Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.


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1 1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights. AMU (Atomic Mass Unit) was derived by taking 1/12 of the carbon12 atom. A carbon12 atom has a mass of x g. 1/12 ( x g) = x g 1 AMU = x g Since it was known that hydrogen was 1/12 the mass of carbon, the mass of a hydrogen atom was designated as 1 amu Carbon is 12 times greater in mass than the hydrogen, therefore, it was given the mass of 12 amu. The masses of all the elements, in amu, were determined by their relative masses to hydrogen or carbon. mass of an p + mass of an n o mass of an e = amu = amu = amu Formula Masses: Calculate the masses of each kind of atom in the formula unit. Add the masses of all atoms in the formula unit. e.g. Calculate the formula mass (amu) of Calcium sulfate (CaSO 4 ) 1 amu = 1.66 x g 1 Ca = 1 x amu = amu 1 S = 1 x amu = amu 4 O = 4 x amu = amu amu Calculate the grams of one formula unit of calcium sulfate amu CaSO x g CaSO amu CaSO 4 = 2.26 x g CaSO 4 The Mole (Avogadro s number) A number of atoms, ions, or molecules that is large enough to see and handle. A mole = number of things Just like a dozen = 12 things One mole = x things Avogadro s number = x Symbol for Avogadro s number is N A.
2 2 How do we know when we have a mole? count it out weigh it out Molar mass  mass in grams numerically equal to the atomic weight of the element in grams. H has an atomic weight of g g of H atoms = x H atoms Mg has an atomic weight of g g of Mg atoms = x Mg atoms Molecules, atoms, protons and electrons 1 molecule of H 2 O = 2 hydrogen atoms and 1 oxygen atom 12 molecules of H 2 O = 2 (12) hydrogen atoms and 1 (12) oxygen atoms doz molecules of H 2 O = 2 (doz) hydrogen atoms and 1 (12) oxygen atoms 6.02 x molecules of H 2 O = 2 (6.02 x ) hydrogen atoms and 1 (6.02 x ) oxygen atoms mol molecules of H 2 O = 2 (mol) hydrogen atoms and 1 (mol) oxygen atoms A mole of hydrogen is equal to its amu in grams. One atom of hydrogen is equal to 1.0 amu 6.02 x atoms of hydrogen are equal to 1.0 gram One mole of hydrogen is equal to 1.0 g Mol of H 2 O = 3 mol atoms 3 mol of H 2 O = 3 x 3 mol atoms = 9 total mol atoms One Mole of Contains Cl 2 or 70.90g x Cl 2 molecules 2(6.022 x ) Cl atoms C 3 H 8 or g x C 3 H 8 molecules 3 (6.022 x ) C atoms 8 (6.022 x ) H atoms A mole of atoms of one element is equal to mass of one atom in amu in grams. e.g. 1 carbon atom = amu (1.994 x g) 6.02 x carbon atoms = grams ( 1 mole of carbon atoms = grams) Molar Mass: A mole of any substance has a mass equal to the summation of the masses of the elements in the compound. A mole of calcium sulfate: (a salt) 6.02 x formula units of CaSO 4 = g 1 mole of formula units of CaSO 4 = g (molar mass)
3 0.5 mole CaSO g CaSO 4 = 68.1 g CaSO mole CaSO x formula units of CaSO 4 = 3.01 x formula units of CaSO g CaSO mole CaSO 4 = moles CaSO If 1 mole of CaSO 4 has a mass of g, what is the mass of 0.5 moles of CaSO 4? 1.0 mole CaSO 4 2. How many formula units are there in 0.5 moles of CaSO 4? 1.0 mole CaSO 4 = 6.02 x formula units of CaSO mole CaSO 4 3. How many moles of CaSO 4 are there in g of CaSO 4? 1.0 mole CaSO 4 = g CaSO g CaSO 4 4. How many formula units of CaSO 4 are there in g of CaSO 4? 1.0 mole CaSO 4 = g CaSO mole CaSO 4 = 6.02 x formula units of CaSO g CaSO mole CaSO x formula units CaSO g CaSO mole CaSO 4 5. How many oxygen atoms are in 2.10 grams of CaSO 4? = x formula units CaSO mole CaSO 4 = g CaSO mole CaSO 4 = 6.0 mole O atoms 1.0 mole O = 6.02 x atoms of O 2.10 g CaSO mol CaSO mol O atoms 6.02 x atoms g CaSO mol CaSO mol O = 5.57 x atoms of oxygen A mole of H 2 O has a molar mass of 18.0 g 2 H = 2 (1.0 g) = 2.0 g 1 O = 1(16.0) g = 16.0g 18.0 g
4 4 If you have 5.00 x molecules of water, what is the mass of the water? If you have 52.5 g of H 2 O, how many hydrogen atoms do you have? Percent Composition and Formulas of Compounds The Law of Definite Proportions (Constant Composition) tells us that a compound has the same kinds of atoms in the same ratio. Therefore, we can determine the percent composition of any substance: part % = x 100 whole For example: What is the percent composition of all elements in CH 3 COOH? 1. Need molar mass of molecule. 2 C = 2 (12.0 g) = 24.0 g 4 H = 4 (1.0 g) = 4.0 g 2 O = 2 (16.0 g) = 32.0 g 2. Take the percentage of each element g % C = x 100 = 40.0 % 4.0 g % H = x 100 = 6.67 % 32.0 g % O = x 100 = 53.3 % 40.0 % % % = % = 100. % What is the percent composition of oxygen in calcium sulfate? part % = whole x 100% % Ca = mass of O x 100 mass of compound = amu x amu = % Empirical and Molecular Formulas: Molecular Empirical formula: Simplest formula
5 5 C 2 H 4 C 4 H 8 CH 2 C 6 H 12 Experimental Determination of Empirical and Molecular Formulas: Example 1 Suppose we know that a compound contains only the elements carbon, hydrogen, and oxygen, and we weigh out a g sample for analysis. We find that this g sample of compound contains g of carbon, g of hydrogen, and g of oxygen. We begin by converting to moles. We now know that g of the compound contains mol of C atoms, mol of H atoms, and mol of O atoms. What is the ratio of the atoms in the empirical formula? Divide by the smallest number of moles = 1 mol C = 2 mol H CH 2 O empirical formula = 1 mol O Suppose we know that the molecular formula mass (molar mass) of this compound is gram/mol. Empirical mass = g 6 (CH 2 O) = C 6 H 12 O g = g Example 2 When a g sample of vanadium metal is heated in air, it reacts with oxygen to achieve a final mass of g. Calculate the empirical formula of this vanadium oxide g vanadium + oxygen g V 1.0 mol V g vanadium = mol V atoms g oxygen g V
6 g O 1.0 mol O g O = mol O atoms mol = 1 mol V x 2 = 2 mol V mol mol = 2.5 mol O x 2 = 5 mol O mol V 2 O 5 Empirical From % composition: The most common form of nylon (Nylon6) is % carbon, 12.38% nitrogen, 9.80% hydrogen, and 14.14% oxygen. Calculate the empirical formula for Nylon6. In 100 grams, you would have g C, g N, 9.80 g H, and g O g C 1.0 mol C = mol C = mol C g C mol g N 1.0 mol N = mol N = mol N g N mol 9.80 g H 1.0 mol H = 9.72 mol H = 11.0 mol H g H mol g O 1.0 mol O = mol O = mol O g O mol C 6 NH 11 O
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