An organic compound with a molecular weight o | vedclass

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(B) $1$. Calculate the empirical formula: $C = 20/12 = 1.67$,$H = 6.67/1 = 6.67$,$N = 46.67/14 = 3.33$,$O = (100 - (20 + 6.67 + 46.67))/16 = 26.66/16

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$(NH_2)_2CO$

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(B) $1$. Calculate the empirical formula: $C = 20/12 = 1.67$,$H = 6.67/1 = 6.67$,$N = 46.67/14 = 3.33$,$O = (100 - (20 + 6.67 + 46.67))/16 = 26.66/16 = 1.67$. $2$. Divide by the smallest value $(1.67)$: $C = 1, H = 4, N = 2, O = 1$. $3$. The empirical formula is $CH_4N_2O$. The empirical formula weight is $12 + 4 + 28 + 16 = 60$. $4$. Since the molecular weight is $60$,the molecular formula is $CH_4N_2O$,which corresponds to urea $(NH_2)_2CO$. $5$. Urea,upon heating,forms biuret,which gives a violet color with alkaline copper sulfate (Biuret test).

An organic compound with a molecular weight of $60$ contains $C = 20\%$,$H = 6.67\%$,and $N = 46.67\%$,with the remainder being oxygen. Upon heating,the solid residue gives a violet color with alkaline copper sulfate solution. Identify the compound.

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