Question

Calculate the mass in grams of each of the following samples. a. \(6.14 \times 10^{-4}\) moles of sulfur trioxide b. \(3.11 \times 10^{5}\) moles of lead(IV) oxide c. 0.495 mole of chloroform, \(\mathrm{CHCl}_{3}\) d. \(2.45 \times 10^{-8}\) moles of trichloroethane, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}\) e. 0.167 mole of lithium hydroxide f. 5.26 moles of copper(I) chloride

Short answer

The mass in grams for each of the samples are: a. 0.0492 g b. \(7.43 \times 10^7\) g c. 59.09 g d. \(3.28 \times 10^{-6}\) g e. 4.00 g f. 520.74 g

Step-by-step solution

a. Mass of 6.14 x 10^{-4} moles of sulfur trioxide (SO3)

1. Find the molar mass of SO3: \(M(SO3) = (1 \times M(S)) + (3 \times M(O))\) Using the periodic table, we find the atomic masses: M(S) = 32.06 g/mol, M(O) = 16.00 g/mol Calculate \(M(SO3)\) \(M(SO3) = (1\times 32.06) + (3 \times 16.00) = 80.06\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 6.14 \times 10^{-4} moles \times 80.06\ g/mol = 0.0492\ g\)

b. Mass of 3.11 x 10^{5} moles of lead(IV) oxide (PbO2)

1. Find the molar mass of PbO2: \(M(PbO2) = (1 \times M(Pb)) + (2 \times M(O))\) Using the periodic table, we find the atomic masses: M(Pb) = 207.2 g/mol, M(O) = 16.00 g/mol Calculate \(M(PbO2)\) \(M(PbO2) = (1 \times 207.2) + (2 \times 16.00) = 239.2\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 3.11 \times 10^5 moles \times 239.2\ g/mol = 7.43 \times 10^7\ g\)

c. Mass of 0.495 mole of chloroform (CHCl3)

1. Find the molar mass of CHCl3: \(M(CHCl3) = (1 \times M(C)) + (1 \times M(H)) + (3 \times M(Cl))\) Using the periodic table, we find the atomic masses: M(C) = 12.01 g/mol, M(H) = 1.01 g/mol, M(Cl) = 35.45 g/mol Calculate \(M(CHCl3)\) \( M(CHCl3) = (1\times 12.01) + (1\times 1.01) + (3 \times 35.45) = 119.37\ g/mol \) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 0.495 moles \times 119.37\ g/mol = 59.09\ g\)

d. Mass of 2.45 x 10^{-8} moles of trichloroethane (C2H3Cl3)

1. Find the molar mass of C2H3Cl3: \(M(C2H3Cl3) = (2 \times M(C)) + (3 \times M(H)) + (3 \times M(Cl))\) Using the periodic table, we find the atomic masses: M(C) = 12.01 g/mol, M(H) = 1.01 g/mol, M(Cl) = 35.45 g/mol Calculate \(M(C2H3Cl3)\) \(M(C2H3Cl3) = (2\times 12.01) + (3\times 1.01) + (3 \times 35.45) = 133.93\ g/mol \) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 2.45 \times 10^{-8} moles \times 133.93\ g/mol = 3.28 \times 10^{-6}\ g\)

e. Mass of 0.167 mole of lithium hydroxide (LiOH)

1. Find the molar mass of LiOH: \(M(LiOH) = (1 \times M(Li)) + (1 \times M(O)) + (1 \times M(H))\) Using the periodic table, we find the atomic masses: M(Li) = 6.94 g/mol, M(O) = 16.00 g/mol, M(H) = 1.01 g/mol Calculate \(M(LiOH)\) \(M(LiOH) = (1\times 6.94) + (1\times 16.00) + (1 \times 1.01) = 23.95\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 0.167 moles \times 23.95\ g/mol = 4.00\ g\)

f. Mass of 5.26 moles of copper(I) chloride (CuCl)

1. Find the molar mass of CuCl: \(M(CuCl) = (1 \times M(Cu)) + (1 \times M(Cl))\) Using the periodic table, we find the atomic masses: M(Cu) = 63.55 g/mol, M(Cl) = 35.45 g/mol Calculate \(M(CuCl)\) \(M(CuCl) = (1\times 63.55) + (1\times 35.45) = 99.00\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 5.26 moles \times 99.00\ g/mol = 520.74\ g\)

Key concepts

Molar Mass Calculation

The concept of molar mass is crucial in chemistry for converting moles to grams, and vice versa. Molar mass essentially represents the mass of one mole of a given substance. Here's how you calculate it:

• Identify the chemical formula of the compound you're dealing with.
• Refer to the periodic table to look up the atomic mass of each element in the compound.
• Multiply the atomic mass of each element by the number of times it appears in the formula.
• Add up all these values to get the total molar mass.

For example, sulfur trioxide (\(\text{SO}_3\)) requires you to find the atomic masses of sulfur and oxygen. Sulfur has an atomic mass of 32.06 g/mol, and oxygen has an atomic mass of 16.00 g/mol. Calculating molar mass would involve:
\(M(\text{SO}_3) = (1 \times 32.06) + (3 \times 16.00) = 80.06\text{ g/mol}\).
This simplicity is key to performing stoichiometric calculations and converting between moles and grams.

Chemical Compounds

Chemical compounds consist of two or more different elements that are chemically bonded together in fixed proportions. Understanding the composition of a compound helps you interpret its chemical formula, which is vital for further calculations.
Here are some things to keep in mind about chemical compounds:

• Each compound has a unique formula reflecting its specific ratio of atoms.
• An example is \(\text{PbO}_2\), lead(IV) oxide, which includes one lead (Pb) atom and two oxygen (O) atoms.
• The type of bond—ionic or covalent—affects the properties of the compound.
• Compounds can be broken down into simpler substances through chemical reactions.

Recognizing the differences between compounds, such as sulfur trioxide and chloroform, \(\text{CHCl}_3\), and their unique formulas, is fundamental for calculating molar masses and performing conversions in stoichiometric problems.

Stoichiometry

Stoichiometry is the area of chemistry that involves the calculation of reactants and products in chemical reactions. It is a handy tool for scientists to predict yields and measure reaction efficiency.
Here’s what makes stoichiometry vital:

• It explores the quantitative relationships using balanced chemical equations.
• Stoichiometry uses the mole concept to convert between mass and number of particles.
• It allows for scaling lab reactions to industrial processes based on reactant or product usages and yields.

To convert moles to grams in stoichiometric ratios, such as with other samples of chloroform or trichloroethane (\(\text{C}_2\text{H}_3\text{Cl}_3\)), you need to follow these steps:

• Determine the molar mass as described earlier.
• Use the molar mass to convert from moles to grams using the formula: \(\text{mass} = \text{moles} \times \text{molar mass}\).
• This conversion is often essential for experimental guidance and proportion calculation in reactions.

A firm grasp of stoichiometry provides a deeper understanding of how much of each compound is consumed or produced in a chemical reaction, aiding in precise chemical analysis and synthesis.