Anonymous

What is the empirical formula & empirical formula mass for each of the following compounds?

(a)C2H4

(b)C2H6O2

(c)N2O5

(d)Ba3(PO4)2

(e)Te4I16

Relevance

Empirical formula is the simplest formula that represents the whole number ratio between the elements in a compound.

a) C2H4: Divide the subscripts by 2 to get the Empirical formula

Empirical formula is CH2

Empirical formula mass = 1(12.01 g/mol) + 2(1.008 g/mol) = 14.03 g/mol

b) C2H6O2 (divide the subscripts by 2)

Empirical formula is CH3O; Empirical formula mass = 1(12.01 g/mole) + 6(1.008 g/mole) + 1(16.0 g/mole) = 34.1 g/mole

c) N2O5 (There is no common factor for 2 and 5). So the empirical formula is N2O5 (the same as molecular formula)

Empirical Formula mass = 2 (14.0 g/mole) + 5(16.0 g/mole) = 108 g/mole

d) Empirical formula is Ba3(PO4)2 (Same as molecular formula)

Empirical formula mass = 3(137.33 g/mole) + 2(30.97 g/mole) + 8(16.0 g/mole) = 601.9 g/mole

e) Te4I16 (4 is the common factor for 4 and 16): Empirical formula = TeI4

Empirical formula mass = 1(127.60 g/mole) + 4(126.90 g/mole) = 635.2 g/mole

• Dr W
Lv 7

the empirical formula is the formula with the lowest whole # ratios of all the elements in the molecule.

(a). C2H4...

both 2 and 4 are divisible by 2. let's do that.. we get 2/2 =1... 4/2 =2... ie..CH2.. that is the empirical formula for C2H4. empirical formula mass = 12.01 + 2x1.008 = 14.03 g / unit

(b) C2H6O2.. divide all by 2.. CH3O is the empirical formula. mass = 12.01 + 3x1.008+ 16 = 31.0 g/unit

(c)N2O5.. cannot be divided into smaller whole numbers. mass = 108 g / unit

(d) Ba3(PO4)2. again, cannot be divided into smaller whole number ratios. mass = 602 g/unit

(e) Te4I16.. 4 and 16 can be divided by 4. TeI4 is the empirical formula. mass = 127.6+4*126.9 = 635 g/unit