Question

Suppose you want to invest in a mining operation. You know the total mass of four ore deposits, the percent of that deposit that is a mineral containing iron (Fe), and the chemical formula for the mineral:

Deposit A contains 31,000,000 tons of ore containing 26% sidererite [FeCO₃]

• Deposit B contains 16,000,000 tons of ore containing 63% magnetite[Fe₃O₄]
• Deposit C contains 67,000,000 tons of ore containing 25% hematite [Fe₂O₃]
• Deposit D contains 46,000,000 tons of ore containing 30% limonite [FeO(OH)]

• Knowing this information, you can determine the amount of the desired element (Fe) in the deposit. I have calculated the amount of Fe in Deposit A as an example. (Hint: You will need to obtain some information from the periodic table of elements (Links to an external site.) and do several calculations.)

• Deposit A contains 31,000,000 tons of ore containing 26% sidererite [FeCO3]

  • 0.26*31,000,000 = 8,060,000 tons of siderite
  • Atomic mass of Fe: 55.8 amu
  • Atomic mass of C: 12.0 amu
  • Atomic mass of O: 16.0 amu
  • Atomic mass of siderite: {(55.8*1) + (12.0*1)+(16.0*3)] = 115.8 amu
  • Fraction of Fe in siderite: (55.8*1)/115.8 = 0.482
  • Amount of Fe in deposit: 0.482* 8,060,000 = 3,883,834 tons of Fe

  Therefore, Deposit A contains 3,883,834 tons of Fe.

Questions 1 – 3: For deposits B, C, and D, calculate the amount of Iron in the deposit.  *Note: when completing your calculations, remember that there is a different % of Fe in each of the different minerals!

Question 4: Which of the 4 hypothetical ore deposits (A, B, C, or D) above contains the most iron by weight, and therefore is worth the most money? For this simple example, assume that all other costs are equal.

Question 5: Next, think about some of the other probable costs associated with a mining operation. Consider all of the steps involved (from discovering the deposit to bringing it to market), all of which must be considered before going into business if you want to make a profit. List as many as you can think of. (Note: Some should be related directly to mining and geology, but others will not.)

Question 6: Finally, consider all of the waste involved in the process of obtaining minerals – both in removing the ore from the ground, and (for at least some materials) in separating the desired element or compound from the rest of the material. What do you think (or know) is done with all of this waste material?

Answer 1

Question 1 :- Amount of Iron in deposits B

• Deposit B contains 16,000,000 tons of ore containing 63% magnetite[Fe3O4]
• Deposit of magnetite in Deposit B = 0.63 x 16,000,000 = 10,080,000 tons
• Atomic mass of Fe: 55.8 amu
• Atomic mass of O: 16.0 amu
• Atomic mass of magnetite = (3*55.8+4*16) = 231.4 amu
• Fraction of Fe in Magnetite = (3*55.8)/231.4 = 0.72
• Amount of Fe in depoits = 0.72*10,080,000 = 7,257,600 tons

Question 2 :- Amount of iron in deposit C

• Deposit C contains 67,000,000 tons of ore containing 25% hematite [Fe2O3]
• Deposit of Hematite in Deposit C = 0.25 x 67,000,000 tons = 16,750,000 tons
• Atomic mass of Fe: 55.8 amu
• Atomic mass of O: 16.0 amu
• Atomic mass of Hematite = (2*55.8+3*16) = 159.6 amu
• Fraction of Fe in Hematite = (2*55.8)/159.6 = 0.69
• Amount of Fe in depoits = 16,750,000 * 0.69 = 11,557,500 tons

Question No 3 :- Amount of Fe in deposit D

• Deposit D contains 46,000,000 tons of ore containing 30% limonite [FeO(OH)]
• Deposit of Limonite in Deposit D = 0.30 * 46,000,000 = 13,800,000 tons
• Atomic mass of Fe: 55.8 amu
• Atomic mass of O: 16.0 amu
• Atomic mass of H: 1 amu
• Atomic mass of Limonite = (55.8+16+1) = 72.8 amu
• Fraction of Fe in Limonite = (55.8)/72.8 = 0.76
• Amount of Fe in depoits = 13,800,000 * 0.76 = 10,488,000 tons

Question 4

Ans:- Deposits C contain Highest amount of Fe = 11,557,500 tons