Mass or weight percent is the percentage of a given element or compound within a solution. Now, we're gonna say that mass percent can be seen as mass component which is usually the mass of our solute divided by the total mass of our solution times 100. Now, for example, if we're given 23% NaOH, this means that we have 23%, which translates to 23 grams of NaOH over 100 grams of solution. Because we're dealing with the percentage, we always assume it's out of 100 grams. Now, we can further expand on this and say that we have 23 grams of NaOH still on top and remember a solution is made up of solute plus solvent. So, that'd be 23 grams still of NaOH plus 100 minus 23, which gives us 77 grams of our solvent. Within our calculations, it's important that you are able to see that mass percent can be broken like this, broken down into this setup and it can further be expanded where we look at both the solute and solvent components individually. Keep this in mind as we approach more questions dealing with mass percent. Now, look at example 1. Take a look at it. Once you do attempt it on your own, if you get stuck, move on to the next video and see how I approach example 1.
- 1. The Chemical World9m
- 2. Measurement and Problem Solving2h 25m
- 3. Matter and Energy2h 15m
- Classification of Matter18m
- States of Matter8m
- Physical & Chemical Changes19m
- Chemical Properties8m
- Physical Properties5m
- Temperature (Simplified)9m
- Law of Conservation of Mass5m
- Nature of Energy5m
- First Law of Thermodynamics7m
- Endothermic & Exothermic Reactions7m
- Heat Capacity16m
- Thermal Equilibrium (Simplified)8m
- Intensive vs. Extensive Properties13m
- 4. Atoms and Elements2h 33m
- The Atom (Simplified)9m
- Subatomic Particles (Simplified)12m
- Isotopes17m
- Ions (Simplified)22m
- Atomic Mass (Simplified)17m
- Periodic Table: Element Symbols6m
- Periodic Table: Classifications11m
- Periodic Table: Group Names8m
- Periodic Table: Representative Elements & Transition Metals7m
- Periodic Table: Phases (Simplified)8m
- Periodic Table: Main Group Element Charges12m
- Atomic Theory9m
- Rutherford Gold Foil Experiment9m
- 5. Molecules and Compounds1h 50m
- Law of Definite Proportions9m
- Periodic Table: Elemental Forms (Simplified)6m
- Naming Monoatomic Cations6m
- Naming Monoatomic Anions5m
- Polyatomic Ions25m
- Naming Ionic Compounds11m
- Writing Formula Units of Ionic Compounds7m
- Naming Acids18m
- Naming Binary Molecular Compounds6m
- Molecular Models4m
- Calculating Molar Mass9m
- 6. Chemical Composition1h 23m
- 7. Chemical Reactions1h 43m
- 8. Quantities in Chemical Reactions1h 16m
- 9. Electrons in Atoms and the Periodic Table2h 32m
- Wavelength and Frequency (Simplified)5m
- Electromagnetic Spectrum (Simplified)11m
- Bohr Model (Simplified)9m
- Emission Spectrum (Simplified)3m
- Electronic Structure4m
- Electronic Structure: Shells5m
- Electronic Structure: Subshells4m
- Electronic Structure: Orbitals11m
- Electronic Structure: Electron Spin3m
- Electronic Structure: Number of Electrons4m
- The Electron Configuration (Simplified)20m
- The Electron Configuration: Condensed4m
- Ions and the Octet Rule9m
- Valence Electrons of Elements (Simplified)5m
- Periodic Trend: Metallic Character4m
- Periodic Trend: Atomic Radius (Simplified)7m
- Periodic Trend: Ionization Energy (Simplified)9m
- Periodic Trend: Electron Affinity (Simplified)7m
- Electron Arrangements5m
- The Electron Configuration: Exceptions (Simplified)12m
- 10. Chemical Bonding2h 10m
- Lewis Dot Symbols (Simplified)7m
- Ionic Bonding6m
- Covalent Bonds6m
- Lewis Dot Structures: Neutral Compounds (Simplified)8m
- Bonding Preferences6m
- Multiple Bonds4m
- Lewis Dot Structures: Multiple Bonds10m
- Lewis Dot Structures: Ions (Simplified)8m
- Lewis Dot Structures: Exceptions (Simplified)12m
- Resonance Structures (Simplified)5m
- Valence Shell Electron Pair Repulsion Theory (Simplified)4m
- Electron Geometry (Simplified)7m
- Molecular Geometry (Simplified)9m
- Bond Angles (Simplified)11m
- Dipole Moment (Simplified)14m
- Molecular Polarity (Simplified)7m
- 11 Gases2h 15m
- 12. Liquids, Solids, and Intermolecular Forces1h 11m
- 13. Solutions3h 1m
- 14. Acids and Bases2h 14m
- 15. Chemical Equilibrium1h 27m
- 16. Oxidation and Reduction1h 33m
- 17. Radioactivity and Nuclear Chemistry53m
Solutions: Mass Percent: Study with Video Lessons, Practice Problems & Examples
Mass percent is the ratio of the mass of a solute to the total mass of a solution, expressed as a percentage. For example, a 23% NaOH solution contains 23 grams of NaOH in 100 grams of solution, with 77 grams being the solvent. Understanding this concept is crucial for calculations involving solutions, as it allows for the breakdown of components into solute and solvent, facilitating further analysis in chemistry, including stoichiometry and concentration calculations.
Mass Percent can be examined as the percentage of solute found within a given amount of solution.
Mass Percent in Solutions
Solutions: Mass Percent Concept 1
Video transcript
Solutions: Mass Percent Example 1
Video transcript
So here it states calculate the amount of water in kilograms that must be added to 12 grams of urea in the preparation of an 18.3% by mass solution. Here the molar mass of urea is given as 60.055 grams per mole. Alright, so we need to determine what the kilograms of water are. And we're dealing here with mass percent. So mass percent here, would equal the grams of our solute. So mass of or grams of solute divided by grams of solution times 100. Pulling in what we know. Well, we know that the percentage is 18.3%, the mass of our solute is this urea here and it's still down here as well. We don't know what the mass of our water is so that's going to be our x and this is getting multiplied by 100. Alright. So what we're going to do here is we're going to realize that we need to cross-multiply, well, we need to multiply both sides here by \(12.0 + x\). So this cancels out with this. So now we have \(18.3 \times (12 + x)\) here in parenthesis. This 12 and this 100 are multiplying each other to give us 1200. Next, what we're gonna do is we can solve this in 2 ways. Either we can divide out the 18.3 right now or we can distribute it. It's up to you which method you want to use. So here when I distribute it, I get \(219.6 + 18.3x = 1200\). Subtract out 219 here. And then I'm gonna get here, \(18.3x = 980.4\). Divide both sides here by 18.3. \(x = \) the mass of our water which comes out to 53.6 grams of \(H_2O\). But we don't want the answer at the end of being grams, we want it in kilograms, so we do one more conversion. For every 1 kilogram, it's equal to 1,000 grams. So, that gives me 0.0536 kilograms of water. So, just remember in this question we're talking about mass percent so use that formula to isolate our variable which in this case is grams of water. Once we do that, convert it to kilograms to get our final answer. Now that you've seen example 1, attempt to do the next example on your own, come back and see if your answer matches up with mine.
Solutions: Mass Percent Example 2
Video transcript
A solution was prepared by dissolving 51 grams of KBr, Potassium Bromide, in 310 ml of water. Calculate the mass percent of potassium bromide in the solution. Alright, so that's going to be mass percent equals the grams of our solute which is KBr divided by grams of solution. Here our solution is made up of our solute KBr and our solvent water. So, we need grams of KBr plus grams of water and this gets multiplied by 100. We already know the grams of KBr so we can immediately plug those in. So, that's 51 grams of KBr here on top and here on the bottom. Now, we need to determine the grams of water, but instead we're given milliliters. Remember that water is roughly about 1 in terms of its density. Water is 1 gram per 1 milliliter. We use that information to help us get the grams of water. Now, this is a piece of information you just have to remember. The density of water is around roughly 1. So that's 310 grams that we have of water. So plug that in and this gets multiplied by 100 and it gives us, at the end, 14.1%. So, this question wasn't that bad, you just had to remember the density of water is 1 and use that to find its grams. Once you find all the components, plug it into the formula to get our final answer.
Here’s what students ask on this topic:
What is mass percent in chemistry?
Mass percent, also known as weight percent, is a way of expressing the concentration of a component in a mixture or solution. It is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100. The formula is:
For example, a 23% NaOH solution means there are 23 grams of NaOH in 100 grams of the solution.
How do you calculate mass percent of a solution?
To calculate the mass percent of a solution, use the formula:
For example, if you have 10 grams of NaCl dissolved in 90 grams of water, the total mass of the solution is 100 grams. The mass percent is:
What is the difference between mass percent and volume percent?
Mass percent and volume percent are both ways to express concentration, but they differ in what they measure. Mass percent is the ratio of the mass of the solute to the total mass of the solution, multiplied by 100. Volume percent, on the other hand, is the ratio of the volume of the solute to the total volume of the solution, multiplied by 100. Mass percent is useful for solid and liquid solutes, while volume percent is often used for liquid-liquid solutions.
How do you convert mass percent to molarity?
To convert mass percent to molarity, follow these steps:
- Calculate the mass of the solute in grams.
- Convert the mass of the solute to moles using its molar mass.
- Determine the volume of the solution in liters.
- Use the formula:
For example, for a 10% NaCl solution with a density of 1.1 g/mL, calculate the moles of NaCl and divide by the volume in liters to find the molarity.
Why is mass percent important in chemistry?
Mass percent is important in chemistry because it provides a straightforward way to express the concentration of a solute in a solution. It is particularly useful in stoichiometry, where precise measurements of reactants and products are crucial. Mass percent allows chemists to easily calculate the amounts of substances needed for reactions, analyze the composition of mixtures, and prepare solutions with specific concentrations. It also helps in quality control and standardization in industrial processes.
Your Introduction to Chemistry tutor
- Use the following table for problems 9.25 to 9.28: A solution containing 80. g of KCl in 200. g of H₂O at 50...
- Use the following table for problems 9.25 to 9.28: A solution containing 80. g of NaNO₃ in 75 g of H₂O at 50...
- How is volume/volume percent concentration defined and for what types of solutions is it typically used?
- In a laboratory experiment, a 10.0-mL sample of NaCl solution is poured into an evaporating dish with a mass o...