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A certain metal oxide has the formula MO where \(\mathrm{M}\) denotes the metal. A \(39.46-\mathrm{g}\) sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, \(31.70 \mathrm{~g}\) of the metal is left over. If \(\mathrm{O}\) has an atomic mass of 16.00 amu, calculate the atomic mass of \(\mathrm{M}\) and identify the element.

Short Answer

Expert verified
The atomic mass of M is approximately 65.36 amu; the metal is zinc (Zn).

Step by step solution

01

Determine Mass of Oxygen Removed

Since the initial mass of the compound MO is 39.46 g and the mass of metal M left after removing oxygen is 31.70 g, the mass of oxygen removed can be calculated as follows:\[\text{Mass of Oxygen} = 39.46 \, \text{g} - 31.70 \, \text{g} = 7.76 \, \text{g}.\]
02

Use Moles of Oxygen to Find Moles of Compound

Using the atomic mass of oxygen, we can find the moles of oxygen in 7.76 g:\[\text{Moles of O} = \frac{7.76 \, \text{g}}{16.00 \, \text{g/mol}} = 0.485 \, \text{mol}.\]Since the formula of the compound is MO, the moles of MO is also 0.485 mol.
03

Calculate the Moles of Metal M

Because the compound formula is MO, the moles of metal M in the compound is equal to the moles of oxygen, which is 0.485 mol.
04

Calculate the Atomic Mass of Metal M

The mass of metal M is given as 31.70 g. Using the moles calculated, find the atomic mass of M:\[\text{Atomic Mass of M} = \frac{31.70 \, \text{g}}{0.485 \, \text{mol}} \approx 65.36 \, \text{g/mol}.\]
05

Identify the Element

Using a periodic table, match the calculated atomic mass to find the closest element. The atomic mass closest to 65.36 is zinc (Zn), with an atomic mass of approximately 65.38 amu.

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Key Concepts

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

Metal Oxides
Metal oxides are compounds that consist of a metal element and oxygen. These compounds occur naturally and can have various applications, including forming protective layers on metal surfaces to prevent rusting. In chemistry, understanding metal oxides is important as they are often involved in reactions that remove or release oxygen, such as in the given problem where the oxide reacts with hydrogen. This reaction helps to isolate the metal by removing oxygen, which typically results in the formation of water. Seeing how metal oxides react can give insights into chemical properties and reactivity of metals under various conditions. Metal oxides can vary widely depending on the specific metal involved, affecting properties such as color, density, and chemical reactivity.
Stoichiometry
Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. By using stoichiometry, you can predict how much of each substance is needed or produced in a reaction. In the case of the metal oxide MO, stoichiometry helps us understand the ratio between metal (M) and oxygen (O) in the compound. It tells us that for every one mole of oxygen, there is one mole of metal. This 1:1 ratio is crucial in determining the exact amounts of substances involved, especially when calculating unknown components like the atomic mass of \(\text{M}\). With stoichiometry, we translate these ratios into measurable amounts using the concept of moles, which simplifies complex chemical equations into manageable calculations.
Mole Concept
The mole concept is a key principle in chemistry that represents a specific number of particles, usually atoms or molecules, in a given substance. A mole is defined as \(6.022 \times 10^{23}\) particles, referring to Avogadro's number. It is as essential in chemistry as dozen is to everyday counting. This concept allows chemists to count atoms by weighing them, making quantitative chemical analysis possible. For calculating the atomic mass of \(\text{M}\) in our problem, we relied on the mole concept to convert grams into moles, making it easier to work with large amounts of particles. By knowing the moles of oxygen used, we directly inferred the moles of the metal involved, showcasing how crucial and practical the mole concept is in solving chemical equations and reactions.
Element Identification
Element identification involves determining which element corresponds to an unidentified atomic mass based on known values. Once you know the molar mass obtained from experimental data, comparing it with the periodic table allows you to match this mass with a specific element. In the given problem, the calculated atomic mass of element \(\text{M}\) was about \(65.36 \, \text{g/mol}\), which closely matched that of zinc (approximately \(65.38 \, \text{amu}\)). This identification process is critical in many scientific fields, as it provides a deeper understanding of the material's composition and its potential reactions and uses. Element identification is a step crucial for confirming theoretical data with experimental outcomes, ensuring that chemical equations, materials, and reactions are correctly interpreted and understood.

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Most popular questions from this chapter

The compound 2,3 -dimercaptopropanol \(\left(\mathrm{HSCH}_{2} \mathrm{CHSHCH}_{2} \mathrm{OH}\right),\) commonly known as British Anti-Lewisite (BAL), was developed during World War I as an antidote to arsenic-containing poison gas. (a) If each BAL molecule binds one arsenic (As) atom, how many As atoms can be removed by \(1.0 \mathrm{~g}\) of BAL? (b) BAL can also be used to remove poisonous heavy metals like mercury \((\mathrm{Hg})\) and lead \((\mathrm{Pb})\). If each \(\mathrm{BAL}\) binds one \(\mathrm{Hg}\) atom, calculate the mass percent of \(\mathrm{Hg}\) in a BAL-Hg complex. (An \(\mathrm{H}\) atom is removed when a BAL molecule binds an \(\mathrm{Hg}\) atom.)

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Give an everyday example that illustrates the limiting reactant concept.

Chemical analysis shows that the oxygen-carrying protein hemoglobin is 0.34 percent Fe by mass. What is the minimum possible molar mass of hemoglobin? The actual molar mass of hemoglobin is about \(65,000 \mathrm{~g}\). How would you account for the discrepancy between your minimum value and the experimental value?

When combined, aqueous solutions of sulfuric acid and potassium hydroxide react to form water and aqueous potassium sulfate according to the following equation (unbalanced): $$ \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{KOH}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) $$ Determine what mass of water is produced when a beaker containing \(100.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(250 \mathrm{~mL}\) water is added to a larger beaker containing \(100.0 \mathrm{~g}\) KOH dissolved in \(225 \mathrm{~mL}\) water. Determine the mass amounts of each substance (other than water) present in the large beaker when the reaction is complete.

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