Question

A 5.00 g sample of an anhydrous Group II metal nitrate loses 3.29g in mass on strong heating.
Which metal is present?
A magnesium
B calcium
C strontium
D barium

Answer 1

When group II metal nitrate is heated, it produces metal oxide and nitrogen dioxide gas, and the ratio between metal nitrate and metal oxide is 1:1 .

From 1 mole nitrate you get 1 mole oxide. As:

\(\mathsf{Number} \space \mathsf{of } \space \mathsf{moles } = \cfrac{\textbf{Given mass}}{\textbf{Molar Mass}} \)

Molar mass of 2 Nitrate ions (anions) = \(\mathsf{2*(14+48)} \) = \(\mathsf{124 gm}\) .
And the Atomic mass of oxygen is 16 gm.
So the oxide has weight of : \(\mathsf{5-3.29 g = 1.71 g}\) .

Now as metal oxide and metal nitrate have equal no. of moles, so :

\(\cfrac{\textbf{Given mass of metal nitrate}}{\textbf{Molar mass of metal nitrate}} = \cfrac{\textbf{Given mass of metal oxide}}{\textbf{Molar mass of metal oxide}} \)

Let us say that the mass of the metal is x grams.

So, molar mass of metal nitrate = x + 124 grams

Molar mass of metal oxide = x + 16 grams.

Therefore , we get :

\(\cfrac{5}{x+124} = \cfrac{1.71}{x+16} \)

\(5(x+16) = 1.71(x+124) \)

\(5x + 80 = 1.71 x + 212.04 \)

\(5x - 1.71 x = 212.04 - 80 \)

\(3.29 x = 132.04\)

\(x = 40 (\mathsf{Approx.} )\)

Therefore, the mass of the metal used is 40 grams.

Therefore it is Calcium as it has mass of 40 grams (atomic mass) .