass of the compound. Empirical & Molecular Formula Questions (additional practice)
7 1. The empirical formula of maleic acid is CHO. Its molecular mass is 116 g/mol. What is its molecular formula? 2. A well known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula C 2 H 4 NO. If its molar mass is 116 g/mol, what is the molecular formula of the compound. 3. A compound contains C, H and O. A g sample contains g carbon and g hydrogen. A gaseous sample of this gas has a mass 5.25 times the mass of an identical volume of oxygen gas under the same conditions. What is the molecular formula of the compound? 4. Cinnamic alcohol is used mainly in perfumery, particularly soaps and cosmetics. Its molecular formula is C 9 H 10 O. (a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol. (b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469g? 5. The molar mass of caffeine (ahhh.sweet caffeine) is g. Caffeine molecules are made up of C, H, N and O. If the empirical formula of caffeine is C 4 H 5 N 2 O, what is the molecular formula? 6. Monosodium glutamate (MSG) has sometimes been suspected as the cause of Chinese restaurant syndrome because this food-flavour enhancer can induce headaches and chest pains. (interestingly, note that the Chinese people have used MSG for at least a thousand years with no ill effects). MSG has the following composition by mass 35.51% C 4.77% H 37.85% O 8.29% N 13.60% Na What is the molecular formula if its molar mass is 169 g? 8. A certain hydrocarbon is found to consist of 83.6% carbon and 16.4% hydrogen (by mass). If one molecule of this compound has a mass of 1.43 x g, calculate its molecular formula. 10. When oleic acid is analysed it is found to consist of 76.6% carbon, 12.1% hydrogen and 11.3% oxygen (by mass). It is also found that there are exactly 18 times as many atoms present in one molecular of oleic acid than there are in one molecule of water! Calculate the molecular formula of oleic acid.
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