Empirical and Molecular Formula Worksheet

SHOW WORK ON A SEPARATE SHEET OF PAPER.

 

Write the empirical formula for the following compounds.

1) C6H6

CH

 

2) C8H18

      C4H9

 

3) WO2

      WO2

 

4) C2H6O2

      CH3O

 

5) X39Y13

       X2Y

6)   A compound with an empirical formula of C2OH4 and a molar mass of 88 grams per mole. What is the molecular formula of this compound?

 

First find the empirical formula mass of the compound.

Carbon:       2 (12.0 g/mol)

Oxygen:      1 (16.0 g/mol)

Hydrogen:   4 (1.0 g/mol)

 

=44 g/mol

 

So this compound has a molar mass of 88 g/mol and an empirical formula mass of 44 g/mol.  So the molecular formula must have twice as many of each atom as the empirical formula.

 

88 / 44 = 2

 

Molecular formula is 2 times the empirical formula

2(C2OH4)  =  C4O2H8

 

7)   A compound with an empirical formula of C4H4O and a molar mass of 136 grams per mole. What is the molecular formula of this compound?

 

Use the same strategy as problem #6

4(12 g/mol) + 4(1 g/mol) + 1(16 g/mol) = 68 g/mol

 

136/68 = 2

 

2(C4H4O)  = C8H8O2

 

8)   A compound with an empirical formula of CFBrO and a molar mass of 254.7 grams per mole. What is the molecular formula of this compound?

 

Use the same strategy at problems 6 & 7.

 

1(12 g/mol) + 1(19 g/mol) + 1(79.9 g/mol) + 1(16 g/mol) = 127 g/mol

254.7/127 = 2

 

2(CFBrO)  = C2F2Br2O

 

9)   A compound with an empirical formula of C2H8N and a molar mass of 46 grams per mole. What is the molecular formula of this compound?

 

C:  2(12) = 24 g/mol

H: 8(1) =      8 g/mol

N:  1(14) =  14 g/mol

= 46 g/mol

46/46=1

So the molecular formula and the empirical formula are the same.

 

10)    A well-known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula C2H4NO. If its molar mass is 116.1 g/mol, what is the molecular formula of the compound?

 

C:  2(12 g/mol)

H: 4(1 g/mol)

N: 1(14 g/mol)

O: 1(16 g/mol)

=24+4+14+16=58 g/mol

116.1/58 = 2

2(C2H4NO) = C4H8N2O2

 

 

 

12)    Nitrogen and oxygen form an extensive series of oxides with the general formula NxOy. One of them is a blue solid that comes apart, reversibly, in the gas phase. It contains 36.84% N. What is the empirical formula of this oxide?

 

Pretend you have 100 grams of this substance.  That way you can convert all the percentages directly into grams.

 

Divide the mass in grams by the atomic mass to determine the number of moles of each atom.  

36.84% N  =  36.84 g / (14g/mol) =2.6 mol

63.16% O  =  63.16 g / (16 g/mol) =3.95 mol

 

This gives us N2.6O3.95

 

But we only want integers, so divide by the smallest number.

 

N2.6/2.6O3.95/2.6

 

This gives us  N1O1.5  .  We can convert 1.5 to an integer if we multiply everything by 2.

This leaves us with an empirical formula of N2O3 .

 

 

 

 

 

13. A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine. What is the empirical formula of the indium compound?

 

The iridium chloride has a mass of 0.5000 g.  Of this, 0.2404 g are chlorine.  Therefore, iridium must have a mass of 0.5000g – 0.2404g = 0.2596 g

 

Cl:        0.2404 g / (35.5 g/mol) = 6.77 x 10-3 mol

Ir:         0.2596 g / (192.2 g/mol) = 1.35 x 10-3 mol

 

So we have Ir1.35E-3Cl6.77E-3

 

To get integers, divide by the smallest of these numbers, 1.35 x 10-3.  This leaves us with IrCl5 .

 

(hint:  when dividing, you can just do 6.77 / 1.35 and forget about the powers of 10.  Since they’re both 10-3 they cancel out.)

 

 

14.     An unknown compound was found to have a percent composition as follows:

47.0 % potassium, 14.5 % carbon, and 38.5 % oxygen. What is its empirical formula? If the true molar mass of the compound is 166.22 g/mol, what is its molecular formula?

 

K:     47 g / (39 g/mol) = 1.21 mol

C:     14.5 g / (12 g/mol) = 1.21 mol

O:     38.5 g / (16 g/mol) = 2.41 mol

Divide each number by the smallest to get

KCO2

 

 

15. Rubbing alcohol was found to contain 60.0 % carbon, 13.4 % hydrogen, and the remaining mass was due to oxygen. What is the empirical formula of rubbing alcohol?

 

The oxygen must be 100% - 60% - 13.4% = 26.6%

Imagine we have 100 g of rubbing alcohol.  Now we can convert these percentages into grams.  Then we can divide by their atomic mass to get moles.

 

C:   60 g / (12.0 g/mol)  = 5 mol

H:   13.4 g / (1.0 g/mol) =13.4 mol

O:   26.6 g / (16.0 g/mol) = 1.6625

 

Divide each number by the smallest number, 1.6625.  This gives us

C3H­8O.