Mole Ratios

Key Questions

  • Question #a106e

    Mole ratios are used as conversion factors between products and reactants in stoichiometry calculations.

    For example, in the reaction

    2H₂(g) + O₂(g) → 2H₂O(g)

    The mole ratio between O₂ and H₂O is (1 mol O₂)/(2 mol H₂O).

    The mole ratio between H₂ and H₂O is (2 mol H₂)/(2 mol H₂O).

    Example:

    How many moles of O₂ are required to form 5.00 moles of H₂O?

    Solution:

    5.00 mol H₂O × (1 mol O₂)/(2 mol H₂O) = 2.50 mol O₂.

    If the question had been stated in terms of grams, you would have had to convert grams of H₂O to moles of H₂O, then moles of H₂O to moles of O₂ (as above), and finally moles of O₂ to grams of O₂.

    Ernest Z. · · Jan 15 2014
  • Question #dc373

    Answer:

    To get the experimental molar ratio, you divide the moles of each reactant that you actually used in the experiment by each other.

    Explanation:

    EXAMPLE 1

    Consider the reaction: "2Al" + "3I"_2 → "2AlI"_3

    What is the experimental molar ratio of "Al" to "I"_2 if 1.20 g "Al" reacts with 2.40 g "I"_2?

    Solution

    Step 1: Convert all masses into moles.

    1.20 cancel("g Al") × "1 mol Al"/(26.98 cancel("g Al")) = "0.044 48 mol Al"

    2.40 cancel("g I₂") × ("1 mol I"_2)/(253.8 cancel("g I₂")) = "0.009 456 mol I"_2

    Step 2: Calculate the molar ratios

    To calculate the molar ratios, you put the moles of one reactant over the moles of the other reactant.

    This gives you a molar ratio of "Al" to "I"_2 of 0.04448/0.009456

    Usually, you divide each number in the fraction by the smaller number of moles. This gives a ratio in which no number is less than 1.

    The experimental molar ratio of "Al" to "I"_2 is then 0.04448/0.009456 = 4.70/1 (3 significant figures)

    The experimental molar ratio of "I"_2 to "Al" is 1/4.70

    Note: It is not incorrect to divide by the larger number and express the above ratios as 1:0.213 and 0.213:1, respectively. It is just a matter of preference.

    EXAMPLE 2

    A student reacted 10.2 g of barium chloride with excess silver nitrate, according to the equation

    "BaCl"_2("aq") + "2AgNO"_3("aq") → "2AgCl(s)" + "Ba(NO"_3)_2("aq")

    She isolated 14.5 g of silver chloride. What was her experimental molar ratio of "AgCl" to "BaCl"_2?

    Solution

    Step 1: Convert all masses into moles

    10.2 cancel("g BaCl₂") × ("1 mol BaCl"_2)/(208.2 cancel("g BaCl₂")) = "0.048 99 mol BaCl"_2

    14.5 cancel("g AgCl") × "1 mol AgCl"/(143.3 cancel("g AgCl")) = "0.1012 mol AgCl"

    Step 2: Calculate the molar ratios

    The experimental molar ratio of "AgCl" to "BaCl"_2 is 0.1012/0.04899 = 2.07/1

    Here is a video example:


    video from: Noel Pauller

    Ernest Z. · 183 · Jan 25 2014

Questions

Stoichiometry