# Guide to Stoich: Glucose and Oxygen

## 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...

1. Balancing Equations
2. Mole to Mole Conversions
3. Mass to Mass Conversions
4. Limiting and Excess Reactants
5. Theoretical Yield
6. 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

1. Balance the Equation if an element is not balanced. This case, all elements are not balanced on both sides
2. 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

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

Balancing Chemical Equations

## Mole to Mole Converstion

How to do Mole to Mole Conversion...

-Glucose and Oxygen-

3 Steps (Below)

Answer: 73.14 Moles of Water (H20)

Stoichiometry Problems: Moles to Moles

## Mass to Mass Converstions

How to do Mass to Mass Conversions...

-Glucose and Oxygen-

Stoichiometry Problems: Grams to Grams

## 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.

## Theoretical Yield

Theoretical Yield is product of the limiting reactant, the smallest amount you got when calculating excess and limiting.

## 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.