Chemical Equations and Calculations

Transcription

1 Chemical Equations and Calculations A chemical equation is a shorthand way of indicating what is going on in a chemical reaction. We could do it the long way Two molecules of Hydrogen gas react with one molecule of Oxygen gas to produce two molecule of liquid water. Or we can use the shorthand H (g) + O (g) H O (l) The second way is easier to read and keep track of everything going on. The equation above uses phase symbols. Phase symbols are used to denote the phase of substances in an equation. (g) (s) (l) (aq) gas solid liquid aqueous (water solution) The chemical equation must represent reality. The symbol,, is sometimes used to denote a reaction that is heated and is written above the arrow. Catalyst symbols or formulas are also placed above the arrow. Balancing equations An equation is balanced when the number of atoms of each type present is the same on both sides of the equation. The chemical formulas CANNOT be changed in the process of balancing an equation. The process of balancing an equation can sometimes seem to be trial and error. In reality, there is a method to the madness. When we balance a chemical equation we are looking at the relationships between the elements on both sides of the equation. This relationship helps us to figure out what needs to be done. Some basic rules for balancing chemical equations are 1. Start with the most complicated chemical first and start with the element that appears the most number of times in that compound.. Save free elements for last. CH 4 (g) + O (g) CO (g) + H O (l)

2 Here CH 4 is the most complicated and Hydrogen appears the most number of times so we will start with that element. O is a free element (not combined with other elements) so we will leave that for last. There are 4 Hydrogens on the reactant side and on the product side. Therefore, we multiply the water on the product side by and the hydrogens balance. CH 4 (g) + O (g) CO (g) + H O (l) The carbons are already balanced. Now we can balance the oxygens. There are on the reactant side and 4 on the product side. If we multiply the O by we balance the oxygen. CH 4 (g) + O (g) CO (g) + H O (l) Now the equation is balanced. Stoichiometry: Molar interpretation vs. Molecular interpretation of a chemical reaction The balanced chemical reaction gives you a series of conversion factors to use in problem solving. CH 4 + O CO + H O The conversion factors are obtained from the coefficients in the balanced chemical reaction. These conversion factors can be used to relate the amounts of reactant to other reactants or to amounts of products. It is this reason that, in order to solve a chemical problem, you first need to have a balanced chemical equation. How many grams of solid Barium Sulfate can be produced from the reaction of g of Barium Nitrate with Sodium Sulfate? The other product of the reaction is Sodium Nitrate. First we need to write the balanced chemical equation: Ba(NO ) + Na SO 4 BaSO 4 + NaNO Now we can proceed with the calculation.

3 1mol Ba(NO ) 1mol BaSO.91 g BaSO g Ba(NO = 4 4 ) 18.1g BaSO g Ba(NO) 1mol Ba(NO) 1mol BaSO4 Solutions Solvent is the substance that does the dissolving. Usually present in the larger amount. Solute is the substance that is dissolved. Usually present in the smaller amount. In solutions involving water, water is usually the solvent, even if it is present in the smaller amount. We can quantify all of this by using concentration units. The unit used most in chemistry is called molarity or the molar concentration. The symbol for molarity is M and is defined as the number of moles of solute in one liter of solution. Molarity (M) = moles of solute liters of solution This concentration unit is just a conversion factor that allows us to convert between the volume of a solution and the number of moles of solute present in that solution or vice versa. Giving the volume of a solution and the concentration of that solution will allow you to calculate the moles of reactant for use in stoichiometry problems. Dilution of solutions the solution to pollution is dilution Not exactly true. Some pollutants are dangerous in the environment even in very small amounts. When solutions are diluted the number of moles of solute remains the same. Only the volume of the solution changed (by adding water). Because the moles of solute remain the same we can start with: This can be rewritten as: initial moles of solute = final moles of solute initial molarity initial volume = final molarity final volume. Three of these values will always be given.

4 Stoichiometry of Reactions in Solution When solutions react, it is very difficult to specify the number of grams of reactant (or solute) present. It is much easier to specify the volume and molarity of the solution(s) used. We can use these two numbers to calculate the number of moles of reactant present and from there continue on with the stoichiometric calculation..51 ml of M Sodium Carbonate reacts with Calcium Nitrate solution. The products are solid Calcium Carbonate and aqueous Sodium Nitrate. What mass of Calcium Carbonate can be produced from this mixture? Answer: First we need to have a balanced chemical equation: Na CO (aq) + Ca(NO ) (aq) NaNO (aq) + CaCO (s) Now we can perform the stoichiometric calculation:.51l Na CO sol' n 10 L Na CO sol' n mol Na CO 1mL Na CO sol' n 1L Na CO sol' n 1mol CaCO g CaCO 1mol Na CO 1mol CaCO = 1.44 g CaCO Limiting Reactants The Limiting Reactant is the reactant that is completely consumed during the course of a reaction. If the reactant is not completely consumed it is known as an excess reactant. The limiting reactant is what determines the amount of product(s). When all of the limiting reactant is consumed, no more product(s) can be produced. Determining the limiting reactant 1. Calculate the amount of product that would result from each of the reactants.. The reactant that corresponds to whichever amount of product is smallest is the limiting reactant. All other reactants are excess reactants.

5 Theoretical Yield is the maximum amount of product that can be produced from a reaction mixture. It is the amount of product determined from the limiting reactant. Actual Yield is the amount of product that results when the reaction is carried out in the lab. Percent yield is the ratio of actual yield to theoretical yield multiplied by 100. % yield = actual yield theoretical yield 100% Why determine the % yield? When a chemist reports his results from a reaction, he/she does so by using the percent yield. This is done because the percent yield should be the same for the same reaction regardless of how much reactant is initially used g of Ethanol (C H 6 O) reacts with 10.5 g of Oxygen to produce Carbon Dioxide and water. How many grams of Carbon Dioxide are produced in this reaction? What is the limiting reactant? If only 0.00 g of Carbon Dioxide are produced, what is the percent yield? Balanced Chemical Equation: Calculations: 0.00 g CO.54 g CO C H 6 O + O CO + H O 1mol CH6O mol CO 44.01g CO g CH6O =.54 g CO g CH6O 1mol CH6O 1mol CO 1mol O mol CO 44.01g CO 10.5 g O = 9.98 g CO.00 g O mol O 1mol CO 100 = 89.45% The limiting reactant is the reactant that gives the smallest amount of product. In this case, that is Ethanol. The theoretical yield is that amount of product,.54 g CO. The percent yield is 89.45%.