To calculate the number of moles of solute in a solution, you need to know the molarity of the solution and the volume of solution. The molarity tells you the number of moles of solute per liter of solution. Once you know the molarity, you can calculate the moles of solute by multiplying the molarity by the volume in liters.
For example, if you have 250 mL of a 0.5 M solution, the molarity is 0.5 mol/L. To convert 250 mL to liters, divide by 1000:
250 mL / 1000 = 0.25 L
So for a 0.5 M solution with a volume of 0.25 L, the number of moles of solute is:
Moles of solute = Molarity x Volume in liters
Moles of solute = 0.5 mol/L x 0.25 L = 0.125 moles
So for 250 mL of a 0.5 M solution, there are 0.125 moles of solute.
Step-By-Step Method
Here is a step-by-step method for calculating the moles of solute in a solution:
1. Identify the molarity (M) of the solution.
2. Convert the volume in mL to liters by dividing by 1000.
3. Use the formula:
Moles of solute = Molarity x Volume in liters
4. Plug in the values and calculate.
Let’s go through an example:
* Molarity = 0.75 M
* Volume = 325 mL
1. Molarity given as 0.75 M
2. Convert volume:
325 mL / 1000 = 0.325 L
3. Use the formula:
Moles of solute = Molarity x Volume in liters
Moles of solute = 0.75 mol/L x 0.325 L
4. Calculate:
Moles of solute = 0.244 mol
So for a 0.75 M solution with a volume of 325 mL, there are 0.244 moles of solute.
Molarity Calculations
To find the molarity of a solution, you need to know the moles of solute and the liters of solution.
* Molarity = Moles of solute / Liters of solution
For example, if you dissolve 5.5 moles of a solute into 2.5 liters of solution, the molarity is:
* Moles of solute = 5.5 mol
* Liters of solution = 2.5 L
* Molarity = Moles / Liters
= 5.5 mol / 2.5 L
= 2.2 M
So the molarity of a solution containing 5.5 moles of solute dissolved in 2.5 liters of solution is 2.2 M.
Let’s go through another example:
* Moles of solute = 1.65 mol
* Liters of solution = 0.425 L
* Molarity = Moles / Liters
= 1.65 mol / 0.425 L
= 3.9 M
For 1.65 moles of solute dissolved in 0.425 liters of solution, the molarity is 3.9 M.
Dilution Calculations
Dilution is when you add more solvent to a solution to decrease the concentration. The number of moles of solute stays constant, but the volume increases, so the molarity decreases.
Dilution calculations can be done using the formula:
M1V1 = M2V2
* M1 = initial molarity
* V1 = initial volume
* M2 = diluted molarity
* V2 = diluted volume
Let’s go through an example:
* M1 = 2 M
* V1 = 150 mL
* M2 = ? (diluted molarity)
* V2 = 250 mL
Plug into the formula:
(2 M)(150 mL) = (M2)(250 mL)
300 mL = 250M2
M2 = 300/250 = 1.2 M
So diluting 150 mL of a 2 M solution to a final volume of 250 mL gives a diluted molarity of 1.2 M.
The dilution formula is very useful for preparing solutions of specific molarity by diluting a stock solution.
Impact of Temperature and Pressure
Temperature and pressure can affect molarity calculations:
Temperature
– As temperature increases, the volume of the solution increases due to thermal expansion. This decreases the molarity since the moles remain constant.
– Conversely, decreasing the temperature can increase the molarity as the volume decreases.
Pressure
– Increasing the pressure will decrease the volume, thereby increasing the molarity.
– Decreasing the pressure allows the volume to expand, lowering the molarity.
To account for temperature and pressure, you can use the combined gas law:
PV = nRT
Where:
P = pressure
V = volume
n = moles
R = ideal gas constant
T = temperature (in Kelvin)
This allows you to calculate the new volume at the different temperature and pressure, then use that adjusted volume in the molarity calculation. Proper lab technique requires monitoring temperature and pressure when preparing and measuring solutions.
Using Molarity
Some key uses of molarity in chemistry include:
– Following recipes for chemical reactions: Many reactions require specific amounts of reactants measured in moles. Molarity allows convenient measurement of reactants.
– Standardizing solutions: Molarity is used to prepare solutions like acids or bases to an exact concentration. These standardized solutions are useful reagents for many experiments.
– Titrations: Molarity is used in acid-base titrations and redox titrations to determine unknown concentrations. By measuring volumes in a titration, the moles and molarity can be calculated.
– Calculating solution stoichiometry: Using molarity, the amount of reactants and products in solution stoichiometry can be determined starting from the balanced chemical equation.
– Determining solute properties: Colligative properties like vapor pressure lowering and osmotic pressure are directly related to molarity. It allows calculation of these solution property changes.
– Electrochemistry: Molarity appears in the Nernst equation to calculate electrode potentials. It provides insight into the energetics of redox reactions.
Molarity is one of the most common units used by chemists to describe concentration. It has widespread applications spanning many areas of chemistry both in the laboratory and in theory. Expertise using molarity is essential for all chemists.
Common Molarity Problems
Here are some common molarity problems and how to solve them:
Problem 1: Calculating Moles and Molarity
If you are given the mass of a compound and asked to find the molarity of a solution containing it, follow these steps:
1. Convert mass of compound to moles using molar mass
2. For solids, assume 100% dissolution. For liquids, use given density and volume to get moles.
3. Divide moles of compound by liters of solution to get molarity.
Example:
* Mass of NaOH (s) = 4.00 g
* Volume of solution = 250 mL
1. Convert NaOH mass to moles:
4.00 g NaOH x (1 mol NaOH / 40.00 g NaOH) = 0.100 mol NaOH
2. Assume 100% dissolution of the solid
3. Convert volume to liters:
250 mL / 1000 mL/L = 0.250 L
Molarity = 0.100 mol NaOH / 0.250 L = 0.400 M
Problem 2: Calculating Dilution
If you need to prepare a diluted solution from a concentrated stock solution, use the dilution formula:
M1V1 = M2V2
Where:
M1 = molarity of stock solution
V1 = volume of stock solution to use
M2 = desired molarity of diluted solution
V2 = desired final volume of diluted solution
Rearrange and solve for the needed variable.
Example:
* Stock solution is 6.0 M HCl
* Need 150 mL of 2.5 M HCl
Plug into dilution formula:
(6.0 M)(V1) = (2.5 M)(150 mL)
V1 = 62.5 mL of stock solution
Use 62.5 mL of 6.0 M HCl and dilute to 150 mL final volume to make 2.5 M HCl.
Problem 3: Limiting Reagent and Reaction Stoichiometry
If you need to find the limiting reagent, reaction yield, or remaining moles when mixing two solutions:
1. Determine moles of each reactant using molarity and volume.
2. Compare reactant moles to the mole ratio in the balanced equation.
3. The reactant with fewer moles is limiting.
4. Use stoichiometry and the limiting reactant to find remaining moles and yield.
Example:
* Mix 100 mL of 0.75 M CuSO4 with 200 mL of 1.5 M NaOH
* Reaction: CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4
1. Moles CuSO4 = (0.75 M)(0.100 L) = 0.075 mol
Moles NaOH = (1.5 M)(0.200 L) = 0.30 mol
2. CuSO4 : NaOH mole ratio is 1:2 from the equation.
3. CuSO4 is limiting since it has fewer moles.
4. Use 1:2 ratio:
Limiting CuSO4 would react with 0.075 mol x 2 = 0.15 mol NaOH
NaOH has 0.30 mol, so 0.30 – 0.15 = 0.15 mol NaOH left.
Yield of Cu(OH)2 is also 0.075 mol.
Practice Problems
Here are some practice problems to test your understanding of molarity calculations:
1) What is the molarity of 245 mL of a 0.55 M NaBr solution?
2) If 36.5 g KCl is dissolved in 425 mL of solution, what is the molarity?
3) What volume of a 2.0 M HCl solution can be diluted with water to make 150 mL of 0.25 M HCl?
4) How many moles of solute are in 125 mL of a 3.4 M solution?
5) If you mix 50.0 mL of 1.5 M FeCl3 with 75.0 mL of 0.50 M NaOH, which is the limiting reagent?
Solutions to Practice Problems
Here are the solutions:
1) Molarity = 0.55 M (Molarity does not change on dilution or mixing)
2) Moles KCl = 36.5 g x (1 mol/74.55 g) = 0.49 mol
Molarity = 0.49 mol/0.425 L = 1.1 M
3) (2.0 M)(V1) = (0.25 M)(150 mL)
V1 = 15 mL of 2.0 M HCl
4) Moles = (3.4 M)(0.125 L) = 0.425 mol
5) Moles FeCl3 = (0.050 L)(1.5 M) = 0.075 mol
Moles NaOH = (0.075 L)(0.50 M) = 0.0375 mol
NaOH is limiting since it has fewer moles.
Practice Makes Perfect
Calculating molarity and performing dilution and stoichiometry calculations takes practice. Work through many examples until the concepts become familiar. Some key tips:
– Memorize molar mass of common compounds
– Convert all volumes to liters
– Understand dilution formula and limiting reactant logic
– Pay attention to mole ratios from balanced equations
– Check your work and units throughout the problem
– Review missed problems and learn from your mistakes
With sufficient practice, molarity and related calculations will become second-nature. It is a fundamental quantitative skill all chemists need to master. Don’t get discouraged, eventually the many repetitions will yield mastery.
Conclusion
In summary, here are the key steps to calculate moles of solute from molarity and volume:
1. Identify the molarity of the solution.
2. Convert volume of solution from mL to liters.
3. Use formula: Moles = Molarity x Volume in liters.
4. Calculate to find moles of solute.
Common calculations also include:
– Determining molarity from moles and volume.
– Dilution of solutions using the dilution formula.
– Reaction stoichiometry problems with limiting reagents.
– The combined gas law to account for temperature and pressure effects.
With diligent practice, chemists develop essential expertise using molarity in all types of calculations. Molarity is the most common concentration unit in chemistry. Familiarity with moles, molarity, and stoichiometry provides a quantitative foundation for chemistry success.
