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?

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Answer 1 · 2016-02-29T10:44:12

$0.423$ $mol$ Al give or take w/ sig figs

$11.2$ $g$ Al x ( $1$ $mol$ / $26.98$ $g$ )= $0.415$ $mol$ Al

$22.3$ $g$ Ni x( $1$ $mol$ / $58.69$ $g$ )= $0.380$ $mol$ Ni

$5.0$ $g$ Co x ( $1$ $mol$ / $58.93$ $g$ )= $0.0848$ $mol$ Co

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

$61.5$ $g$ Fe $xx((1 mol)/55.84)=1.10$ $mol$ Fe

Divide each one by the smallest number of moles: cobalt at $0.0848$ $mol$ .

Al = $0.415/0.0848 = 4.89$ $mol$
Ni = $0.380/0.0848 = 4.48$ $mol$
Co = $0.0848/0.0848 = 1.00$ $mol$
Fe= $1.10/0.0848= 12.97$ $mol$

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

Al = $4.89$ $mol xx 2= 10$ $mol$ Al
Ni = $4.48$ $mol xx 2= 9$ $mol$ Ni
Co = $1.00$ $mol xx 2= 2$ $mol$ Co
Fe= $12.97$ $mol xx 2= 26$ $mol$ Fe

This gives us an empirical formula of $Al_10Co_2Fe_26Ni_9$ .

If we add up the Fe, Co, Ni, and Al in the empirical formula, we get $2366.7$ $gmol^-1$ of Alnico

$100$ $g$ Alnico x ( $1$ $mol$ Alnico/ $2366.7$ $g$ Alnico) x ( $10$ $mol$ Al/ $1$ $mol$ Alnico) = $0.423$ $mol$ Al