Determine the empirical formula of each of the following compounds if a sample contains (a) 0.104 mol k, 0.052 mol c, and 0.156 mol o; (b) 5.28 g sn and 3.37 g f; (c) 87.5% n and 12.5% h by mass.

The empirical formula of the compounds are:

A. K₂CO₃

B. SnF₄

C. NH₂

The empirical formula of a compound is defined as the simplest formula which gives an expression of the ratio of the atoms of the different elements present in the compound.

The empirical formula of the compounds given in question above can be obtained as follow:

K = 0.104 mole

C = 0.052 mole

O = 0.156 mole

Divide by the smallest

K = 0.104 / 0.052 = 2

C = 0.052 / 0.052 = 1

O = 0.156 / 0.052 = 3

Therefore, the empirical formula of the compound is K₂CO₃

Sn = 5.28 g

F = 3.37 g

Sn = 5.28 / 118.71 = 0.044

F = 3.37 / 19 = 0.177

Sn = 0.044 / 0.044 = 1

F = 0.177 / 0.044 = 4

Therefore, the empirical formula of the compound is SnF₄

N = 87.5%

H = 12.5%

N = 87.5 / 14 = 6.25

H = 12.5 / 1 = 12.5

N = 6.25 / 6.25 = 1

H = 12.5 / 6.25 = 2

Therefore, the empirical formula of the compound is NH₂

Learn more: link

The answer to your question is:

a) K₂CO₃

b) SnF₄

c) NH₂

Explanation:

a)  

0.104 mol K

0.052 mol C

0.156 mol O

Divide mol by the lowest number (0.052)

K = 0.104/0.052 = 2

C = 0.052/0.052 = 1

O = 0.156/0.052 = 3

Then        empirical formula =  K₂CO₃

b)

5.28 g of Sn   AW Sn = 119 g

3.37 g F          AW F = 19 g

    119 g of Sn 1 mol                    19 g of F 1 mol

     5.28g      x                         3.37 g     x

      x = 5.28(1)/119 = 0.044 mol               x = 3.37(1)/19 = 0.117

Divide by the lowest number (0.044)

Sn     0.044 / 0.044 = 1                            F = 0.117/0.044 = 4

Then SnF₄

c)

87.5% N     = 87.5 g

12.5% H     = 12.5 g

AW N = 28g

AW H = 2 g

   28 g  1 mol                           2 g 1 mol

   87.5g   x                               12.5g  x    

    x = 87.5(1)/28 = 3.125                         x = 12.5(1)/2 = 6.25

The lowest number is 3.125

Then

N      3.125/3.125 = 1                           H = 6.25/3.125 = 2

                        NH₂