Question

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.

Answer 1

Answer:

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

       

Answer 2

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:

A. Determination of the empirical formula

K = 0.104 mole

C = 0.052 mole

O = 0.156 mole

Empirical formula =?

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

B. Determination of the empirical formula

Sn = 5.28 g

F = 3.37 g

Empirical formula =?

Divide by their molar mass

Sn = 5.28 / 118.71 = 0.044

F = 3.37 / 19 = 0.177

Divide by the smallest

Sn = 0.044 / 0.044 = 1

F = 0.177 / 0.044 = 4

Therefore, the empirical formula of the compound is SnF₄

C. Determination of the empirical formula.

N = 87.5%

H = 12.5%

Empirical formula =?

Divide both side by their molar mass

N = 87.5 / 14 = 6.25

H = 12.5 / 1 = 12.5

Divide by the smallest

N = 6.25 / 6.25 = 1

H = 12.5 / 6.25 = 2

Therefore, the empirical formula of the compound is NH₂

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