Question

One mole of a compound M(HCO3)2 has a mass of 162g. Calculate the relative atomic mass of M. [H = 1, C =12, O = 16]

Answer 1

Answer:One of the mole of M(HCO3)2 = 162g; M +(1 + 12 + 16 x 3) x 2 = 162; M +122 = 162,:. M = 162 - 122 = 40g

Answer 2

Given:

Molar Mass of M(HCO₃)₂=162g

To Find:

Relative atomic mass of 'M'

Solution:

On Finding the Molar Formula Unit mass of "M(HCO₃)₂"

i.e., M+2[1+12+(16)3]

M+2[13+48]

=(M+122)g

But, we are given that the molar mass of M(HCO₃)₂ is 162g

» M+122=162

or M=162–122

» M=40g/mol

Therefore, the Atomic Mass of Element(Metal) 'M' is 40u (Ca-Calcium).

And, Thus;

$Relative \: Atomic \: Mass = \frac{mass \: in \: 'u'}{1 u}$

"A r" of M=40u/1u=40

.

.

∴ The relative atomic mass (A r) of M is 40.

This statement explains that the atomic mass of M is 40 times the mass of 1 u