# Basic Stoichiometry

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• What is Stoichiometry? Generally, stoichiometry involves using a balanced chemical equation to determine the amounts of products produced during a chemical reaction.
Example of stoichiometry:

1 cup flour + 3 eggs + 2 sticks butter + 4 cups milk 1 cake

• What does a chemical equation tell us? A chemical equation tells us the ratios in which the reactants combine to form products. These are mole ratios. For example, from the equation above, we need three eggs to make at least one cake. This can be mathematically written as:  3 eggs/ 1 cake which means “3 eggs per 1 cake”.
Say we only have 2 eggs. How many cakes can we form? Set up a proportion using our “mole ratio” we just wrote:
3 eggs   = 2 eggs
1 cake     x cakes

Solving for x, we find that we can make 0.66 cakes with 2 eggs. Although as any baker would know, you can’t make 0.66 cakes. But unlike baking a cake, you can form 0.66 moles of a certain product in a chemical reaction. Now we’re going to take things to the next level.

• Solving  stoichiometry problems

Many problems give background information that is good to know, but for our purposes is irrelevant. The following problem seems complex, but we’ll break it down step-by-step so that it will be a piece of cake.

Problem: In photosynthesis, plants convert carbon dioxide and water into glucose (C6H12O6) according to the following chemical equation:

6 CO2 (g) + 6H2O (l) 6O2 (g) + C6H12O6 (aq)

Suppose you determine that a particular plant consumes 37.8 g CO2 in one week. Assuming that there is more than enough water present to react with all the CO2, what mass of glucose (in grams) can the plant synthesize from the CO2?

Solution

The chemical equation is already given, which is a bonus. Whenever doing stoichiometry always write and balance the chemical equation first (if not given)

• Identify key words-usually the word “excess” means that there is no limiting reactant to deal with, greatly simplifying the work load. In this case, we are told that there is “more than enough water present” which is more or less code for excess water.

Given: mass of CO2 = 37.8 g
Unknown: mass of C6H12O6 (glucose)

• It’s usually always a good idea to immediately convert all your quantities into moles right away:
37.8 g CO2            =    0.86 moles CO2
44.0 g/mol CO2

• Look at the balanced chemical equation and use the mole ratio between CO2 and C6H12O6 just like the cake example. The mol ratio on the left is taken from the chemical equation

6 moles CO2    =    0.860 moles CO2
1 mole C6H12O6      x moles C6H12O6

X = 0.14 moles of glucose (C6H12O6)

Since we were asked to find the mass of glucose, multiply the moles of glucose produced by its molecular weight:
Moles of glucose X molecular weight of glucose = mass of glucose
0.143 moles X 180.2 g/mol ~ 25.8 grams of glucose