A compound with the empirical formula CH2 has a molar mass of 98 g/mol. What is

Question

A compound with the empirical formula CH2 has a molar mass of 98 g/mol. What is the molecular formula for this compound? 2-A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula? 3-The molecular mass of the compound is 132 amu. What is the molecular formula? 4What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass? 5What is the empirical formula of a compound composed of 36.5 g potassium (K) and 7.47 g oxygen (O)?

Explanation / Answer

1)

n = MF mass / EF mass
  
Where MF and EF are the respective molecular formula mass and empirical formula mass.

CH2 empirical formula mass = 12 + 2*1 = 14 gmol-1
  
n = 98 / 14 = 7

So molecular formula =(empirical formula)n
=(CH2)7
=C7H14

2) let there be 100 g of the sample, then.

NUMBER OF MOLES OF OXYGEN = 36.32/16 =2.27 moles
NUMBER OF MOLES OF CARBON = 54.53/12 =4.544 moles
NUMBER OF MOLES OF HYDROGEN = (9. 15/1 =9.15moles
  
Moles of oxygen are the least so,
Oxygen = 2.27 / 2.27 = 1
Carbon = 4.54 / 2.27 = 2
Hydrogen =9.15 / 2.27 = 4.030 approx 4

So empirical formula would be C2H4O.

3) Now, n = MF mass / EF mass

Where MF and EF are the respective molecular formula mass and empirical formula mass.

Empirical formula mass of C2H4O = 44 amu

n = 132 / 44 = 3

So molecular formula =(empirical formula)n

=(C2H4O) 3

=C6H12O3

4) let there be 100 g of the sample, then

Number of moles of hydrogen =3.25/ 1 = 3.25

Number of moles of oxygen = 77.39/16 = 4.83

Number of moles of carbon = 19.36/12=1.613

Moles of carbon are the least

So, H = 3.25 / 1.613 = 2.01

C = 1.613 / 1.613 = 1

O = 4.83 / 1.613 = 3

So empirical formula would be CH2O3.

5) number of moles of potassium =36.5 / 39 = 0.935

Number of moles of oxygen = 7.47 / 16 = 0.466

  Moles of oxygen are the least

So, O = 0.466 / 0.466 = 1

K = 0.935 / 0.466 = 2

So empirical formula would be K2O

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