Stoichiometry: Reaction Calculations 3.14

Calculating Quantities in Reactions

• Stoichiometry uses proportional reasoning to find mole ratios from balanced equations.
• Mole ratios help to solve the stoichiometry problems
• Stoichiometry problems can involve mass, volume, or the number of particles

Balanced Equations Show Proportions

• Balanced chemical equations show the proportions of reactants and products
• These equations are similar to muffin recipes that show how much of each ingredient is needed
• For the synthesis of water, the balanced equation is: 2H2 + O2 → 2H2O
• In this reaction two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water
• Coefficients in a balanced equation show the number of particles for each substance in the reaction
• Calculations use proportions from balanced chemical equations to determine the quantity of each reactant and product
• Stoichiometry assumes reactions go to completion, with all of the given reactant turning changing into the product
• It's also assumed that every reaction happens perfectly, with no loss of product during collection

Relative Amounts in Equations

• Equations can be interpreted in terms of moles
• The coefficients represent the moles of each substance
• For example, 2C8H18 + 25O2 → 16CO2 + 18H2O shows that two moles of C8H18 react with 25 moles of O2 to form 16 moles of CO2 and 18 moles of H2O
• Stoichiometry deals with the quantities of substances in chemical reactions

Mole Ratio

• The mole is the unit that bridges the gap between one substance and another in stoichiometry problems
• Coefficients in a balanced chemical equation show the relative numbers of moles
• Mole ratios are conversion factors that convert from moles of one substance to moles of another

Converting Between Amounts in Moles

• Identify the amount in moles that is known from the problem
• Set up the mole ratio using coefficients from the balanced equation
• Put the known substance on the bottom and unknown substance on top
• Multiply the original amount by the mole ratio

Using Mole Ratios

• To prepare 312 moles of ammonia (NH3) in the reaction N2 + 3H2 → 2NH3, the steps are the following:
• The known amount of NH3 is 312 mol with the unknown amount of H2
• The mole ratio is:

3 mol H2/2 mol NH3

• Calculating the moles:

? mol H2 = 312 mol NH3 * (3 mol H2/2 mol NH3) = 468 mol H2

Moles, Mass, and Volume

• Substances are often measured by mass or volume
• It may be necessary to convert between mass, volume, and moles
• There are three basic steps when solving stoichiometry problems
• Change the given units into moles, use the mole ratio to determine moles of the desired substance, and change out of moles to the required units
• Conversion factors that involve moles will remind the steps that need to be followed

Solving Stoichiometry Problems

• Gather information and make sure the equation is balanced, write down the known information for the given substance, and identify the units one needs to find for the unknown substance
• Plan work by thinking through the 3 steps: change to moles, use the mole ratio, and change out of moles
• Determine what the conversion factor needed is in each step
• Write the mole ratio in the form:

moles of unknown substance/moles of given substance

• Calculate by writing a question mark followed by the units of the answer, and the quantity of the given substance
• Write the conversion factors, including the mole ratio, in order so that change the units of the given substance to the units needed for the answer
• Cancel units to ensure that the remaining units are for the unknown substance
• Round the answer to the correct number of significant figures and report with correct units and the name/formula of the substance
• Verify the result by estimating to determine if the answer should be larger or smaller than the initial value, and double-check the work

Mass Calculations

• The conversion factor for converting between mass and moles is the molar mass of the substance
• The molar mass is the sum of atomic masses from the periodic table for the atoms in a substance
• The problem is a three-step process: convert mass in grams to moles, use the mole ratio, and convert moles to grams

Solving Mass-Mass Problems

• Use molar mass to convert from mass of known to amount of known
• Use mole ratio to convert from amount of known to amount of unknown
• Use molar mass to convert from amount of unknown to mass of unknown

Problems Involving Mass

• Given 1221 g of H2 and excess N2, and the equation N2 + 3H2 → 2NH3, To find the mass of NH3 that can be made, follow these steps:
• Mass of H2 = 1221 g and molar mass of H2 = 2.02 g/mol. The unknown is the mass of NH3, and the molar mass of NH3 is 17.04 g/mol
• Use the molar mass of H2 to change grams of H2 to moles. The mole ratio is 2 mol NH3/3 mol H2 which will leave the units of mol NH3
• Convert moles of NH3 to grams using the molar mass of NH3

Volume Calculations

• When reactants are liquids, they are measured by volume; doing calculation involving liquids adds steps such as volume to mass/mass to volume
• There are 5 conversion factors: volume to mass/mass to volume, 2 densities, 2 molar masses, and a mole ratio
• Use the density of the substance as the conversion factor to convert from volume to mass or from mass to volume
• Other ways of including volume: if a substance is a gas at standard temperature and pressure (STP), use the molar volume of a gas to change directly between volume of the gas and moles

Solving Volume-Volume Problems

• Use density to convert from volume of known to mass of known, and from volume of unknown to mass of unknown
• Use molar mass to convert from mass of known to amount of known, and amount of unknown to mass of unknown
• Use the mole ratio to convert from amount of known to amount of unknown

Volume Problems

• For the reaction POCl3(l) + 3H2O(l) → H3PO4(l) + 3HCl(g), given 56 mL POCl3, the density of POCl3 is 1.67 g/mL, and the density of H3PO4 is 1.83 g/mL., the following steps can find the volume of H3PO4 that forms:
  • The molar mass for POCI3 is 153.32 g/mol, molar mass for H3PO4 is 98.00 g/mol. Given is 1 mol POCI3 = 1 mol H3PO4
• Use the density of POCl3 followed by the molar mass of POCl3 to change milliliters of POCl3 to moles
• Multiply molar mass and density of H3PO4 to transfer moles of H3PO4 into milliliters

Particle Problems

• Avogadro's number, 6.022 × 1023 particles/mol, is how stoichiometry problems are solved

Solving Particle Problems

• Use Avogadro's number to convert from particles of a known substance to the amount in moles of the known substance
• Use the mole ratio between the known to amount of unknown
• Use Avogadro's number to convert from amount of unknown to particles of the unknown substance

Amount of Particles

• Given is 1.89 × 1024 molecules of C5H12, The amount of grams of C5H8, and the reaction C5H12(l) → C5H8(l) + 2H2(g) is performed by this:
• Avogadro's number of 6.022 × 1023 molecules/mol is used to change to moles, then the mole ratio, and the molar mass of C5H8 is 68.13 g/mol to change to grams

Limiting Reactants and Percentage Yield

• When gasoline runs out in a car, the car cannot go any further even though there is oxygen; gasoline will limit the distance traveled because it stops the engine
• 100% of the reactants change into products
• The real world can limit the yield of reaction

Limiting Reactant

• When reactants of a reaction are seldom present in ratios equal to the mole ratio in the balanced equation, one of the reactants is used up first
• Zinc limits the amount of product that can form in the reaction Zn + 2HCl → ZnCl2 + H2. Zinc is used up which makes it the limiting reactant
• The HCl is the excess reactant because there is more than enough HCI present to react with all of the Zn

Determine Theoretical Yield

• The theoretical yield of a reaction should always be calculated based on the limiting reactant.
• The maximum quantity of product that a reaction could theoretically make if everything about the reaction works perfectly is called the theoretical yield

Limiting Reactant Problems

• To identify the limiting reactant and the theoretical yield of phosphorous acid, H3PO3, if 225 g of PCl3 is mixed with 125 g of H2O, in this reaction: PCl3 + 3H2O → H3PO3 + 3HCl:
• The mass of PCl3 is 225 g and the mass of H2O is 125 g, looking for the mass of H3PO3 (unknown),
• The molar mass of PCl3 = 137.32 g/mol, the molar mass of H2O = 18.02 g/mol, and the molar mass of H3PO3 = 82.00 g/mol
• Problems calculating the mass of H3PO3 are set up that are expected to form from each reactant

Limiting Reactants Issues

• In industry, the cheapest reactant is often used as the excess reactant
• In this way, the expensive reactant is more completely used up

Actual Yield

• The mass of product actually formed can be called the actual yield, which is the measured amount of a product
• It can be less than expected
• There are several reasons why the actual yield is usually less that the theoretical yields in chemical reactions

Percentage Yield

• Percentage yield describes the efficiency of a reaction whose formula,

percentage yield = actual yield/theoretical yield * 100%

Find Percentage Yield

• It is possible to determine the limiting reactant, theoretical yield, and percentage yield if given 14.0 g N2 mixed with 9.0 g H2 results in 16.1 g NH3 formed. Using our known reaction N2 + 3H2 → 2NH3:
• The mass of N2 = 14.0 g and mass of H2 = 9.0 g, looking for the theoretical yield of NH3 that has an actual yield = 16.1 g
• The molar mass of N2 = 28.02 g/mol, molar mass of H2 = 2.02 g/mol, molar mass of NH3 = 17.04 g/mol with 1 mol N2 = 2 mol NH3 and 3 mol H2 = 2 mol NH3
• Set up problems that will calculate the mass of NH3 that will form from the reactant

Actual Yield

• If the percentage yield for a reaction is known, a close estimate can be found if its actual field value.
• Percentage yield in a particular reaction is fairly consistent

Calculate Actual Yield

• To find the amount of grams of CH3COOC5H11 should form if 4808 g are theoretically possible and the percentage yield for the reaction is 80.5%
• Know the theoretical yield of CH3COOC5H11 = 4808 g and that the percentage yield = 80.5%, find the actual yield of CH3COOC5H11
• Then the value can be calculated with the equation

80.5% = actual yield/4808 g * 100