Q3: Calculating Ore Processing for Thallium Extraction

In summary: If you have a fraction a/b and you divide both parts of that fraction by the same value, let's say c, you get a/c / b/c = a/b. It's not the same as a/c / b/c = 1/b.
  • #1
Maxwell Kraft
6
0
Hello. I just started a first year chemistry course after switching to the sciences from the arts, and wanted to double-check some equations with you because this is a little new to me. I'm good with the qualitative stuff, but the quantitative parts come a little harder.

Q1: A sheet of metal is 93.3 mm wide and 40.6 mm long. If it weighs 4.877 g and the density of the metal is 1.74 g/cm3, what is the thickness of the sheet (in mm)?

My answer: 4.877g / 1.74g/cm^3 = 2.80cm^3

93.3mm x 40.6mm = 3780mm (3 sig figures)

2.80cm^3 / 3780mm = 0.740mm

93.3mm x 40.6mm x 0.740mm = 2.80cm^3

Thus, the sheet is 0.740mm thick.

Q2: If PV = [gR(T+273.15)]/M, solve for M when P = 334, V = 0.350, g = 0.274, R = 62.37, and T = 39.

My answer:

PV = [gR(T+273.15)] / M

(334)(0.350) = (0.274) (62.37) (312.15) / M

116 = 5330 (3 sig figures) / M

116 / 5330 = M

0.0217 = M

Q3: An ore contains 42.3 % of the mineral ilmenite, FeTiO3, which is a source of the element Ti. How much ore must be processed in order to obtain 41.0 kg of Ti?

Molar masses

Fe = 55.85g
Ti = 2004.4g
O = 16.00g x 3 = 48.00g

total = 308.2g

308.2g / 204.4g = 1.50

41.0kg x 1.50 = 61.5kg

100 / 42.3 = 2.36

61.5 x 2.36 = 145 kg of ore are needed to obtain 41.0kg of thallium.

Thank you all in advance.

M.
 
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  • #2
First looks OK. Second looks wrong (don't round down intermediate values, check your math). Third looks wrong (check molar mass of Ti).
 
  • #3
Opps. Got the element wrong for number three. Should be...

Q3: An ore contains 42.3 % of the mineral ilmenite, FeTiO3, which is a source of the element Ti. How much ore must be processed in order to obtain 41.0 kg of Ti?

Molar masses

Fe = 55.85g
Ti = 48.77
O = 16.00g x 3 = 48.00g

total = 151.7g

308.2g / 151.7g = 3.168

41.0kg x 3.168 = 129.9kg

100 / 42.3 = 2.36

129.5 x 2.36 = 300 kg of ore are needed to obtain 41.0kg of titanium.

For number 2, is my mistake that I rounded before I got to a final answer?
 
  • #4
Maxwell Kraft said:
For number 2, is my mistake that I rounded before I got to a final answer?

One of mistakes. Check your math.
 
  • #5
Maxwell Kraft said:
116 = 5330 (3 sig figures) / M

116 / 5330 = M

0.0217 = M

QUOTE]

Look over these three lines again.
 
  • #6
Ok, let me work through this and see if I can figure out where I went wrong.

Q2: If PV = [gR(T+273.15)]/M, solve for M when P = 334, V = 0.350, g = 0.274, R = 62.37, and T = 39.

My answer:

PV = [gR(T+273.15)] / M

(334)(0.350) = (0.274) (62.37) (312.15) / M

The left side actually works out to 116.9, and I won't round the right side.

116.9 = 5334.5 / M

Moving 5334.5 to the left side, it goes from the numerator to the demoninator.

116.9 / 5334.5 = M

0.0219 = M (3 signifigant figures)

I hope I got that right, because I have another one =P

Q4: A piece of nickel wire has a diameter of 0.505 mm. If nickel has a density of 8.90 g/cc, how long (in meters) should you cut a piece of wire to obtain 0.0247 moles of nickel?

My answer: since a wire is a cyllinder, then the volume = πr2h

r = 0.505 / 2 = 0.252

Nickel weighs 58.69 g/mole, so there are 58.69g/mole x 0.0247 moles = 1.45g of the wire needed to obtain the correct length.

So v = π (0.252)2h

Since densty is 8.90 g/cc, then its volume is 1.45g / 8.90g/cc = 0.163 cc

So 0.163 cc = π (0.252)2h

h = 0.163 cc / π (0.252) 2

h = 0.817cm of wire is needed

I’d like to thank you all for giving me a hand with this. I know this is something you have to practice, but I’m afraid of practicing it wrong and forming bad habits.
 
  • #7
Maxwell Kraft said:
116.9 = 5334.5 / M

Moving 5334.5 to the left side, it goes from the numerator to the demoninator.

116.9 / 5334.5 = M

Why have you moved M from denominator to numerator? What you are doing you are dividing both sides by 5334.5, it cancels out on the right - but M stays where it was!

Ni wire looks OK to me.
 
  • #8
But, if 116.9 = 5334.5 / M, then doesn't 116.9 / 5334.5 = M? Both sides are divided by 5334.5, so the left side becomes 116.9 / 5334.5 and the right side becomes (5334.5 / M) / 5334.5, which cancels out to just M.
 
  • #9
No, it cancels out to 1/M.
 

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The purpose of basic chemistry calculations is to determine the quantities of substances involved in a chemical reaction or process. These calculations are essential for understanding the composition, properties, and behavior of matter.

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