Problem 136
Question
The stainless steel used in the Gateway Arch in St. Louis contains \(72.0 \% \mathrm{Fe}, 19.0 \% \mathrm{Cr}\), and the remainder is nickel. A 10.0 -g sample of this stainless steel is treated to convert the metals to their oxides: \(10.3 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}, 2.71 \mathrm{~g}\) \(\mathrm{Cr}_{2} \mathrm{O}_{3},\) and \(1.14 \mathrm{~g} \mathrm{NiO} .\) Calculate the mass percent of each metal in the sample.
Step-by-Step Solution
Verified Answer
Fe: 72.1%, Cr: 18.54%, Ni: 8.96%
1Step 1: Calculate total oxygen mass in oxide compounds
First, we calculate the masses of oxygen present in each of the metal oxides.- For \(\text{Fe}_2\text{O}_3\): - Molar mass of \(\text{Fe}_2\text{O}_3\) is \(159.7\,\text{g/mol}\) (56 for Fe and 16 for O; **2 Fe and 3 O**), so weight % of O is \(\frac{48}{159.7} \approx 0.3\). - Mass of O in \(10.3 \text{ g} \text{Fe}_2\text{O}_3\) is \(10.3 \times 0.3 = 3.09\text{ g}\).- For \(\text{Cr}_2\text{O}_3\): - Molar mass of \(\text{Cr}_2\text{O}_3\) is \(152\,\text{g/mol}\) (52 for Cr and 16 for O; **2 Cr and 3 O**), so weight % of O is \(\frac{48}{152} \approx 0.3158\). - Mass of O in \(2.71 \text{ g} \text{Cr}_2\text{O}_3\) is \(2.71 \times 0.3158 = 0.856\text{ g}\).- For \(\text{NiO}\): - Molar mass of \(\text{NiO}\) is \(74.7\,\text{g/mol}\) (58.7 for Ni and 16 for O), so weight % of O is \(\frac{16}{74.7} \approx 0.2143\). - Mass of O in \(1.14 \text{ g} \text{NiO}\) is \(1.14 \times 0.2143 = 0.244\text{ g}\).
2Step 2: Calculate the mass of each metal
Subtract the mass of oxygen from each oxide compound to find the mass of the metal alone.- Mass of Fe in \(\text{Fe}_2\text{O}_3\): - Metal mass = \(10.3 - 3.09 = 7.21\text{ g}\).- Mass of Cr in \(\text{Cr}_2\text{O}_3\): - Metal mass = \(2.71 - 0.856 = 1.854\text{ g}\).- Mass of Ni in \(\text{NiO}\): - Metal mass = \(1.14 - 0.244 = 0.896\text{ g}\).
3Step 3: Calculate mass percent of each metal
Use the mass of each metal and the total mass of the sample (10.0 g) to find the percentage by mass.- Mass percent of Fe: - \(\text{Percent of Fe} = \left(\frac{7.21}{10.0}\right) \times 100\% = 72.1\%\).- Mass percent of Cr: - \(\text{Percent of Cr} = \left(\frac{1.854}{10.0}\right) \times 100\% = 18.54\%\).- Mass percent of Ni: - \(\text{Percent of Ni} = \left(\frac{0.896}{10.0}\right) \times 100\% = 8.96\%\).
Key Concepts
Mass Percent CalculationMolar MassOxidation of Metals
Mass Percent Calculation
Mass percent calculation is an essential concept in stoichiometry. It helps determine the proportion of each constituent element within a compound or mixture. In the problem, we are asked to find the percentage mass of metals in a stainless steel sample treated to convert them into oxides. This requires understanding how much of each metal was present before it reacted with oxygen.
Then, to calculate the mass percent of a specific metal, divide the mass of that metal by the total mass of the sample, and multiply by 100. For example, if the mass of iron (Fe) in the sample is 7.21 g: \[ \text{Percent of Fe} = \left( \frac{7.21}{10.0} \right) \times 100\% \]This way, we can express the metal content as a percentage of the total sample mass. Mass percent thus provides insight into the distribution of elements in metal alloys, crucial for quality control and analysis.
- First, calculate the total mass of each metal oxide.
- Subtract the mass of oxygen within each oxide to find the mass of the metal alone.
Then, to calculate the mass percent of a specific metal, divide the mass of that metal by the total mass of the sample, and multiply by 100. For example, if the mass of iron (Fe) in the sample is 7.21 g: \[ \text{Percent of Fe} = \left( \frac{7.21}{10.0} \right) \times 100\% \]This way, we can express the metal content as a percentage of the total sample mass. Mass percent thus provides insight into the distribution of elements in metal alloys, crucial for quality control and analysis.
Molar Mass
The molar mass concept plays a pivotal role in determining the composition of compounds and reacting masses. It is the mass of a single mole of a compound, typically expressed in grams per mole (g/mol). For chemical compounds like \(\text{Fe}_2\text{O}_3\), \(\text{Cr}_2\text{O}_3\), and \(\text{NiO}\), their molar masses are essential in calculating the amount of each element within the respective oxides.
Understanding molar mass is critical for manipulating chemical equations and converting between mass and moles, particularly for processes involving multiple reactions or substances.
- To find the molar mass of \(\text{Fe}_2\text{O}_3\): Sum up the atomic masses of 2 Fe (56) and 3 O (16), resulting in 159.7 g/mol.
- For \(\text{Cr}_2\text{O}_3\): Add 2 Cr (52) and 3 O (16), leading to 152 g/mol.
- In \(\text{NiO}\): Combine Ni (58.7) and O (16), arriving at 74.7 g/mol.
Understanding molar mass is critical for manipulating chemical equations and converting between mass and moles, particularly for processes involving multiple reactions or substances.
Oxidation of Metals
Oxidation of metals is a chemical process where metals react with oxygen to form metal oxides. This reaction fundamentally involves the transfer of electrons, where the metal atoms lose electrons, becoming oxidized, and the oxygen gains electrons, forming oxides. In the Gateway Arch stainless steel problem, each metal in the alloy reacts with oxygen to form its respective oxide.
These oxides have differing compositions and properties from the pure metals. Metal oxidation is significant in both industrial applications and natural processes, as it impacts material durability, appearance, and chemical resistance. Additionally, understanding the oxidation process is crucial for techniques like corrosion prevention and metallurgy.
- Iron (Fe) forms \(\text{Fe}_2\text{O}_3\).
- Chromium (Cr) forms \(\text{Cr}_2\text{O}_3\).
- Nickel (Ni) forms \(\text{NiO}\).
These oxides have differing compositions and properties from the pure metals. Metal oxidation is significant in both industrial applications and natural processes, as it impacts material durability, appearance, and chemical resistance. Additionally, understanding the oxidation process is crucial for techniques like corrosion prevention and metallurgy.
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