Stoichiometry

Look back at Moles

Practice

• 45.3 g of carbon is how many moles?• 5.00 liters of carbon dioxide at

STP is how many moles?• How many molecules of

hydrogen are there in 40.0 L of hydrogen gas at STP?

Arithmetic of Equations

• Stoichiometry – calculation of quantities in chemical reactions

• A chemical reaction tells us a few things: (recipe)– N2 + 3H2 → 2NH3

– We can compare number of atoms, number of molecules, moles, mass and volume

– Which of these are conserved? (same on both side)

How to interpret a reaction

• Given the reaction: N2 + 3H2 → 2NH3

• We can easily get number of each atom, each molecule, and volume of each gas.

• The tricky one is the mass of the reactants and products.

• By the law of conservation of mass, we know that each side of an equation needs to have an equal mass, but we need to find the mass of each side to show that they are equal.

Mass of Reactants and Products

• N2 + 3H2 → 2NH3

• Reactants: 1 mol N2 = 28 g

• 3 mol H2 = 3 X 2 g = 6 g• Total = 34 g• Products: 2 mol NH3 = 2 X 17 g = 34 g

• Both sides are 34 g, which is a good sign the calculation was done right

Now try this out

• Given the reaction 2Mg + O2 → 2MgO• Find the number of moles and mass of

reactants and products

• Next do the same for this unbalanced reaction– That means YOU need to balance it!

• H2SO4 + NaOH → Na2SO4 + H2O

Mole Ratios

• Mole ratio: conversion factor derived from the coefficients of a balanced chemical reactions in terms of moles

• N2 + 3H2 → 2NH3

• This equation gives us three mole ratios• Nitrogen to hydrogen, nitrogen to ammonia, and

hydrogen to ammonia or• 1 mol N2 1 mol N2 3 mol H2

• 3 mol H2 2 mol NH3 2 mol NH3

Get all mole ratios for this reaction

•2Mg + O2 → 2MgO

Using mole ratios

• We can use mole ratios to convert from moles of one substance to moles of a different one.

• If we have 3.50 mol N2 and excess H2 in the reaction: N2 + 3H2 → 2NH3

• How many moles of NH3 will be formed?

• What if we had 3.50 mol H2 and excess N2?

Now you try these

• Given the equation: 2Mg + O2 → 2MgO• How many moles of oxygen will react with

5.00 mol Mg?

• How many moles of Mg and O2 reacted if we ended up with 3.70 mol MgO?

Mass-Mass Calculations

• With our new tool, the mole ratio, we can calculate the mass of any other substance in a reaction if we are given the mass of one.

It will take THREE steps:1. Convert given grams to given moles (molar mass)2. Convert given moles to wanted moles(Mole Ratio)3. Convert wanted moles to wanted grams (molar

mass)

Example

• Given the reaction: N2 + 3H2 → 2NH3

• How many grams of ammonia will be formed if 50 g of nitrogen react?

• Given the same amount of nitrogen, how much hydrogen reacted?

Your turn

• Given the reaction: 2Mg + O2 → 2MgO• How many grams of Mg reacted if 75.3 g MgO

were formed?

• How many grams of O2 reacted if 75.3 g MgO were formed?

Other Stoichiometric Calculations

• Instead of doing a mass-mole conversion, we can do the two other conversions from Ch 10.

• Instead of giving a mass, we could get a volume or number of particles

• Also, instead of calculating mass at the end, we can get volume or number of particles

• Look at Figure 12.8 on page 363

Modified Steps

1. Convert given grams, liters or particles to given moles

2. Convert given moles to wanted moles(Mole Ratio)

3. Convert wanted moles to wanted grams, liters or particles

Examples

• Given the reaction: 2Mg + O2 → 2MgO• How many atoms of Mg reacted if 75.3 g MgO

were formed?

• Given the reaction: N2 + 3H2 → 2NH3

• How many liters of ammonia will be formed if 3.50 L of nitrogen react?

Limiting Reagent and Percent Yield

• Limiting reagent is the more practical application of stoichiometry that can be used in the laboratory

• Percent yield compares the amount of product formed in an experiment compared to what the maximum would be

Limiting and Excess Reactants

• Limiting Reactant- reactant that is used up in a reaction, which limits that amount of product

• Excess Reactant- reactant that remains after the reaction has stopped

Determining a limiting reactant• For these problems, we will be given the starting

amounts of two reactants• We need to use both and use each to get the

amount of one of the products formed

N2 + 3H2 → 2NH3

• If 5.3 moles of nitrogen gas react with 5.3 moles of hydrogen gas, how many moles of ammonia is produced? Which reactant is limiting?

N2 + 3H2 → 2NH3

• If 30.0 g of nitrogen gas react with 30.0 g of hydrogen gas, how many grams of ammonia is produced? Which reactant is limiting?

Your turn

• 2Mg + O2 → 2MgO• Which reactant is limiting if 20.0 g of

magnesium react with 20.0 g of oxygen? How much product will be formed?

Yield Definitions

• Theoretical Yield- maximum amount of product that can be formed by the given amount of reactants

• Actual Yield- amount of product actually formed when an experiment is performed

• Percent Yield- ratio of the actual yield to the theoretical yield expressed as a percent

Percent Yield• Tells us how efficient an experiment was• Can be as a result of an error made or may just be a

result of the procedure of the experiment• Actual yield is determined in the laboratory or given

in the problem• Theoretical yield is determined with a calculation

Example

• N2 + 3H2 → 2NH3

• If 10.0 g nitrogen react with 3.00 g hydrogen and 10.0 g ammonia is formed, what is the percent yield?

Practice

• 2Mg + O2 → 2MgO• If 15.0 g magnesium react with 10.0 g oxygen

to form 16.5 g magnesium oxide, what is the percent yield?