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Earth's mass is estimated to be \(5.98 \times 10^{24} \mathrm{~kg}\), and titanium represents \(0.05 \%\) by mass of this total. (a) How many moles of Ti are present? (b) If half of the Ti is found as ilmenite (FeTiO \(_{3}\) ), what mass of ilmenite is present? (c) If the airline and auto industries use \(1.00 \times 10^{5}\) tons of Ti per year, how many years will it take to use up all the Ti ( 1 ton \(=2000\) lb)?

Short Answer

Expert verified
Moles of Ti: approx. 6.26 x 10^{25} mol. Mass of ilmenite: approx. 5.4 x 10^{23} kg. Years to use up Ti: approx. 5.98 \times 10^{22} years.

Step by step solution

01

- Find the mass of titanium (Ti) in Earth's mass

Compute the mass of titanium by multiplying Earth's total mass by the percentage of titanium.\[ \text{Mass of Ti} = 5.98 \times 10^{24} \text{ kg} \times \frac{0.05}{100} \]
02

- Calculate the moles of titanium

Using the molar mass of titanium (\text{Ti}), which is approximately 47.87 g/mol, we can find the number of moles of titanium. Convert the mass from kg to g first.\[ \text{mass of Ti in g} = \text{mass of Ti in kg} \times 10^3 \]Then,\[ \text{moles of Ti} = \frac{\text{mass of Ti in g}}{ \text{molar mass of Ti}} \]
03

- Determine the mass of titanium in ilmenite

Since half of the titanium is in the form of ilmenite (FeTiO_3), find out the mass of titanium that corresponds to ilmenite.\[ \text{Mass of Ti in ilmenite} = \frac{ \text{mass of Ti} }{2} \]
04

- Calculate the mass of ilmenite

Ilmenite has a molar mass given by:\[ \text{Molar mass of FeTiO}_3 = \text{molar mass of Fe (55.85 g/mol)} + \text{molar mass of Ti (47.87 g/mol)} + 3 \times \text{molar mass of O (16.00 g/mol)} \]First, find the moles of ilmenite using the proportion of titanium in it. Then,\[ \text{moles of ilmenite} = \frac{\text{mass of Ti in ilmenite in g}}{47.87} \]Finally, convert back to mass:\[ \text{mass of ilmenite} = \text{moles of ilmenite} \times \text{molar mass of FeTiO}_3 \]
05

- Convert consumption rate to kg per year

Given that the airline and auto industries use 1.00 \times 10^5 tons of Ti per year and 1 ton = 2000 lbs, convert the usage rate to kilograms:\[ \text{mass of Ti per year in kg} = 1.00 \times 10^5 \text{ tons/year} \times 2000 \text{ lbs/ton} \times 0.453592 \text{ kg/lb} \]
06

- Calculate how many years it will take to use up all titanium

Determine the total number of years by dividing the total mass of titanium by the annual usage rate.\[ \text{years} = \frac{\text{mass of Ti in kg}}{\text{mass of Ti used per year in kg}} \]

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Key Concepts

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

mass percentage
When dealing with large quantities of substances, like the mass of titanium (Ti) in the Earth, we often use mass percentage to express the concentration of one component in a mixture. Here, titanium makes up 0.05% of Earth's mass. Calculating the mass of Ti from its percentage means multiplying Earth's total mass by the percentage (converted to a decimal). To find the mass of titanium, you use:

\[ \text{Mass of Ti} = 5.98 \times 10^{24} \text{ kg} \times \frac{0.05}{100} \]

This step converts the percentage to a workable mass and is integral in proceeding with further calculations related to titanium within Earth's mass. Mass percentage helps convert massive or microscopic scale quantities into more tangible data.
molar mass
Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance (moles). It's expressed in g/mol. For this exercise, titanium has a molar mass of approximately 47.87 g/mol. To find out how many moles are present in a certain mass of titanium, you need to know its molar mass:

\[ \text{moles of Ti} = \frac{\text{mass of Ti in g}}{\text{molar mass of Ti}} \]

Since Earth's Ti mass is derived in kilograms, convert kg to grams by multiplying by 1000. Molar mass connects the mass of a substance with the amount of substance and is particularly useful for converting between these two measures in chemical calculations.
mole calculations
Mole calculations, a fundamental concept in chemistry, involve converting between mass, moles, and number of entities (atoms, molecules). Starting with the known mass of Ti, we first convert the mass from kilograms to grams, and then we can calculate the moles of Ti:

\[ \text{moles of Ti} = \frac{\text{mass of Ti in g}}{47.87} \]

Once you have the moles of titanium, you can further calculate the mass proportion of titanium in other compounds. For instance, calculating the mass of ilmenite (FeTiO\textsubscript{3}), given titanium's role in it, involves using the formula of ilmenite to find its molar mass and then converting moles back to mass. This process shows the practical application of mole calculations in compound formation and analysis.
chemical formulas
Chemical formulas like FeTiO\textsubscript{3} (ilmenite) provide a clear picture of the composition of a compound. They indicate the types and numbers of atoms involved. In our exercise, ilmenite consists of iron (Fe), titanium (Ti), and oxygen (O). To solve for the mass of ilmenite, we need to know its molar mass:

\[ \text{Molar mass of FeTiO}_3 = 55.85 \text{ g/mol (Fe)} + 47.87 \text{ g/mol (Ti)} + 3 \times 16.00 \text{ g/mol (O)} \]

Understanding chemical formulas allows us to understand the weight and proportion of each element in the compound. This understanding is crucial for calculating mass ratios and compositions in various chemical reactions and compounds, such as converting the mass of titanium into the mass of ilmenite.

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Most popular questions from this chapter

Selenium is prepared by the reaction of \(\mathrm{H}_{2} \mathrm{SeO}_{3}\) with gascous \(\mathrm{SO}_{2}\) (a) What redox process does the sulfur dioxide undergo? What is the oxidation state of sulfur in the product? (b) Given that the reaction occurs in acidic aqueous solution, what is the formula of the sulfur-containing species? (c) Write the balanced redox equation for the process.

The key step in the manufacture of sulfuric acid is the oxidation of sulfur dioxide in the presence of a catalyst, such as \(\mathrm{V}_{2} \mathrm{O}_{5}\) At \(727^{\circ} \mathrm{C}, 0.010 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) is injected into an empty \(2.00-\mathrm{L}\) container \(\left(K_{p}=3.18\right)\) (a) What is the equilibrium pressure of \(\mathrm{O}_{2}\) that is needed to maintain a \(1 / 1\) mole ratio of \(\mathrm{SO}_{3}\) to \(\mathrm{SO}_{2} ?\) (b) What is the equilibrium pressure of \(\mathrm{O}_{2}\) needed to maintain a \(95 / 5\) mole ratio of \(\mathrm{SO}_{3}\) to \(\mathrm{SO}_{2} ?\)

Hydrogen is by far the most abundant element cosmically. In interstellar space, it exists mainly as \(\mathrm{H}_{2}\). In contrast, on Earth, it exists very rarely as \(\mathrm{H}_{2}\) and is ninth in abundance in the crust. Why is hydrogen so abundant in the universe? Why is hydrogen so rare as a diatomic gas in Earth's atmosphere?

A key part of the carbon cycle is the fixation of \(\mathrm{CO}_{2}\) by photosynthesis to produce carbohydrates and oxygen gas. (a) Using the formula \(\left(\mathrm{CH}_{2} \mathrm{O}\right)_{n}\) to represent a carbohydrate, write a balanced equation for the photosynthetic reaction. (b) If a tree fixes \(48 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) per day, what volume of \(\mathrm{O}_{2}\) gas measured at 1.0 atm and \(78^{\circ} \mathrm{F}\) does the tree produce per day? (c) What volume of air ( \(0.040 \mathrm{~mol} \% \mathrm{CO}_{2}\) ) at the same conditions contains this amount of \(\mathrm{CO}_{2} ?\)

Because of their different molar masses, \(\mathrm{H}_{2}\) and \(\mathrm{D}_{2}\) effuse at different rates (Section 5.5 ). (a) If it takes 16.5 min for \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to effuse, how long does it take for \(0.10 \mathrm{~mol}\) of \(\mathrm{D}_{2}\) to do so in the same apparatus at the same \(T\) and \(P ?\) (b) How many effusion steps does it take to separate an equimolar mixture of \(\mathrm{D}_{2}\) and \(\mathrm{H}_{2}\) to \(99 \mathrm{~mol} \%\) purity?

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