13.1: Units of Concentration


Introduction

What does concentration mean?

In the sciences, there are basically two different meanings for concentration; first, how close things are to each other, and second, the composition of substances within a mixture. In General Chemistry 1 we used concepts like density and molarity to describe the first concept, and mass percent and mole fraction to describe the second.

• How dense (packed) things are
• Density ($$d=\frac{mass}{volume}$$)
• Molarity ($$M=\frac{moles}{volume}$$),

Note, the above can be used to describe a pure substance or a mixture. Molarity was very useful for identifying the number of solute particles in a solution as you could only measure the total solution (either its mass or volume), but could not directly measure its components.
• Relative composition of a mixture
• Mass Percent $$\left (Mass \; \%_{A}=100(\frac{Mass_{A}}{Mass_{Total}}) \right )$$
• Mole Fraction $$\left (X_{A}=(\frac{moles_{A}}{moles_{Total}}) \right )$$

Note, the above can be used to describe the relative composition of homogenous mixtures (solutions), but don't really tell you about how close things are to each other. For example, we used the concept of mole fraction to describe the partial pressure of a gas, where the partial pressure was the mole fraction times the total pressure, but that told you nothing about how close the molecules were to each other

It should also be noted that "percent" is simply the "fraction times 100", and so they are two different ways of stating the same relationship, and that it is a matter of convention to use percent for mass and fraction for the number of molecules, but you could also have "mass fraction" and "mole percent".

Note

When it comes to units of measurement, there are also two fundamentally different ways of mathematically describing the concentrations of things

• One Part to the Whole
• This is the only kind we have dealt with, and describes things like molarity (mole of a compound to the total volume) and fractions (mole or mass of one compound to the total moles or mass of all compounds)
• One Part to Another Part
• We will introduce several new measurements, which fit this, like molality (moles of solute to kg solvent), and g solute/100 g solvent. Each of these is representing one entity (the solute) to another entity (the solvent), which is different than relating to the whole (which is the solute and solvent).

Units of Concentration

Molarity M:

$\text{molarity (M)} =\dfrac{\text{moles solute}}{\text{liter solution}}$

Molarity is useful because by measuring the volume of a solution you can determine the number of solute particles (n=MV, where n=moles solute). Since many reactions occur in solution phases and solutes are often reactants, we used molarity in stoichiometric calculations. We will also learn in the next chapter that the rate of reactions is also dependent on their concentration, and molarity can express that.

NOTE: This is a Part-to-Whole type of calculation, that is, a measurement of the volume (the whole: as the solution is the solvent plus all solutes allows you to determine the moles of a substance (the part). n = MV, where n=moles, M=Molarity and V=Volume. So for example, when making solutions of a specific molarity, you dilute the solute to a desired volume.

Molality m:

$\text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}}$

Molality is the ratio of moles of solute to the mass of solute in Kg. This is often used in calculating how a solute affects a physical property of a solvent, like the boiling or melting point of a solvent. For example, we could determine how much salt is required to lower the freezing point of a specific amount of water by 1 degree.

NOTE: This is a Part-to-Part type calculation, that is, moles solute per kilogram solvent (not solution). So for example, when making solutions of a desired molality, you add solute to a specific amount (mass) of solvent (you do not dilute to volume).

Mole Fraction

$Mole \: Fraction_{A}=\left ( \frac{mole_{A}}{moles_{total}} \right )$
$\sum_{n=1}^{all}X_{n}=1$

Mole fraction of substance A is often denoted by XA and the sum of the mole fractions = 1. Although you can have a mass fraction, chemists usually use this for mole fractions.

NOTE: This is a Part-to-Whole type calculation, that is mole of a species divided by the total moles of all species, this is dimensionless.

Mass Percent

$Mass\%_{A}=\left ( \frac{mass_{A}}{mass_{total}} \right )100$
$\sum_{n=1}^{all}Mass\%_n=100$

Percentage is related to fraction, it is simply the fraction times 100. Chemists usually use percentage to describe mass fractions. Note, the sum of the mass percents of all substances within a sample = 100%.

Note: This is a Part-to-Whole type calculation, That is, this is 100 times the mass fraction.

Ratio

The ratio of two substances is the amount of one compared to the amount of another. If one of these is the solvent, then the ratio of solute to solven becomes an indicator of how concentrated the solute is in the solvent. This is not normally used, but may help you understand grams solute per 100 g water (see below).

Note: This is a Part-to-Part type calculation: So mass ratio of a solute in a solvent would be mass solute divided by mass solvent (not solution, as in mass fraction)

Grams Solute per 100 g H2O (S)

This is often given the symbol (S) and used in graphing properties of solvents, like how a solute concentration effects the boiling point (temperature) of water. It tells you how concentrated a solute is in terms of g solute per 100 g water. This is simply the ratio times 100, and has units g solute/100 g solvent. Note: a rule-of-thumb, is a soluble substance has a solubility greater than 0.1g/100 g solvent, and an insoluble substance is less, but of course, that is all relevant.

$S=100(mass\; ratio)=\frac{mass\; solute}{mass\; solvent}(100)$

If you dissolve Xg solute into Yg solvent, S = (X/Y)100g

Note: This is a Part-to-Part type calculation. Simply speaking it is the mass ration times 100.

Parts-per notation

These are often called "pseudo-units" and are dimensionless quantities used to describe very dilute solute concentrations. The meaning can be ambiguous, but they are often used in engineering.

Parts per million

Often expressed as g solute/106g solvent or "medium", where the medium is a sample like dirt or body mass. Note, mass is not a measure of the parts like a number of entities, but this is the convention)

$ppm=\frac{g\; solute}{10^{6}g\; solvent}=\frac{g\; solute}{10^{6}g\; medium}$
for aqueous solutions we note that the density of water is 1g/ml, so the above equation can be converted to
$ppm=\frac{mg_{solute}}{L_{ water}}$

Exercise $$\PageIndex{1}$$

Show how to convert g solute/106g solvent = mg solute/L water (when the solvent is water)

$$\displaystyle ppm=\left (\frac{g \; solute}{10 ^6 \; g \;H_2 O} \right )\left ( \frac{10^3 \;g \; H_2 O}{L} \right )\left ( \frac{10^3mg \; solute}{g} \right )=\frac{mg \; solute}{L \; H_2 O}$$

Exercise $$\PageIndex{2}$$

Calculate the concentration of selenium in ppm if 5.4 grams are found in 250 kg of water?

note 1 kg is 103g so 250 kg is 2.5x105 g.

$$\left (\frac{5.4\, g \; solute}{2.5 x10 ^5 \; g \;H_2 O} \right )\left ( \frac{10^3 \;g \; H_2 O}{L} \right )\left ( \frac{10^3mg \; solute}{g} \right )=\frac{22\, mg \; solute}{L \; H_2 O}=22 \,ppm$$

Parts per billion

Often expressed as g solute/109g solvent (note, mass is not a unit of quantity, but this is the convention)

$ppb=\frac{g\; solute}{10^{9}g\; solvent}$
for aqueous solutions we note that the density of water is 1g/ml, so the above equation can be converted to
$ppb=\frac{\mu g_{solute}}{L_{ water}}$

Exercise $$\PageIndex{3}$$

Show how to convert g solute/109g solvent = $$\mu$$g solute/L water (when the solvent is water)

$$\displaystyle ppb=\left (\frac{g \; solute}{10 ^9 \; g \;H_2 O} \right )\left ( \frac{10^3 \;g \; H_2 O}{L} \right )\left ( \frac{10^6 \mu g \; solute}{g} \right )=\frac{\mu g \; solute}{L \; H_2 O}$$

Worksheet

Concentration worksheet

Concentration worksheet key

Note:

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Concentration Conversions

The following problem have Video Solutions in the Answers

Exercise $$\PageIndex{4}$$

Concentrated aqueous ammonia is 28% by mass NH3 and has a density of 0.90g/ml. What is the mole fraction ammonia in concentrated ammonia? Note: Answer contains a video.

XAmmonia = 0.29 , note how this is close to the mass %, which is because ammonia and water have similar molar masses.

Exercise $$\PageIndex{5}$$

Concentrated aqueous ammonia is 28% by mass NH3 and has a density of 0.90g/ml. What is the molarity of ammonia in concentrated ammonia? Note: Answer contains a video.

MAmmonia = 15M.

Exercise $$\PageIndex{6}$$

Concentrated aqueous ammonia is 28% by mass NH3 and has a density of 0.90g/ml. What is the molality of ammonia in concentrated ammonia? Note: Answer contains a video.