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Chapter 3: Problem 37

(a) What is the mass, in grams, of \(2.50 \times 10^{-3} \mathrm{~mol}\) of ammonium phosphate? (b) How many moles of chloride ions are in \(0.2550 \mathrm{~g}\) of aluminum chloride? (c) What is the mass, in grams, of \(7.70 \times 10^{20}\) molecules of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\) (d) What is the molar mass of cholesterol if \(0.00105 \mathrm{~mol}\) has a mass of \(0.406 \mathrm{~g}\) ?

Short Answer

Expert verified
(a) The mass of \(2.50 \times 10^{-3} \mathrm{~mol}\) of ammonium phosphate is \(0.373 \mathrm{~g}\). (b) There are \(5.74 \times 10^{-3} \mathrm{~mol}\) of chloride ions in \(0.2550 \mathrm{~g}\) of aluminum chloride. (c) The mass of \(7.70 \times 10^{20}\) molecules of caffeine is \(0.248 \mathrm{~g}\). (d) The molar mass of cholesterol is \(386.67 \mathrm{~g/mol}\).

Step by step solution

01

Find the molar mass of ammonium phosphate

First, we need to determine the chemical formula of ammonium phosphate, which is (NH4)3PO4. Then, the molar mass can be calculated by finding the atomic masses of each element and adding them up: Molar mass = 3 × (1 × 14.01 + 4 × 1.01) + 1 × (15.99 + 1 × 15.99 + 4 × 1.008) = 149.09 g/mol.
02

Calculate the mass of given moles

Next, we need to calculate the mass of given moles using the molar mass. We know that mass = moles × molar mass: Mass = (2.50 × 10⁻³ mol) × (149.09 g/mol) = 0.373 g. Question (b):
03

Find the molar mass of aluminum chloride

First, we need to determine the chemical formula of aluminum chloride, which is AlCl3. Then, we can calculate the molar mass by finding the atomic masses of each element and adding them up: Molar mass = 1 × 26.98 + 3 × 35.45 = 133.33 g/mol.
04

Calculate the moles of aluminum chloride

We can calculate the moles of aluminum chloride using the given mass and molar mass. We know that moles = mass/molar mass: Moles = 0.2550 g / 133.33 g/mol = 1.91 × 10⁻³ mol.
05

Calculate the moles of chloride ions

For each molecule of aluminum chloride, there are three chloride ions. So, the moles of chloride ions can be found by multiplying the moles of aluminum chloride with 3: Moles of Cl⁻ = 1.91 × 10⁻³ mol × 3 = 5.74 × 10⁻³ mol. Question (c):
06

Find the molar mass of caffeine

First, we need to determine the chemical formula of caffeine, which is C8H10N4O2. Then, we can calculate the molar mass by finding the atomic masses of each element and adding them up: Molar mass = 8 × 12.01 + 10 × 1.01 + 4 × 14.01 + 2 × 16.00 = 194.19 g/mol.
07

Calculate moles of caffeine

We can calculate the moles of caffeine using Avogadro's number (6.022 × 10²³ molecules/mol) and the given number of molecules: Moles = (7.70 × 10²⁰ molecules) / (6.022 × 10²³ molecules/mol) = 1.28 × 10⁻³ mol.
08

Calculate the mass of the given number of caffeine molecules

We can calculate the mass using the molar mass and the moles of caffeine. We know that mass = moles × molar mass: Mass = (1.28 × 10⁻³ mol) × (194.19 g/mol) = 0.248 g. Question (d):
09

Convert mass and moles to molar mass

To find the molar mass, we can use the mass and moles of cholesterol. We know that molar mass = mass / moles: Molar mass = 0.406 g / 0.00105 mol = 386.67 g/mol.

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Most popular questions from this chapter

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: $\mathrm{O}_{3}(g)+2 \mathrm{NaI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ $$ \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q) $$ (a) How many moles of sodium iodide are needed to remove $5.95 \times 10^{-6} \mathrm{~mol}\( of \)\mathrm{O}_{3} ?(\mathbf{b})$ How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3}\) ?

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol \((\mathrm{THC}),\) the active ingredient in marijuana, is required to produce intoxication. The molecular formula of \(\mathrm{THC}\) is $\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2} .\( How many moles of THC does this \)25 \mu \mathrm{g}$ represent? How many molecules?

Calculate the following quantities: (a) mass, in grams, of \(1.50 \times 10^{-2} \mathrm{~mol} \mathrm{CdS}\) (b) number of moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(86.6 \mathrm{~g}\) of this substance (c) number of molecules in $8.447 \times 10^{-2} \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{6}$ (d) number of \(\mathrm{O}\) atoms in $6.25 \times 10^{-3} \mathrm{~mol} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of $\mathrm{CO}, 0.733 \mathrm{~g}\( of \)\mathrm{CO}_{2},\( and \)0.450 \mathrm{~g}\( of \)\mathrm{H}_{2} \mathrm{O}$ were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing $3.50 \mathrm{~g}\( of sodium carbonate is mixed with one containing \)5.00 \mathrm{~g}$ of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?
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