Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) \(\mathrm{CH}_{4},\) (b) \(\mathrm{NO}_{2}\), (c) \(\mathrm{SO}_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (e) NaI, (f) \(\mathrm{K}_{2} \mathrm{SO}_{4},(\mathrm{~g}) \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Short Answer

Expert verified
(a) 16.05 amu, (b) 46.01 amu, (c) 80.07 amu, (d) 78.12 amu, (e) 149.89 amu, (f) 174.27 amu, (g) 310.18 amu.

Step by step solution

01

Determining Atomic Masses

To calculate the molecular mass of the substances, we first need to know the atomic masses of each element: - C (Carbon) = 12.01 amu - H (Hydrogen) = 1.01 amu - N (Nitrogen) = 14.01 amu - O (Oxygen) = 16.00 amu - S (Sulfur) = 32.07 amu - Na (Sodium) = 22.99 amu - I (Iodine) = 126.90 amu - K (Potassium) = 39.10 amu - Ca (Calcium) = 40.08 amu - P (Phosphorus) = 30.97 amu
02

Calculate Molecular Mass of CH4

For \(\mathrm{CH}_{4}\): - 1 Carbon atom = 12.01 amu- 4 Hydrogen atoms = 4 \times 1.01 = 4.04 amu- Total = 12.01 + 4.04 = 16.05 amu
03

Calculate Molecular Mass of NO2

For \(\mathrm{NO}_{2}\): - 1 Nitrogen atom = 14.01 amu- 2 Oxygen atoms = 2 \times 16.00 = 32.00 amu- Total = 14.01 + 32.00 = 46.01 amu
04

Calculate Molecular Mass of SO3

For \(\mathrm{SO}_{3}\): - 1 Sulfur atom = 32.07 amu- 3 Oxygen atoms = 3 \times 16.00 = 48.00 amu- Total = 32.07 + 48.00 = 80.07 amu
05

Calculate Molecular Mass of C6H6

For \(\mathrm{C}_{6} \mathrm{H}_{6}\): - 6 Carbon atoms = 6 \times 12.01 = 72.06 amu- 6 Hydrogen atoms = 6 \times 1.01 = 6.06 amu- Total = 72.06 + 6.06 = 78.12 amu
06

Calculate Molecular Mass of NaI

For NaI: - 1 Sodium atom = 22.99 amu - 1 Iodine atom = 126.90 amu - Total = 22.99 + 126.90 = 149.89 amu
07

Calculate Molecular Mass of K2SO4

For \(\mathrm{K}_{2} \mathrm{SO}_{4}\):- 2 Potassium atoms = 2 \times 39.10 = 78.20 amu- 1 Sulfur atom = 32.07 amu- 4 Oxygen atoms = 4 \times 16.00 = 64.00 amu- Total = 78.20 + 32.07 + 64.00 = 174.27 amu
08

Calculate Molecular Mass of Ca3(PO4)2

For \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\):- 3 Calcium atoms = 3 \times 40.08 = 120.24 amu- 2 Phosphorus atoms = 2 \times 30.97 = 61.94 amu- 8 Oxygen atoms = 8 \times 16.00 = 128.00 amu- Total = 120.24 + 61.94 + 128.00 = 310.18 amu

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Atomic Mass
To begin any molecular mass calculation, it's essential to understand the concept of atomic mass. Atomic mass refers to the mass of a single atom of a chemical element. It is usually expressed in atomic mass units (amu). This unit provides a convenient way to express the large numbers associated with the mass of individual atoms.
Each element has a specific atomic mass based on the weighted average of all its isotopes. For example, carbon ( C) has an atomic mass of 12.01 amu, while hydrogen ( H) has an atomic mass of 1.01 amu.
The atomic masses are vital as they are used in calculating the molecular or formula mass of compounds. By adding up the atomic masses of each constituent atom in the chemical formula, you can determine the overall mass of a molecule.
Chemical Formula
A chemical formula provides a simple notation to represent the elements in a compound and the number of each type of atom present. It serves as a map that shows how atoms combine and in what proportions. For example, the chemical formula CH₄ tells us that one molecule contains one carbon atom and four hydrogen atoms.
When using a chemical formula, it's important to accurately note the numbers of each type of atom, as this directly influences the molecular mass calculation. Even a small change in a compound's chemical formula can result in a different substance. For instance, H₂O (water) and H₂O₂ (hydrogen peroxide) are completely different, even though they share similar elements. Thus, a chemical formula is crucial for stoichiometry, as it provides the basis for calculating reactant and product quantities.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). The concept is an extension of molecular mass, which is measured in amu. To find the molar mass, you add together the atomic masses of the individual atoms that make up a molecule, just like in molecular mass calculations.
For example, the molar mass of methane ( CH₄) can be calculated by summing the atomic masses: one carbon atom (12.01 amu) plus four hydrogen atoms (4 × 1.01 amu = 4.04 amu), resulting in a molecular mass of 16.05 amu.
Although molar mass and molecular mass are numerically identical for a molecule (since they sum to 16.05), molar mass is typically expressed in g/mol to reflect its relevance in laboratory measurements and stoichiometry.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is essential for understanding how much of each reactant is needed to produce a desired amount of product. This concept uses a balanced chemical equation along with the molar masses of the substances involved.
When dealing with stoichiometry, you'll convert between moles and masses to ensure you have the correct proportions for a reaction to occur efficiently. This calculation helps chemists understand how much of each chemical is required or produced in a reaction. For instance, if you were to use the molar masses calculated, you could determine how many grams of a reactant are necessary to produce a specific amount of product.
Understanding stoichiometry allows students and chemists to predict yields, optimize reactions, and reduce waste, making it a fundamental aspect of chemistry education.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the empirical formulas of the compounds with the following compositions: (a) 2.1 percent \(\mathrm{H}\), 65.3 percent \(\mathrm{O}, 32.6\) percent \(\mathrm{S} ;\) (b) 20.2 percent \(\mathrm{Al}\), 79.8 percent \(\mathrm{Cl}\).

Explain why, in combustion analysis, we cannot determine the amount of oxygen in the sample directly from the amount of oxygen in the products \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\)

A reaction having a 90 percent yield may be considered a successful experiment. However, in the synthesis of complex molecules such as chlorophyll and many anticancer drugs, a chemist often has to carry out multiple-step syntheses. What is the overall percent yield for such a synthesis, assuming it is a 30 -step reaction with a 90 percent yield at each step?

Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is 2: 1 . A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is \(178 \mathrm{~g}\).

Ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), an important industrial organic chemical, can be prepared by heating hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) at \(800^{\circ} \mathrm{C}\) : $$ \mathrm{C}_{6} \mathrm{H}_{14} \stackrel{\Delta}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{4}+\text { other products } $$ If the yield of ethylene production is 42.5 percent, what mass of hexane must be used to produce \(481 \mathrm{~g}\) of ethylene?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free