AP Chemistry 1.4: Composition of Mixtures

Pure substance

A pure substance contains only one type of element or compound and has a fixed composition.

Mixture

A mixture contains two or more substances that are physically combined. Its composition can vary, and its properties depend partly on the relative amounts of its components.

How does the composition of a mixture affect its properties?

The relative amounts of the components influence the mixture's properties. For example, the amount of sugar affects a beverage's sweetness, while the proportions of elements in an alloy affect its strength and corrosion resistance.

Solution

A solution is a homogeneous mixture whose composition and properties are uniform throughout. It contains a solvent and one or more solutes.

Concentration

Concentration quantitatively describes the relative amount of a component in a mixture or solution. Terms such as dilute and concentrated provide qualitative descriptions of relatively low and high concentration.

Solvent

The solvent is the solution component present in the greatest relative amount and serves as the medium in which other substances dissolve.

Solute

A solute is a component dissolved in a solvent, typically present in a smaller amount or lower concentration.

Aqueous solution

An aqueous solution is a solution in which water is the solvent.

Molarity

Molarity, $M$, is the number of moles of solute per liter of solution: $M = \frac{n_{\mathrm{solute}}}{V_{\mathrm{solution}}}$. Its units are typically $\mathrm{mol\,L^{-1}}$ or M.

Why must solution volume be converted to liters when using molarity?

Molarity is defined as moles of solute per liter of solution, so a volume given in milliliters must be converted using $1\ \mathrm{L} = 1000\ \mathrm{mL}$ before substitution.

A beverage contains $0.133\ \mathrm{mol}$ sucrose in $355\ \mathrm{mL}$ of solution. What is its molarity?

$355\ \mathrm{mL} = 0.355\ \mathrm{L}$, so $M = \frac{0.133\ \mathrm{mol}}{0.355\ \mathrm{L}} = 0.375\ \mathrm{M}$.

How can the number of moles of solute be calculated from molarity and solution volume?

Rearrange the molarity equation: $n_{\mathrm{solute}} = M V_{\mathrm{solution}}$, with $V$ expressed in liters.

How many moles of solute are present in $10.0\ \mathrm{mL}$ of a $0.375\ \mathrm{M}$ solution?

Convert the volume to $0.0100\ \mathrm{L}$ and use $n = MV$: $n = (0.375)(0.0100) = 0.00375\ \mathrm{mol}$.

How can the volume of solution containing a specified amount of solute be found from molarity?

Rearrange $M = \frac{n}{V}$ to obtain $V = \frac{n}{M}$. The resulting volume is in liters when molarity is in mol/L.

How is molarity calculated when the solute mass, rather than moles, is given?

First convert mass to moles using molar mass, then divide by solution volume: $M = \frac{m_{\mathrm{solute}}/MM_{\mathrm{solute}}}{V_{\mathrm{solution}}}$.

A $0.500\ \mathrm{L}$ vinegar sample contains $25.2\ \mathrm{g}$ acetic acid, $\mathrm{CH_3CO_2H}$, with molar mass $60.052\ \mathrm{g/mol}$. What is its molarity?

$M = \frac{25.2\ \mathrm{g}/60.052\ \mathrm{g\,mol^{-1}}}{0.500\ \mathrm{L}} = 0.839\ \mathrm{M}$.

How can the mass of solute in a solution be calculated from molarity and volume?

Use $n = MV$ and then convert moles to mass: $m_{\mathrm{solute}} = M V_{\mathrm{solution}} MM_{\mathrm{solute}}$.

What mass of NaCl is present in $0.250\ \mathrm{L}$ of a $5.30\ \mathrm{M}$ solution? Use $MM_{\mathrm{NaCl}} = 58.44\ \mathrm{g/mol}$.

$m = (5.30\ \mathrm{mol/L})(0.250\ \mathrm{L})(58.44\ \mathrm{g/mol}) = 77.4\ \mathrm{g\ NaCl}$.

How can the volume of solution containing a specified solute mass be calculated directly?

Convert mass to moles and divide by molarity: $V = \frac{m_{\mathrm{solute}}}{MM_{\mathrm{solute}} M}$. The volume is in liters if $MM$ is in g/mol and $M$ is in mol/L.

What volume of a $0.839\ \mathrm{M}$ acetic acid solution contains $75.6\ \mathrm{g}$ acetic acid? Use $MM = 60.05\ \mathrm{g/mol}$.

$V = \frac{75.6\ \mathrm{g}}{(60.05\ \mathrm{g/mol})(0.839\ \mathrm{mol/L})} = 1.50\ \mathrm{L}$.

Why should intermediate values generally not be rounded during solution-concentration calculations?

Rounding intermediate results can introduce cumulative error in the final answer. Retain extra digits as guard digits and round only the final result according to significant figures.

Mass percentage concentration

Mass percentage is the mass of a component divided by the total mass of solution, multiplied by $100\%$: $\%\,\mathrm{mass} = \frac{m_{\mathrm{component}}}{m_{\mathrm{solution}}} \times 100\%$.

What does a $7.4\%$ by-mass NaOCl solution contain per $100.0\ \mathrm{g}$ of solution?

It contains $7.4\ \mathrm{g}$ of NaOCl per $100.0\ \mathrm{g}$ of total solution. The remaining mass consists primarily of solvent and other components.

A $5.0\ \mathrm{g}$ sample contains $3.75\ \mathrm{mg}$ glucose. What is the glucose mass percentage?

Convert $3.75\ \mathrm{mg}$ to $0.00375\ \mathrm{g}$, then calculate $\frac{0.00375}{5.0} \times 100\% = 0.075\%$.

How can solution density be used with mass percentage to determine solute mass from solution volume?

First convert solution volume to solution mass using $m_{\mathrm{solution}} = \rho V$. Then multiply by the mass fraction: $m_{\mathrm{solute}} = m_{\mathrm{solution}}\left(\frac{\%\,\mathrm{mass}}{100}\right)$.

A hydrochloric acid solution is $37.2\%$ HCl by mass and has density $1.19\ \mathrm{g/mL}$. What mass of HCl is in $0.500\ \mathrm{L}$ of solution?

$m_{\mathrm{solution}} = (500\ \mathrm{mL})(1.19\ \mathrm{g/mL}) = 595\ \mathrm{g}$; $m_{\mathrm{HCl}} = 595(0.372) = 221\ \mathrm{g}$.

Volume percentage concentration

Volume percentage is the volume of liquid solute divided by the total solution volume, multiplied by $100\%$: $\%\,\mathrm{vol} = \frac{V_{\mathrm{solute}}}{V_{\mathrm{solution}}} \times 100\%$.

A $355\ \mathrm{mL}$ bottle contains rubbing alcohol that is $70\%$ isopropanol by volume. What volume of isopropanol does it contain?

$V_{\mathrm{isopropanol}} = 0.70(355\ \mathrm{mL}) = 248.5\ \mathrm{mL}$.

How can the mass of a liquid solute be found from volume percentage and solute density?

Find solute volume using $V_{\mathrm{solute}} = \left(\frac{\%\,\mathrm{vol}}{100}\right)V_{\mathrm{solution}}$, then use $m_{\mathrm{solute}} = \rho_{\mathrm{solute}} V_{\mathrm{solute}}$.

A $70\%$ by-volume isopropanol solution has a total volume of $355\ \mathrm{mL}$. If isopropanol has density $0.785\ \mathrm{g/mL}$, what mass of isopropanol is present?

$m = (355\ \mathrm{mL})(0.70)(0.785\ \mathrm{g/mL}) = 195\ \mathrm{g}$.

Mass-volume percentage concentration

Mass-volume percentage is the mass of solute divided by the total solution volume, expressed as a percentage: $\%\,(m/v) = \frac{m_{\mathrm{solute}}}{V_{\mathrm{solution}}} \times 100\%$. For example, $0.9\%\,(m/v)$ means $0.9\ \mathrm{g}$ solute per $100\ \mathrm{mL}$ solution.

How is blood glucose commonly reported using mass-volume units?

Blood glucose is often reported in milligrams per deciliter, $\mathrm{mg/dL}$, where $1\ \mathrm{dL} = 100\ \mathrm{mL}$.

Parts per million (ppm)

For mass-based concentration, $\mathrm{ppm} = \frac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 10^6$. It is useful for very dilute concentrations such as trace pollutants.

Parts per billion (ppb)

For mass-based concentration, $\mathrm{ppb} = \frac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 10^9$. Like ppm, ppb may also be defined using volume or mass-volume ratios when specified.

What is the relationship between ppm and ppb?

$1\ \mathrm{ppm} = 1000\ \mathrm{ppb}$, so a concentration in ppb can be converted to ppm by dividing by $1000$.

A water sample contains lead at $15\ \mathrm{ppb}$. What is this concentration in ppm?

$15\ \mathrm{ppb}\left(\frac{1\ \mathrm{ppm}}{1000\ \mathrm{ppb}}\right) = 0.015\ \mathrm{ppm}$.

How can the mass of a trace solute be calculated from a mass-based ppm or ppb value?

Rearrange the definitions: $m_{\mathrm{solute}} = \frac{\mathrm{ppm}\,m_{\mathrm{solution}}}{10^6}$ or $m_{\mathrm{solute}} = \frac{\mathrm{ppb}\,m_{\mathrm{solution}}}{10^9}$.

A $300\ \mathrm{mL}$ glass of water has a lead concentration of $15\ \mathrm{ppb}$. Assuming water has density $1.00\ \mathrm{g/mL}$, what mass of lead is present?

The water mass is approximately $300\ \mathrm{g}$. Thus $m_{\mathrm{Pb}} = \frac{(15)(300)}{10^9}\ \mathrm{g} = 4.5 \times 10^{-6}\ \mathrm{g} = 4.5\ \mu\mathrm{g}$.

A $50.0\ \mathrm{g}$ wastewater sample contains $0.48\ \mathrm{mg}$ mercury. What are its mercury concentrations in ppm and ppb?

Convert $0.48\ \mathrm{mg}$ to $0.00048\ \mathrm{g}$: $\mathrm{ppm} = \frac{0.00048}{50.0} \times 10^6 = 9.6\ \mathrm{ppm}$. Since $1\ \mathrm{ppm} = 1000\ \mathrm{ppb}$, the concentration is $9600\ \mathrm{ppb}$.

When calculating a mass percentage, ppm, or ppb, what unit requirement must be satisfied?

The numerator and denominator must use the same type of quantity and compatible units, such as grams of solute per grams of solution. Unit conversion is required before forming the ratio.

How does molarity differ from mass percentage and ppm/ppb?

Molarity uses moles of solute per liter of solution, whereas mass percentage and mass-based ppm/ppb use mass ratios. Molarity is especially useful for stoichiometric calculations because chemical equations relate substances in moles.

Dilution

Dilution is the process of lowering a solution's concentration by adding solvent. The amount of solute remains constant while the solution volume increases.

What happens to solute moles, solution volume, and concentration during dilution?

The moles of solute remain unchanged, the total solution volume increases, and the concentration decreases because $M = \frac{n}{V}$.

Derive the dilution equation from the definition of molarity.

Before dilution, $n_1 = M_1V_1$; after dilution, $n_2 = M_2V_2$. Because $n_1 = n_2$, $M_1V_1 = M_2V_2$, commonly written as $C_1V_1 = C_2V_2$.

What conditions must be met when applying $C_1V_1 = C_2V_2$?

The equation applies when a solution is diluted without changing the amount of solute. Concentration units must be consistent, and the two volume units must match or cancel.

How is the final concentration calculated when a stock solution is diluted to a known final volume?

Rearrange the dilution equation: $C_2 = \frac{C_1V_1}{V_2}$.

A $0.850\ \mathrm{L}$ sample of $5.00\ \mathrm{M}$ solution is diluted to $1.80\ \mathrm{L}$. What is the final concentration?

$C_2 = \frac{(5.00\ \mathrm{M})(0.850\ \mathrm{L})}{1.80\ \mathrm{L}} = 2.36\ \mathrm{M}$.

How is the final volume calculated when the initial stock volume and both concentrations are known?

Rearrange $C_1V_1 = C_2V_2$ to obtain $V_2 = \frac{C_1V_1}{C_2}$.

What volume of $0.12\ \mathrm{M}$ HBr can be prepared from $11\ \mathrm{mL}$ of $0.45\ \mathrm{M}$ HBr?

$V_2 = \frac{(0.45)(0.011\ \mathrm{L})}{0.12} = 0.041\ \mathrm{L} = 41\ \mathrm{mL}$.

How is the volume of concentrated stock solution required for a dilution calculated?

Use $V_1 = \frac{C_2V_2}{C_1}$, where $C_1$ is the stock concentration and $C_2,V_2$ describe the desired diluted solution.

What volume of $1.59\ \mathrm{M}$ KOH is needed to prepare $5.00\ \mathrm{L}$ of $0.100\ \mathrm{M}$ KOH?

$V_1 = \frac{(0.100\ \mathrm{M})(5.00\ \mathrm{L})}{1.59\ \mathrm{M}} = 0.314\ \mathrm{L}$, or $314\ \mathrm{mL}$.