# Chemistry percentage composition HARD. (1 Viewer)

#### imxprt

##### New Member
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

##### Moderator
Moderator
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

##### Active Member
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

##### New Member
Yeah this is the question

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#### Eagle Mum

##### Active Member
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.

Last edited:

#### CM_Tutor

##### Moderator
Moderator

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

\bg_white \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

\bg_white \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

\bg_white \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

##### Active Member
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.