Question

0.45g of an organic compound gave 0.44g of CO2 and 0.09g of H2O. the molecular mass of compound is 90amu calculate the molecular formula​

Answer 1

Answer:

C2H2O4.

Explanation:

To calculate the % of Carbon we have :-

=12/44  * (weight of CO2 /weight of organic compound)* 100.                       ==12/44 * 0.440.45 * 100 which on solving we will get 26.67 % of carbon.

Now, % of Hydrogen will be:-

=2/18* (weight of H2O/weight of organic compound)* 100.                      

= 218 * 0.090.45 * 100 which on solving we will get 2.22 % of hydrogen.

Again % of Oxygen will be  = 100 − (% of C+ % of H)= 100 − (26.67 + 2.22) which on solving we will get 71.11 % of the oxygen.

Since the atomic masses of the C, H and O is 12 1 and 16 so the % atomic mass will be 26.67/12 = 2.22 for C, 2.22/1 = 2.22 for H and 71.11/16= 4.44 for O.So, the simplest ratio will be 2.22 hence C will be 1 , H will be 1 and O will be 2.

So, the empirical formula of the compound will be = CHO2

Empirical formula mass is = 12 + 1 +32 = 45

Now, n = Molecular mass / Empirical formula mass which is 90 /45 =2.

So, the molecular formula is Empirical formula * n = CHO​2 x 2 = C2H2O4.

Answer 2

Explanation:

Step I. Percentage of elements <br> Percentage of carbon

<br>

<br> Percentage of hydrogen

<br>

<br> Percentage of oxygeb

<br> Step II. Empirical formula of the compound <br>

<br> Empirical formula of the compound

<br> Step III. Molecular formula of the compound <br> Empirical formula mass

<br> Molecular mass = 90 amu (Given) <br>

<br> Molecular formula

Empirical formula