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Ch7-p1 Chapter 7 - Chemical Quantities The Mole In everyday life, we use counting units to deal with large quantities. For example, we buy eggs by the dozen (12), paper by the ream (500), and pop by the case (24). In chemistry, we count molecules, atoms, and ions by the mole. A mole contains 6.02 X 1023 items. This very large number is called Avogadro’s number, after the Italian physicist Amedeo Avogadro. One mole of any element, molecule, or ionic compound contains Avogadro’s number of atoms. Therefore: 1mol of C contains 6.02 X 1023 carbon atoms. 1mol of Al contains 6.02 X 1023 aluminum atoms. 1mol of Br contains 6.02 X 1023 bromine atoms. 1mol of H2O contains 6.02 X 1023 H2O molecules. 1mol of NaCl contains 6.02 X 1023 NaCl formula units. Ch7-p2 The subscripts in a chemical formula also represent moles. For example, the chemical formula for aspirin is C9H8O4. Therefore, 1 mole of aspirin contains 9 moles of carbon atoms, 8 moles of hydrogen atoms, and 4 moles of oxygen atoms. How many C atoms are in 0.350 mol of C6H12O6? How many Cl atoms in 2.0 mole of PCl3? How many moles of C2H6O contain 5.0 x 10 24 atoms of H? Ch7-p3 Molar Mass The mass in grams of 1 mole of substance is called its molar mass. The molar mass (in grams) of any substance is always equal to its formula weight (in amu). Determine the molar masses for the following substances: a) b) c) d) e) C Ag H2O NaCl NO3- What is the mass of 1 mole of glucose, C6H12O6? Ch7-p4 How many moles of glucose are in 5.38 g? How many grams of glucose are in 2.74 moles? The concept of the mole provides a bridge between masses and numbers of particles. Using molar mass and Avogadro’s Number as conversion factors, we can convert grams moles atoms. For example, calculate the number of Cu atoms in a 3g copper penny (assuming the penny is 100% Cu). Ch7-p5 How many glucose molecules are in 5.23 g of C6H12O6? Percent Composition and Empirical Formula Recall that the subscripts in a molecular formula represent a definite proportion of the elements within a compound. Therefore, a mole of any compound contains a definite proportion by mass of its elements. By using the molar mass of a compound, we can determine its percent composition. For example, determine the percent composition of K2CO3. The molecular formula of a compound represents the true formula of a compound. A formula that gives the Ch7-p6 lowest whole-number ratio of atoms in the compound is called the empirical formula. For example, hydrazine with molecular formula of N2H4, has an empirical formula of NH2. The mole concept can be used to determine empirical formulas. e.g. A sample of a compound contains 3.24 g Na, 2.26 g S, and 4.51 g O. What is its empirical formula? e.g. Calculate the empirical formula of a compound that has percent composition 36% Al and 64% S. Ch7-p7 Stoichiometry Consider the following reaction: 2H2 + O2 2H2O The coefficients from the balanced equation tell us that we need two molecules of hydrogen to react with one molecule of oxygen to give two molecules of water. From the mole concept, we can interpret the equation in terms of moles. Thus, 2 moles of hydrogen react with 1 mole of oxygen to give 2 moles of water. The coefficients in the above equation are called stoichiometrically equivalent quantities. The relationship between these quantities can be expressed as follows: 2 mol H2 K 1 mol O2 K 2 mol H2O These stoichiometric relations can be used to give conversion factors for relating quantities of reactants and products in a chemical reaction. e.g. Calculate the number of moles of H2O produced from 1.57 moles of O2. Ch7-p8 e.g. Using the equation below, calculate the mass of CO2 that is produced in burning 1.00 g of butane, C4H10. 2C4H10 + 13O2 8CO2 + 10H2O Ch7-p9 e.g. Nitrogen gas reacts with hydrogen gas to produce ammonia by the following equation: N2 + 3H2 2NH3 a) If you have 1.8 moles of H2, how many grams of NH3 can be produced? b) How many grams of H2 are needed to react with 2.80 g of N2? c) How many grams of NH3 can be produced from 12 g of H2? Ch7-p10 Percent Yield. All of the equations we’ve seen so far have assumed that 100% of product was formed. In reality, this is rarely the case. Side reactions or the equilibrium can prevent reactions from going to completion. When chemists perform reactions, there are two things they want to know; a) what is the theoretical yield of the reaction and b) what is the actual yield of the reaction. • The theoretical yield is the calculated amount of product if the reaction proceeds to completion. • The actual yield is the amount of product obtained after the reaction is complete. If we know the values of the theoretical yield and actual yield, then we can express these values as a percentage, called the percent yield. e.g. What is the percent yield if 40.0 g of CO are produced from the reaction of 30.0 g O2? 2C + O2 2CO Ch7-p11 e.g. From the following equation, calculate the mass of CO2 that can be produced if the reaction of 45.0 g of propane (C3H8) and sufficient oxygen has a 60% yield. C3H8 + 5O2 3CO2 + 4H2O Ch7-p12 Equilibrium Constants. We’ve learned that when a reversible reaction has reached equilibrium, the forward and reverse reaction occur at the same rate. We’ve also learned that at equilibrium, the amounts of reactants and products remain constant. Because of this, we can express the ratio of products to reactants. This relationship between reactants and products is called an equilibrium constant. equilibrium constant = Keq = products reactants Consider the general reaction: aA + bB cC + dD In this equation, the capital letters A and B are reactants and C and D are products. The lower case letters represent their coefficients in the balanced equations. In the equilibrium constant, the products are multiplied in the numerator and divided by the reactants in the denominator. [C]c[D]d Keq = [A]a[B]b Ch7-p13 The square brackets indicate that the concentrations (moles per 1 litre of solution) of the products and reactants are used. Also note that the concentrations of reactants and products are raised to the power of their respective coefficients. e.g. Write the expression for the equilibrium constant Keq for the following reactions: 2HBr CO + 2H2 CH4 + 2H2S H2 + Br2 CH3OH CS2 + 4H2 Write the balanced chemical equation that would give the Ch7-p14 following equilibrium constant expression. [NO2]2 Keq = [NO]2[O2] When Keq > 1, the equilibrium favours the products. When Keq < 1, the equilibrium favours the reactants. When Keq = 1, the concentration of reactants and products are equal.