Crash Course Video on solutions
Another crash course video on solutions
A Stoichiometry Problem using Molarity.
In the chemical equation above, the numbers in red are called the coefficients and represent the number of moles needed of each molecule to have a balanced equation. These numbers also establish the ratio of moles that are used and produced.
The equation is the recipe which describes the relationship between the reactants on the left side of the arrow to the products on the right side of the arrow in terms of moles.
If I am told to use 125 mL of 0.5 M Hydrogen Peroxide (H2O2) with an excess of the rest of the reactants, I can calculate how many grams of oxygen are produced (O2).
The first equation that we use is for Molarity (M).
Molarity (M) = moles/liter
We know the Molarity of hydrogen peroxide is 0.5M.
We also know that volume we are using is 125 mL. However, this needs to be converted to liters.
(125mL/1)(1L/1000mL) = 0.125 L
Now we are ready to use the Molarity equation:
0.5 M = moles/0.125 mL
(0.5M)(0.125mL) = moles of H2O2
0.0625 = moles of H2O2 However, 0.5 M only has 1 significant figure so we must round...
0.6 moles of H2O2 are being used. Now we can go back to the chemical equation and use the ratio of moles to determine how many moles of oxygen are produced.
 |
| Picture 1 |
As shown in the picture on the left, for every 5 moles of hydrogen peroxide reacted, 5 moles of oxygen are produced. (Click on the picture to enlarge the image.)
 |
| Picture 2 |
Picture 2 shows the calculation of the moles of hydrogen peroxide used is 0.06 moles. The next step shows that relationship of moles of hydrogen peroxide to moles of oxygen is 5/5 which is equal to one. Therefore, 0.06 moles of oxygen are going to be produced.
Also shown in this picture, is the calculation of mass of oxygen in one mole of oxygen. 32.0 g of oxygen equals 1 mole of oxygen. We can use this information to determine how many grams of oxygen are in 0.06 moles of oxygen.
 |
| Picture 3 |
Picture 3 shows how to convert moles of oxygen to grams. Remember, since we are multiplying, we must use the least number of significant figures. In this case, in the number 0.06 moles of oxygen there is only one significant figure. Therefore, our answer must have only 1 significant figure.
But what if we are given the concentration of substance in molality instead?
Molality is the number of moles of solute per kilogram of solvent.
Molality (m) = moles ÷ kg of solvent
No comments:
Post a Comment