Chemistry percentage composition HARD. (1 Viewer)

imxprt · Feb 23, 2021

Hey can anyone help me with this question.
A steel company requires 2 tonnes of Fe. the company decided to mine from a site containing goethite. FeO(OH). which contains 31% of the mineral. how much of the goethite needs to be mined to obtain 2 tonnes of Fe.

CM_Tutor · Feb 23, 2021

Goethite, FeO(OH), being 31% Fe means that the m(Fe) makes up 31% of m(goethite):

m(Fe) = 31% x m(goethite)

So, if I wanted 500 g Fe (say), then I would need to solve:

500 = 0.31 x m(goethite)

m(goethite) = 500 / 0.31 = 1612.9 ... g = 1.61 kg

Now, you want m(Fe) to be 2 tonne = 2000 kg, so ...

Eagle Mum · Feb 28, 2021

The question seems somewhat ambiguous - have you typed it correctly?

The percentage by mass of Fe in the mineral goethite is fixed at 62.85% (=55.845/(55.845+2x15.999+1.008)), so it doesn’t matter what percentage of the rock is goethite, since you would need 2000 / 0.6285 kg (3182 kg) of goethite to extract 2000 kg of iron. Therefore, I don’t see how the 31% is relevant.

You would need to mine 2000 / (0.6285 x 0.31) = 10 265 kg of iron ore (mixture of iron compounds with sand & clay) to extract 2000 kg of iron.

imxprt · Feb 28, 2021

Yeah this is the question

Eagle Mum · Feb 28, 2021

imxprt said:
Yeah this is the question

In your working out, x (=2000/(0.6285x0.31)) is the required mass of iron ore, whereas 2000/0.6285 is the required mass of goethite.
My sympathies - textbook/test/exam questions ought to be rigorously set.

CM_Tutor · Mar 1, 2021

My answer above is wrong.

I originally took the question to mean that you were given that the %Fe in goethite was 31%, but I did not stop to consider whether this is reasonable - and it isn't.

As @Eagle Mum has correctly pointed out, the %Fe in goethite is

$\begin{align*} %\text{Fe in goethite}&=\frac{\text{M}(\text{Fe})}{\text{M}(\text{FeO}(\text{OH}))} \times \frac{100}{1} \\ &= \frac{55.85}{55.85+2 \times 15.9994 + 1.0080} \times \frac{100}{1} \\ &= \frac{55.85}{88.8568} \times \frac{100}{1} \\ &= 62.853...% \\ &\approx 62.85% \quad (\text{4 sig. fig.}) \end{align*}$

So, if you need 2000 kg of Fe, the mass of goethite required is

$\begin{align*} %\text{Fe in goethite}&=\frac{\text{m(Fe) required}}{\text{m(FeO(OH)) required}} \times \frac{100}{1} \\ \frac{62.853...}{100} &= \frac{2000\text{ kg}}{\text{m(FeO(OH)) required}} \\ \text{m(FeO(OH)) required}&= \frac{2000}{0.62853...} \\ &= 3181.98...\text{ kg} \\ &\approx 3182\text{ kg} \quad (\text{4 sig. fig.}) \end{align*}$

And if the ore is 31% goethite, then the mass of ore required is

$\begin{align*} \text{%goethite in ore}&=\frac{\text{m(goethite) required}}{\text{m(ore) required}} \times \frac{100}{1} \\ \frac{31}{100} &= \frac{3181.98...\text{ kg}}{\text{m(ore) required}} \\ \text{m(ore) required}&= \frac{3181.98...}{0.31} \\ &= 10264.45...\text{ kg} \\ &\approx 10.26\text{ tonne} \quad (\text{4 sig. fig.}) \end{align*}$

So, 10.26 tonne of ore is required to obtain 3182 kg of goethite, and thus to obtain 2000 kg of Fe after extraction and purification, etc.

Eagle Mum · Mar 1, 2021

Yes, I agree with CM_Tutor in such circumstances as this question it’s best to ATQ with 3182 kg of goethite AND add that 10.26 tonnes of ore are required to obtain the required amount of goethite and therefore Fe.

Chemistry percentage composition HARD. (1 Viewer)

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