You are watching: What is the same on the left and right side of a balanced equation
Even though nlinux.orgical compounds are broken up and new compounds are formed during a nlinux.orgical reaction, atoms in the reactants do not disappear, nor do new atoms appear to form the products. In nlinux.orgical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products—they are merely reorganized into different arrangements. In a complete nlinux.orgical equation, the two sides of the equation must be present on the reactant and the product sides of the equation.
Coefficients and Subscripts
There are two types of numbers that appear in nlinux.orgical equations. There are subscripts, which are part of the nlinux.orgical formulas of the reactants and products; and there are coefficients that are placed in front of the formulas to indicate how many molecules of that substance is used or produced.
Steps in Balancing a nlinux.orgical Equation
Identify the most complex substance. Beginning with that substance, choose an element(s) that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element(s) on both sides. Balance polyatomic ions (if present on both sides of the nlinux.orgical equation) as a unit. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients. Count the numbers of atoms of each kind on both sides of the equation to be sure that the nlinux.orgical equation is balanced.Example \(\PageIndex{1}\): Combustion of Heptane
Balance the nlinux.orgical equation for the combustion of Heptane (\(\ce{C_7H_{16}}\)).
\<\ce{C_7H_{16} (l) + O_2 (g) → CO_2 (g) + H_2O (g) } \nonumber\>
Solution
| 1. Identify the most complex substance. | The most complex substance is the one with the largest number of different atoms, which is \(C_7H_{16}\). We will assume initially that the final balanced nlinux.orgical equation contains 1 molecule or formula unit of this substance. |
| 2. Adjust the coefficients. | a. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side: \<\ce{C7H16 (l) + O2 (g) → } \underline{7} \ce{CO2 (g) + H2O (g) } \nonumber \> 7 carbon atoms on both reactant and product sidesb. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: \<\ce{C7H16 (l) + O2 (g) → 7 CO2 (g) + } \underline{8} \ce{H2O (g) } \nonumber \> 16 hydrogen atoms on both reactant and product sides |
| 3. Balance polyatomic ions as a unit. | There are no polyatomic ions to be considered in this reaction. |
| 4. Balance the remaining atoms. | The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side: \<\ce{C7H16 (l) + }\underline{11} \ce{ O2 (g) → 7 CO2 (g) + 8H2O (g) } \nonumber\> 22 oxygen atoms on both reactant and product sides |
| 5. Check your work. | The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a nlinux.orgical equation is balanced. |
Example \(\PageIndex{2}\): Combustion of Isooctane
Combustion of Isooctane (\(\ce{C_8H_{18}}\))
\<\ce{C8H18 (l) + O2 (g) -> CO_2 (g) + H_2O(g)} \nonumber\>
Solution
The assumption that the final balanced nlinux.orgical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. The combustion of any hydrocarbon with oxygen produces carbon dioxide and water.
| 1. Identify the most complex substance. | The most complex substance is the one with the largest number of different atoms, which is \(\ce{C8H18}\). We will assume initially that the final balanced nlinux.orgical equation contains 1 molecule or formula unit of this substance. |
| 2. Adjust the coefficients. | a. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products: \<\ce{C8H18 (l) + O2 (g) -> }\underline{8} \ce{ CO2 (g) + H2O(g)}\nonumber\> 8 carbon atoms on both reactant and product sidesb. 18 hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products: \<\ce{C8H18 (l) + O2 (g) -> 8CO2 (g) + }\underline{9} \ce{ H2O(g)} \nonumber\> 18 hydrogen atoms on both reactant and product sides |
| 3. Balance polyatomic ions as a unit. | There are no polyatomic ions to be considered in this reaction. |
| 4. Balance the remaining atoms. | The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient (\(\dfrac{25}{2}\)) to balance the oxygen atoms: \<\ce{C8H18 (l) + } \underline{ \dfrac{25}{2} } \ce{O2 (g)→ 8CO2 (g) + 9H2O(g) }\nonumber\> 25 oxygen atoms on both reactant and product sidesThe equation is now balanced, but we usually write equations with whole number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the nlinux.orgical equation by 2: \< \underline{2} \ce{C8H18(l) + } \underline{25} \ce{O2(g) ->} \underline{16} \ce{CO2(g) + } \underline{18} \ce{H2O(g)} \nonumber\> |
| 5. Check your work. | The balanced nlinux.orgical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side. Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly. |
Summary
To be useful, nlinux.orgical equations must always be balanced. Balanced nlinux.orgical equations have the same number and type of each atom on both sides of the equation. The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in nlinux.orgical reactions.See more: Why Was Your Measured Ph Level Of The Buffer Solutions Different From The Calculated Values?