Balancing Chemical Equations What goes in must come out!

Law of Conservation of Mass When a chemical reaction occurs, new compounds are created, BUT… – Matter cannot be destroyed or createdatomsare just rearranged to form new compounds. – Matter cannot be destroyed or created; atoms are just rearranged as the atoms change partners to form new compounds. 3 – If there are 3 atoms of oxygen in the reactants, there MUST be 3 atoms of oxygen in the products. – Number of each atom in reactants = number of each atom in products. – Balanced Equation (c) McGraw Hill Ryerson 2007 If you could collect and measure all of the exhaust from this car, you would find that mass of reactants (gas + O 2 ) = mass of products (exhaust).

Subscripts vs. Coefficients The subscripts tell you how many atoms of a particular element are in a compound. The coefficient tells you about the quantity, or number, of molecules of the compound. The subscripts tell you how many atoms of a particular element are in a compound. The coefficient tells you about the quantity, or number, of molecules of the compound.

Subscripts vs. Coefficients Coefficient is the only thing that can change Subscripts must remain the same

5 If a chemical equation does not obey the law of conservation of mass the equation is said to be what? NOT BALANCED So Let’s look at the steps we need to take to BALANCE chemical equations H 2 + O 2  H 2 O Let’s work with the following equation:

6 #R atom #P 2 H 2 2 O 1 For example: H 2 + O 2  H 2 O Step 1. Create a table A table that shows us what atoms are present in this reaction, how many there are and are they reactants or products?

7 #R atom #P 2 H2 2O1 Step 2: Go to the first atom that’s not balanced and balance it! Since the O atoms are not balanced what do we need to do to balance it? Right! Multiply it by 2 (Only multiply) x2

8 #R atom #P 2 H2 2 O1 H 2 + O 2  2H 2 O In step 2 we balanced the number of O atoms by multiplying the product side by 2. This now becomes the new coefficient in the chemical equation. Modify the equation to reflect the change Is the equation balanced? x2

9 #R atom #P 2 H 4 2 O 2 2H 2 + O 2  2H 2 O Step 3. Move to the next unbalanced atom. What is it? How can we balance the Hydrogen? #R atom #P 2 H 4 2 O 2 Adjust the equation to reflect your changes Multiply Reactants by 2 2x Do we have a balanced Chemical Equation now? Yes we do!

Polyatomics When an equation has a polyatomic in it, such as in this balanced equations – 2AgNO 3 + MgCl 2  2AgCl + Mg(NO 3 ) 2 If the polyatomic appears on BOTH the reactant and product side of the equation, then count the polyatomic as an “ATOM”.

Polyatomics So the above reactant atoms would be: If the same polyatomic does not appear on both sides, break the Polyatomic down into atoms! #R atom #P 2 Ag 2 2 NO Mg 1 2Cl 2

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13 #R atom #P 1 NA 1 1 OH 2 1 Ca 1 2 Br 1 Is this equation balanced? NaOH + CaBr 2  Ca(OH) 2 + NaBr What atoms do we have in This equation? 1) Count atoms & Start the table 2) Do the #Reactant atoms = the # of Product atoms? 3) So pick the 1st unbalanced atom & begin balancing

14 #R atom #P 1 Na 1 1 OH 2 1 Ca 1 2 Br 1 We’ll start with balancing Hydroxide NaOH + CaBr 2  Ca(OH) 2 +NaBr How can we make both Hydroxides equal? Sure we’ll multiply #R OH by 2 2x Hydroxide is now balanced so let’s move to the next Unbalanced atom, which is? … Next step> rewrite the modified eqn. 2NaOH + CaBr 2  Ca(OH) 2 + NaBr

15 #R atom #P 2 Na 1 2 OH 2 1 Ca 1 2 Br 1 What can we do to balance the Bromine? 2NaOH + CaBr 2  Ca(OH) 2 + 2NaBr Now adjust the table to reflect The changes and then rewrite the Eqn. Sure! Multiply the #P Bromine by 2 x2

16 #R atom #P 2Na 2 2 OH 2 1 Ca 1 2 Br 2 Let’s update the table with the new #’s Based on our updated equation. 2NaOH + CaBr 2  Ca(OH) 2 + 2NaBr Are we now balanced? Sure!

17 #R atom #P 2 C 1 6 H 2 2 O 3 Ok Try Balancing this equation: C 2 H 6 + O 2  CO 2 + H 2 O Step 2. Balance the #P Carbon #R atom #P 2 C 2 6 H 2 2 O 5 C 2 H 6 + O 2  2CO 2 + H 2 O Step 1. Total up the atoms & Re-write the equation Are we done?

18 #R atom #P 2 C 2 6 H 2 2 O 5 C 2 H 6 + O 2  2CO 2 + 3H 2 O #R atom #P 2 C 2 6 H 6 2 O 7 Step 3. Carbons are balanced now but Hydrogen isn’t. So, balance Hydrogen atoms next Carbon and Hydrogen are now balanced but oxygen isn’t. Multiply #P Hydrogen by 3 Step 4. Re-write the eqn. & Retotal the number of atoms x3

19 C 2 H 6 + 3½ O 2  2CO 2 + 3H 2 O #R atom #P 2 C 2 6 H 6 2 O 7 Step 5. To balance Oxygen multiply O by 3½ Step 6. Re-write the eqn. & Retotal the number of atoms It looks like we’re balanced. But, are we? No! We can’t have 3 ½ Oxygen molecules! Only whole Numbers are allowed. So what do we need to do to fix this? 3 ½ x

20 C 2 H 6 + 3½ O 2  2CO 2 + 3H 2 O #R atom #P 4 C 4 12 H O 14 Step 7. Let’s clean this up by Multiply everything by 2 Step 8. Retotal #R and the #P atoms 2C 2 H O 2  4CO 2 + 6H 2 O Are we balanced? YES! x 2

21 Try this problem NH 4 OH + FeCl 3  Fe(OH) 3 + NH 4 Cl #R atom #P 1 NH OH 3 1 Fe 1 3 Cl 1 Start here. Recognize we Have polyatomics but they Appear on both sides of the Equation. OK … Now finish it up

22 Answer to previous problem 3NH 4 OH + FeCl 3  Fe(OH) 3 + 3NH 4 Cl