Balancing Chemical Equations
Balancing Chemical Equations
The most important point to remember when dealing with chemical equations is that the amount of an element on one side of the equation must equal the amount of that element on the other side of the equation. Thus, the equation is said to be, “balanced,” in compliance with the Law of Conservation of Mass. This scientific principle states that matter can neither be created nor destroyed in a chemical reaction.
Example: See if you can identify whether the following equation is balanced, and therefore in accordance with the Law of Conservation of Mass.
H2 + O2 → H2O
There are two hydrogen atoms on both sides of the equation (H2), however, the equation is not balanced, because the number of oxygen atoms on the left side of the arrow does not equal the number of oxygen atoms on the right side of the arrow. It is useful to make a chart listing the number of each atoms in a chemical equation so you can keep track of what needs to be balanced.
Example:
Start Finish
H 2 2
O 2 1
One essential rule in balancing equations is that you can never change the subscripts, because that would result in a change of the chemical composition of a molecule. This would create an entirely different reaction and create something completely different.
Subscripts are the little numbers that come after an atom, such as in the 2 in H2O.
Only the coefficients may be manipulated.
Coefficients are the large numbers in front of the molecules that tell how many of each molecule are present. When there are no coefficients in front of a molecule, it is implied that there is only one of that molecule as in the example we’re using now.
Example:
H2 + O2 → 2H2O
Placing a coefficient of “2” in front of “H2O” is a step in the right direction, because this denotes that there are now 2 oxygen atoms on the right side of the equation, balancing with the 2 oxygen atoms on left side of the equation. Update your table.
Start Finish
H 2 4
O 2 2
As you can see, there are now an equal amount of oxygen atoms, but now too many hydrogen atoms on the right. The coefficient of “2” applies to all of the atoms in the molecule it is placed in front of. Two H2’s produces four hydrogen atoms. You multiply the H’s subscript by the coefficient of the whole molecule to achieve the total number of H atoms.
Example:
coefficient x subscript
2H2O: 2 x 2 = 4
You may also place coefficients on the left side of an equation. To match the number of H atoms on the right, place a 2 in front of the H2 on the left.
Example:
2H2 + O2 → 2H2O
The same coefficient x subscript rule applies.
Example:
coefficient x subscript
2H2: 2 x 2 = 4
Add up the total number of atoms on both sides. Consult your table.
Start Finish
H 4 4
O 2 2
The equation is now balanced. The number of atoms on one side equals the number of atoms on the opposite side. Many complex chemical equations are harder to balance, but don’t give up. It’s like solving a puzzle; It takes practice. Oftentimes the coefficients you assign initially will be the wrong amount or require readjusting, but through trial and error you will achieve the right result.
A more complex example demonstrates this point.
Example:
Cu(OH)2 + HNO3 → Cu(NO3)2 + H2O
Diatomic ions are positively or negatively charged compounds made up of more than 1
element, in this case, (OH) and (NO3), hydroxide and nitrate respectively.
Example:
Start Finish
Cu 1 1
O 5 7
H 3 2
N 1 2
While the Cu atoms are easy to count, the rest are much trickier. It is important to note that the 2 beside the OH in Cu(OH)2 counts for both the oxygen and the hydrogen atom, thus meaning there are two of each. The same applies for the NO3 in the Cu(NO3)2 molecule on the right side of the equation. There are three oxygen atoms times the 2 subscript outside the parentheses, yielding a total of 6. Also, there are two total nitrogen atoms in this molecule, due to the 2 subscript outside of the parentheses.
When balancing equations, it is usually best to balance the elements that are present in the fewest molecules, and then balance the elements that are present in more molecules. Since Cu and N only occur once on each side of the equation, it is wise to balance them first. By putting a coefficient of 2 in front of 2HNO3, there are then 2 N atoms on both sides of the equation.
Example:
Cu(OH)2 + 2HNO3 → Cu(NO3)2 + H2O
Start Finish
Cu 1 1
O 8 7
H 4 2
N 2 2
All that needs balancing is the amount of H and O atoms. This can simply be done by adding a 2 coefficient to H2O on the right.
Example:
Cu(OH)2 + 2HNO3 → Cu(NO3)2 + 2H2O
Start Finish
Cu 1 1
O 8 8
H 4 4
N 2 2
Try to balance the following equation:
Example:
NH4NO3 → N2O + H2O
Answer:
Your first observation should be that there is an imbalance of hydrogen and nitrogen.
Start Finish
O 3 2
H 4 2
N 2 2
By placing a 2 coefficient in front of the H20 molecule on the right side of the equation, both hydrogen and oxygen attain balanced levels, and mass conservation is satisfied.
Example:
NH4NO3 → N2O + 2H2O
The hardest types of balancing equations are those such as the following combustion in which a fractional coefficient is required for balancing. The following describes a hydrocarbon and oxygen gas reacting to form carbon dioxide and water vapor. The (g) means that they are in a gaseous state.
Example:
C2H6 (g) + O2 (g) → CO2 (g) + H2O (g)
Start Finish
C 2 1
H 6 2
O 2 3
In any reaction, it is advisable to save molecules such as N2 or H2 or in this case, O2, for last, first tackling the molecules including C and H. By balancing CO2 with a coefficient of 2 and H2O with a coefficient of 3, the number of H and C atoms are now in equilibrium.
Example:
C2H6 (g) + O2 (g) → 2CO2 (g) + 3H2O (g)
Start Finish
C 2 2
H 6 6
O 2 7
It is at this point in the process of balancing the equation that a fraction is necessary in order to make the O atoms on the left side equal the O atoms on the right side. This is because there is an odd number on the right and an even number on the left. A whole number coefficient can not be used.
Multiplying the left O2 by 7/2 , which is equal to 3.5, will allow the O2 on the left to total 7.
However, the equation is not yet balanced. Fractional coefficients of molecules are not acceptable.
The final step is multiplying all molecules by the denominator of the fraction to make every coefficient a whole number. This means multiplying them by 2.
2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g)
The equation is at last, balanced. However, there is room for one more error in certain instances, even after all the balancing has been completed. An important thing to remember is that all coefficients in a balanced equation should be the smallest whole numbers they can be. Observe how all the coefficients can be simplified dividing by 2 in the following example.
Example:
4NaN3 → 4Na + 6N2
2NaN3 → 2Na + 3N2
You should now have a sufficient understanding of how to balance most chemical equations. A few additional general steps that may be of use are as follows:
Consider the first element listed in the first molecule in the equation.
If the element is mentioned in two or more formulas on the same side of the arrow, skip it until after the other elements are balanced.
If this element is mentioned in one formula on each side of the arrow, balance it by placing coefficients in front of one or both of these formulas.
Moving from left to right, repeat the process for each element.
When you place a number in front of a formula that contains an element you tried to balance previously, recheck that element and put its atoms back in balance.
Continue this process until the number of atoms of each element is balanced.
Often, an element can be balanced by using the subscript for this element on the left side of the arrow as the coefficient in front of the formula containing this element on the right side of the arrow and vice versa (using the subscript of this element on the right side of the arrow as the coefficient in front of the formula containing this element on the left side).
Sometimes it’s easiest to first balance pure nonmetallic elements (H2, O2, N2, F2, Cl2, Br2, I2, S8, Se8, and P4 ) with a fractional coefficient (1/2, 3/2, 5/2, etc.). If a fraction is used, you eliminate it later by multiplying each coefficient in the equation by the denominator as previously stated.
If polyatomic ions don’t change in the reaction, and therefore appear in the same form on both sides of the chemical equation, they can be balanced as if they were single atoms.
Finally, if you find an element difficult to balance, leave it and come back to it.
The most important point to remember when dealing with chemical equations is that the amount of an element on one side of the equation must equal the amount of that element on the other side of the equation. Thus, the equation is said to be, “balanced,” in compliance with the Law of Conservation of Mass. This scientific principle states that matter can neither be created nor destroyed in a chemical reaction.
Example: See if you can identify whether the following equation is balanced, and therefore in accordance with the Law of Conservation of Mass.
H2 + O2 → H2O
There are two hydrogen atoms on both sides of the equation (H2), however, the equation is not balanced, because the number of oxygen atoms on the left side of the arrow does not equal the number of oxygen atoms on the right side of the arrow. It is useful to make a chart listing the number of each atoms in a chemical equation so you can keep track of what needs to be balanced.
Example:
Start Finish
H 2 2
O 2 1
One essential rule in balancing equations is that you can never change the subscripts, because that would result in a change of the chemical composition of a molecule. This would create an entirely different reaction and create something completely different.
Subscripts are the little numbers that come after an atom, such as in the 2 in H2O.
Only the coefficients may be manipulated.
Coefficients are the large numbers in front of the molecules that tell how many of each molecule are present. When there are no coefficients in front of a molecule, it is implied that there is only one of that molecule as in the example we’re using now.
Example:
H2 + O2 → 2H2O
Placing a coefficient of “2” in front of “H2O” is a step in the right direction, because this denotes that there are now 2 oxygen atoms on the right side of the equation, balancing with the 2 oxygen atoms on left side of the equation. Update your table.
Start Finish
H 2 4
O 2 2
As you can see, there are now an equal amount of oxygen atoms, but now too many hydrogen atoms on the right. The coefficient of “2” applies to all of the atoms in the molecule it is placed in front of. Two H2’s produces four hydrogen atoms. You multiply the H’s subscript by the coefficient of the whole molecule to achieve the total number of H atoms.
Example:
coefficient x subscript
2H2O: 2 x 2 = 4
You may also place coefficients on the left side of an equation. To match the number of H atoms on the right, place a 2 in front of the H2 on the left.
Example:
2H2 + O2 → 2H2O
The same coefficient x subscript rule applies.
Example:
coefficient x subscript
2H2: 2 x 2 = 4
Add up the total number of atoms on both sides. Consult your table.
Start Finish
H 4 4
O 2 2
The equation is now balanced. The number of atoms on one side equals the number of atoms on the opposite side. Many complex chemical equations are harder to balance, but don’t give up. It’s like solving a puzzle; It takes practice. Oftentimes the coefficients you assign initially will be the wrong amount or require readjusting, but through trial and error you will achieve the right result.
A more complex example demonstrates this point.
Example:
Cu(OH)2 + HNO3 → Cu(NO3)2 + H2O
Diatomic ions are positively or negatively charged compounds made up of more than 1
element, in this case, (OH) and (NO3), hydroxide and nitrate respectively.
Example:
Start Finish
Cu 1 1
O 5 7
H 3 2
N 1 2
While the Cu atoms are easy to count, the rest are much trickier. It is important to note that the 2 beside the OH in Cu(OH)2 counts for both the oxygen and the hydrogen atom, thus meaning there are two of each. The same applies for the NO3 in the Cu(NO3)2 molecule on the right side of the equation. There are three oxygen atoms times the 2 subscript outside the parentheses, yielding a total of 6. Also, there are two total nitrogen atoms in this molecule, due to the 2 subscript outside of the parentheses.
When balancing equations, it is usually best to balance the elements that are present in the fewest molecules, and then balance the elements that are present in more molecules. Since Cu and N only occur once on each side of the equation, it is wise to balance them first. By putting a coefficient of 2 in front of 2HNO3, there are then 2 N atoms on both sides of the equation.
Example:
Cu(OH)2 + 2HNO3 → Cu(NO3)2 + H2O
Start Finish
Cu 1 1
O 8 7
H 4 2
N 2 2
All that needs balancing is the amount of H and O atoms. This can simply be done by adding a 2 coefficient to H2O on the right.
Example:
Cu(OH)2 + 2HNO3 → Cu(NO3)2 + 2H2O
Start Finish
Cu 1 1
O 8 8
H 4 4
N 2 2
Try to balance the following equation:
Example:
NH4NO3 → N2O + H2O
Answer:
Your first observation should be that there is an imbalance of hydrogen and nitrogen.
Start Finish
O 3 2
H 4 2
N 2 2
By placing a 2 coefficient in front of the H20 molecule on the right side of the equation, both hydrogen and oxygen attain balanced levels, and mass conservation is satisfied.
Example:
NH4NO3 → N2O + 2H2O
The hardest types of balancing equations are those such as the following combustion in which a fractional coefficient is required for balancing. The following describes a hydrocarbon and oxygen gas reacting to form carbon dioxide and water vapor. The (g) means that they are in a gaseous state.
Example:
C2H6 (g) + O2 (g) → CO2 (g) + H2O (g)
Start Finish
C 2 1
H 6 2
O 2 3
In any reaction, it is advisable to save molecules such as N2 or H2 or in this case, O2, for last, first tackling the molecules including C and H. By balancing CO2 with a coefficient of 2 and H2O with a coefficient of 3, the number of H and C atoms are now in equilibrium.
Example:
C2H6 (g) + O2 (g) → 2CO2 (g) + 3H2O (g)
Start Finish
C 2 2
H 6 6
O 2 7
It is at this point in the process of balancing the equation that a fraction is necessary in order to make the O atoms on the left side equal the O atoms on the right side. This is because there is an odd number on the right and an even number on the left. A whole number coefficient can not be used.
Multiplying the left O2 by 7/2 , which is equal to 3.5, will allow the O2 on the left to total 7.
However, the equation is not yet balanced. Fractional coefficients of molecules are not acceptable.
The final step is multiplying all molecules by the denominator of the fraction to make every coefficient a whole number. This means multiplying them by 2.
2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g)
The equation is at last, balanced. However, there is room for one more error in certain instances, even after all the balancing has been completed. An important thing to remember is that all coefficients in a balanced equation should be the smallest whole numbers they can be. Observe how all the coefficients can be simplified dividing by 2 in the following example.
Example:
4NaN3 → 4Na + 6N2
2NaN3 → 2Na + 3N2
You should now have a sufficient understanding of how to balance most chemical equations. A few additional general steps that may be of use are as follows:
Consider the first element listed in the first molecule in the equation.
If the element is mentioned in two or more formulas on the same side of the arrow, skip it until after the other elements are balanced.
If this element is mentioned in one formula on each side of the arrow, balance it by placing coefficients in front of one or both of these formulas.
Moving from left to right, repeat the process for each element.
When you place a number in front of a formula that contains an element you tried to balance previously, recheck that element and put its atoms back in balance.
Continue this process until the number of atoms of each element is balanced.
Often, an element can be balanced by using the subscript for this element on the left side of the arrow as the coefficient in front of the formula containing this element on the right side of the arrow and vice versa (using the subscript of this element on the right side of the arrow as the coefficient in front of the formula containing this element on the left side).
Sometimes it’s easiest to first balance pure nonmetallic elements (H2, O2, N2, F2, Cl2, Br2, I2, S8, Se8, and P4 ) with a fractional coefficient (1/2, 3/2, 5/2, etc.). If a fraction is used, you eliminate it later by multiplying each coefficient in the equation by the denominator as previously stated.
If polyatomic ions don’t change in the reaction, and therefore appear in the same form on both sides of the chemical equation, they can be balanced as if they were single atoms.
Finally, if you find an element difficult to balance, leave it and come back to it.
Balancing Chemical Equations (practice) | |
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