Abstract
Lab: Chemistry
Published: Not Published
Chemical equations represent how a chemical reaction proceeds from reactants to products through physical or chemical change using chemical formulas.
Stoichiometry is a term that describes the relative quantities of reactants and products in a chemical reaction. It is based on the Law of Conservation of Mass, which is a fundamental law that states that matter is neither created nor destroyed. Quite simply, the number and identity of reactant atoms must equal the number and identity of product atoms. Reactions rearrange atoms but do not create or destroy them. This requires that a proposed reaction must be balanced, meaning that the number of atoms for each element are equal on the reactant and product sides.
For example, in the chemical equation below, the left side (the reactants) includes one copper atom, one hydrogen atom, one nitrogen atom, and three oxygen atoms.
Cu + HNO3 → Cu(NO3)2 + 4 H2O + NO
On the right side, notice the water product has a number preceding it. This is a coefficient and represents the number of molecules in the reaction. Using this information, we can tally the number of atoms on the product side. There is one copper atom and eight hydrogen atoms (4 x 2). Tallying the nitrogen and oxygen atoms requires a bit more math. There are two nitrogen atoms in the first product and one nitrogen atom in the third product, which gives a total of three nitrogen atoms. For oxygen, there are six oxygen atoms in the first product, four oxygen atoms in the second product, and one oxygen atom in the third product for a total of 11 oxygen atoms.
If left in this form, the reaction wouldn’t be feasible because it defies the Law of Conservation of Masses. There are more hydrogen, nitrogen, and oxygen atoms on the product side. Therefore, the equation needs to be balanced.
Balancing an equation is an iterative process that requires adding coefficients to each side until the numbers become equal. There are several approaches to balance a chemical equation. One approach uses a table to visualize the numbers and a bit of trial and error.
# of atoms on the reactant side | # of atoms on the product side | ||||||
Copper | Hydrogen | Nitrogen | Oxygen | Copper | Hydrogen | Nitrogen | Oxygen |
1 | 1 | 1 | 3 | 1 | 8 | 3 | 11 |
Since there is only one hydrogen atom on the reactant side but eight hydrogen atoms on the product side, multiplying the compound containing the nitrogen on the reactant side by eight would balance hydrogen. The second row in the new table reflects this change to the number of atoms.
# of atoms on the reactant side | # of atoms on the product side | |||||||
Copper | Hydrogen | Nitrogen | Oxygen | Copper | Hydrogen | Nitrogen | Oxygen | |
1 | 1 | 1 | 3 | 1 | 8 | 3 | 11 | |
8 HNO3 | 1 | 8 | 8 | 24 | 1 | 8 | 3 | 11 |
Next, to increase the number of nitrogens on the product side, multiplying the product Cu(NO3)2 by three would raise the number of nitrogen atoms from three to seven. The number of nitrogen atoms is not balanced yet, but there are still coefficients for one reactant and one product to consider.
# of atoms on the reactant side | # of atoms on the product side | |||||||
Copper | Hydrogen | Nitrogen | Oxygen | Copper | Hydrogen | Nitrogen | Oxygen | |
1 | 1 | 1 | 3 | 1 | 8 | 3 | 11 | |
8 HNO3 | 1 | 8 | 8 | 24 | 1 | 8 | 3 | 11 |
3 Cu(NO3)2 | 1 | 8 | 8 | 24 | 3 | 8 | 7 | 23 |
If there are two molecules of NO produced, this adds one more nitrogen atom and one more oxygen atom to the product side, balancing these two species with the reactant side.
# of atoms on the reactant side | # of atoms on the product side | |||||||
Copper | Hydrogen | Nitrogen | Oxygen | Copper | Hydrogen | Nitrogen | Oxygen | |
1 | 1 | 1 | 3 | 1 | 8 | 3 | 11 | |
8 HNO3 | 1 | 8 | 8 | 24 | 1 | 8 | 3 | 11 |
3 Cu(NO3)2 | 1 | 8 | 8 | 24 | 3 | 8 | 7 | 23 |
2 NO | 1 | 8 | 8 | 24 | 3 | 8 | 8 | 24 |
The only atom left unbalanced now is copper. Increasing the number to three copper atoms on the reactant side balances the equation.
# of atomson the reactant side | # of atoms on the product side | |||||||
Copper | Hydrogen | Nitrogen | Oxygen | Copper | Hydrogen | Nitrogen | Oxygen | |
1 | 1 | 1 | 3 | 1 | 8 | 3 | 11 | |
8 HNO3 | 1 | 8 | 8 | 24 | 1 | 8 | 3 | 11 |
3 Cu(NO3)2 | 1 | 8 | 8 | 24 | 3 | 8 | 7 | 23 |
2 NO | 1 | 8 | 8 | 24 | 3 | 8 | 8 | 24 |
3 Cu | 3 | 8 | 8 | 24 | 3 | 8 | 8 | 24 |
The balanced equation is written as follows:
3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 4 H2O + 2 NO
Balancing the equation is also essential for determining the limiting reactant because the coefficient of the compounds is used to calculate how much product is produced by each reactant (product yield). From this quantity, the reactant producing the least amount of product is considered the limiting reactant--which is completely consumed in the reaction and therefore limits the total amount of product generated. This calculated quantity also represents the theoretical yield of the reaction, which is needed to calculate the percent yield.
The balanced equation is more than a simple accounting of atoms. The coefficients describe the molar relationship between products and reactants, i.e., how much product is produced by each reactant. The number of moles of reactant is used to calculate the number of moles of another product or reactant. The reactant that produces the least amount of product is considered the limiting reactant.
The limiting reactant is completely consumed in the reaction and therefore limits the total amount of product generated. Once the limiting reactant is entirely consumed, no more product will form. The possible amount of product that could be formed based on the limiting reactant is the theoretical yield of the reaction.
The actual yield is compared to the theoretical yield, resulting in the ‘percent yield’. A percent yield of 100% means that, based on the reactants used, the maximum possible amount of product was produced. Percent yields less than 100% are common and indicate that there was some product loss during the reaction. The percent yield is never greater than the theoretical yield. If this is the case, experimental or calculation errors occurred.
Explore More Videos
ABOUT JoVE
Copyright © 2024 MyJoVE Corporation. All rights reserved