Guide to Stoich: Glucose and Oxygen
By Madeline Hermogenes, 7th pd.
What is Stoichiometry?
Stoichiometry: is a relationship between relative qualities of substances taking part in a reaction or making of a compound, typically in a ratio of whole integers.
Below are each components to Stoich (Stoichiometry), I will explain on how to find each of the following by using Glucose and Oxygen as my example...
- Balancing Equations
- Mole to Mole Conversions
- Mass to Mass Conversions
- Limiting and Excess Reactants
- Theoretical Yield
- Percent Yield
Importance of Balancing Equations
Importance of Balancing Equations
When you are writing your equation, you must balance the equation. It needs to follow the Law of Conservation of Mass and when you do Mole to Mole or Mass to Mass conversions, they serve necessary so that the correct amount of reactants can be determined and the amount of products can be predicted.
We are reacting Glucose and Oxygen, which yields to make Carbon Dioxide and Water. How did I know the product of Glucose and Oxygen? This type of reaction is combustion, it's where you take a Hydrocarbon burned in excess Oxygen. The product is always carbon dioxide and water.
How to Balance?
When you balance the equation, you want to have the same amount of each element balance with it's corresponding element, we do this by adding coefficients.
Example: Balancing Glucose and Oxygen
Equation of Glucose and Oxygen before balancing:
C6H12O6 + 02 --> C02 + H20
***Tip: Balance combustion reactions in this order, CHO/ Carbon, Hydrogen, Oxygen
- Balance the Equation if an element is not balanced. This case, all elements are not balanced on both sides
- Start balancing Carbon first...
Carbon-6 | Carbon-1
Multiply each number to get the lowest common product
Carbon-6x1 | Carbon-1x6 = Carbon-6 | Carbon-6
3. Start balancing Hydrogen... check if balancing carbon effected Hydrogen's number; did not affect.
Hydrogen-12 | Hydrogen-2
Multiply each number to get the lowest common product
Hydrogen-12x1 | Hydrogen-2x6 = Hydrogen-12 | Hydrogen-12
4. Balance Oxygen... see if balancing other elements effected Oxygen's numbers; affected oxygen.
Oxygen-6 Oxygen-2 | Oxygen-12 Oxygen-6
***Note: Oxygen is separate from itself, find the lowest product of one Oxygen on both sides; two of the oxygens are already balanced...
Oxygen-6 Oxygen-2x6 | Oxygen-12x1 Oxygen-6 = Oxygen-18 | Oxygen 18
Answer:
6 Carbons | 6 Carbons
18 Oxygen | 18 Oxygen
12 Hydrogen | 12 Hyrdogen
C6H1206 (s) + 602 (g) --> 6C02 (g) + 6H20 (l)
D-glucose + Molecular Oxygen --> Carbon Dioxide + Water/Oxidane
Glucose: 180.156g/mol
Oxygen: 31.998g/mol
Water: 18.015g/mol
Carbon Dioxide: 44.008 g/mol
Mole to Mole Converstion
How to do Mole to Mole Conversion...
-Glucose and Oxygen-
3 Steps (Below)
Answer: 73.14 Moles of Water (H20)
Step 1: Write out Equation
| Step 2: Set up conversion
| Step 3: Solve Solve: Multiple the Top (12.19 x 6) and divide by the bottom (1) |
Step 1: Write out Equation
- Write out Equation *Make sure you balance it, and label the states of matter*
- Identify what type of reaction it is
Step 2: Set up conversion
- Top left-corner, put your given (Ex: 12.19 moles of Glucose)
- Want to get out of the Given (Glucose), bottom right-corner: use the coefficient of your given (1 mole); mol. ratio
- Top right-corner, what it's asking for (Moles of Water)
- Put the coefficient of Water (6 mole); mol. ratio
Mass to Mass Conversions
Mass to Mass Converstions
How to do Mass to Mass Conversions...
-Glucose and Oxygen-
Step 1: Write out Equation
| Step 2: Set up conversion
| Step 3: SolveYou multiply the Top... 12.1x6x18.015= 1307.889gCO2 and Divide it by the bottom... 1307.889/180.15 = 17.7gCO2 |
Step 1: Write out Equation
- Write out Equation *Make sure you balance it, and label the states of matter*
- Identify what type of reaction it is
Step 2: Set up conversion
- Top left-corner, put your given... 12.1 grams of Glucose
- Diagonal to Given: use the molar mass of Glucose, found on the periodic table and calculated, 180.156 g/mol C6H12O6, to get out of grams of the Glucose
- Above molar mass of Glucose, the molar mass of the Glucose always = 1 mol. C6H12O6
- Diagonal to 1 mol. of Glucose, you want to get out of mol. of Glucose/Given, and become a mol. of what the question is asking for... CO2 To do that, you must use coefficients, or mol. ratio of what you are asked for... 1 mol. CO2
- Above 1 mol. CO2, what you are asked for, CO2, is the second part of the mol. ratio of what you are asked for 6 mol. CO2
- Diagonal is 6 mol. CO2, can equal 1 mol. CO2
- Above 1 mol. CO2, equals the molar mass of 1 mol. of an element 44.009g/mol CO2
Limiting and Excess Reactant
Limiting and Excess Reactant
Limiting and Excess reactants, your limiting reactant is the reactant that will run out first in the Equation. While your Excess is the reactant you will still have left over. To solve for Limiting and Excess, you start with your given, 12.3gGlucose. And you two Mass to Mass conversions to find the common product, which is H20.
Find Mass to Mass Coversion of first GivenDo the exact same thing in Mass to Mass Conversion, except use the numbers corresponding with the elements used... Start with your Given of 12.3gGlucose,and diagonal to it should be it's molar mass, 180.156g/mol. Above should always be 1 mole of the given, and diagonal to that should be one of the mol. ratios or coefficient of your given. The number above that should be 2nd part of the mol. ratio, or coefficient of what your being asked for, 6 mol.H20. Diagoal to that should be 1 mol. of H20, and above that should be the molar mass of H20, 18.015g/mol H20. | Find Mass to Mass Coversion on Second GivenDo the exact same thing in Mass to Mass Conversion, except use the numbers corresponding with the elements used... Start with your Given of 12.3gC02,and diagonal to it should be it's molar mass, 44.009g/mol. Above should always be 1 mole of the given, and diagonal to that should be one of the mol. ratios or coefficient of your given, 6 mol. The number above that should be 2nd part of the mol. ratio, or coefficient of what your being asked for, 6 mol.CO2. Diagoal to that should be 1 mol. of CO2, and above that should be the molar mass of CO2, 18.015g/molH20. | SolveMultiply the Top, and divide by the bottom... Answers: 7.38gH20-excess: 7.38gGlucose 5.03gH20-limiting: 12.3gCO2 |
Find Mass to Mass Coversion of first Given
Do the exact same thing in Mass to Mass Conversion, except use the numbers corresponding with the elements used...
Start with your Given of 12.3gGlucose,and diagonal to it should be it's molar mass, 180.156g/mol. Above should always be 1 mole of the given, and diagonal to that should be one of the mol. ratios or coefficient of your given. The number above that should be 2nd part of the mol. ratio, or coefficient of what your being asked for, 6 mol.H20. Diagoal to that should be 1 mol. of H20, and above that should be the molar mass of H20, 18.015g/mol H20.
Find Mass to Mass Coversion on Second Given
Do the exact same thing in Mass to Mass Conversion, except use the numbers corresponding with the elements used...
Start with your Given of 12.3gC02,and diagonal to it should be it's molar mass, 44.009g/mol. Above should always be 1 mole of the given, and diagonal to that should be one of the mol. ratios or coefficient of your given, 6 mol. The number above that should be 2nd part of the mol. ratio, or coefficient of what your being asked for, 6 mol.CO2. Diagoal to that should be 1 mol. of CO2, and above that should be the molar mass of CO2, 18.015g/molH20.
Theoretical Yield
Theoretical Yield is product of the limiting reactant, the smallest amount you got when calculating excess and limiting.
Answer: 5.03gH20
Percent Yield
Percent Yield
Percent yield is used to see if your reactant came out with want it was suppose to be, this case, we are to expect 5.04gH20. But, there are factors where you might get more or less than 100% H20. It could be more limiting reactant, another substances is still there, or maybe you spilt your product.
To find percent yield...
You take the Actual Yield: 2.98gH20, and divide it by the Theoretical: 5.03gH20
And than multiply that number by 100.
Answer: 59.24%
It ended up being less than 100%, that means there wasn't enough Water than there should be. That means an action caused a change in outcome, like spilling.