Molar Mass of element = atomic mass (bottom number on PT)
Molar Mass of compound =
AaBbCc = a x molar mass A + b * molar mass of B + c * molar mass C
MgCl2 = 1 x 24.3 + 2 * 35.5 = 95.3 g/ mol
Percent by mass = molar mass element
x 100
molar mass of compound
% O in Mg(OH)2 = molar mass oxygen
x 100 =
2 (16)
x 100
molar mass of Mg(OH)2
24.3 + 2 (16) + 2(1)
% O in Mg(OH)2 = (32 / 58.3) x 100 = 54.9%
Empirical Formula
1. change percent to grams, change grams to moles (grams ÷ molar mass = moles)
2. pick smallest number of moles
3. divide all moles by smallest number of moles should get whole numbers which become subscripts
a. if numbers are not whole numbers multiply each number by a number to make it a whole
number
4. write empirical formula with subscripts
Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur
step 1
step 2 &3
step 3a
step 4
Al = 35.98g / 27.0 = 1.33 / 1.33 = 1
x2=2
Al2S3
S = 64.02 g / 32.1 = 1.99 / 1.33 = 1.50 x 2 = 3
Molecular Formula
1. find empirical formula (see above)
2. find molar mass of empirical formula (see above)
3. find x = molar mass of molecular formula (given in problem)
molar mass of empirical formula
4. multiply subscripts of empirical formula by x and write molecular formula
A compound was found to contain 49.98g carbon and 10.47g hydrogen. The molar mass of the
compound is 58.12 g/mol. What is the molecular formula?
Step 1: find empirical formula
C = 49.98 / 12 = 4.165 / 4.165 = 1 x 2 = 2
C2H5
H = 10.47 / 1 = 10.47 / 4.165 = 2.5 x 2 = 5
Step 2: find molar mass of empirical formula
C2H5 = 2(12) + 5(1) = 29
Step 3: x = molar mass of molecular / molar mass of empirical
x = 58.12 / 29 = 2
Step 4: multiply subscripts by x
C2 H5
x2 x2
C4H10
Hydrate
1. find moles of water and moles of anhydrous compound
Hydrate = Anhydrous + H2O
2. find x = moles of water
moles of anhydrous
3. write formula of hydrate
anhydrous · x H2O
The anhydrous mass of CuSO4 is 25.42g and the mass of water is 14.3 g. What is the hydrate formula?
H2O = 14.3 / (2(1) + 1(16) = 14.3 / 18 = 0.794 mol H2O
CuSO4 = 25.42 / (1(63.5) + 1(32.1) + 4(16)) = 25.42 / 159.6 = 0.159 mol anhydrous
x = mol water / mol anhydrous = 0.794 / 0.159 = 5
CuSO4 · 5 H2O