CHS Chemistry
Chapter 8 – MOLES – Notes & Procedures
Counting Atoms in a Formula:
SUBSCRIPTS are the small numbers that appear below and to the right of an element symbol or
a parenthesis in a chemical formula. The subscript multiplies whatever it comes after.
Fe2O3 means there are 2 iron atoms for each 3 oxygen atoms in a crystal of Fe2O3.
If a subscript comes after a parenthesis, it multiplies everything inside the parentheses.
There are 2 nitrogens and 6 oxygens in Ca(NO3)2.
COEFFICIENTS are the large numbers that come before a formula. A coefficient multiplies
everything in the entire formula.
4 Na2CO3 means there are 8 sodiums, 4 carbons and 12 oxygens in 4 sodium carbonates.
Calculating Molar Mass:
The molar mass is the mass (in grams) of one mole of a substance. It is equivalent to the
formula mass, but with units of grams instead of a.m.u. To calculate the MOLAR MASS of a
compound, you simply add up the masses of all the atoms of all the elements in the formula.
Example: calcium hydroxide – Ca(OH)2
1 calcium:
2 oxygen:
2 hydrogen:
MOLAR MASS
1 x 40.08 =
2 x 16.00 =
2 x 1.01 =
=
40.08
32.00
2.02
74.10 g/mol
Converting Moles and Mass (grams):
To convert between moles and grams you need the Molar Mass as a conversion factor.
1. Start with the value (and units) given in the problem.
2. Decide whether to multiply or divide (if you have moles, MOLE-tiply)
3. Take the given value and multiply it by the molar mass or divide it BY the molar mass.
Example: Calculate the mass of 2.55 moles of potassium bromide (KBr).
1. Calculate the molar mass of KBr = 119.00 g/mol
2. You are given moles so multiply.
3. 2.55 mol x 119.00 g/mol = 303.45 g
Example: How many moles of magnesium fluoride (MgF2) are in 822 g?
1. Calculate the molar mass of MgF2 – 62.31 g/mol.
2. You are given g (not moles) so divide BY the molar mass.
3. ___822 g___ = 13.19 mol
62.31 g/mol
Percent Composition: ( % = (part ÷ total) x 100)
1. Calculate the molar mass (writing down the total mass of each element as it’s calculated)
2. For the requested element (or all elements if directed), element mass x 100 = %
total mass
3. Check that the percents add up to 100%.
Example: Calculate the percent composition for each element in silicon dioxide (SiO2).
1. Molar mass: Si = 28.09; O = 2x16.00 = 32.00
2. % Si = 28.09 x 100 = 46.75%
60.09
%O = 32.00 x 100 = 53.25 %
60.09
3. 46.75 + 53.25 = 100%
MM = 60.09 g/mol
Empirical Formula: (the Lowest-Terms formula)
In this type of problem you will be given the percentage or grams of each element in the
compound (you can use either % or g without converting anything).
1.
2.
3.
4.
Take the given value for each element and divide each by the atomic mass for that element.
Look at the results for each element from #1. See which result is the lowest (smallest) value
Divide all of the results from #1 by the smallest one. These will be the subscripts.
If any of the subscripts is not close enough to obviously round to a whole number, you must
multiply all the subscripts by a number that will make them whole numbers.
Molecular Formula:
The molecular formula is a multiple of the empirical formula. You must calculate the
multiplier and multiply the empirical formula by it to get the molecular formula.
1. Calculate the formula mass of the empirical formula (empirical mass).
2. Formula mass of the compound or __Molar Mass_ = multiplier
Empirical Mass
3. Multiply the empirical formula subscripts by the “multiplier”.
Example: Analysis of a 10.150 g sample of a compound known to contain only phosphorus and
oxygen, indicates a phosphorus content of 4.433 g. What is the empirical formula of the
compound.
(Subtract to find oxygen content. 10.15 g – 4.433g = 5.717g oxygen)
1. P: 4.433/30.97 = 0.143 O: 5.717/16.00 = 0.357 (0.143 is the smallest of these)
2. P: 0.143/0.14.3 = 1
O: 0.357/0.143 = 2.5 (multiply by 2 to get whole numbers)
3.
P1O2.5 x 2 = P2O5 (empirical mass = 141.94)
The molar mass of this compound is found to be 283.89 g/mol. What is the compound's
molecular formula?
1. 283.89 / 141.94 = 2
2. P2O5 x 2 =
P4O10