Question

Calculate the percentage composition of each element in ammonium carbonate. [N=14 , H = 1 , C = 12; O = 16​

Answer 1

Answer:-

Molecular mass of Ammonium carbonate :-

$\\ \sf\longmapsto (NH_4)_2CO_3$

$\\ \sf\longmapsto 2(14u+4\times 1u)+12u+3(16u){}$

$\\ \sf\longmapsto 2(14u+4u)+12u+48u$

$\\ \sf\longmapsto 2(28u)+60u$

$\\ \sf\longmapsto 56u+60u$

$\\ \sf\longmapsto 106u$

Formula:-

$\boxed{\sf Mass\℅of\;x=\dfrac{Mass\:of\:x\:in\;1mol\:of\:the\:compound}{Molecular\:Mass\:of\:the\;compound}\times 100}$

Nitrogen:-

• Atomic mass=28u

$\\ \sf\longmapsto \dfrac{28}{106}\times 100$

$\\ \sf\longmapsto 26.4\℅$

Hydrogen:-

• Atomic mass=8u

$\\ \sf\longmapsto \dfrac{8}{106}\times 100$

$\\ \sf\longmapsto 7.5\℅$

Carbon:-

• Atomic mass=12u

$\\ \sf\longmapsto \dfrac{12}{106}\times 100$

$\\ \sf\longmapsto 11.3\℅$

Oxygen:-

• Atomic mass=48u

$\\ \sf\longmapsto \dfrac{48}{106}\times 100$

$\\ \sf\longmapsto 54.8\℅$