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(a) What is the mass, in grams, of \(0.0714 \mathrm{~mol}\) of iron(III) sulfate? (b) How many moles of ammonium ions are in \(8.776 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(6.52 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{\mathrm{g}} \mathrm{O}_{4} ?\) (d) What is the molar mass of diazepam (Valium \(^{8}\) ) if \(0.05570\) mol weighs \(15.86 \mathrm{~g}\) ?

Short Answer

Expert verified
(a) The mass of 0.0714 mol of iron(III) sulfate is 28.56 g. (b) There are 0.1826 mol of ammonium ions in 8.776 g of ammonium carbonate. (c) The mass of \(6.52 \times 10^{21}\) molecules of aspirin, C鈧塇鈧圤鈧, is 0.195 g. (d) The molar mass of diazepam (Valium) is 284.74 g/mol.

Step by step solution

01

Calculate the molar mass of iron(III) sulfate

Iron(III) sulfate has the chemical formula Fe鈧(SO鈧)鈧. To determine the molar mass, we need to sum the atomic masses of all atoms in the compound. Fe: 55.85 g/mol S: 32.06 g/mol O: 16.00 g/mol Molar mass of Fe鈧(SO鈧)鈧 = 2(55.85) + 3(32.06*4 + 16.00*3) = 111.70 + 3(128.24 + 48) = 399.88 g/mol
02

Calculate the mass of iron(III) sulfate

Use the formula mass = moles 脳 molar mass: Mass of Fe鈧(SO鈧)鈧 = 0.0714 mol 脳 399.88 g/mol = 28.56 g (b) How many moles of ammonium ions are in 8.776 g of ammonium carbonate?
03

Calculate the molar mass of ammonium carbonate

Ammonium carbonate has the chemical formula (NH鈧)鈧侰O鈧. To determine the molar mass, we need to sum the atomic masses of all atoms in the compound: N: 14.01 g/mol H: 1.01 g/mol C: 12.01 g/mol O: 16.00 g/mol Molar mass of (NH鈧)鈧侰O鈧 = 2(14.01 + 4(1.01)) + 12.01 + 3(16) = 2(18.05) + 12.01 + 48 = 96.11 g/mol
04

Convert the mass of ammonium carbonate to moles

Use the formula moles = mass / molar mass Moles of (NH鈧)鈧侰O鈧 = 8.776 g / 96.11 g/mol = 0.0913 mol
05

Convert the moles of ammonium carbonate to moles of ammonium ions

There are two moles of ammonium ions (NH鈧勨伜) for each mole of ammonium carbonate: Moles of NH鈧勨伜 = 0.0913 mol 脳 2 = 0.1826 mol (c) What is the mass, in grams, of \(6.52 \times 10^{21}\) molecules of aspirin, C鈧塇鈧圤鈧?
06

Calculate the molar mass of aspirin

Aspirin has the chemical formula C鈧塇鈧圤鈧. To determine the molar mass, we need to sum the atomic masses of all atoms in the compound: C: 12.01 g/mol H: 1.01 g/mol O: 16.00 g/mol Molar mass of C鈧塇鈧圤鈧 = (9 脳 12.01) + (8 脳 1.01) + (4 脳 16) = 180.16 g/mol
07

Convert the number of molecules to moles

Use Avogadro's number (6.022 脳 10虏鲁 particles/mol) to find the moles of aspirin: Moles of C鈧塇鈧圤鈧 = (6.52 脳 10虏鹿 molecules) / (6.022 脳 10虏鲁 molecules/mol) = 0.001083 mol
08

Calculate the mass of aspirin

Use the formula mass = moles 脳 molar mass: Mass of C鈧塇鈧圤鈧 = 0.001083 mol 脳 180.16 g/mol = 0.195 g (d) What is the molar mass of diazepam (Valium) if 0.05570 mol weighs 15.86 g?
09

Calculate the molar mass of diazepam

Use the formula molar mass = mass / moles Molar mass = 15.86 g / 0.05570 mol = 284.74 g/mol

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Key Concepts

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

Molar Mass Calculation
Understanding the concept of molar mass is fundamental in mastering stoichiometry problems. Molar mass, often expressed in grams per mole (g/mol), is the mass of one mole of a substance. It can be calculated by summing the atomic masses of all atoms present in a formula unit of the compound, as found on the periodic table.For example, in part (a) of the exercise, the molar mass of iron(III) sulfate, Fe鈧(SO鈧)鈧, is calculated. First, you identify the number of each type of atom in the compound, and then multiply each by their respective atomic masses: Iron (Fe) is 55.85 g/mol, Sulfur (S) is 32.06 g/mol, and Oxygen (O) is 16.00 g/mol. With this, we consider the stoichiometric coefficients from the compound's formula, leading to the equation: Molar mass of Fe鈧(SO鈧)鈧 = 2(55.85) + 3(32.06*4 + 16.00*3) = 399.88 g/mol.
The exercise audibly demands slow, careful addition to avoid common mistakes like missing an atom or miscalculating the total mass.
Moles to Grams Conversion
The conversion from moles to grams and vice versa is a pivotal part of stoichiometry, allowing the translation of chemical quantities into measurable units. The formula for this conversion relies on the substance's molar mass: Mass (in grams) = Number of Moles 脳 Molar Mass (in g/mol).

For instance, in the exercise, once the molar mass of iron(III) sulfate was determined as 399.88 g/mol, the mass of 0.0714 moles of the compound could be found by simply multiplying:Mass of Fe鈧(SO鈧)鈧 = 0.0714 mol 脳 399.88 g/mol = 28.56 g.
Reversing the process allows for the determination of the number of moles from a given mass, as with ammonia, Moles of (NH鈧)鈧侰O鈧 = 8.776 g / 96.11 g/mol = 0.0913 mol.

It is essential to work these calculations with precision, ensuring that the correct molar mass is used for the substance in question.
Avogadro's Number
Avogadro's number, approximately 6.022 脳 10虏鲁, is the number of particles (atoms, molecules, ions, etc.) in one mole of a substance. It acts as a bridge between the microscopic world of atoms and the macroscopic world we measure in the laboratory.

Converting Particles to Moles to Grams

For instance, to find the mass of a specific number of molecules, as in part (c) of the exercise involving aspirin, C鈧塇鈧圤鈧, you first convert the number of molecules to moles using Avogadro's number:Moles of C鈧塇鈧圤鈧 = (6.52 脳 10虏鹿 molecules) / (6.022 脳 10虏鲁 molecules/mol) = 0.001083 mol.
Then, convert the moles to grams using the molar mass:Mass of C鈧塇鈧圤鈧 = 0.001083 mol 脳 180.16 g/mol = 0.195 g.

Understanding Avogadro's number is essential for interpreting various quantities in chemical reactions and processes. The calculations often involve exponential numbers, which is why careful attention to detail and the use of scientific notation is crucial to ensuring accuracy.

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Most popular questions from this chapter

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\); but other products containing \(\mathrm{Cl}\). such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\), are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\). (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and \(\mathrm{HCl}\). (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\)

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