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Balancing equations 1.
Always check the equation by counting atoms on each side of the equation. For each element, the number of atoms on the left has to be equal to the number of atoms on the right.
2. Start by putting a "1" in front of the most complex compound (the one containing the greatest number and variety of atoms). H2SO4(aq) + Ba(OH)2(aq)
2 H2O(l) + Na2SO4(aq)
4 hydrogens, 1 sulfur, 1 barium, 6 oxygens
4 hydrogens, 1 sulfur, 1 barium, 6 oxygens
2(NH4)3PO4(aq) + 3 CaCl2
Ca3(PO4)2(s) + 6 NH4Cl(aq)
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
3. Start by putting a "1" in front of the most complex compound (the one containing the greatest number and variety of atoms). 1 C6H12O6(s) + O2(g)
CO2(g) + H2O(l)
4. Balance elements that appear in multiple reactants or multiple products last. In combustion equations, this means that the oxygen is always balanced last. ...in the equation above, balance the O last, because it appears in more than one compound on each side of the equation. Balancing C's: 1 C6H12O6(s) + O2(g)
6 CO2(g) + H2O(l)
Balancing H's: 1 C6H12O6(s) + O2(g)
6 CO2(g) + 6 H2O(l)
Balancing O's: 1 C6H12O6(s) + 6 O2(g)
6 CO2(g) + 6 H2O(l)
5. Whole number coefficients are preferred. But use fractions during the balancing process. The equation can be scaled to remove them later. 6. Be certain that the total charge is the same on each side after you have a massbalanced equation.
7. Don't touch the subscripts! Changing the subscripts in a formula results in an entirely
different chemical compound (and often one that doesn't exist). Remember that the coefficients show how much compound is involved in the reaction, while the subscripts show how much element is in the compound. Try a "1" for most 1 C3H7S(l) + O2(g) complex compound Balancing C's: Balancing S's:
CO2 + H2O(l) + SO2
1 C3H7S(l) + O2(g)
3 CO2 + H2O(l) + SO2
1 C3H7S(l) + O2(g)
3 CO2 + H2O(l) + 1 SO2
Balancing H's:
Balancing O's:
1 C3H7S(l) + O2(g)
3 CO2 + 7/2 H2O(l) + 1 SO2
2 C3H7S(l) + O2(g)
6 CO2 + 7 H2O(l) + 2 SO2
2 C3H7S(l) + 23/2 O2(g) 4 C3H7S(l) + 23 O2(g)
6 CO2 + 7 H2O(l) + 2 SO2 12 CO2 + 14 H2O(l) + 4 SO2
8. Be certain that the total charge is the same on each side after you have a massbalanced equation.
9. Don't touch the subscripts! Changing the subscripts in a formula results in an entirely
different chemical compound (and often one that doesn't exist). Remember that the coefficients show how much compound is involved in the reaction, while the subscripts show how much element is in the compound. Mass balanced but not charge balanced: Both mass and charge balanced
Zn(s) + Fe3+(aq) Zn(s) + 2 Fe3+(aq)
Zn2+(aq) + Fe2+(aq) Zn2+(aq) + 2 Fe2+(aq)
10. Don't touch the subscripts! Changing the subscripts in a formula results in an entirely
different chemical compound (and often one that doesn't exist). Remember that the coefficients show how much compound is involved in the reaction, while the subscripts show how much element is in the compound. NaOH(aq) + Al2(SO4)3(aq)
Al(OH)3(s) + Na2SO4(aq)
Wrong: Na2(OH)3(aq) + Al2(SO4)3(aq)
Al(OH)3(s) + Na2(SO4)3(aq)
Correct: 6 NaOH(aq) + Al2(SO4)3(aq)
2 Al(OH)3(s) + 3 Na2SO4(aq)
11. When you notice that the same polyatomic ions appear on both sides of the equation, it's much easier to balance the equation as though the polyatomic ions were 'atoms'. For example: 12. A missing coefficient is understood to be a "1". 2(NH4)3PO4(aq) + 3 CaCl2
Ca3(PO4)2(s) + 6 NH4Cl(aq)
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
6 ammoniums, 2 phosphates, 3 calciums, 6 chlorides
6 ammoniums, 2 phosphates, 3 calciums, 6 chlorides
13. A missing coefficient is understood to be a "1". 1 HCl(aq) + 1 NaOH 1 NaCl(aq) + 1 H2O(l) is more compactly written as HCl(aq) + NaOH NaCl(aq) + H2O(l) but both are technically correct!
14. Write the equation so that the coefficients are the smallest set of integers possible. For example, if all of the coefficients are divisible by 2, divide them all by 2. 2 H3O+(aq) + 2 OH 4 H2O(l) is more compactly written as H3O+(aq) + OH 2 H2O(l) but both are technically correct!
15. Equations that involve electron transfer (redox reactions) are sometimes difficult to
balance by inspection. When you notice that the oxidation numbers for an element aren't identical on both sides of the equation, you have a redox equation. You can sometimes balance such equations by inspection if the equation is simple (and if you're stubborn enough) but there is a special recipe for balancing redox equations that relies on balancing the changes in oxidation number. We'll show you the recipe in second semester general chemistry, when redox reactions are discussed in more detail. If you're ready for the recipe now, here's a sketchy outline of the procedure for balancing redox reactions: 1.
Assign oxidation numbers to atoms in all species.
2. Identify which reactant is being oxidized (the oxidation number increases when it reacts) and which is being reduced (the oxidation number goes down). 3. Write the reaction as two separate "half reactions": one for the oxidation, and one for the reduction. 4. Balance each separate half reaction as follows:
1. Balance missing oxygens by adding H2O to the opposite side the equation.
2. Balance missing hydrogens by adding H+ to the opposite side. 3. Balance the charge by adding electrons (remembering that oxidation involves electron loss, and reduction involves electron gain). 5. Multiply each half reaction by an integer so that the number of electrons in both half reactions is the same. 6. Add the two half reactions together.
7. If the reaction is taking place in alkaline solution, and the resulting equation contains H+, add the reaction for the dissociation of water H2O(l)
H+(aq) + OH(aq)
to the equation to eliminate hydrogen ions.
Always check the equation by counting atoms on each side of the equation. For each element, the number of atoms on the left has to be equal to the number of atoms on the right.
2. Start by putting a "1" in front of the most complex compound (the one containing the greatest number and variety of atoms). H2SO4(aq) + Ba(OH)2(aq)
2 H2O(l) + Na2SO4(aq)
4 hydrogens, 1 sulfur, 1 barium, 6 oxygens
4 hydrogens, 1 sulfur, 1 barium, 6 oxygens
2(NH4)3PO4(aq) + 3 CaCl2
Ca3(PO4)2(s) + 6 NH4Cl(aq)
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
3. Start by putting a "1" in front of the most complex compound (the one containing the greatest number and variety of atoms). 1 C6H12O6(s) + O2(g)
CO2(g) + H2O(l)
4. Balance elements that appear in multiple reactants or multiple products last. In combustion equations, this means that the oxygen is always balanced last. ...in the equation above, balance the O last, because it appears in more than one compound on each side of the equation. Balancing C's: 1 C6H12O6(s) + O2(g)
6 CO2(g) + H2O(l)
Balancing H's: 1 C6H12O6(s) + O2(g)
6 CO2(g) + 6 H2O(l)
Balancing O's: 1 C6H12O6(s) + 6 O2(g)
6 CO2(g) + 6 H2O(l)
5. Whole number coefficients are preferred. But use fractions during the balancing process. The equation can be scaled to remove them later. 6. Be certain that the total charge is the same on each side after you have a massbalanced equation.
7. Don't touch the subscripts! Changing the subscripts in a formula results in an entirely
different chemical compound (and often one that doesn't exist). Remember that the coefficients show how much compound is involved in the reaction, while the subscripts show how much element is in the compound. Try a "1" for most 1 C3H7S(l) + O2(g) complex compound Balancing C's: Balancing S's:
CO2 + H2O(l) + SO2
1 C3H7S(l) + O2(g)
3 CO2 + H2O(l) + SO2
1 C3H7S(l) + O2(g)
3 CO2 + H2O(l) + 1 SO2
Balancing H's:
Balancing O's:
1 C3H7S(l) + O2(g)
3 CO2 + 7/2 H2O(l) + 1 SO2
2 C3H7S(l) + O2(g)
6 CO2 + 7 H2O(l) + 2 SO2
2 C3H7S(l) + 23/2 O2(g) 4 C3H7S(l) + 23 O2(g)
6 CO2 + 7 H2O(l) + 2 SO2 12 CO2 + 14 H2O(l) + 4 SO2
8. Be certain that the total charge is the same on each side after you have a massbalanced equation.
9. Don't touch the subscripts! Changing the subscripts in a formula results in an entirely
different chemical compound (and often one that doesn't exist). Remember that the coefficients show how much compound is involved in the reaction, while the subscripts show how much element is in the compound. Mass balanced but not charge balanced: Both mass and charge balanced
Zn(s) + Fe3+(aq) Zn(s) + 2 Fe3+(aq)
Zn2+(aq) + Fe2+(aq) Zn2+(aq) + 2 Fe2+(aq)
10. Don't touch the subscripts! Changing the subscripts in a formula results in an entirely
different chemical compound (and often one that doesn't exist). Remember that the coefficients show how much compound is involved in the reaction, while the subscripts show how much element is in the compound. NaOH(aq) + Al2(SO4)3(aq)
Al(OH)3(s) + Na2SO4(aq)
Wrong: Na2(OH)3(aq) + Al2(SO4)3(aq)
Al(OH)3(s) + Na2(SO4)3(aq)
Correct: 6 NaOH(aq) + Al2(SO4)3(aq)
2 Al(OH)3(s) + 3 Na2SO4(aq)
11. When you notice that the same polyatomic ions appear on both sides of the equation, it's much easier to balance the equation as though the polyatomic ions were 'atoms'. For example: 12. A missing coefficient is understood to be a "1". 2(NH4)3PO4(aq) + 3 CaCl2
Ca3(PO4)2(s) + 6 NH4Cl(aq)
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
24 hydrogens, 6 nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines
6 ammoniums, 2 phosphates, 3 calciums, 6 chlorides
6 ammoniums, 2 phosphates, 3 calciums, 6 chlorides
13. A missing coefficient is understood to be a "1". 1 HCl(aq) + 1 NaOH 1 NaCl(aq) + 1 H2O(l) is more compactly written as HCl(aq) + NaOH NaCl(aq) + H2O(l) but both are technically correct!
14. Write the equation so that the coefficients are the smallest set of integers possible. For example, if all of the coefficients are divisible by 2, divide them all by 2. 2 H3O+(aq) + 2 OH 4 H2O(l) is more compactly written as H3O+(aq) + OH 2 H2O(l) but both are technically correct!
15. Equations that involve electron transfer (redox reactions) are sometimes difficult to
balance by inspection. When you notice that the oxidation numbers for an element aren't identical on both sides of the equation, you have a redox equation. You can sometimes balance such equations by inspection if the equation is simple (and if you're stubborn enough) but there is a special recipe for balancing redox equations that relies on balancing the changes in oxidation number. We'll show you the recipe in second semester general chemistry, when redox reactions are discussed in more detail. If you're ready for the recipe now, here's a sketchy outline of the procedure for balancing redox reactions: 1.
Assign oxidation numbers to atoms in all species.
2. Identify which reactant is being oxidized (the oxidation number increases when it reacts) and which is being reduced (the oxidation number goes down). 3. Write the reaction as two separate "half reactions": one for the oxidation, and one for the reduction. 4. Balance each separate half reaction as follows:
1. Balance missing oxygens by adding H2O to the opposite side the equation.
2. Balance missing hydrogens by adding H+ to the opposite side. 3. Balance the charge by adding electrons (remembering that oxidation involves electron loss, and reduction involves electron gain). 5. Multiply each half reaction by an integer so that the number of electrons in both half reactions is the same. 6. Add the two half reactions together.
7. If the reaction is taking place in alkaline solution, and the resulting equation contains H+, add the reaction for the dissociation of water H2O(l)
H+(aq) + OH(aq)
to the equation to eliminate hydrogen ions.
