Calculating Percent by Mass/volume: Name - Chem Worksheet 15-2 ... - Free Printable

Problem: Calculating Percent by Mass and Volume

The worksheet provides problems related to calculating percent by mass and percent by volume of solutions. Below, I will solve each problem step-by-step using the provided equations and concepts.

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#### Useful Equations
1. Percent by Mass:
\[
\text{Percent by Mass} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100
\]
where:
\[
\text{mass of solution} = \text{mass of solute} + \text{mass of solvent}
\]

2. Percent by Volume:
\[
\text{Percent by Volume} = \left( \frac{\text{volume of solute}}{\text{volume of solution}} \right) \times 100
\]
where:
\[
\text{volume of solution} = \text{volume of solute} + \text{volume of solvent}
\]

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Problem 1: Percent by Mass of Iron(II) Sulfate

Question: What is the percent by mass of 5.0 g of iron(II) sulfate dissolved in 75.0 g of water?

Solution:
1. Identify the mass of solute and solvent:
- Mass of solute (iron(II) sulfate) = 5.0 g
- Mass of solvent (water) = 75.0 g

2. Calculate the mass of the solution:
\[
\text{Mass of solution} = \text{Mass of solute} + \text{Mass of solvent} = 5.0 \, \text{g} + 75.0 \, \text{g} = 80.0 \, \text{g}
\]

3. Use the percent by mass formula:
\[
\text{Percent by Mass} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 = \left( \frac{5.0 \, \text{g}}{80.0 \, \text{g}} \right) \times 100
\]

4. Perform the calculation:
\[
\text{Percent by Mass} = \left( \frac{5.0}{80.0} \right) \times 100 = 0.0625 \times 100 = 6.25\%
\]

Answer:
\[
\boxed{6.25\%}
\]

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Problem 2: Percent by Volume of Benzene

Question: A solution is made by adding 25 mL of benzene to 80 mL of toluene. What is the percent by volume of benzene?

Solution:
1. Identify the volume of solute and solvent:
- Volume of solute (benzene) = 25 mL
- Volume of solvent (toluene) = 80 mL

2. Calculate the volume of the solution:
\[
\text{Volume of solution} = \text{Volume of solute} + \text{Volume of solvent} = 25 \, \text{mL} + 80 \, \text{mL} = 105 \, \text{mL}
\]

3. Use the percent by volume formula:
\[
\text{Percent by Volume} = \left( \frac{\text{Volume of solute}}{\text{Volume of solution}} \right) \times 100 = \left( \frac{25 \, \text{mL}}{105 \, \text{mL}} \right) \times 100
\]

4. Perform the calculation:
\[
\text{Percent by Volume} = \left( \frac{25}{105} \right) \times 100 \approx 0.2381 \times 100 \approx 23.81\%
\]

Answer:
\[
\boxed{23.81\%}
\]

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Problem 3: Percent by Mass of Ammonium Nitrate

Question: A solution is formed by adding 35 g of ammonium nitrate to 250 g of water. What is the percent by mass of ammonium nitrate?

Solution:
1. Identify the mass of solute and solvent:
- Mass of solute (ammonium nitrate) = 35 g
- Mass of solvent (water) = 250 g

2. Calculate the mass of the solution:
\[
\text{Mass of solution} = \text{Mass of solute} + \text{Mass of solvent} = 35 \, \text{g} + 250 \, \text{g} = 285 \, \text{g}
\]

3. Use the percent by mass formula:
\[
\text{Percent by Mass} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 = \left( \frac{35 \, \text{g}}{285 \, \text{g}} \right) \times 100
\]

4. Perform the calculation:
\[
\text{Percent by Mass} = \left( \frac{35}{285} \right) \times 100 \approx 0.1228 \times 100 \approx 12.28\%
\]

Answer:
\[
\boxed{12.28\%}
\]

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Problem 4: Percent by Volume of Isopropanol

Question: What is the percent by volume of a solution formed by mixing 25 mL of isopropanol with 45 mL of water?

Solution:
1. Identify the volume of solute and solvent:
- Volume of solute (isopropanol) = 25 mL
- Volume of solvent (water) = 45 mL

2. Calculate the volume of the solution:
\[
\text{Volume of solution} = \text{Volume of solute} + \text{Volume of solvent} = 25 \, \text{mL} + 45 \, \text{mL} = 70 \, \text{mL}
\]

3. Use the percent by volume formula:
\[
\text{Percent by Volume} = \left( \frac{\text{Volume of solute}}{\text{Volume of solution}} \right) \times 100 = \left( \frac{25 \, \text{mL}}{70 \, \text{mL}} \right) \times 100
\]

4. Perform the calculation:
\[
\text{Percent by Volume} = \left( \frac{25}{70} \right) \times 100 \approx 0.3571 \times 100 \approx 35.71\%
\]

Answer:
\[
\boxed{35.71\%}
\]

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Problem 5: Mass Percent of Each Component in a Mixture

Question: What is the mass percent of each component in the mixture formed by adding 12 g of calcium sulfate, 18 g of sodium nitrate, and 25 g of potassium chloride to 500 g of water?

Solution:
1. Identify the masses of all components:
- Mass of calcium sulfate = 12 g
- Mass of sodium nitrate = 18 g
- Mass of potassium chloride = 25 g
- Mass of water = 500 g

2. Calculate the total mass of the solution:
\[
\text{Total mass of solution} = 12 \, \text{g} + 18 \, \text{g} + 25 \, \text{g} + 500 \, \text{g} = 555 \, \text{g}
\]

3. Calculate the mass percent of each component:
- Calcium sulfate:
\[
\text{Mass percent of CaSO}_4 = \left( \frac{12 \, \text{g}}{555 \, \text{g}} \right) \times 100 \approx 2.16\%
\]
- Sodium nitrate:
\[
\text{Mass percent of NaNO}_3 = \left( \frac{18 \, \text{g}}{555 \, \text{g}} \right) \times 100 \approx 3.24\%
\]
- Potassium chloride:
\[
\text{Mass percent of KCl} = \left( \frac{25 \, \text{g}}{555 \, \text{g}} \right) \times 100 \approx 4.50\%
\]
- Water:
\[
\text{Mass percent of H}_2\text{O} = \left( \frac{500 \, \text{g}}{555 \, \text{g}} \right) \times 100 \approx 90.09\%
\]

Answer:
\[
\boxed{2.16\%, 3.24\%, 4.50\%, 90.09\%}
\]

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Problem 6: Percent by Mass of Sodium Chloride

Question: A solution is made by dissolving 125 g of sodium chloride in 1.5 kg of water. What is the percent by mass?

Solution:
1. Convert the mass of water to grams:
\[
1.5 \, \text{kg} = 1500 \, \text{g}
\]

2. Identify the mass of solute and solvent:
- Mass of solute (sodium chloride) = 125 g
- Mass of solvent (water) = 1500 g

3. Calculate the mass of the solution:
\[
\text{Mass of solution} = \text{Mass of solute} + \text{Mass of solvent} = 125 \, \text{g} + 1500 \, \text{g} = 1625 \, \text{g}
\]

4. Use the percent by mass formula:
\[
\text{Percent by Mass} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 = \left( \frac{125 \, \text{g}}{1625 \, \text{g}} \right) \times 100
\]

5. Perform the calculation:
\[
\text{Percent by Mass} = \left( \frac{125}{1625} \right) \times 100 \approx 0.0769 \times 100 \approx 7.69\%
\]

Answer:
\[
\boxed{7.69\%}
\]

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Problem 7: Percent by Volume of Acetone

Question: What is the percent by volume of a solution formed by adding 15 L of acetone to 28 L of water?

Solution:
1. Identify the volume of solute and solvent:
- Volume of solute (acetone) = 15 L
- Volume of solvent (water) = 28 L

2. Calculate the volume of the solution:
\[
\text{Volume of solution} = \text{Volume of solute} + \text{Volume of solvent} = 15 \, \text{L} + 28 \, \text{L} = 43 \, \text{L}
\]

3. Use the percent by volume formula:
\[
\text{Percent by Volume} = \left( \frac{\text{Volume of solute}}{\text{Volume of solution}} \right) \times 100 = \left( \frac{15 \, \text{L}}{43 \, \text{L}} \right) \times 100
\]

4. Perform the calculation:
\[
\text{Percent by Volume} = \left( \frac{15}{43} \right) \times 100 \approx 0.3488 \times 100 \approx 34.88\%
\]

Answer:
\[
\boxed{34.88\%}
\]

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Problem 8: Volume of Methyl Alcohol

Question: An experiment requires a solution that is 80% methyl alcohol by volume. What volume of methyl alcohol should be added to 200 mL of water to make this solution?

Solution:
1. Let the volume of methyl alcohol added be \( V \) mL.
2. The total volume of the solution will be:
\[
\text{Total volume} = V + 200 \, \text{mL}
\]

3. According to the problem, the solution is 80% methyl alcohol by volume:
\[
\text{Percent by Volume} = \left( \frac{\text{Volume of methyl alcohol}}{\text{Total volume of solution}} \right) \times 100 = 80\%
\]

4. Substitute the known values into the equation:
\[
80 = \left( \frac{V}{V + 200} \right) \times 100
\]

5. Simplify the equation:
\[
0.8 = \frac{V}{V + 200}
\]

6. Solve for \( V \):
\[
0.8(V + 200) = V
\]
\[
0.8V + 160 = V
\]
\[
160 = V - 0.8V
\]
\[
160 = 0.2V
\]
\[
V = \frac{160}{0.2} = 800 \, \text{mL}
\]

Answer:
\[
\boxed{800 \, \text{mL}}
\]

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Final Answers

1. \(\boxed{6.25\%}\)
2. \(\boxed{23.81\%}\)
3. \(\boxed{12.28\%}\)
4. \(\boxed{35.71\%}\)
5. \(\boxed{2.16\%, 3.24\%, 4.50\%, 90.09\%}\)
6. \(\boxed{7.69\%}\)
7. \(\boxed{34.88\%}\)
8. \(\boxed{800 \, \text{mL}}\)