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

Mole (mol): The unit of measurement for the amount of a substance.
- Used for measuring large amounts of tiny things.
- 1 mole of something = 602,214,076,000,000,000,000,000 of it = 6.022 \times 10^{23}.
- Avogadro’s number: The number of particles in exactly one mole of a pure substance, which is 6.022 \times 10^{23}.
- Example: 1 mole of gold contains 6.022 \times 10^{23} atoms of gold.

Problem: If you have 4.0 moles of helium, how many atoms do you have?
Solution:
4.0 \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}

How many atoms are in 7.1 moles of silver?
- Answer: 4.27 \times 10^{24} atoms
If you have 6.25 \times 10^{21} atoms of lithium, how many moles of lithium do you have?
- Answer: 0.0104 moles

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 \text{ g of Al} = 1 \text{ mol of Al}
- Conversion factors:
  - \frac{26.98 \text{ g of Al}}{1 \text{ mol of Al}}
  - \frac{1 \text{ mol of Al}}{26.98 \text{ g of Al}}

If you have 3.50 mol of gold, how many grams do you have?
- Solution:
  3.50 \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}
You perform a chemical reaction and produce 12.6 g of aluminum. How many moles of Al did you make?
- Solution:
  12.6 \text{ g of Al} \times \frac{1 \text{ mol}}{26.98 \text{ g of Al}} = 0.467 \text{ mol of Al}

What is the mass, in grams, of 0.0185 mol of sodium?
- Answer: 0.43 grams
How many moles of silver are in 3.60 \times 10^{-5} g of silver?
- Answer: 3.34 \times 10^{-7} moles
How many atoms of sulfur are in 2.00 g of it?
- Answer: 3.75 \times 10^{22} atoms

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

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 \text{ mol of LiF} \times \frac{25.94 \text{ g of LiF}}{1 \text{ mol}} = 64.9 \text{ g of LiF}
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 \text{ g KMnO4} \times \frac{1 \text{ mol}}{158.04 \text{ g KMnO4}} = 0.791 \text{ mol KMnO4}

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

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 = \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 = \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}

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.
- \frac{54.87}{100} = \frac{\text{g of O}}{225 \text{ g total}}
- 12345.75 = 100(\text{g of O})
- \text{g of O} = 123.4 \text{ g}

Find the percent composition of Ca(NO3)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?