Back To CourseChemistry 101: General Chemistry
14 chapters | 131 lessons | 11 flashcard sets
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Nikki has a master's degree in teaching chemistry and has taught high school chemistry, biology and astronomy.
It's party time in the chemistry lab, and to celebrate, we're filling up balloons with different gases. But we're not just any party planners here, we're scientists, so we're being meticulous with our work. We want all of our balloons to have the exact same volume: 22.4 liters.
Before we inflated anything, we made sure the temperature was freezing (0 degrees Celsius or 273 Kelvin (K)) and the air pressure was 1 atmosphere (atm). These temperature and pressure conditions are known as Standard Temperature and Pressure or STP. At STP, one mole of any gas will fill the balloon to 22.4 liters. This quantity is known as molar volume.
We can represent molar volume as 1 mol = 22.4 L or as a ratio 22.4 L/1 mol. As long as we fill our balloons with exactly one mole of gas, they will all have the same volume.
Because molar volume holds true for all gases at STP, stoichiometric calculations involving gas are quite simple. The volume of the gas can be determined by multiplying the number of moles of gas by 22.4 L.
Imagine you have 2 moles of oxygen gas at STP and one last balloon to blow up. If you inflate the balloon with all 2 moles of gas, how big will the balloon be? To find out, multiply the number of moles of oxygen, 2, by molar volume, 22.4 L/1 mol.
(2 mol O2 x 22.4 L/1 mol = 44.8 L)
Moles cancel out, and you are left with 44.8 L, the answer!
If you have a volume of gas, the number of moles of gas present can be determined using the reverse operation. In other words, multiply the volume of our gas by the reciprocal of molar volume, 1 mol/22.4 L.
Let's practice! How many moles of nitrogen are present in a 2 L sample held at STP? To begin, we take our volume, 2 L, and multiply by the reciprocal of molar volume, 1 mol/22.4 L.
(2 L x 1 mol/22.4 L = 0.089 moles)
Liters cancel out, leaving us with 0.089 moles.
Let's go back to our laboratory and create some gases. We are doing a reaction that breaks apart hydrogen peroxide into water vapor and oxygen gas. The balanced reaction for this process is:
2H2O2 (liquid) -> 2H2O (gas) + O2 (gas)
We have approximately 68.02 g of liquid hydrogen peroxide. How many liters of oxygen gas will be produced in this reaction if carried out at STP?
We'll approach solving this problem like any stoichiometry problem and follow these four basic steps:
Step 1: Balance the equation. This equation is already balanced. On to step two!
Step 2: Find the moles of the reactant. My reactant is in grams, so I must convert grams to moles. I take the given mass of hydrogen peroxide, 68.02 g, and divide by the molar mass of hydrogen peroxide, 34.01 g/ 1 mol.
68.02 g H2O2 x 1 mol H2O2/34.01 g = 2 moles H2O2
Grams cancel out. I'm left with 2 moles of hydrogen peroxide.
Step 3: Use the mole ratio between the reactant and product to find the moles of product made. Remember to look at your balanced equation in order to find the mole ratio. For this problem, the mole ratio is 2 moles H2O2 (the reactant) to 1 mole oxygen gas (the product). I'm looking to determine moles of oxygen, so I make sure that part of the ratio is the numerator.
2 moles H2O2 x 1 mole O2/2 moles H2O2 = 1 mole O2
Moles H2O2 cancel out. The reaction will produce 1 mole of oxygen gas.
Step 4: Convert moles of product to the desired unit. I'm interested in the volume of oxygen, so I multiply the number of moles of oxygen produced by molar volume.
1 mole O2 x 22.4 L/1 mol = 22.4 L O2
Moles cancelled out, and my final product was 22.4 liters of oxygen gas.
Solutions are homogenous mixtures made of a solute and a solvent. Examples of solutions include salt water, apple juice or Gatorade. One of the most important properties of a solution is its concentration, or the ratio of solute to solvent. Most of the time, the concentration of a solution is expressed using molarity, which is represented by a capital M.
Molarity (M) is the ratio of moles of solute to liters of solution. Molarity is calculated by dividing the number of moles of solute by the liters of solution.
Molarity = moles solute /liters solution
The units for molarity can be capital M or mol/L.
Let's say we have 0.5 moles of sugar dissolved in 1 liter of water. The molarity of this solution is 0.5 M. Molarity is found by dividing 0.5 moles sugar (our solute) by 1 liter of water (our solution).
0.5 M = 0.5 moles sugar/1 liter water
Sometimes we know the molarity of a solution and the volume, but we do not know the moles of solute present. To determine this, we use our molarity equation (molarity = moles solute/liters of solution), plug in the information we do know (molarity and liters of solution) and solve for the unknown (moles of solute). When solving molarity problems, it is helpful to represent the units of molarity as mol/L.
In the lab, I have 0.10 liters of a sodium chloride solution. I know that the molarity of the solution is 2.5, but I don't know how many moles of sodium chloride are present. To solve, I take my molarity equation: molarity = moles solute/liters solution. I plug in the information I do know (2.5 molarity and 0.10 L) and solve for the unknown (moles of solute). When solving, I will use mol/L for the units for molarity.
2.5 mol/L = moles solute/ 0.10 L solution
To get moles solute by itself, I multiply each side by 0.10 L. Voila! After crossing liters, I am left with 0.25 moles of solute.
Solutions are everywhere in chemistry labs, not just as cool lab decorations, but as essential ingredients to chemical reactions.
One very cool reaction happens between lead ions (Pb2+) and iodine ions (I-) when they are in solution together. They react to form a bright yellow solid called lead (II) iodide (PbI2).
Pb2+ + 2I- --> PbI2
In the lab, we want to make as much lead (II) iodide as possible. We have an unlimited amount of iodide ions, but we only have a small amount of lead ions, limiting the amount of lead (II) iodide that we can make.
We have 0.250 L of a lead solution that is 1.0 M. Using this, how many grams of lead (II) iodide can we possibly make in this reaction?
To solve, we'll approach solving this like any stoichiometry problem and follow the four basic steps:
Step 1: Balance the equation. My equation (Pb2+ + 2I- --> PbI2) is already balanced, so it's on to step 2!
Step 2: Find the moles of the reactant. To find the number of moles of lead ion present, I need to use my molarity equation, molarity = moles solute/liter solution. I know that my molarity is 1.0 and that I have 0.250 L of solution. I plug this information into my equation and solve for moles of solute.
1.0 mol/L = moles solute /0.250 L solution
moles solute = 0.250 moles
Step 3: Use the mole ratio between the reactant and product to find the moles of product made. Remember to look at the balanced equation to find the mole ratio. The mole ratio for this problem is 1 mole lead ions to 1 mole lead (II) iodide. I multiply my moles lead ions (0.250 moles) by my molar ratio (1 mole lead (II) iodide/ 1mol lead ions) to determine the number of moles of lead (II) iodide that will be produced.
0.250 moles Pb2+ x 1 mol PbI2/1 mol Pb2+ = 0.250 moles PbI2
Moles of lead ions will cancel out, and I am left with 0.250 moles PbI2.
Step 4: Convert moles of product to the desired unit. I want to know how many grams of lead (II) iodide I am producing, so I multiply my moles of lead (II) iodide (0.250 moles) by the molar mass of lead (II) iodide (461.00 g/mol). This reaction should produce 115.25 grams of lead (II) iodide.
0.250 moles PbI2 x 461.00 g/mol = 115.25 g PbI2
Molar Volume at STP means that one mole of any gas will occupy 22.4 liters of space.
When performing stoichiometry involving gas, remember that at STP, 1 mole of gas = 22.4 L, or that there is 1 mol gas/22.4 L.
Molarity is used to describe the concentration of a solution. Molarity is the ratio of moles solute to liters of solution. The units for molarity are M or mol/L. When performing stoichiometry involving solutions, remember that molarity = moles solute/liters solution.
Approach stoichiometry problems for gases or solutions with the same four steps:
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Back To CourseChemistry 101: General Chemistry
14 chapters | 131 lessons | 11 flashcard sets