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Unformatted text preview: Empirical and Molecular Formulas
A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY PERCENT Percent Composition
Consider some of the family of nitrogenoxygen compounds: NO2, nitrogen dioxide and closely related, NO, nitrogen monoxide (or nitric oxide) Percent Composition
Consider some of the family of nitrogenoxygen compounds: NO2, nitrogen dioxide and closely related, NO, nitrogen monoxide (or nitric oxide) WEIGHT
Ethanol, C2H6O 52.13% C 13.15% H 34.72% O
Structure of NO2 Structure of NO2 Chemistry of NO, nitrogen monoxide Percent Composition
Consider NO2, Molar mass = ? What is the weight percent of N and of O?
Wt. % N = 14.0 g N • 1 00% = 30.4 % 46.0 g NO 2 Determining Formulas
In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. or SIMPLEST formula. A compound of B and H is 81.10% B. What is its empirical formula? • Because it contains only B and H, it must contain 18.90% H. • In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. • Calculate the number of moles of each constitutent. constitutent. Wt. % O = 2 ( 16.0 g O per mole) x 1 00% = 6 9.6% 46.0 g PROBLEM : A compound of B and H is
81.10% B. What is its empirical formula? What are the weight percentages of N and O in NO? Page 1 A compound of B and H is 81.10% B. What is its empirical formula? A compound of B and H is 81.10% B. What is its empirical formula? A compound of B and H is 81.10% B. What is its empirical formula? Calculate the number of moles of each element in 100.0 g of sample. Now, recognize that atoms combine in atoms the ratio of small whole numbers.
1 atom B + 3 atoms H --> 1 molecule BH 3 or 1 mol B atoms + 3 mol H atoms ---> 1 mol BH 3 molecules molecules Find the ratio of moles of elements in the compound. Take the ratio of moles of B and H. Always divide by the smaller number. 1 mol 81.10 g B • = 7 .502 mol B 10.81 g 18.90 g H • 1 mol = 1 8.75 mol H 1.008 g 18.75 mol H 2.499 mol H 2.5 mol H = = 1.000 mol B 1.0 mol B 7.502 mol B
But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B 2H5 A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula? molecular
Is the molecular formula B 2H5, B4H 10, B6H15, B8 H20, etc.? A compound of B and H is 81.10% B. Its empirical formula is B2H 5. What is its molecular formula? We need to do an EXPERIMENT to find EXPERIMENT to the MOLAR MASS. Here experiment gives 53.3 g/ mol Compare with the mass of B 2H5 = 26.66 g/unit 26.66 Find the ratio of these masses.
2 units of B2H5 53.3 g/mol = 1 mol 26.66 g/unit of B2H5 Determine the formula of a Determine the formula of a compound of Sn and II using the compound of Sn and using the following data. following data.
• Reaction of Sn and I2 is done using is excess Sn. Sn. • Mass of Sn in the beginning = 1.056 g • Mass of iodine (I2 ) used = 1.947 g • Mass of Sn remaining = 0.601 g • See p. 133 B2H6 B2H6 is one example of this class of compounds. is Molecular formula = B 4H10 Page 2 Tin and Iodine Compound
Find the mass of Sn that combined with 1.947 g I2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used: Tin and Iodine Compound
Now find the number of moles of I 2 that that combined with 3.83 x 10 -3 mol Sn. Mass mol Sn. of I2 used was 1.947 g. used Tin and Iodine Compound
Now find the ratio of number of moles of moles of I and Sn that combined.
1.534 x 10 -2 mol I 3.83 x 10-3 mol Sn = 4.01 mol I 1.00 mol Sn 1.947 g I2 • 1 mol = 7.671 x 10-3 mol I2 253.81 g 0.455 g Sn • 1 mol = 3.83 x 10-3 mol Sn 118.7 g This is equivalent to 2 x 7.671 x 10 -3 or 1.534 x 10-2 mol iodine atoms mol Empirical formula is SnI4 SnI Page 3 ...
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This note was uploaded on 01/11/2011 for the course ENGINEERIN MAE 107 taught by Professor Pozikrizdis during the Fall '08 term at San Diego.
- Fall '08