Question

A compound on analysis gave the following percentage
composition: C - 54.54%, H = 9.09%, 0 = 36.36%

Answer 1

c=54.54%=54.54/12=4.545            4.545/2.2725=2
H=9.09%=9.09/1=9.09                    9.09/2.2725=4
O=36.36%=36.36/16=2.2725          2.2725/2.2725=1
  so,
      the compound is C2H4O (CH3CHO)

Answer 2

Question : A compound on analysis gave the following percentage composition by mass.

1, Hydroge = 9.09%

2, Oxygen = 35.35%

3,Carbon = 54.55%

5,Molar mass of compound = 88.88%

Find out molecular formula of compound

Answer:

Step 1 : = H + O + C

= 9.09+36.36+54.55

= 100

Step 2: = H = 9.09% 9.09g

O = 36.36% 36.36g

C =54.55% 54.55g

( here we change % into g)

Step 3: Convertion g into mol

Mole of H = 9.09÷1.008

=9.17 mol

Mole of O = 36.36÷16

= 2.272 mol

Mole of C = 54.55÷12.000

= 4.545 mol

Step 4 : H = 9.017 ÷ 2

= 4.5085

O = 2.272 ÷ 2

= 1.136

C = 4.545 ÷ 2

= 0.2725

therefore

4.5085 : 1.126 : 0.2725

Ratio = 4 : 1 : 2

emeeical formula = H4 OC2

Step 5 : emerical into molecular formula

Molecular formula = CH3 - CHO

hope it help you friend ^_^ .