Key Terms
Example
25.2 g acetic acid (CH3CO2H, molar mass 60.05 g/mol) in 0.500 L Moles acetic acid = 25.2 / 60.05 = 0.420 mol Molarity =
Molar mass
Mass in grams of one mole of a substance. Units = g/mol.
Going from particles to mass requires two steps
Particles --> moles --> mass
Moles of saccharin
0.0400 g / 183.18 g/mol = 2.18 x 10^-4 mol
Molecules
2.18 x 10^-4 x 6.022 x 10^23 = 1.31 x 10^20 molecules Carbon atoms: 1.31 x 10^20 x 7 = 9.18 x 10^20 atoms C
Formula mass
Sum all (atomic mass x atom count) for every element
Steps from mass data
1. Convert mass of each element to moles (divide by molar mass).
Example (from source)
27.29% C and 72.71% O In 100 g: 27.29 g C and 72.71 g O Moles C: 27.29 / 12.01 = 2.272 mol Moles O: 72.71 / 16.00 = 4.54
You need two pieces of information
1. Empirical formula (from percent composition) 2.
Steps
1. Calculate empirical formula mass (sum atomic masses in empirical formula).
Formula
N = (molar mass) / (empirical formula mass) Molecular formula = (empirical formula) x n
Solution
Homogeneous mixture. Composition is uniform throughout.
Solvent
The component present in greatest amount; the dissolving medium. Solute: the component dissolved in the solvent; present
Concentration
Quantitative measure of how much solute is in a given amount of solution. Dilute: low solute concentration.
Units
Mol/L, written as M.