# Alnico is a strongly magnetic alloy (solid mixture) of Co, Ni, Al and Fe. A 100.0 g sample of Alnico has 11.2g Al, 22.3g Ni, and 5.0g Co in it; the rest is Fe. How many moles of aluminum are there?

Feb 29, 2016

$0.423$ $m o l$ Al give or take w/ sig figs

#### Explanation:

$11.2$ $g$ Al x ($1$ $m o l$/$26.98$ $g$)=$0.415$ $m o l$ Al

$22.3$ $g$ Ni x($1$ $m o l$/$58.69$ $g$)=$0.380$ $m o l$ Ni

$5.0$ $g$ Co x ($1$ $m o l$/$58.93$ $g$)=$0.0848$ $m o l$ Co

$100$ $g - 11.2$ $g$ Al $- 22.3 g$ $N i - 5.0$ $g$ Co = $61.5$ $g$ Fe

$61.5$ $g$ Fe $\times \left(\frac{1 m o l}{55.84}\right) = 1.10$ $m o l$ Fe

Divide each one by the smallest number of moles: cobalt at $0.0848$ $m o l$.

Al = $\frac{0.415}{0.0848} = 4.89$ $m o l$
Ni = $\frac{0.380}{0.0848} = 4.48$ $m o l$
Co =$\frac{0.0848}{0.0848} = 1.00$ $m o l$
Fe= $\frac{1.10}{0.0848} = 12.97$ $m o l$

Since Ni is $4.48 \approx 4.50$, we multiply everything by $2$ to get a whole number, $9$, for Ni.

Al = $4.89$ $m o l \times 2 = 10$ $m o l$ Al
Ni = $4.48$ $m o l \times 2 = 9$ $m o l$ Ni
Co = $1.00$ $m o l \times 2 = 2$ $m o l$ Co
Fe= $12.97$ $m o l \times 2 = 26$ $m o l$ Fe

This gives us an empirical formula of $A {l}_{10} C {o}_{2} F {e}_{26} N {i}_{9}$.

If we add up the Fe, Co, Ni, and Al in the empirical formula, we get $2366.7$ $g m o {l}^{-} 1$ of Alnico

$100$ $g$ Alnico x ($1$ $m o l$ Alnico/$2366.7$ $g$ Alnico) x ($10$ $m o l$ Al/$1$ $m o l$ Alnico) = $0.423$ $m o l$ Al