Question #2694b

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Question #2694b

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Answer 1 · 2017-06-17T18:00:55

The formula mass is 399.86 u.

Explanation:

The calculation might be easier if you first multiply all atoms inside the parentheses by the subscript 3.

Then ${Fe}_{2} {({SO}_{4})}_{3}$ $"Fe"_2("SO"_4)_3$ becomes ${Fe}_{2} S_{3} O_{12}$ $"Fe"_2"S"_3"O"_12$ .

Now, we can calculate the formula mass.

$No. m Atom m m A_{r} m m Mass$ $bb("No."color(white)(m)"Atom"color(white)(mm)A_text(r)color(white)(mm) "Mass")$
$l 2 m m m l Fe m m l 55.84 m l l 111.68$ $color(white)(l)2color(white)(mmml)"Fe"color(white)(mml)55.84color(white)(mll)111.68$
$l 3 m m m l S m m m 32.06 m m 96.18$ $color(white)(l)3color(white)(mmml)"S"color(white)(mmm) 32.06color(white)(mm)96.18$
$12 m m m O m m l l 16.00 m l l 192.00$ $12color(white)(mmm)"O"color(white)(mmll)16.00color(white)(mll)192.00$
$m m m m m m m l l Total = 399.86$ $color(white)(mmmmmmmll)"Total = 399.86"$

The formula mass is 399.86 u.

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Question #2694b

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