Question

(a) Define atomic mass unit.

(b) Calculate the molecular mass of the following :

(i) $CuS{O}_4.5H_{2}O$ (ii) $(N{H}_4{)}_{2}C{O}_3$

(iii) $(N{H}_2{)}_{2}CO$ (iv) $M{g}_3{N}_2$

Given atomic mass of Cu = 63.5, H = 1, O = 16, C = 12, N = 14, Mg = 24, S = 32

Answer 1

(a) a unit of mass for expressing masses of atoms, molecules, or nuclear particles equal to ¹/₁₂ the mass of a single atom of the most abundant carbon isotope 12C — called also dalton

An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus. The AMU is used to express the relative masses of, and thereby differentiate between, various isotopes of elements.


(b)
(i) Atomic mass of H = 1 g
atomic mass of O = 16 g
atomic mass of Cu =63.5 g
atomic mass of S = 32 g

now ,
molar mass of CuSO4.5H2O = atomic mass of Cu + atom mass of S + 4× atomic mass of O +5× {2 × atomic mass of H + atomic mass of O }

= 63.5 + 32 + 4× 16 +5 (2 + 16) g

=63.5 +32 + 64 + 90 g

= 249.5 g

(ii) Add atomic mass of all the elements in the following compound..
atomic mass of N =14 ; of H =1 ; of C=14 ; of O =16
NH4 = 18 but there are two NH4 so total 36
+ 12(of C) + 48 (of 3 O) =96

(iii) Molecular mass of (NH2)2CO= 14*2+4+12+16
=60g

(iv) Add all the atomic mass of following èlements in the following compound..
atomic mass of Mg=24 ; òf N= 14
there are 3 Mg so 3×24 =72 and 2 N so 2×14 =28 so moleculr mass =28+72 =100