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Chapter 3: Problem 35

(a) What is the mass, in grams, of \(2.50 \times 10^{-3}\) 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 \(0.00105 \mathrm{~mol}\) weighs \(0.406 \mathrm{~g}\) ?

Short Answer

Expert verified
(a) The mass of \(2.50 \times 10^{-3}\) mol of ammonium phosphate is 0.373 g. (b) There are \(5.73 \times 10^{-3}\) moles of chloride ions in 0.2550 g of aluminum chloride. (c) The mass of \(7.70 \times 10^{20}\) molecules of caffeine is 0.249 g. (d) The molar mass of cholesterol is 386.67 g/mol.

Step by step solution

01

(a) Find molar mass of Ammonium Phosphate

To find the mass, we first have to find the molar mass of ammonium phosphate, which has the formula (NH4)3PO4. The molar mass of N = 14.01 g/mol The molar mass of H = 1.01 g/mol The molar mass of P = 30.97 g/mol The molar mass of O = 16.00 g/mol The molar mass of ammonium phosphate is: \((3 * 14.01) + (12 * 1.01) + 30.97 + (4 * 16.00) = 149.09 \mathrm{~g/mol}\)
02

(a) Find mass

Now we can find the mass, by multiplying the given moles with the molar mass of ammonium phosphate. Mass = moles * molar mass Mass = \(2.50 \times 10^{-3}\) mol * 149.09 g/mol = 0.373 g
03

(b) Moles of Aluminum Chloride

First find the molar mass of aluminum chloride, AlCl3. The molar mass of Al = 26.98 g/mol The molar mass of Cl = 35.45 g/mol Molar mass of AlCl3 = 26.98 + (3 * 35.45) = 133.33 g/mol Now find the moles of aluminum chloride given the mass. Moles = mass / molar mass Moles = \(0.2550 \mathrm{~g} / 133.33 \mathrm{~g/mol}\) = \(1.91 \times 10^{-3}\) mol
04

(b) Moles of Chloride

In 1 mol of AlCl3, there are 3 moles of chloride ions (Cl-). Therefore, the moles of chloride ions in \(1.91 \times 10^{-3}\) mol of AlCl3 is: \((1.91 \times 10^{-3} \mathrm{~mol}) * 3 = 5.73 \times 10^{-3} \mathrm{~mol}\)
05

(c) Molar mass and moles of Caffeine

First, find the molar mass of caffeine, C8H10N4O2. Molar mass of C = 12.01 g/mol Molar mass of N = 14.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of caffeine = \((8 * 12.01) + (10 * 1.01) + (4 * 14.01) + (2 * 16.00) = 194.19 \mathrm{~g/mol}\) Now, find the moles of caffeine from the given number of molecules, using Avogadro's number (\(6.022 \times 10^{23} \mathrm{~mol^{-1}}\)). Moles = number of molecules / Avogadro's number Moles = \(7.70 \times 10^{20}\) molecules / \(6.022 \times 10^{23} \mathrm{~mol^{-1}}\) = \(1.28 \times 10^{-3} \mathrm{~mol}\)
06

(c) Mass of Caffeine

Now, find the mass from the moles and molar mass of caffeine. Mass = moles * molar mass Mass = \(1.28 \times 10^{-3} \mathrm{~mol} * 194.19 \mathrm{~g/mol}\) = 0.249 g
07

(d) Molar Mass of Cholesterol

Use the provided mass and moles of cholesterol to find its molar mass. Molar mass = mass / moles Molar mass = \(0.406 \mathrm{~g} / 0.00105 \mathrm{~mol}\) = 386.67 g/mol

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Most popular questions from this chapter

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(1.50 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Glycine, an amino acid used by organisms to make proteins, is represented by the molecular model below. (a) Write its molecular formula. (b) Determine its molar mass. (c) Calculate the mass of 3 moles of glycine. (d) Calculate the percent nitrogen by mass in glycine. [Sections \(3.3\) and \(3.5]\)

A particular coal contains \(2.5 \%\) sulfur by mass. When this coal is burned at a power plant, the sulfur is con- verted into sulfur dioxide gas, which is a pollutant. To reduce sulfur dioxide emissions, calcium oxide (lime) is used. The sulfur dioxide reacts with calcium oxide to form solid calcium sulfite. (a) Write the balanced chemical equation for the reaction. (b) If the coal is burned in a power plant that uses 2000 tons of coal per day, what mass of calcium oxide is required daily to eliminate the sulfur dioxide? (c) How many grams of calcium sulfite are produced daily by this power plant?

(a) Diamond is a natural form of pure carbon. How many moles of carbon are in a \(1.25\) -carat diamond (1 carat \(=0.200 \mathrm{~g}\) )? How many atoms are in this diamond? (b) The molecular formula of acetylsalicylic acid (aspirin), one of the most common pain relievers, is \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\). How many moles of \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) are in a \(0.500-\mathrm{g}\) tablet of aspirin? How many molecules of \(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{O}_{4}\) are in this tablet?

Determine the empirical formula of each of the following compounds if a sample contains (a) \(0.104 \mathrm{~mol} \mathrm{~K}\). \(0.052 \mathrm{~mol} \mathrm{C}\), and \(0.156 \mathrm{~mol} \mathrm{O} ;\) (b) \(5.28 \mathrm{~g} \mathrm{Sn}\) and \(3.37 \mathrm{~g} \mathrm{~F}\); (c) \(87.5 \% \mathrm{~N}\) and \(12.5 \% \mathrm{H}\) by mass.
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