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Chalcopyrite, the principal ore of copper (Cu), contains 34.63 percent Cu by mass. How many grams of Cu can be obtained from \(7.35 \times 10^{3} \mathrm{~kg}\) of the ore?

Short Answer

Expert verified
2,545,305 grams of copper can be obtained.

Step by step solution

01

Convert Ore from Kilograms to Grams

First, you need to convert the mass of the ore from kilograms to grams because the percentage is given in terms of grams. We know that 1 kilogram is equal to 1000 grams, so multiply the mass of the ore by 1000: \[ 7.35 \times 10^3 \text{ kg} \times 1000 \text{ g/kg} = 7.35 \times 10^6 \text{ g}. \]
02

Calculate Mass of Copper

Now, use the percentage by mass to find the mass of copper in the ore. Multiply the total grams of ore by the percentage of copper:\[ 7.35 \times 10^6 \text{ g} \times \frac{34.63}{100} = 2.545305 \times 10^6 \text{ g}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Copper Extraction
Copper extraction is a vital process in the mining industry. It involves extracting copper from its ore, chalcopyrite ( CuFeS_2 ). Chalcopyrite is a copper iron sulfide mineral and is among the primary ores from which copper is extracted. This process is crucial because copper is an essential material used in electrical wiring, plumbing, and various industrial machines due to its excellent electrical conductivity and malleability. Understanding the copper extraction process allows industries to efficiently extract this valuable metal.

During extraction, the raw ore undergoes several physical and chemical processes. Typically, the ore is first crushed and then concentrated via froth flotation. This step separates the chalcopyrite from other minerals. Once a high concentration of chalcopyrite is obtained, it is then subjected to pyrometallurgical methods, involving heating and melting, where techniques like smelting are used to isolate copper from sulfur and iron.
Percentage Composition
Percentage composition in chemistry refers to the proportion of a specific element within a compound, relative to the total mass of the compound, expressed as a percentage.

For chalcopyrite, knowing the percentage composition of copper is crucial in determining how much copper can be extracted. In this exercise, chalcopyrite contains 34.63% copper by mass. To find this, we divide the mass of copper by the total mass of the chalcopyrite and then multiply by 100 to get the percentage:
  • Mass of copper in chalcopyrite.
  • Total mass of chalcopyrite.
  • Formula: \(\frac{\text{mass of Cu}}{\text{total mass of ore}} \times 100 = \text{percentage composition}\).
Knowing this percentage allows us to calculate the extractable copper from any given amount of chalcopyrite ore, as shown in our exercise.
Mass Conversion
Mass conversion is an important step often needed in chemistry to standardize units for calculation. It simplifies calculations by ensuring the quantities involved are consistently expressed in the same units.

In our exercise, we need to convert the mass of chalcopyrite ore from kilograms to grams before determining the mass of copper. Here's why:
  • 1 kilogram equals 1000 grams.
  • Given: \(7.35 \times 10^3\) kilograms of ore.
  • Conversion: Multiply by 1000, resulting in \(7.35 \times 10^6\) grams.
Converted into grams, it aligns perfectly with the percentage composition, which dictates how much of that mass is copper.
Ore Processing
Ore processing is a multidisciplinary procedure for turning ores into raw materials that can be used by various industries. It encompasses a series of stages, one of which is crucially discussed in our exercise: knowing the amount of a valuable metal like copper present in your ore.

The steps involved in ore processing, particularly for chalcopyrite, typically include:
  • Crushing and grinding the ore to increase the surface area.
  • Froth flotation to concentrate the ore.
  • Smelting and refining to isolate the copper.
Through these processes, we optimize the yield of copper. Understanding the entire processing chain is crucial not just for industrial efficiency but for economic valuation of a mining endeavor.

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Most popular questions from this chapter

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