Strategies in Solving Chemistry Problems

Problem solving Step 1: Analyze the problem: Read the problem carefully. Write down the givens and what it is that you need to solve for

Strategies in Solving Chemistry Problems Problem solving (cont) Step 2: Develop a plan for solving the problem. Consider the possible paths between the given and the unknown. What principles or equations relate the given with the unknown. Your plan may involve a single step or several steps

Strategies in Solving Chemistry Problems Problem solving (cont) Step 3: Solve the problem. Use the known information and suitable equations or relationships to solve for the unknown (dimensional analysis will solve a great number of problems)

Strategies in Solving Chemistry Problems Problem solving (cont) Step 4: Check the solution. Read the problem again to make sure you found all the solutions asked for in the problem

Strategies in Solving Chemistry Problems Step 5 Check sig figs. Check the answer to make sure that the correct number of sig figs are in the answer. Standard rounding rules apply, 5 and up round up, 4 and below round down

Strategies in Solving Chemistry Problems Step 6 ensure that each answer has the correct type of units. Keeping track of units is difficult. Using dimensional analysis will help in keeping track of units. 3–73–7

A Problem Solving Method GERC G = Given, list all the givens and what is missing E = Equation, determine the correct equation to use R = Algebraically rearrange the equation to solve for what is missing C = Calculate, do the math, ensure proper units and sig figs

Chemical Stoichiometry Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Stoichiometry Given an amount of either starting material or product, determining the other quantities. use conversion factors from –molar mass (g - mole) –balanced equation (mole - mole) keep track

Also called Quantitative Chemistry

Quick Review Atomic mass The mole Mass sample of atoms Determining moles-atoms Calculate number of atoms Number of moles in mass Molar mass and calculating molar mass Calculating percent of composition of compounds

Summary of all Mole calculations: moles Number of particles Avogadro’s Constant = Mass of Substance Molar Mass = Pressure x Gas Volume Gas Constant x Temperature = Volume of Solution x Concentration =

Determining the Formula of a Compound

Formulas molecular formula = (empirical formula) n [n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH

Determining Empirical Formulas There is a 5 step process 1.Convert the percent of each element to mass (if we use 100g then percent equals mass) Example: Butane is 82.66% C and 17.34% H. Use a 100g sample to change %’s into mass 82.66g C and 17.34g of H

Determining Empirical Formulas Step 2: Convert the mass of each element to moles Example: 82.66g X 1/12.01 = moles of Carbon X 1/ = moles of Hydrogen

Determining Empirical Formulas Step 3: Use number of moles as subscripts Example: C H 17.20

Determining Empirical Formulas Step 4: Attempt to get inegers as subscripts by dividing by the smallest subscript Example: C 6.882/6.882 H 17.20/6.882 C 1 H 2.499

Determining Empirical Formulas Step 5: If any subscripts are fractional quantities multiply all the subscripts by the smallest integer that will convert all subscripts to integers Example: C 1 H (The subscript for H is real close to 5/2=2.5, 2 is the smallest integer. C 1X2 H 2.50X2 = C 2 H 5

Copyright © Houghton Mifflin Company. All rights reserved.3–22 Determining Molecular Formulas Molar Mass ÷ Empirical Formula Mass = multiplier per element empirical formula X multiplier = molecular formula

Sample Problem Determine the Empirical formula and the molecular formula for a compound that has the following: 71.65%Cl 24.27%C 4.07%H And it’s molar mass is g/mol

Chemical Equations

Copyright © Houghton Mifflin Company. All rights reserved.3–25 Chemical Equations Chemical equations consist of 2 basic parts reactants + reactants → products The reactants and products have various physical states (states of matter). There are 4 basics states of matter

Copyright © Houghton Mifflin Company. All rights reserved.3–26 Physical States in a Chemical Reaction Solid (s) Liquid (l) Gas (g) Dissolved in water (aq) Example: HCl (aq) + NaHCO 3(s) → CO 2(g) + H 2 O + NaCl (aq)

Chemical Equation A representation of a chemical reaction: C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O reactantsproducts

Chemical Equation C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Balancing Chemical Equations Rules for Balancing equations: 1.Determine the reaction that is occurring a. What are the reactants b. What are the Products c. What is the physical state 2. Write the unbalanced equation

Balancing Chemical Equations Rules for Balancing equations: 3.Balance the equation starting with the most complicated molecule and then determine what coefficients are necessary a. DO NOT Change identities b. DO NOT Change reactants c. DO NOT Change products

Notice The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts must not be changed to balance an equation. A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction. Coefficients can be fractions, although they are usually given as lowest integer multiples. Trial and error is a valid method to balance a chemical equation.

Stoichiometric Calculations: Amounts of Reactants and Products

Stoichiometric Calculations Calculating masses of reactants and products in chemical reactions is a 5 step process 1.Balance the equation for the reaction 2.Convert the known mass of the reactant or product to moles of that substance 3.Use the balanced equation to set up the appropriate mole ratios 4.Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product 5.Convert from moles back to grams if required

Sample Problem Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming lithium carbonate and liquid water. What is the mass of gaseous carbon dioxide can be absorbed by 1.00Kg of lithium hydroxide? 1.Step 1 write the equation and balance it LiOH(s) + CO 2 (g) → Li 2 CO 3 (s) + H 2 O(l) Balanced: 2LiOH (s) + CO 2 (g) → Li 2 CO 3 (s) + H 2 O(l)

Sample Problem Step 2: convert given mass of LiOH to moles 1.00kg LiOH X 1000g/1kg X 1mol LiOH/23.95 = 41.8 mol LiOH

Sample Problem Step 3: use the appropriate mole ratio 1 mol CO 2 /2 mol LiOH

Sample Problem Step 4: calculate moles of CO 2 needed to react with the given mass of LiOH using the mole ratio 41.8 mol LiOH x 1 mol CO 2 /2 mol LiOH = 20.9 mol CO 2

Copyright © Houghton Mifflin Company. All rights reserved.3–38 Sample Problem Step 5: calculate the mass of CO 2 using the molar mass 44.0g/mol 20.9 mol CO 2 X 44.0g CO 2 /1mol = 9.20 X10 2 g CO 2 It takes 920 g CO 2 (g) will be absorbed by 1.00Kg LiOH (s)

Limiting Reagent Reactant that determines the amount of product formed. The one you run out of first. Makes the least product.

Limiting reagent To determine the limiting reagent requires that you do two stoichiometry problems. Figure out how much product each reactant makes. The one that makes the least is the limiting reagent.

5 steps in limiting reactants 1.Balance the equation for the reaction 2.Convert the known mass of the reactant or product to moles of that substance 3.Determine which reactant is limiting 4.Use the amount of limiting reactant and the appropriate mole ratios to calculate the number of moles of the desired reactant or product 5.Convert from moles back to grams if required

Excess Reagent The reactant you don’t run out of. The amount of stuff you make is the yield. The theoretical yield is the amount you would make if everything went perfect. The actual yield is what you make in the lab.

Yield How much you get from an chemical reaction

Percent Yield % yield = Actual x 100% Theoretical % yield = what you got x 100% what you could have got

The End Copyright © Houghton Mifflin Company. All rights reserved.3–45