Question

An organic compound contains carbon, hydrogen oxygen and nitrogen in the weight
ratio 3 : 1 : 8 : 3.5 calculate its empirical formula

Answer 1

Answer:

% of C given = 20%

% of N, H, O combined =

[math]100-20=80%[/math]

Ratio of N:H:O = 7:1:4

Let the respective percentages be 7x, x and 4x.

According to the sum,

[math]7x + x + 4x = 80[/math]

[math]=》12x = 80[/math]

[math]=》 x = 80/12[/math]

% of N = [math]80/12 * 7 = 46.67%[/math]

% of H = [math]80/12 * 1 = 6.67%[/math]

% of O = [math]80/12 * 4 = 26.68%[/math]

So, the % of C, N, H, O are 20%, 46.67%, 6.67% and 26.68%.

Explanation:

So, the minimum molecular mass of the compound is 72.06 amu.

Hope this helps.

Answer 2

Given,

$\qquad \qquad \sf C \: \: \: \: H \: \: \: \:O \: \: \: \: N$

To Calculate the empirical formula :

Step-1 :

Weight percentage of elements taken as weight

$\qquad\qquad \sf 3 \: \: \: \: 1 \: \: \: \:8 \: \: \: \: 3.5$

Step-2 :

Weight of element divided with atomic weight

$\displaystyle \qquad\qquad \sf \frac{3}{12} \: \: \: \: \frac{1}{1} \: \: \: \: \frac{8}{16} \: \: \: \: \frac{3.5}{14}$

$\displaystyle \qquad\qquad \sf 0.25 \: \: \: \: 1 \: \: \: \: 0.5 \: \: \: \: 0.25$

Step-3 :

Find a simple ratio of atoms of elements

(This can be done by divided with smallest number)

Divide with 4

$\displaystyle \qquad\qquad \sf 0.25 \times 4 \: \: \: \: 1 \times 4 \: \: \: \: 0.5 \times 4 \: \: \: \: 0.25 \times 4$

$\displaystyle \qquad\qquad \sf 1\: \: \: \: 4 \: \: \: \: 2 \: \: \: \: 1$

Step-4 :

Write simple ratio of atom aganist of each element as prefix

$\qquad \qquad \sf C_{1} \: \: \: \: H_{4}\: \: \: \:O_{2} \: \: \: \: N_{1}$

The empirical formula is

$\qquad \qquad \longmapsto \: \: \underline{\boxed{ \: \bf CH_{4}O_{2}N \: } }$

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