Molarity and Stoichiometry

Stoichiometry Review

• Transferring from one chemical to another requires using moles.
• Methods to convert to moles:
  • Particles: Use Avogadro's number.
  • Mass: Use molar mass.
  • Volume of a gas: Use 22.4 liters (This may be covered later).
  • Molarity: Dealing with a solution.

Molarity to Moles Conversion

• Molarity is defined as moles of solute per liter of solution.
• To get moles from molarity, multiply molarity by the volume of the solution in liters: $\text{Moles} = \text{Molarity} \times \text{Liters of Solution}$

Example Problem 1

• Problem: How many moles of calcium chloride (CaCl2) form when 0.2 liters of 1 molar HCl react with excess calcium hydroxide?
• Given:
  • Volume of HCl = 0.2 liters
  • Molarity of HCl = 1 M (1 mole/liter)
• Unknown: Moles of CaCl2
• Solution:
  • The excess calcium hydroxide is extraneous information.
  • Multiply the volume and molarity to get moles of HCl: $0.2 \text{ liters} \times 1 \frac{\text{mole}}{\text{liter}} = 0.2 \text{ moles of HCl}$
  • Use the mole ratio from the balanced equation:
    $2 \text{ moles of HCl} : 1 \text{ mole of CaCl}_2$
  • Convert moles of HCl to moles of CaCl2:$0.2 \text{ moles HCl} \times \frac{1 \text{ mole CaCl}<em>2}{2 \text{ moles HCl}} = 0.1 \text{ moles CaCl}</em>2$
  • Therefore, 0.1 moles of CaCl2 are formed.

Example Problem 2

• Problem: What mass of magnesium iodide (MgI2) is required to fully react with 100 milliliters of 0.25 molar silver nitrate (AgNO3)?
• Givens:
  • Volume of AgNO3 = 100 mL
  • Molarity of AgNO3 = 0.25 M
• Unknown: Mass of MgI2
• Solution:
  • Write the half equation: $MgI<em>2 + AgNO</em>3 \rightarrow$
  • Complete the double replacement reaction:
  • Swap metals: Silver pairs with iodine, and magnesium pairs with nitrate.
  • Check charges: Silver is +1, iodide is -1, magnesium is +2, nitrate is -1.
  • Write the correct formulas:$AgI \text{ and } Mg(NO<em>3)</em>2$
  • The balanced equation is: $MgI<em>2 + 2AgNO</em>3 \rightarrow 2AgI + Mg(NO<em>3)</em>2$
  • Convert volume to liters: 100 mL = 0.1 liters.
  • Moles of $AgNO_3$: $0.1 \text{ liters} \times 0.25 \frac{\text{moles}}{\text{liter}} = 0.025 \text{ moles}$
  • Use the mole ratio to convert moles of $AgNO<em>3$ to moles of $MgI</em>2:$ $\frac{1 \text{ mole }MgI<em>2}{2 \text{ moles }AgNO</em>3}$
  • Convert moles of $MgI<em>2$ to grams using molar mass: $Mg = 24.3 \text{ and } I = 126.9 \text{ (so 2 iodine atoms is 253.8)}$, molar mass of $MgI</em>2$ is $24.3 + 253.8 = 278.1 \text{ g/mol}$
  • Moles of $MgI<em>2$ to mass of $MgI</em>2:$ $0.025 \text{ moles }AgNO<em>3 \times \frac{1 \text{ mole }MgI</em>2}{2 \text{ moles }AgNO<em>3} \times \frac{278.1 \text{ grams }MgI</em>2}{1 \text{ mole }MgI_2} = 3.47625 \text{ grams}$
  • Sig Figs: The least number of significant figures in the givens is two (0.25 M), so round the answer to two significant figures.
  • Final answer: 3.5 grams of $MgI_2$.