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Chapter 3: Problem 66

Prevacid is used to treat gastroesophageal reflux disease \((\mathrm{GERD})\) . The chemical formula of Prevacid is $\mathrm{C}_{16} \mathrm{H}_{14} \mathrm{F}_{3} \mathrm{N}_{3} \mathrm{O}_{2} \mathrm{S}$ . a. What is the molar mass of Prevacid? b. What mass of fluorine is in 0.75 mol of Prevacid? c. What number of carbon atoms is in 0.75 mol of Prevacid? d. What is the mass of \(4.25 \times 10^{21}\) molecules of Prevacid?

Short Answer

Expert verified
a. The molar mass of Prevacid is approximately 369.372 g/mol. b. The mass of fluorine in 0.75 mol of Prevacid is 42.75 g. c. There are \(7.227 \times 10^{24}\) carbon atoms in 0.75 mol of Prevacid. d. The mass of \(4.25 \times 10^{21}\) molecules of Prevacid is approximately 1.054 g.

Step by step solution

01

Calculate the molar mass of Prevacid

To calculate the molar mass of Prevacid, we'll need to take the sum of the molar masses of all the atoms in the formula. Molar mass of Prevacid = (16 × molar mass of carbon) + (14 × molar mass of hydrogen) + (3 × molar mass of fluorine) + (3 × molar mass of nitrogen) + (2 × molar mass of oxygen) + molar mass of sulfur Here are the molar masses of each element: C = 12.01 g/mol, H = 1.008 g/mol, F = 19.00 g/mol, N = 14.01 g/mol, O = 16.00 g/mol, and S = 32.07 g/mol.
02

Calculate the total molar mass

Using the molar masses and the formula from Step 1, we can now find the molar mass of Prevacid. Molar mass of Prevacid = (16 × 12.01) + (14 × 1.008) + (3 × 19.00) + (3 × 14.01) + (2 × 16.00) + 32.07 Molar mass of Prevacid = 192.16 + 14.112 + 57.00 + 42.03 + 32.00 + 32.07 = 369.372 g/mol (rounded to 3 decimal places) a. The molar mass of Prevacid is approximately 369.372 g/mol.
03

Calculate the mass of fluorine in 0.75 mol of Prevacid

Given that there are 3 fluorine atoms in one molecule of Prevacid and we have 0.75 mol of Prevacid, we can calculate the mass of fluorine: Mass of fluorine = Number of moles × Molar mass of fluorine Mass of fluorine in 0.75 mol of Prevacid = 0.75 × 3 × 19.00 = 42.75 g b. The mass of fluorine in 0.75 mol of Prevacid is 42.75 g.
04

Calculate the number of carbon atoms in 0.75 mol of Prevacid

There are 16 carbon atoms in one molecule of Prevacid. We will use Avogadro's number (\(6.022 \times 10^{23}\)) to convert from moles to the number of carbon atoms: Number of carbon atoms = 0.75 mol × 16 carbon atoms/mol of Prevacid × Avogadro's number Number of carbon atoms = 0.75 × 16 × 6.022 × 10^{23} Number of carbon atoms = 7.227 × 10^{24} c. There are \(7.227 \times 10^{24}\) carbon atoms in 0.75 mol of Prevacid.
05

Calculate the mass of \(4.25 \times 10^{21}\) molecules of Prevacid

Now we'll need to determine the mass of a specific number of molecules, in this case, \(4.25 \times 10^{21}\) molecules. To do this, we can use the molar mass of Prevacid and Avogadro's number to convert from molecules to mass: Mass of \(4.25 \times 10^{21}\) molecules = \(\dfrac{4.25 \times 10^{21}\mathrm{molecules}}{6.022\times10^{23} \mathrm{molecules/mol}}\) × 369.372 g/mol Mass of \(4.25 \times 10^{21}\) molecules = 2.854 × 10^{-3} mol × 369.372 g/mol Mass of \(4.25 \times 10^{21}\) molecules = 1.054 g (rounded to 3 decimal places) d. The mass of \(4.25 \times 10^{21}\) molecules of Prevacid is approximately 1.054 g.

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Most popular questions from this chapter

Adipic acid is an organic compound composed of 49.31\(\% \mathrm{C}\) , $43.79 \% \mathrm{O},$ and the rest hydrogen. If the molar mass of adipic acid is 146.1 \(\mathrm{g} / \mathrm{mol}\) , what are the empirical and molecular formulas for adipic acid?

Give the balanced equation for each of the following chemical reactions: a. Glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron((III) chloride and hydrogen sulfide gas. c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\)

Hexamethylenediamine $\left(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{N}_{2}\right)$ is one of the starting materials for the production of nylon. It can be prepared from adipic acid $\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\right)$ by the following overall equation: $$ \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}(l)+2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{6} \mathrm{H}_{16} \mathrm{N}_{2}(l)+4 \mathrm{H}_{2} \mathrm{O}(l) $$ What is the percent yield for the reaction if 765 g of hexamethylenediamine is made from \(1.00 \times 10^{3} \mathrm{g}\) of adipic acid?

One of relatively few reactions that takes place directly between two solids at room temperature is $$ \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow $$ $$ \mathrm{Ba}(\mathrm{SCN})_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g) $$ In this equation, the \(\cdot 8 \mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate. a. Balance the equation. b. What mass of ammonium thiocyanate (NH_sCN) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?

Natural rubidium has the average mass of 85.4678 \(\mathrm{u}\) and is composed of isotopes \(^{85} \mathrm{Rb}(\mathrm{mass}=84.9117 \mathrm{u})\) and $^{87} \mathrm{Rb}\( . The ratio of atoms \)^{85} \mathrm{Rb} /^{87} \mathrm{Rb}$ in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\) .
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