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Chapter 7: Problem 12

Calculate the molar mass for each of the following: a. \(\mathrm{O}_{2}\) b. \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) c. \(\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}\)

Short Answer

Expert verified
a. 32.00 u b. 136.09 u c. 354.20 u

Step by step solution

01

- Identify the atomic masses

Look up the atomic masses of all elements involved in each compound. For example: - O (Oxygen) has an atomic mass of 16.00 u - K (Potassium) has an atomic mass of 39.10 u - H (Hydrogen) has an atomic mass of 1.01 u - P (Phosphorus) has an atomic mass of 30.97 u - Fe (Iron) has an atomic mass of 55.85 u - Cl (Chlorine) has an atomic mass of 35.45 u - O (Oxygen) has an atomic mass of 16.00 u
02

- Calculate the molar mass for \(\mathrm{O}_{2}\)

The molecule \(\mathrm{O}_{2}\) has two oxygen atoms: \[ \text{Molar mass of } \mathrm{O}_{2} = 2 \times 16.00\, \text{u} = 32.00\, \text{u} \]
03

- Calculate the molar mass for \(\mathrm{KH}_{2} \mathrm{PO}_{4}\)

The molecule \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) has 1 potassium atom, 2 hydrogen atoms, 1 phosphorus atom, and 4 oxygen atoms: \[ \text{Molar mass of } \mathrm{KH}_{2} \mathrm{PO}_{4} = 39.10\, \text{u} + 2 \times 1.01\, \text{u} + 30.97\, \text{u} + 4 \times 16.00\, \text{u} = 136.09\, \text{u} \]
04

- Calculate the molar mass for \(\mathrm{Fe(ClO}_{4})_{3}\)

The molecule \(\mathrm{Fe} \left( \mathrm{ClO}_{4} \right)_{3} \) has 1 iron atom, 3 chlorine atoms, and 12 oxygen atoms: \[ \text{Molar mass of } \mathrm{Fe} \left( \mathrm{ClO}_{4} \right)_{3} = 55.85\, \text{u} + 3 \times 35.45\, \text{u} + 12 \times 16.00\, \text{u} = 354.20\, \text{u} \]

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Key Concepts

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

Atomic Mass
Atomic mass, also known as atomic weight, is a fundamental concept in chemistry. It represents the mass of a single atom, usually measured in atomic mass units (u). This value is approximately equal to the number of protons and neutrons in the atom. For example, Oxygen (O) has an atomic mass of 16.00 u because its nucleus contains 8 protons and 8 neutrons.
If you look up the atomic masses in the periodic table, you will find the relative mass of each element. This value is crucial when calculating the molar mass of compounds.
Molecular Mass
The molecular mass is the sum of the atomic masses of all atoms in a molecule. To calculate it, you simply add together the atomic masses of each element in the molecule. For instance, in the molecule \(\text{O}_2\), there are two oxygen atoms, so the molecular mass is \[2 \times 16.00 \text{ u} = 32.00 \text{ u}\].
For more complex compounds like \(\text{KH}_2\text{PO}_4\), you sum all the atomic masses of potassium (K), hydrogen (H), phosphorus (P), and oxygen (O) atoms in the compound.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It involves the calculation of how much of each reactant is needed and how much product can be formed. When calculating molar masses, stoichiometry helps in understanding the proportions of elements in compounds. For example, in \(\text{Fe(ClO}_4)_3\), knowing that there is 1 iron atom, 3 chlorine atoms, and 12 oxygen atoms is crucial for calculating its molar mass.
Chemical Formulas
A chemical formula is a way to represent a substance using symbols for its constituent elements. It gives insight into the number and type of atoms in the smallest unit of the substance. For example, \(\text{H}_2\text{O}\) indicates that each molecule of water contains 2 hydrogen atoms and 1 oxygen atom. Chemical formulas are essential for determining molar mass, as seen in the examples given.
Understanding the chemical formula allows us to break down the compound into its individual elements and then sum their atomic masses.
Elements
Elements are pure substances that consist of only one kind of atom. They are the building blocks of all matter. Every element has a unique atomic number and atomic mass. Elements combine in fixed ratios to form compounds. For instance, iron (Fe), chlorine (Cl), and oxygen (O) are elements that combine to form \(\text{Fe(ClO}_4)_3\).
Knowing about elements helps in understanding chemical reactions and compounds. By looking at the periodic table, we get valuable information about the properties and atomic masses of the elements.

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Most popular questions from this chapter

Fatty acids undergo reaction with oxygen gas and form gaseous carbon dioxide and liquid water when utilized for energy in the body. a. Write and balance the equation for the combustion of the fatty acid aqueous capric acid, \(\mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O}_{2}\). b. Write and balance the equation for the combustion of the fatty acid aqueous myristic acid, \(\mathrm{C}_{14} \mathrm{H}_{28} \mathrm{O}_{2}\).

In the mitochondria of human cells, energy is provided by the oxidation and reduction reactions of the iron ions in the cytochromes in electron transport. Identify each of the following as an oxidation or a reduction: a. \(\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+}\) b. \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}\)

Calcium cyanamide, \(\mathrm{CaCN}_{2}\), reacts with water to form calcium carbonate and ammonia. $$ \mathrm{CaCN}_{2}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CaCO}_{3}(s)+2 \mathrm{NH}_{3}(g) $$ a. How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are needed to react with \(75.0 \mathrm{~g}\) of \(\mathrm{CaCN}_{2} ?\) b. How many grams of \(\mathrm{NH}_{3}\) are produced from \(5.24 \mathrm{~g}\) of \(\mathrm{CaCN}_{2} ?\) c. How many grams of \(\mathrm{CaCO}_{3}\) form if \(155 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) reacts?

How would each of the following change the rate of the reaction shown here? $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ a. adding some \(\mathrm{SO}_{2}(g)\) b. increasing the temperature c. adding a catalyst d. removing some \(\mathrm{O}_{2}(g)\)

The equation for the formation of silicon tetrachloride from silicon and chlorine is (7.9) $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(g)+157 \mathrm{kcal} $$ a. Is the formation of \(\mathrm{SiCl}_{4}\) an endothermic or exothermic reaction? b. Is the energy of the product higher or lower than the energy of the reactants?
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