Question

Chromium is obtained by heating chromium(III) oxide with carbon. Calculate the mass percent of chromium in the oxide, and then use this value to calculate the quantity of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) required to produce \(850 \mathrm{kg}\) of chromium metal.

Short answer

Cr constitutes 68.42% of mass in Cr extsubscript{2}O extsubscript{3}, and 1242.88 kg of Cr extsubscript{2}O extsubscript{3} is required to produce 850 kg of chromium.

Step-by-step solution

Determine Molar Masses

First, calculate the molar mass of chromium (Cr) and chromium(III) oxide (Cr extsubscript{2}O extsubscript{3}). The atomic mass of Cr is approximately 52.00 g/mol, and oxygen (O) is about 16.00 g/mol. Thus, \( \text{Molar mass of Cr} = 52.00 \text{ g/mol} \times 2 = 104.00 \text{ g/mol} \). \( \text{Molar mass of } \text{O}_{3} = 16.00 \text{ g/mol} \times 3 = 48.00 \text{ g/mol} \). Thus, total molar mass of Cr extsubscript{2}O extsubscript{3} is \( 104.00 + 48.00 = 152.00 \text{ g/mol} \).

Calculate Mass Percent of Chromium

The mass percent of an element in a compound is given by the formula: \( \text{Mass percent of Cr} = \left( \frac{\text{Mass of Cr in the compound}}{\text{Total mass of the compound}} \right) \times 100 \% \). Insert values: \( \left( \frac{104.00 \text{ g/mol}}{152.00 \text{ g/mol}} \right) \times 100\% = 68.42\% \). Thus, Cr constitutes 68.42% of Cr extsubscript{2}O extsubscript{3} by mass.

Mass of Cr2O3 Needed

To produce 850 kg of chromium, you need to calculate the amount of Cr extsubscript{2}O extsubscript{3} based on the mass percent of chromium. Use the formula: \( \text{Mass of Cr extsubscript{2}O extsubscript{3}} = \frac{\text{Mass of Cr}}{\text{Mass percent of Cr}} \times 100\% \). Substitute the known values: \( \frac{850 \text{ kg}}{68.42\%} \times 100\% = 1242.88 \text{ kg} \). Therefore, you need 1242.88 kg of Cr extsubscript{2}O extsubscript{3}.

Key concepts

Molar Mass Calculation

Molar mass calculation is an essential part of understanding chemical composition. The molar mass of an element is the weight of one mole of its atoms, often given in grams per mole (g/mol). When calculating the molar mass of a compound like chromium(III) oxide (\(\text{Cr}_2\text{O}_3\)), it is crucial to add up the molar masses of all the constituent atoms.

For chromium, the atomic mass is approximately 52.00 g/mol. Since there are two chromium atoms in CrₓOₓ, we multiply: 52.00 g/mol × 2 = 104.00 g/mol.

Oxygen has an atomic mass of about 16.00 g/mol. With three oxygen atoms in the compound, we get: 16.00 g/mol × 3 = 48.00 g/mol.

Adding these gives the total molar mass: 104.00 g/mol + 48.00 g/mol = 152.00 g/mol for Cr₂O₃.

• This shows that finding molar mass involves calculating individual components.
• Such detailed breakdowns are crucial for understanding complex compounds.

Mass Percent Calculation

Mass percent calculation helps in determining the proportion of a particular element within a compound. This is done by comparing the molar mass of the element with the molar mass of the entire compound, then multiplying by 100 to convert it to a percentage.

For chromium in Cr₂O₃, its mass percent is:\[\left( \frac{104.00 \, \text{g/mol}}{152.00 \, \text{g/mol}} \right) \times 100\% = 68.42\%\]Hence, 68.42% of Cr₂O₃ by mass is chromium.

This calculation is vital. It informs us how much chromium is present and thus aids in figuring out how much of the compound is required for chemical reactions involving chromium.

• Mass percent provides useful information about component ratios.
• It assists in calculating necessary quantities for the desired yield from reactions.

Chemical Reactions

Chemical reactions are processes where reactants are transformed into products through rearrangements of atoms. The reaction of chromium(III) oxide with carbon to produce chromium and carbon dioxide is a classic example.

In this example, Cr₂O₃ is reduced by carbon, leaving chromium as the product. The understanding of molar masses and mass percent is crucial here. It allows one to derive the required amounts of reactants to achieve a certain mass of product, specifically in industrial applications like metal extraction.

The reduction involves breaking the bond between chromium and oxygen and forming a new bond between oxygen and carbon.

• Chemical equations represent such reactions and are balanced by ensuring equal atoms on both sides.
• The role of stoichiometry is crucial for these calculations.

Compound Composition

Understanding compound composition involves knowing the types and numbers of atoms involved in a chemical compound, such as chromium(III) oxide (Cr₂O₃). This compound is composed of: - 2 chromium atoms - 3 oxygen atoms

Such insights into the elemental makeup are crucial for chemical analysis, calculation of molar masses, and determining the mass percent of each element in a compound.

Knowing the composition helps in various applications, like predicting the amount of a specific element available in a chemical reaction. It defines the interaction characteristics of the compound and is essential for accurately projecting yields from reactions like the extraction of chromium metal from its oxide.

• Understanding a compound's composition helps with precise calculations and expectations in chemical processes.
• It forms the basis for all further chemical analysis.