The percentage concentration it is a way of expressing the ratio of the solute in one hundred parts of mixture or solution. It should be noted that these "parts" can be expressed in units of mass or volume. Thanks to this concentration, the composition of a solution is known which, unlike that of a pure compound, is not constant..
Likewise, not only its composition varies, but also its organoleptic properties. The tea jar in the image below takes on more intense flavors (and colors) as more spices are dissolved in the ice water. However, although their properties change, the concentration of these spices remains constant..
If we assume that 100 grams of these are dissolved in water and then stirred enough to homogenize the solution, the grams will be distributed throughout the jar. The percentage concentration of the tea remains constant even if the liquid content is divided into different containers.
This will vary only if more water is added to the jar, which, although it does not modify the original mass of the dissolved spices (solute), does modify its concentration. For the example of the rate of tea, this concentration can be conveniently expressed in mass of solute divided by volume of water.
Thus, it opens the way to the infinite cases where these concentration calculations play a crucial role..
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An understanding of the term "solution" is necessary before addressing the percentage expressions of its concentration..
A solution is a homogeneous or uniform mixture of two or more substances whose particles are of atomic or molecular size..
The components of this are the solute and the solvent. The solute is the material dissolved in a solution, which is found to a lesser extent. Solvent is the dispersion medium in a solution and is found in a higher proportion (like water in a tea jar).
- Percent concentration presents the convenience of avoiding calculations of molarity and other concentration units. In many cases, it is sufficient to know the amount of solute dissolved in the solution. However, for chemical reactions the molar concentration is left aside..
- Facilitates the verification of the law of conservation of mass.
- It is expressed in parts per hundred of solution, within which the solute is counted.
- The relationship between solute and solution can be expressed in units of mass (grams) or volume (milliliters).
The way to calculate it depends on the units in which you want to express it. However, the mathematical calculation is essentially the same.
% (m / m) = (grams of solute / grams of solution) ∙ 100
The weight percent of a solution indicates the number of grams of solute in each 100 grams of solution.
For example, a 10% m / m solution of NaOH contains 10 grams of NaOH per 100 grams of solution. It can also be interpreted this way: 10 g of NaOH dissolve in 90 g of water (100-10).
% (m / v) = (grams of solute / milliliters of solution) ∙ 100
Percent in milligrams is a unit of concentration often used in clinical reports to describe extremely low concentrations of the solute (for example, trace minerals in the blood)..
As a concrete case, we have the following example: the level of nitrogen in a person's blood is 32 mg%, which means that there are 32 mg of dissolved nitrogen per 100 ml of blood.
% (v / v) = (milliliters of solute / milliliters of solution) ∙ 100
The volume percent volume of a solution indicates the number of milliliters of solute in each 100 milliliters of solution.
For example, a 25% v / v solution of alcohol in water contains 25 milliliters of alcohol per 100 milliliters of solution, or what is the same: 75 mL of water dissolves 25 mL of alcohol.
If you have 7 g of KIO3, How many grams of 0.5% m / m solution can be prepared with this amount of salt?
A 0.5% m / m solution is very dilute, and is interpreted as follows: for every 100 grams of solution there are 0.5 grams of KIO3 dissolved. So, to determine the grams of this solution that can be prepared, the conversion factors are used:
7 g KIO3 ∙ (100 g Sun / 0.5 g KIO3) = 1400 g or 1.4 Kg of solution.
How is it possible? Obviously, the large amount of mass came from the water; thus, the 7 grams of KIO3 dissolved in 1393 grams of water.
If you want to prepare 500 grams of a CuSO solution4 at 1%, how many grams of cupric salt are necessary?
Conversion factors are applied to solve for g of CuSO4 desired:
500 g of Sol CuSO4 ∙ (1 g of CuSO4/ 100 g of Sol CuSO4) = 5 g of CuSO4
That is, 5 g of CuSO dissolve4 (a bright bluish-colored salt) in 495 g of water (approximately 495 mL)
If 400 mL of water, 37 grams of sugar, 18 grams of salt and 13 grams of sodium sulfate (NatwoSW4), What is the percentage concentration by mass for each of the components of the mixture?
If the density of water is assumed to be 1g / mL, then the mixture has 400 g of water available. Adding the total mass of the components of the solution we have: (400 + 37 + 18 + 13) = 468 g of solution.
Here the calculation is straightforward and simple:
% Water m / m = (400 g water / 468 g Sun) ∙ 100 = 85.47
% Sugar m / m = (37 g sugar / 468 g Sol) ∙ 100 = 7.90
% Salt m / m = (18 g salt / 468 g Sol) ∙ 100 = 3.84
% NatwoSW4 m / m = (13 g NatwoSW4/ 468 g Sun) ∙ 100 = 2.77
Adding all the individual mass percentages we have: (85.47 + 7.90 + 3.84 + 2.77) = 99.98% ≈ 100%, the total mixture.
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