Q3: Calculating Ore Processing for Thallium Extraction

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Discussion Overview

The discussion revolves around calculating ore processing for thallium extraction, specifically focusing on the quantitative aspects of chemistry problems related to molar mass, density, and volume. Participants are working through multiple questions involving calculations for thickness, molar mass, and the amount of ore needed to extract a specific quantity of titanium.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents calculations for the thickness of a metal sheet, density, and volume, arriving at a thickness of 0.740 mm.
  • Another participant challenges the second question's calculations, suggesting that intermediate rounding may have led to an incorrect answer for M.
  • A correction is made regarding the molar mass of titanium, with a revised total mass for ilmenite leading to a new calculation of 300 kg of ore needed to obtain 41.0 kg of titanium.
  • Participants discuss the importance of not rounding intermediate values in calculations, with one participant attempting to clarify their approach to solving for M in the second question.
  • There is a debate about the manipulation of equations, particularly regarding the placement of M in the equation for the second question, with differing interpretations of how to rearrange the equation.
  • One participant expresses concern about forming bad habits in calculations and seeks reassurance about their approach to a new problem involving nickel wire.

Areas of Agreement / Disagreement

Participants do not reach consensus on the correctness of the calculations for the second question, with ongoing debate about the proper handling of equations and rounding. There are multiple competing views on the calculations presented, particularly regarding the molar mass of titanium and the manipulation of the equation for M.

Contextual Notes

Some calculations depend on the accuracy of molar masses and the handling of significant figures, which remain unresolved. There are also concerns about the implications of rounding during intermediate steps in calculations.

Who May Find This Useful

This discussion may be useful for students in introductory chemistry courses, particularly those focusing on quantitative problem-solving and the importance of precision in calculations.

Maxwell Kraft
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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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First looks OK. Second looks wrong (don't round down intermediate values, check your math). Third looks wrong (check molar mass of Ti).
 
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?
 
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.
 
Maxwell Kraft said:
116 = 5330 (3 sig figures) / M

116 / 5330 = M

0.0217 = M

QUOTE]

Look over these three lines again.
 
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 significant 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.
 
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.
 
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.
 
No, it cancels out to 1/M.