Law of Definite proportions

Elements in a pure compound are always present in the same fixed ratio by mass.
Examples:
H₂O; 11% hydrogen, 89% oxygen
NaHCO₃;  27.3% Na,  1.2% H,  14.3% C,  57.1% O

The formula mass (formula weight) of a substance is the sum of all the atomic masses (atomic weights) of all the atoms in the formula unit.
Examples:
CO₂; 12.0 + 2(16.0) = 44.0 amu / molecule
(NH₄)₂S;   2(14.0) + 8(1.0) + 32.1= 68.1 amu / formula unit

Since atomic masses have the units amu / atom, formula units (sums of atomic masses) have the units amu/formula unit. The formula masses of molecular compounds are often called molecular masses (molecular weight).

Percent Composition - shows the % by mass of the elements in a compound.

Consider lactic acid, C₃H₆O₃

   3C     +   6H    +   3 O
3(12.0) + 6(1.0) + 3(16.0) = 90.0 amu / molecule of lactic acid

%C ₌ 36 amu x 100 = 40.0% C
          90 amu      1

%H = 6 amu  x 100 = 6.7% H
          90 amu     1

%O = 48 amu  x 100 = 53.3% O
           90 amu      1

1.000 gram = 6.02 x 10²³ amu just as 12 troy ounces = 1 troy pound. In this case the ratio of one mole equal to Avogadro's number of amu is a mass conversion factor.

6.022 x 10²³ cats/mol cats
6.022 x 10²³ amu/g

1 formula unit SnF₂ = 156.7 amu
1 mol SnF₂ = 156.7 g
1 molecule C₃H₆O₃ = 90.0 amu
1 mol C₃H₆O₃ = 90.0 g
1 atom Cl = 35.45 amu
1 mol Cl atoms = 35.45 g
1 molecule Cl₂ = 70.90 amu
1 mol Cl₂ molecules = 70.90 g

Why molar mass = formula mass in number?

16 amu CH₄ x 1.00 g              x  6.02 x 10²³ molecules CH₄  = 16 g CH₄
 molecule          6.02 x 10²³ amu      1 mole CH₄                                1 mole CH₄

Note, CH₄ can represent the formula mass = 16 amu/molecule
          or the molar mass = 16 g/mole

atomic mass = atomic weight
formula mass = formula weight (FW)
molecule mass = molecular weight (MW)
molar mass (of an element) = gram atomic weight
molar mass (of a molecular compound) = gram molecular weight (GMW)
molar mass (of an ionic or molecular compound) = gram formula weight (GFW)

One can use (FW & GFW) for ionic or molecular compounds but (MW &
GMW) should be terms applied to molecular compounds only.

222g CO₂ x 1 mol CO₂ = mol CO₂
    1               44.0 g CO₂

222g CO₂ x 1 mol CO₂ x  6.02 x 10²³ molecules CO_{2 =}   molecules CO₂
      1            44.0 g CO₂            1 mol CO₂
 

Calculate the mass in grams of 9.9 x 10²³ molecules of ethylene glycol ( C₂H₆O₂₎.

    2C    +  6H    +    2 O
2(12.0) + 6(1.0) + 2(16.0) = 620 g / mole
 

9.9 x 10²⁶ molecules   _x  1 mol                        _x  62.0 g    =    g C₂H₆O₂
       1                            6.02 x 10²³ molecules    1 mol
 

Empirical formula = smallest whole number ratio of atoms in the formula unit ( molecule ).
 

Compound	Molecular  Formula	Empirical Formula
benzene	C6H6	CH
acetylene	C2H2	CH
ammonia	NH3	NH3
glucose	C6H12O6	CH2O
octane	C8H18	C4H9

Calculate the empirical formula of a compound having the following percent composition:

46.2% C, 7.7% H, 46.2% O
 

46.2 gC x 1 mol C = 3.85 mol C ;  3.85 / 2.89 = 1.33
    1          12.0 gC
 

7.7 g H x 1 mol H = 7.7 mol H;        7.7 / 2.89 = 2.60
   1           1.0 g H
 

46.2 g O x 1 mol O = 2.89 mol O;      2.89 / 2.89 = 1
    1            16.0 g O

C_1.33H_2.66O₁ leads to an empirical formula of C₄H₈O₃

6.2 P are burned in O₂ to produce 14.2 g product. Calculate the empirical formula

       14,2g P_xO_y - 6.2g P  =   8.0g oxygen

6.2 g P x 1 mol P = .20 mol P;     .20 /.20 =  1
  1           31 g P
 

8.0g O x 1 mol O = .50 mol O;       2.5 / 1 = 2.5
    1         16.0 g O

P₁O_2.5 ---> P₂O₅

Let the # emp. formulas in a molecular formula = x
 

x = molecular mass
      emp. formula mass
 

If the emp. formula is P₂O₅ (FW = 142) and the molecular mass is given as 284grams per mole,
then  x = 284/142 = 2 and the molecular formula = P₄O₁₀.