2 Improved formatting edit approved Feb 16 '16 at 14:45 JM97 1,81311 gold badge2222 silver badges5353 bronze badges This isn't a textbook, but there is a way to balance pretty much any chemical equation algebraically. Consider the reaction C4H10 + O2 --> CO2 + H2O$$\ce{C4H10 + O2-> CO2 + H2O}$$ To start, you would put a variable in front of each molecule here. That would become: aC4H10+ bO2 --> cCO2 + dH2O$$\ce{aC4H10 + bO2 -> cCO2 + dH2O}$$ Now, count the amount of the same elements as they appear in the different molecules. In this example, there are 4 carbons in the 'a' molecule, and 1 carbon in the 'c' molecule. I would thus write 4a=1c Repeating this process for the other elements, I get: 10a=2d and 2b=2c+d (because oxygen appears in both molecules on the right side). After this step, you have your three equations. 4a=c 5a=d (This is simplified from 10a=2d) 2b=2c+d Now, choose one of those variables, and make it equal 1. It can be any of your choosing, the end outcome will be the same. In this case, making 'a' 1 will make the whole process much easier, so let's do that. a=1. From this, it follows that c=4, d=5, and b= (8+5)/2 or 13/2. Now if there are any fractions, multiply every single number by the denominator to make everything a whole number. Multiplying by 2 in this case yields: a=2 b=13 c=8 and d=10 Now put those values exactly where you originally put the letters. It becomes (2)C4H10 + (13)O2 --> (8)CO2 + (10)H2O$$\ce{ (2)C4H10 + (13)O2 -> (8)CO2 + (10)H2O}$$ So the steps are: Assign variables to each molecule in the reaction. Write out those variables as equations corresponding to the amount of certain elements in each molecule. Make one variable equal 1. Solve for the rest of the variables. Multiply by the denominators of any fractions to leave whole numbers. Those numbers are your answers. You just plug them in :) If you want any clarification, or another example done, just comment and I'll be happy to. Note: If someone could edit my post, or comment how to use the molecular writing thing, that would be awesome. This isn't a textbook, but there is a way to balance pretty much any chemical equation algebraically. Consider the reaction C4H10 + O2 --> CO2 + H2O To start, you would put a variable in front of each molecule here. That would become: aC4H10+ bO2 --> cCO2 + dH2O Now, count the amount of the same elements as they appear in the different molecules. In this example, there are 4 carbons in the 'a' molecule, and 1 carbon in the 'c' molecule. I would thus write 4a=1c Repeating this process for the other elements, I get: 10a=2d and 2b=2c+d (because oxygen appears in both molecules on the right side). After this step, you have your three equations. 4a=c 5a=d (This is simplified from 10a=2d) 2b=2c+d Now, choose one of those variables, and make it equal 1. It can be any of your choosing, the end outcome will be the same. In this case, making 'a' 1 will make the whole process much easier, so let's do that. a=1. From this, it follows that c=4, d=5, and b= (8+5)/2 or 13/2. Now if there are any fractions, multiply every single number by the denominator to make everything a whole number. Multiplying by 2 in this case yields: a=2 b=13 c=8 and d=10 Now put those values exactly where you originally put the letters. It becomes (2)C4H10 + (13)O2 --> (8)CO2 + (10)H2O So the steps are: Assign variables to each molecule in the reaction. Write out those variables as equations corresponding to the amount of certain elements in each molecule. Make one variable equal 1. Solve for the rest of the variables. Multiply by the denominators of any fractions to leave whole numbers. Those numbers are your answers. You just plug them in :) If you want any clarification, or another example done, just comment and I'll be happy to. Note: If someone could edit my post, or comment how to use the molecular writing thing, that would be awesome. This isn't a textbook, but there is a way to balance pretty much any chemical equation algebraically. Consider the reaction $$\ce{C4H10 + O2-> CO2 + H2O}$$ To start, you would put a variable in front of each molecule here. That would become: $$\ce{aC4H10 + bO2 -> cCO2 + dH2O}$$ Now, count the amount of the same elements as they appear in the different molecules. In this example, there are 4 carbons in the 'a' molecule, and 1 carbon in the 'c' molecule. I would thus write 4a=1c Repeating this process for the other elements, I get: 10a=2d and 2b=2c+d (because oxygen appears in both molecules on the right side). After this step, you have your three equations. 4a=c 5a=d (This is simplified from 10a=2d) 2b=2c+d Now, choose one of those variables, and make it equal 1. It can be any of your choosing, the end outcome will be the same. In this case, making 'a' 1 will make the whole process much easier, so let's do that. a=1. From this, it follows that c=4, d=5, and b= (8+5)/2 or 13/2. Now if there are any fractions, multiply every single number by the denominator to make everything a whole number. Multiplying by 2 in this case yields: a=2 b=13 c=8 and d=10 Now put those values exactly where you originally put the letters. It becomes $$\ce{ (2)C4H10 + (13)O2 -> (8)CO2 + (10)H2O}$$ So the steps are: Assign variables to each molecule in the reaction. Write out those variables as equations corresponding to the amount of certain elements in each molecule. Make one variable equal 1. Solve for the rest of the variables. Multiply by the denominators of any fractions to leave whole numbers. Those numbers are your answers. You just plug them in :) If you want any clarification, or another example done, just comment and I'll be happy to. Note: If someone could edit my post, or comment how to use the molecular writing thing, that would be awesome. 1 answered Feb 16 '16 at 13:05 N A 39122 silver badges1212 bronze badges This isn't a textbook, but there is a way to balance pretty much any chemical equation algebraically. Consider the reaction C4H10 + O2 --> CO2 + H2O To start, you would put a variable in front of each molecule here. That would become: aC4H10+ bO2 --> cCO2 + dH2O Now, count the amount of the same elements as they appear in the different molecules. In this example, there are 4 carbons in the 'a' molecule, and 1 carbon in the 'c' molecule. I would thus write 4a=1c Repeating this process for the other elements, I get: 10a=2d and 2b=2c+d (because oxygen appears in both molecules on the right side). After this step, you have your three equations. 4a=c 5a=d (This is simplified from 10a=2d) 2b=2c+d Now, choose one of those variables, and make it equal 1. It can be any of your choosing, the end outcome will be the same. In this case, making 'a' 1 will make the whole process much easier, so let's do that. a=1. From this, it follows that c=4, d=5, and b= (8+5)/2 or 13/2. Now if there are any fractions, multiply every single number by the denominator to make everything a whole number. Multiplying by 2 in this case yields: a=2 b=13 c=8 and d=10 Now put those values exactly where you originally put the letters. It becomes (2)C4H10 + (13)O2 --> (8)CO2 + (10)H2O So the steps are: Assign variables to each molecule in the reaction. Write out those variables as equations corresponding to the amount of certain elements in each molecule. Make one variable equal 1. Solve for the rest of the variables. Multiply by the denominators of any fractions to leave whole numbers. Those numbers are your answers. You just plug them in :) If you want any clarification, or another example done, just comment and I'll be happy to. Note: If someone could edit my post, or comment how to use the molecular writing thing, that would be awesome.