H O N O R S C H E M I S T RY – S E M E S T E R 2

Concentration and Molarity

Concentration of a Solution

𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝑶𝑹

𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

More dissolved solute = higher concentration

Calculating Concentration

Parts per million (ppm) = g solute / 106 g solution

Molarity (M) = mol solute / liter solution

Molality (m) = mol solute / kg solvent

One mol of 8 different elements

1 M CuSO4

Parts per million (ppm)

Used for VERY small quantities (e.g., lead levels in

drinking water)

Even smaller levels given in parts per billion (ppb)

Mathematical expression: 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛× 1,000,000 𝑝𝑝𝑚

Sample Problem – Calculating ppm

A chemical analysis shows that there are 2.2 mg of lead in

5.00 x102 g of a water sample. Convert the measurement

to ppm.

Calculate:

Convert mg to grams of lead

Divide by grams of water

Multiply by 106 PPM

= 4.4 ppm

Practice

3.0 x 10-4 g of Helium gas are dissolved in 200.0 g of

water. Express this concentration in ppm.

1.5 ppm

A 365.0 g sample of water contains 23 mg Au. How much

gold is present in the sample in ppm?

63 ppm

Additional ppm practice p 461: 2-4, 6-7

Calculating Concentration – Molality (m)

Used when solvent volume changes with temperature

Calculated as: m = mol solute / kg solvent

Example:

You mix 0.75 mol NaCl with 2.5 kg of distilled water. What is the molality of your solution?

0.30 m NaCl

Molality (m)

How many moles of sucrose are required to make a

3.5 m solution in 1.75 kg of water?

6.1 mol sucrose

What is the molality of a solution of 52.0 g baking

soda (NaHCO3) in 125 g of water?

4.95 m NaHCO3

Calculating Molarity

…when given mass of solute and volume of solution

If you are given “mass,” use the

molar mass of the compound to

convert to “moles”

If not already in liters, convert

volume to L

1000 mL = 1000 cm3 = 1 L

Calculating Concentration – Molarity (M)

Calculated as:

M = mol solute / L solution

Example:

You are told that 0.30 mol KBr are present in a 0.40 L

sample solution. What is the molarity?

0.75 M KBr

Sample Problem – Calculating Molarity

What is the molarity of a potassium chloride solution

that has a volume of 400.0 mL and contains 85.0 g KCl?

2.85 M

Practice

Vinegar contains 5.0 g of acetic acid, CH3COOH, in

100.0 mL of solution. Calculate the molarity of acetic acid

in vinegar.

0.83 M acetic acid

A solution of AgNO3 contains 29.66 g of solute in

100.0 mL of solution. What is the molarity of the solution?

1.746 M AgNO3

What mass of KBr is present in 25 mL of 0.85 M solution

of potassium bromide?

2.5 g KBr

Diluting Solutions

You have a concentrated solution but need a weaker

solution for your lab work.

Molarity makes it easy!

V1M1 = V2M2

Diluting Solutions

You have a large beaker filled with 10.0 M HCl in your

stock room. Your experiment needs 2.5 L of 0.10 M

HCl. How do you prepare this?

V1M1 = V2M2 V1 = V2M2/M1

0.025 L = 25 mL of 10.0 M acid diluted to 2.5 L

Diluting Solutions

Given 25 mL of 3.0 M CuSO4•5H20, what is the

molarity of the solution if you dilute it to a final

volume of 1.0 L?

0.075 M CuSO4•5H20

How many mL of 5.50 M NaCl solution do you need

to make 0.450 L of 2.00 M NaCl?

164 mL of 5.5 M NaCl

Solution Stoichiometry

How much of a chemical solution do you need for the

experiment?

Solution Stoichiometry uses molarity as an additional

conversion factor

Just like regular stoichiometry, only now you have one more

conversion factor available!

M = mol/L

Solution Stoichiometry

Conversion Review

moles A ↔ moles B

use mole ratio (mol A/ mol B or mol B/mol A)

mass ↔ # mol

use molar mass (g/mol or mol/g)

# mol ↔ volume

use molarity (mol/L or L/mol)

Sample Problem - Solution Stoichiometry

What volume (in mL) of a 0.500 M solution of copper (II) sulfate is

needed to react with an excess of aluminum to provide 11.0 g of

copper?

1. Write and balance the chemical equation (if not already given)

Unbalanced reaction CuSO4 (aq) + Al (s) Cu (s) + Al2(SO4)3 (aq)

Balanced reaction 3CuSO4 (aq) + 2Al (s) 3Cu(s) + Al2(SO4)3 (aq)

Sample Problem - Solution Stoichiometry

What volume (in mL) of a 0.500 M solution of copper (II) sulfate is

needed to react with an excess of aluminum to provide 11.0 g of

copper?

3CuSO4 (aq) + 2Al (s) 3Cu(s) + Al2(SO4)3 (aq)

3. Plan your approach

Convert mass Cu to mol Cu with molar mass

Use mole ratio of Cu : CuSO4 to get moles of CuSO4

Convert mol CuSO4 into volume with molarity

= 346 mL CuSO4 solution

Practice

An excess of zinc metal is added to 125 mL of 0.100 M HCl

solution. What mass of zinc chloride (aq) is formed?

Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)

0.853 g ZnCl2 (aq)

Yellow CdS pigment is prepared by reacting ammonium sulfide

with cadmium nitrate. What mass of CdS can be prepared by

mixing 2.50 L of a 1.25 M Cd(NO3)2 solution with an excess of

(NH4)2S?

Cd(NO3)2 (aq) + (NH4)2S (aq) CdS (s) + 2NH4NO3 (aq)

452 g CdS (s)