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}2}{2 \text{ moles HCl}} = 0.1 \text{ moles CaCl}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: MgI2 + AgNO3 \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(NO3)2
    • The balanced equation is: MgI2 + 2AgNO3 \rightarrow 2AgI + Mg(NO3)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 AgNO3 to moles of MgI2: \frac{1 \text{ mole }MgI2}{2 \text{ moles }AgNO3}
    • Convert moles of MgI2 to grams using molar mass: Mg = 24.3 \text{ and } I = 126.9 \text{ (so 2 iodine atoms is 253.8)}, molar mass of MgI2 is 24.3 + 253.8 = 278.1 \text{ g/mol}
    • Moles of MgI2 to mass of MgI2: 0.025 \text{ moles }AgNO3 \times \frac{1 \text{ mole }MgI2}{2 \text{ moles }AgNO3} \times \frac{278.1 \text{ grams }MgI2}{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.