Problem 6

A sample of 13.0 g of an unknown metal M was treated with an excess of dilute nitric acid. Excess hot alkaline solution was then added to the resulting mixture, which evolved 1.12 dm

^3

of a gas measured at STP. Identify the metal M.

Step 3 of 3: Valence search

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A_r(\ce{M}) = \dfrac{13.0\ \text{g}}{n(\ce{M})} = \dfrac{13.0 \times n}{8 \times 0.050} = 32.5\,n

Analysis

Since $n(\ce{M}) = \frac{8}{n} n(\ce{NH3}) = \frac{0.40}{n}$ mol, the atomic mass is $A_r = 13.0 / (0.40/n) = 32.5\,n$ . Testing valences: $n=1 \Rightarrow 32.5$ (none); $n=2 \Rightarrow 65.0$ (zinc, $\ce{Zn}$ ); $n=3 \Rightarrow 97.5$ (none); $n=4 \Rightarrow 130$ (none). Therefore the metal is zinc.

Common pitfall. Do not assume the gas is NO or NO2: those gases evolve during acid attack, whereas here gas only evolves upon adding excess base (NH3 from NH4+).