Chemistry · General chemistry II · Worked example

Molar mass of a protein from osmotic pressure

A solution of $0.500 g$ of a protein in enough water to make $100.0 mL$ has an osmotic pressure of $1.86 torr$ at $25.0 ° C$ . Find the protein’s molar mass.

Π = M R T \Rightarrow n = \frac{Π V}{R T}

Convert the pressure to atmospheres

The first examples went from a known solute to a property. This one runs backward, from a measured property to an unknown molar mass. $R = 0.08206 L \cdot atm / (mol \cdot K)$ needs the pressure in atmospheres, and $760 torr$ is exactly $1 atm$ .

Π = \frac{1.86}{760} atm = 2.447 \times 10^{- 3} atm

Write the law

The protein is a nonelectrolyte, so $i = 1$ and $Π = M R T$ .

Π = M R T

Solve for the molarity

Divide both sides by $R T$ before putting in numbers.

M = \frac{Π}{R T}

Name the pressure

Use the value in atmospheres from the first step.

Π = 2.447 \times 10^{- 3} atm

Name the gas constant

The units of $R$ decide the units of everything else.

R = 0.08206 L atm {mol}^{- 1} K^{- 1}

\begin{array}{c} R = 0.08206 \\ L atm {mol}^{- 1} K^{- 1} \end{array}

Convert the temperature

$R$ is per kelvin, so add $273.15$ to the Celsius temperature.

T = 25.0 \circ C + 273.15 = 298.2 K

Substitute

Atmospheres and kelvin cancel, leaving $mol / L$ . $(0.08206) (298.2) = 24.47$ , and $2.447 \times 10^{- 3} \div 24.47$ $= 1.000 \times 10^{- 4}$ .

M = \frac{2.447 \times 10^{- 3} atm}{(0.08206 L atm {mol}^{- 1} K^{- 1}) (298.2 K)} = 1.000 \times 10^{- 4} mol/L

\begin{array}{c} M = \frac{2.447 \times 10^{- 3}}{(0.08206) (298.2)} mol/L \\ = 1.000 \times 10^{- 4} mol/L \end{array}

Moles of protein

The solution’s volume is $100.0 mL$ , which is $0.1000 L$ . Moles are the molarity times the volume.

n = (1.000 \times 10^{- 4} mol/L) (0.1000 L) = 1.000 \times 10^{- 5} mol

\begin{array}{c} n = (1.000 \times 10^{- 4} mol/L) \\ \times (0.1000 L) \\ = 1.000 \times 10^{- 5} mol \end{array}

Molar mass

Divide the mass of the sample by the moles. A protein of $5.00 \times 10^{4} g / mol$ is a medium-sized one.

\frac{0.500 g}{1.000 \times 10^{- 5} mol} = 5.00 \times 10^{4} g/mol

\begin{array}{c} \frac{0.500 g}{1.000 \times 10^{- 5} mol} \\ = 5.00 \times 10^{4} g/mol \end{array}

Result

The protein’s molar mass is about $5.00 \times 10^{4} g / mol$ .

Your turn

1. One like the example

$1.00 g$ of another protein in enough water to make $50.0 mL$ has an osmotic pressure of $18.6 torr$ at $25.0 ° C$ . Find its molar mass.

Show the answer and explanation

$2.00 \times 10^{4} g / mol$ .

$18.6 torr$ is $0.02447 atm$ . $M = Π \div R T$ $= 0.02447 \div (0.08206 \times 298.2)$ $= 1.000 \times 10^{- 3} mol / L$ . In $0.0500 L$ that is $5.00 \times 10^{- 5} mol$ , and $1.00 g \div (5.00 \times 10^{- 5} mol)$ $= 2.00 \times 10^{4} g / mol$ .

\begin{array}{l} 18.6 torr \times \frac{1 atm}{760 torr} = 0.02447 atm \\ Π = M R T \\ M = \frac{Π}{R T} \\ Π = 0.02447 atm \\ R = 0.08206 L atm {mol}^{- 1} K^{- 1} \\ T = 298.2 K \\ M = \frac{0.02447 atm}{(0.08206 L atm {mol}^{- 1} K^{- 1}) (298.2 K)} = 1.000 \times 10^{- 3} mol/L \\ n = (1.000 \times 10^{- 3} mol/L) (0.0500 L) = 5.00 \times 10^{- 5} mol \\ \frac{1.00 g}{5.00 \times 10^{- 5} mol} = 2.00 \times 10^{4} g/mol \end{array}

\begin{array}{c} 18.6 torr \times \frac{1 atm}{760 torr} \\ = 0.02447 atm \\ Π = M R T \\ M = \frac{Π}{R T} \\ Π = 0.02447 atm \\ R = 0.08206 \\ L atm {mol}^{- 1} K^{- 1} \\ T = 298.2 K \\ M = \frac{0.02447}{(0.08206) (298.2)} mol/L \\ = 1.000 \times 10^{- 3} mol/L \\ n = (1.000 \times 10^{- 3} mol/L) \\ \times (0.0500 L) \\ = 5.00 \times 10^{- 5} mol \\ \frac{1.00 g}{5.00 \times 10^{- 5} mol} \\ = 2.00 \times 10^{4} g/mol \end{array}

2. A variation

What osmotic pressure does a $0.150 M NaCl$ solution exert at body temperature, $37.0 ° C$ , assuming ideal dissociation?

Show the answer and explanation

About $7.64 atm$ .

This time the pressure is the unknown and the solute is ionic, so $i = 2$ . $T = 37.0 + 273.15$ $= 310.2 K$ . $Π$ $= (2) (0.150) (0.08206) (310.2)$ $= 7.64 atm$ . A little membrane holds back a large pressure, which is why cells must match the salt of their surroundings.

\begin{array}{l} Π = i M R T \\ i = 2 \\ M = 0.150 mol/L \\ R = 0.08206 L atm {mol}^{- 1} K^{- 1} \\ T = 37.0^{\circ} C + 273.15 = 310.2 K \\ Π = (2) (0.150 mol/L) (0.08206 L atm {mol}^{- 1} K^{- 1}) (310.2 K) = 7.64 atm \end{array}

\begin{array}{c} Π = i M R T \\ i = 2 \\ M = 0.150 mol/L \\ R = 0.08206 \\ L atm {mol}^{- 1} K^{- 1} \\ T = 37.0^{\circ} C + 273.15 = 310.2 K \\ Π = (2) (0.150 mol/L) \\ \times (0.08206 L atm {mol}^{- 1} K^{- 1}) \\ \times (310.2 K) = 7.64 atm \end{array}

3. Somewhere new

Why would a freezing-point measurement be a poor way to find the first protein’s molar mass? Estimate its $Δ T_{f}$ .

Show the answer and explanation

$Δ T_{f}$ would be only about $1.86 \times 10^{- 4} ° C$ , far too small to measure.

The $100.0 mL$ of dilute solution holds about $100.0 g$ of water, so the molality is about $1.000 \times 10^{- 5} mol \div 0.1000 kg$ $= 1.000 \times 10^{- 4} mol / kg$ . $Δ T_{f} \approx 1.86 \times 1.000 \times 10^{- 4}$ $= 1.86 \times 10^{- 4} ° C$ . The same solution gives an easily measured $1.86 torr$ of osmotic pressure.

\begin{array}{l} m = \frac{1.000 \times 10^{- 5} mol}{0.1000 kg} = 1.000 \times 10^{- 4} mol/kg \\ (1.86^{\circ} C kg/mol) (1.000 \times 10^{- 4} mol/kg) = 1.86 \times 10^{- 4}^{\circ} C \end{array}

\begin{array}{c} m = \frac{1.000 \times 10^{- 5} mol}{0.1000 kg} \\ = 1.000 \times 10^{- 4} mol/kg \\ (1.86^{\circ} C kg/mol) \\ \times (1.000 \times 10^{- 4} mol/kg) \\ = 1.86 \times 10^{- 4}^{\circ} C \end{array}

Keep exploring

In the Colligative studio, enter $1.00 \times 10^{- 5} mol$ in $100.0 mL$ at $25 ° C$ and confirm the osmotic pressure is $1.86 torr$ .

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Convert the pressure to atmospheres

Write the law

Solve for the molarity

Name the pressure

Name the gas constant

Convert the temperature

Substitute

Moles of protein

Molar mass

Result

Your turn

1. One like the example

2.00×104 g/⁢mol2.00×104 g/mol.

2. A variation

About 7.64 atm7.64 atm.

3. Somewhere new

ΔTfΔTf​ would be only about 1.86×10−4 °⁢C1.86×10−4 ∘C, far too small to measure.

Keep exploring

Study it, then try it

$2.00 \times 10^{4} g / mol$ .

About $7.64 atm$ .

$Δ T_{f}$ would be only about $1.86 \times 10^{- 4} ° C$ , far too small to measure.