Chemistry Exam Notes: Moles, Molar Mass, and Percent Composition

The Mole

  • Mole (mol): The unit of measurement for the amount of a substance.

    • Used for measuring large amounts of tiny things.

    • 1 mole of something = 602,214,076,000,000,000,000,000602,214,076,000,000,000,000,000 of it = 6.022×10236.022 \times 10^{23}.

    • Avogadro’s number: The number of particles in exactly one mole of a pure substance, which is 6.022×10236.022 \times 10^{23}.

    • Example: 1 mole of gold contains 6.022×10236.022 \times 10^{23} atoms of gold.

Mole Calculation Example

  • Problem: If you have 4.0 moles of helium, how many atoms do you have?

  • Solution:
    4.0 moles of He×6.022×1023 atoms1 mole=2.4×1024 atoms of helium4.0 \text{ moles of He} \times \frac{6.022 \times 10^{23} \text{ atoms}}{1 \text{ mole}} = 2.4 \times 10^{24} \text{ atoms of helium}

Practice Problems

  1. How many atoms are in 7.1 moles of silver?

    • Answer: 4.27×10244.27 \times 10^{24} atoms

  2. If you have 6.25×10216.25 \times 10^{21} atoms of lithium, how many moles of lithium do you have?

    • Answer: 0.0104 moles

Molar Mass

  • Molar mass: The mass of one mole of a pure substance.

    • Units: grams/mole (g/mol)

    • The molar mass value is the same number as the atomic mass (found on the periodic table) but with different units.

    • Atomic mass: mass of one atom in atomic mass units (amu).

    • Molar mass: mass of one mole of particles in grams/mole (g/mol).

  • Molar mass can be used as a conversion factor.

    • Conversion factor: a ratio of equivalent values (equal to 1).

    • Example: The molar mass of aluminum (Al) is 26.98 g/mol.

    • 26.98 g of Al=1 mol of Al26.98 \text{ g of Al} = 1 \text{ mol of Al}

    • Conversion factors:

      • 26.98 g of Al1 mol of Al\frac{26.98 \text{ g of Al}}{1 \text{ mol of Al}}

      • 1 mol of Al26.98 g of Al\frac{1 \text{ mol of Al}}{26.98 \text{ g of Al}}

Molar Mass Examples

  1. If you have 3.50 mol of gold, how many grams do you have?

    • Solution:
      3.50 mol of Au×196.97 g of Au1 mol=689.40 g of Au=689 g of Au3.50 \text{ mol of Au} \times \frac{196.97 \text{ g of Au}}{1 \text{ mol}} = 689.40 \text{ g of Au} = 689 \text{ g of Au}

  2. You perform a chemical reaction and produce 12.6 g of aluminum. How many moles of Al did you make?

    • Solution:
      12.6 g of Al×1 mol26.98 g of Al=0.467 mol of Al12.6 \text{ g of Al} \times \frac{1 \text{ mol}}{26.98 \text{ g of Al}} = 0.467 \text{ mol of Al}

Practice Problems

  1. What is the mass, in grams, of 0.0185 mol of sodium?

    • Answer: 0.43 grams

  2. How many moles of silver are in 3.60×10−53.60 \times 10^{-5} g of silver?

    • Answer: 3.34×10−73.34 \times 10^{-7} moles

  3. How many atoms of sulfur are in 2.00 g of it?

    • Answer: 3.75×10223.75 \times 10^{22} atoms

Molar Mass of Compounds

  • Formula mass: The sum of the average atomic masses of a compound’s elements (in amu).

    • Numerically the same as molar mass but in g/mol (more practical).

  • Example: Find the molar mass for H2O.

    • H molar mass = 1.008 g/mol

    • O molar mass = 16.00 g/mol

    • H2O molar mass = 2(1.008) + 1(16.00) = 2.016 + 16.00 = 18.02 g/mol

Practice Problems

  1. Find the molar mass for NaCl.

  2. Find the molar mass for Mg(OH)2.

  3. Find the molar mass for potassium chloride.

Molar Mass Examples

  1. What is the mass, in grams, of 2.50 mol of LiF?

    • Molar mass of LiF = 1(6.94) + 1(19.00) = 6.94 + 19.00 = 25.94 g/mol

    • Solution:
      2.50 mol of LiF×25.94 g of LiF1 mol=64.9 g of LiF2.50 \text{ mol of LiF} \times \frac{25.94 \text{ g of LiF}}{1 \text{ mol}} = 64.9 \text{ g of LiF}

  2. How many moles are in 125 g of KMnO4?

    • Molar mass of KMnO4 = 1(39.10) + 1(54.94) + 4(16.00) = 39.10 + 54.94 + 64.00 = 158.04 g/mol

    • Solution:
      125 g KMnO4×1 mol158.04 g KMnO4=0.791 mol KMnO4125 \text{ g KMnO4} \times \frac{1 \text{ mol}}{158.04 \text{ g KMnO4}} = 0.791 \text{ mol KMnO4}

Practice Problems

  1. Find the mass of 0.525 mol of BaI2.

  2. Determine the number of moles in 17.9 g of SO2.

  3. How many molecules of NH3 are in 340.5 g of NH3?

Percent Composition

  • Percent composition: The percentage (based on mass) of each element in a compound.

    • % of an element in a compound=Mass of element in 1 mol of compoundMolar mass of compound×100\% \text{ of an element in a compound} = \frac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}} \times 100

    • To solve, start by finding the molar mass of the compound.

Percent Composition Example

  • Find the percent composition of CO2.

    • Molar mass of CO2 = 1(12.01) + 2(16.00) = 12.01 + 32.00 = 44.01 g/mol

    • % composition of C = 12.01 g of C in 1 mol of CO244.01 g total in 1 mol of CO2×100=27.29% C in CO2\frac{12.01 \text{ g of C in 1 mol of CO2}}{44.01 \text{ g total in 1 mol of CO2}} \times 100 = 27.29\% \text{ C in CO2}

    • % composition of O = 2(16.00 g of O in 1 mol of CO2)44.01 g total in 1 mol of CO2×100=72.71% O in CO2\frac{2(16.00 \text{ g of O in 1 mol of CO2})}{44.01 \text{ g total in 1 mol of CO2}} \times 100 = 72.71\% \text{ O in CO2}

Percent Composition Example

  • Magnesium hydroxide is 54.87% oxygen by mass. How many grams of oxygen would you expect to see in 225 g of the compound?

    • Solution: Find what 54.87% of 225 g is.

    • 54.87100=g of O225 g total\frac{54.87}{100} = \frac{\text{g of O}}{225 \text{ g total}}

    • 12345.75=100(g of O)12345.75 = 100(\text{g of O})

    • g of O=123.4 g\text{g of O} = 123.4 \text{ g}

Practice Problems

  1. Find the percent composition of Ca(NO3)2.

  2. Potassium chlorate is 31.91% potassium by mass. How many grams of potassium would you expect to see in 150 g of the compound?