expressing

CONCENTRATIONS

Concentration may be expressed several diff erent ways…

• Percent Composition by Mass (% m/m)

• Percent Composition by Volume (% v/v)

• Mass Volume % (m/v)• Mole Fraction or Mole Ratio (X)• Molarity (M)• Molality (m)• Normality (N)

•Normality (N)

Concentration = Amount of SoluteAmount of Soluti on

Percent Composition by Mass (%m/m)

This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100.

% by Mass = Mass of SoluteMass of Soluti on X 100

NOTE: To find a concentration in parts per million, divide the mass of solute by the mass of solution and multiply by one million, 106.

To find a concentration in parts per billion, divide the mass of theSolute by the mass of the solution and multiply by one billion, 109.

% by Mass (m/m) EXAMPLE• Determine the percent composition by

mass of a 100 g salt solution which contains 20 g salt.

% by Mass = Mass of SoluteMass of Soluti on X 100

% by Mass = 20g NaCl100g Soluti on X 100

m/m= 20% NaCl solution

% by Mass (m/m) EXAMPLE• 10g of KCl and 40g of water would

make 50g of solution. What is the % by mass (m/m) of KCl in concentration.

% by Mass = Mass of SoluteMass of Soluti on X 100

% by Mass = 10g KCl50g Soluti on X 100

m/m= 20% KCl solution

% by Mass (m/m) EXAMPLE• According to the organization SeaFriends,

seawater has 904 ppm of sulfur. What is the mass percent of sulfur in seawater?

% by Mass = Mass of SoluteMass of Soluti on in ppmX 100

% by Mass = 904g Sulfur1x106g Soluti on X 100

m/m= 0.0904% Sulfur in seawater

Answer: Assume one million grams of seawater. The number of grams of solute in one million grams is 904. This gives

% by Mass (m/m) EXAMPLE

• A 4 g sugar cube (Sucrose: C12H22O11) is dissolved in a 350 ml teacup of 80 °C water. What is the percent composition by mass of the sugar solution?Given: Density of water at 80 °C = 0.975 g/ml

% by Mass = Mass of SoluteMass of Soluti on X 100

Step 1 - Determine mass of solute Solute is 4 g of C12H22O11

Step 2 - Determine mass of solventdensity = mass/volumemass = density x volumemass = 0.975 g/ml x 350 mlmass = 341.25 g

Step 3 - Determine the total mass of the solutionmsolution = msolute + msolvent

msolution = 4 g + 341.25 g

msolution = 345.25 g Step 4 - Determine percent composition by mass of the sugar solution.

percent composition = (msolute / msolution) x 100percent composition = ( 4 g / 345.25 g) x 100percent composition = ( 0.0116 ) x 100

percent composition = 1.16%

Percent Composition by Volume (%v/v)

• Volume percent or volume/volume percent most often is used when preparing solutions of liquids. • Volume percent is defined as:

% by Volume = Volume of SoluteVolume of Soluti on X 100

Note that volume percent is relative to volume of solution, not volume of solvent.

% by Volume (v/v) EXAMPLE

• A solution containing 25mL of alcohol in 100mL of solution. What is the % by volume (v/v) of alcohol in the solution?

v/v= 25% alcohol in the solution

% by Volume = Volume of SoluteVolume of Soluti on X 100

% by Volume = 25mL of Alcohol100mL of Soluti on X 100

% by Volume (v/v) EXAMPLE• What is the % volume (v/v) of a solution

prepared by adding 12 mL of bromine (Br2) to enough carbon tetrachloride to give 250 mL of solution?

v/v= 4.8 % Bromine in the solution

% by Volume = Volume of SoluteVolume of Soluti on X 100

% by Volume = 12mL of Bromine250mL of Soluti on X 100

% by Volume (v/v) EXAMPLE• A solution containing 700mL of isopropyl

alcohol in 1000mL of solution. What is the % by volume (v/v) of isopropyl alcohol in the solution?

v/v= 0.7% isopropyl alcohol in the solution

% by Volume = Volume of SoluteVolume of Soluti on X 100

% by Volume =7 0 0 m L o f I s o p r p y l A l c o h o l

1000mL of Soluti on X 100

Percent Composition by Mass/Volume (%m/v)

• The mass volume percent is generally used for dilute solutions, so that 100mL of solution has essentially the same mass as 100g of water. (g/mL)

% Mass/Volume =Grams of Solute

M i l l i l i t e r s o f S o l u ti o n X 100

% by Mass/Volume (m/v) EXAMPLE• Suppose we mix 25g of glucose and enough

water to prepare 500mL of glucose solution. The mass/volume percent concentration is calculated as:

m/v = 5 % glucose in the solution

% Mass/Volume =Grams of Solute

M i l l i l i t e r s o f S o l u ti o n X 100

% Mass/Volume =25g of Glucose

250 mL Soluti on X 100

% by Mass/Volume (m/v) EXAMPLE• Calculate the mass/volume percent (m/v)

concentration of solution prepared by mixing 50g KI with enough water to make 250 mL of KI solution.

m/v = 20 % KI in the solution

% Mass/Volume =Grams of Solute

M i l l i l i t e r s o f S o l u ti o n X 100

% Mass/Volume =50g of KI

250 mL Soluti on X 100

% by Mass/Volume (m/v) EXAMPLE

• A laboratory technician must prepare 400mL of a 5% (m/v) glucose solution. How many grams of glucose must be measured out?

400 mL =5g of glucose

100 mL Soluti on = 20g glucose

Using the 5% (m/v) concentration as a conversion factor, we can calculate the mass of glucose needed for the solution:

% by Mass/Volume (m/v) EXAMPLE

• How many grams of NaCl are needed to prepare 3.0 L of a 1% (m/v) NaCl solution?

3000 mL =1g of glucose

100 mL Soluti on = 30g NaCl

Using the 1% (m/v) concentration as a conversion factor, we can calculate the mass of Sodium Chloride needed for the solution:

% by Mass/Volume (m/v) EXAMPLE

• An antifreeze mixture is made with ethylene glycol and water. How many liters of a 15% (m/v) antifreeze solution can be prepared from 60.0 mL of ethylene glycol?

60 g =1L

150 g anti freeze= 0.40 L soln.

Using the 15% (m/v) concentration as a conversion factor, we can calculate the volume of antifreeze solution:

CALCULATE THE % CONCENTRATION• (m/v) 2.0g sucrose in 100mL of solution• (m/m) 20.0g KCl in 150g of solution• (m/v) 75.0g of Na2SO4 in 0.50 L of

solution• (v/v) 3.0mL of acetone in 40mL of

solution• (m/v) 0.300 kg of glucose in 5.00L of

solution• (m/m) 40.0g of CaCl2 in 250g of solution