Question #fa394

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Question #fa394

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Answer 1 · 2017-02-28T17:43:12

$In one mole of ferric oxide?$ $"In one mole of ferric oxide?"$

Explanation:

A $mole$ $"mole"$ specifies a quantity of $6.022 \times 10^{23}$ $6.022xx10^23$ individual items of stuff. We use it in the same way as we would use a $dozen$ $"dozen"$ , or a $bakers' dozen$ $"bakers' dozen"$ , or a $gross$ $"gross"$ . It is simply another (admittedly large) collective number.

We have $1 \cdot m o l$ $1*mol$ , $6.022 \times 10^{23}$ $6.022xx10^23$ formula units of $F e_{2} O_{3}$ $Fe_2O_3$ , i.e. $2 \times N_{A}$ $2xxN_A$ $iron atoms$ $"iron atoms"$ , and $3 \times N_{A}$ $3xxN_A$ $oxygen atoms$ $"oxygen atoms"$ , where $N_{A} \equiv 6.022 \times 10^{23} \cdot m o l^{- 1}$ $N_A-=6.022xx10^23*mol^-1$ . And thus, there are $2 \times 6.022 \times 10^{23}$ $2xx6.022xx10^23$ iron atoms in such a molar quantity.

Now I know $N_{A}$ $N_A$ iron atoms have a mass of $55.8 \cdot g$ $55.8*g$ , and $N_{A}$ $N_A$ oxygen atoms have a mass of $15.999 \cdot g$ $15.999*g$ . How do I know this?

And thus $1 \cdot m o l$ $1*mol$ of $ferric oxide$ $"ferric oxide"$ has a mass of $1 \cdot m o l \times (2 \times 55.8 \cdot g \cdot m o l^{- 1} + 3 \times 16.00 \cdot g \cdot m o l^{- 1}) = 159.60 \cdot g$ $1*molxx(2xx55.8*g*mol^-1+3xx16.00*g*mol^-1)=159.60*g$ .

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Question #fa394

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