Guide to Stoich: Student edition
Bella Gonzalez
What is stoichiometry? Why is it useful?
The relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers.
Knowing Stoichiometry can help you find the composition of molecules and proportions of substances to combine them and cause them to react completely.
What do we do first?Say you have a given reaction such as; Calcium Chloride reacts with oxygen. This equation would have to be solved and balanced as shown below:
The equation is a synthesis equation. When balancing, both elements on each side must be equal to each other. | Converting to MolesAfter we have our balanced equation it is necessary to find the moles of certain elements or compounds. Say you have 2.12g of CaCl2 and you are trying to find the mole for O2. You start with 2.12g of CaCl2 and find the coefficient (used in the beginning to balance the equation) to cancel each other out. Then using the coefficient O2 (3), you multiply it by the number of grams (2.12) and divide by the coefficient of CaCl2 (1).
The answer comes out to be 6.63 moles of O2 | Converting to MassYour given is 12.1g of CaCl2 and you must find the mass of Ca(ClO3)2. We start by finding the molar mass of CaCl2 which is 110.984g. Next we multiply by the mole of CaCl2 (1), multiply again by the coefficient of Ca(ClO3)2 (1), and finally multiply once again by the molar mass of Ca(ClO3)2 (171.525).
Next we divide by the molar mass of CaCl2 (110.984), the coefficient CaCl2 (1), the mole of Ca(ClO3)2 (1).
The mass for Ca(ClO3)2 is 18.70g. |
What do we do first?
Say you have a given reaction such as; Calcium Chloride reacts with oxygen. This equation would have to be solved and balanced as shown below:
- CaCl2+ O2 --> Ca(ClO3)2
- [1]CaCl2+[3]O2 --> [1]Ca(ClO3)2
The equation is a synthesis equation. When balancing, both elements on each side must be equal to each other.
Converting to Moles
After we have our balanced equation it is necessary to find the moles of certain elements or compounds. Say you have 2.12g of CaCl2 and you are trying to find the mole for O2. You start with 2.12g of CaCl2 and find the coefficient (used in the beginning to balance the equation) to cancel each other out. Then using the coefficient O2 (3), you multiply it by the number of grams (2.12) and divide by the coefficient of CaCl2 (1).
- (2.12 x 3) / 1 = 6.63
The answer comes out to be 6.63 moles of O2
Converting to Mass
Your given is 12.1g of CaCl2 and you must find the mass of Ca(ClO3)2. We start by finding the molar mass of CaCl2 which is 110.984g. Next we multiply by the mole of CaCl2 (1), multiply again by the coefficient of Ca(ClO3)2 (1), and finally multiply once again by the molar mass of Ca(ClO3)2 (171.525).
- 12.1 x 1 x 1 x 171.525= 2075.45
Next we divide by the molar mass of CaCl2 (110.984), the coefficient CaCl2 (1), the mole of Ca(ClO3)2 (1).
- 2075.45/ 110.984 / 1 / 1 = 18.70
The mass for Ca(ClO3)2 is 18.70g.
Limiting and Excess ReactantsYou are given two problems to solve,and you're trying to figure out how to find amount of the element/compound you have to make the final product.
Starting with 12.3g of CaCl2 we multiply by the mole of CaCl2 (1), the coefficient Ca(ClO3)2 (1), and the molar mass Ca(ClO3)2 (171.525).
Divide 2109.75 by The molar mass of CaCl2 (110.984), the co efficient of CaCl2 (1), and the mole of Ca(ClO3)2.
The total number of grams for Ca(ClO3)2 is 19.00g. Repeat the same process when solving for number
The answer for the second problem is 21.97g of O2. The limiting reactant is 19.00g because that is the only amount that Ca(ClO3)2 can make. The excess reactant is 21.97g of O2 because there is more than needed, wanted, or asked for. | What is the theoretical yield? The theoretical yield is the number of a product that can be made over and over again in theory of course. Unless something were to happen to one of the reactants or products. To get the theoretical yield you would use find the smallest number in the final results shown before. In this case it would be 19.00g Ca(ClO3)2. In theory You can only use 19.00g Ca(ClO3)2, unless you were to spill some, or leave some in the testing tube, graduated cylinder, etc. | What is the percent yield?The amount of product that could possibly be produced in a given reaction, calculated according to the starting amount of the limiting reactant. To find this is very simple. The first step is to divide the actual yield by the theoretical yield. The actual yield is 21.2g (the largest number), and divide by the theoretical yield; 19.00g (the smallest number). The product should be then multiplied by 100.
The percent yield is 111.57g. |
Limiting and Excess Reactants
You are given two problems to solve,and you're trying to figure out how to find amount of the element/compound you have to make the final product.
- 12.3g CaCl2 --> __ g Ca(ClO3)2
- 12.3g O2 --> __ g Ca(ClO3)2
Starting with 12.3g of CaCl2 we multiply by the mole of CaCl2 (1), the coefficient Ca(ClO3)2 (1), and the molar mass Ca(ClO3)2 (171.525).
- 12.3 x 1 x 1 x 171.525 = 2109.75
Divide 2109.75 by The molar mass of CaCl2 (110.984), the co efficient of CaCl2 (1), and the mole of Ca(ClO3)2.
- 2109.75/ 110.984 / 1 / 1 = 19.00
The total number of grams for Ca(ClO3)2 is 19.00g.
Repeat the same process when solving for number
- Multiply Mass A, Mole of A (always 1), Coefficient of B, and molar mass of B.
- Then divide the answer from above by the molar mass of A, Coefficient of A, And mole B.
The answer for the second problem is 21.97g of O2.
The limiting reactant is 19.00g because that is the only amount that Ca(ClO3)2 can make. The excess reactant is 21.97g of O2 because there is more than needed, wanted, or asked for.
What is the theoretical yield?
What is the percent yield?
The amount of product that could possibly be produced in a given reaction, calculated according to the starting amount of the limiting reactant. To find this is very simple.
The first step is to divide the actual yield by the theoretical yield. The actual yield is 21.2g (the largest number), and divide by the theoretical yield; 19.00g (the smallest number). The product should be then multiplied by 100.
- (21.2/19.00) x 100 =111.57
The percent yield is 111.57g.