Balancing Chemical Equations What goes in must come out!

Balancing Chemical Equations Balancing a chemical equation is much like the work of an accountant who has to show every penny that comes in and where it has gone to.

Objectives Learn the steps to balancing chemical equations. Take notes to help you understand. Test yourself with a set of equations to balance. Enter your own equations to see if they balance.

Law of Conservation of Mass Law of Conservation of Mass You need to remember this law! The Law of Conservation of Mass states: that mass is neither created nor destroyed in any chemical reaction. Therefore balancing of equations requires the same number of atoms on both sides of a chemical reaction. The number of atoms in the Reactants must equal the Number of atoms in the Products

Because of the principle of the Conservation of Matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Lavoisier, 1788 Chemical Equations

Law of Conservation of Mass The mass of all the reactants (the substances going into a reaction) must equal the mass of the products (the substances produced by the reaction). Reactant + Reactant = Product

A simple equation, such as the synthesis of Iron (II) sulfide, iron + sulfur Iron (II) sulfide Fe + S FeS Note that in a chemical equation, by convention, we use the arrow “ " instead of the equals “ = ".

The last stage is to put in state of matter symbols, (s, l, g, aq), as appropriate (solid, liquid, gas, aqueous or dissolved in water) Fe(s) + S(s) FeS(s)

Balancing Equations ___ Al(s) + ___ Br 2 (l) ---> ___ Al 2 Br 6 (s) 23

Steps to Balancing a Chemical Equation 1. Write all reactants on the left and all products on the right side of the equation arrow. Make sure you write the correct formula for each element 2. Use coefficients in front of each formula to balance the number of atoms on each side.

Steps to Balancing a Chemical Equation 3. Multiply the coefficient of each element by the subscript of the element to count the atoms. Then list the number of atoms of each element on each side. 4. It is often easiest to start balancing with an element that appears only once on each side of the arrow. These elements must have the same coefficient. Next balance elements that appear only once on each side but have different numbers of atoms. Finally balance elements that are in two formulas in the same side.

Re-cap of steps from rule 4: Balance elements that appear only once on each side of the arrow. Next balance elements that appear only once on each side but have different numbers of atoms. Finally balance elements that are in two formulas in the same side.

Balancing Chemical Equations An easier way

First you need an equation with the correct “formulae” ………. You’ll probably be given this in the question Just like this one Mg + O 2  MgO Then all you do is list the atoms that are involved on each side of the arrow Mg + O 2  MgO Mg O Mg O

[1] Just count up the atoms on each side Then start balancing: Mg + O 2  MgO Mg O [2] The numbers aren’t balanced so then add “BIG” numbers to make up for any shortages And adjust totals Mg + O 2  MgO Mg O

Mg + O 2  MgO Mg O But the numbers still aren’t equal, so add another “BIG” number 2 And adjust totals again NOW BOTH SIDES HAVE EQUAL NUMBERS OF ATOMS WE SAY THAT THE EQUATION IS BALANCED!! 2

Try to balance these equations using the same method: [1] Na + Cl 2  NaCl [2] CH 4 + O 2  CO 2 + H 2 O [4] Al + O 2  Al 2 O 3 [3] Li + HNO 3  LiNO 3 + H 2

How did you get on?? [1] 2 Na + Cl 2  2 NaCl [2] CH O 2  CO H 2 O [4] 4 Al + 3 O 2  2 Al 2 O 3 [3] 2 Li + 2 HNO 3  2 LiNO 3 + H 2 Here are the answers: HOPE YOU’VE GOT THE IDEA… REMEMBER TO CHECK THAT YOU CAN DO ELECTROLYSIS EQUATIONS TOO

Example NH 3 + O 2 NO + H 2 O Reactants Products N appears once on both sides in equal numbers, so the coefficient for NH 3 is the same as for NO.

Example: NH 3 + O2 NO + H 2 O Next look at H which appears only once on each side but has different numbers of atoms, 3 on the left and 2 on the right. The least common multiple of 3 and 2 is 6, so rewrite the equation to get 6 atoms of H on both sides: 2NH 3 + O 2 NO + 3H 2 O

Example: 2NH 3 + O2 2 NO + 3H 2 O There are 2 oxygen atoms on the left and 5 on the right — the least common multiple of 2 and 5 is 10, so rewrite the equation as: 2NH 3 + 5O 2 4NO + 6H 2 O

Now count the atoms on each side: 2NH3 + 5O2 4NO + 6H2O Write them out keeping them on the appropriate side of the chemical equation 2 N (nitrogen atoms) 4 N (nitrogen atoms) 6 H (hydrogen atoms) 12 H (hydrogen atoms) 10 O (oxygen atoms) 10 O (oxygen atoms) “YET” This shows the equation not to be balanced “YET”

Check the number again: If you double the N and H on the left the equation will be balanced: 4NH3 + 5O2 4NO + 6H2O

Double-check: 4NH3 + 5O2 4NO + 6H2O 4 N (nitrogen atoms) 4 N (nitrogen atoms) 12 H (hydrogen atoms) 12 H (hydrogen atoms) 10 O (oxygen atoms) 10 O (oxygen atoms) The equation is Balanced

Stoichiometry By combining our abilities to balance equations and do simple unit conversions, we can now complete stoichiometry problems (mass to mass conversions)

Particle and Mole Relationships Chemical reactions stop when one of the reactants is used up. Stoichiometry is the study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemical reaction.Stoichiometry

Particle and Mole Relationships (cont.) Stoichiometry is based on the law of conservation of mass. The mass of reactants equals the mass of the products.

Particle and Mole Relationships (cont.)

A mole ratio is a ratio between the numbers of moles of any two substances in a balanced equation. Refer to coefficients in the balanced equation to determine the mole ratio.mole ratio

Stoichiometry Stoichiometry problems consist of predicting the amount of grams of product that will form from a reaction if you know how many grams of reactant you have, or predicting the number of grams of reactant that will be needed to produce a certain amount of product.

Using Stoichiometry All stoichiometric calculations begins with a balanced chemical equation. 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)

Steps to solve mole-to-mole, mole-to-mass, and mass-to-mass stoichiometric problems 1.Complete Step 1 by writing the balanced chemical equation for the reaction. 2.To determine where to start your calculations, note the unit of the given substance. If mass (in grams) of the given substance is the starting unit, begin your calculations with Step 2. If amount (in moles) of the given substance is the starting unit, skip Step 2 and begin your calculations with Step 3.

Using Stoichiometry (cont.) 3. The end point of the calculation depends on the desired unit of the unknown substance. If the answer must be in moles, stop after completing Step 3. If the answer must be in grams, stop after completing Step 4.

Stoichiometry We will solve these problems using a “T” chart just like we did for unit conversions, but we will add additional cells Given infoUnits to convert intoUnits to convert into etc Units to cancelUnits to cancel etc and we use additional conversion factors

Now for the key part of the lesson… Take notes from what Mr. Ahmad writes on the white board for challenge problems.