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Chapter 7: Problem 63

A mixture contains \(0.250\) mol of \(\mathrm{Mn}_{2} \mathrm{O}_{3}\) and \(20.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\). \((7.1,7.2,7.3)\) a. How many atoms of oxygen are present in the mixture? b. How many grams of manganese are in the mixture?

Short Answer

Expert verified
a) 7.29 脳 10^{23} atoms of oxygen. b) 40.08 g of manganese.

Step by step solution

01

Convert grams of MnO2 to moles

Given: 20.0 g of MnO2. Find the molar mass of MnO2: \[ \text{Molar mass of MnO}_2 = 54.94 \text{ (Mn)} + 2 \times 16.00 \text{ (O)} = 86.94 \text{ g/mol} \] Now, convert grams to moles: \[ \text{Moles of MnO}_2 = \frac{20.0 \text{ g}}{86.94 \text{ g/mol}} = 0.230 \text{ mol} \]
02

Calculate total moles of oxygen atoms

Mn2O3 has 3 oxygen atoms per molecule, and MnO2 has 2 oxygen atoms per molecule. We have:\[ 0.250 \text{ mol of Mn}_2\text{O}_3 \times 3 \text{ atoms O} = 0.750 \text{ mol O} \] \[ 0.230 \text{ mol of MnO}_2 \times 2 \text{ atoms O} = 0.460 \text{ mol O} \] Total moles of oxygen: \[ 0.750 \text{ mol} + 0.460 \text{ mol} = 1.210 \text{ mol O} \]
03

Convert moles of oxygen to atoms

Use Avogadro's number to find the number of oxygen atoms:\[ 1.210 \text{ mol O} \times 6.022 \times 10^{23} \text{ atoms/mol} = 7.29 \times 10^{23} \text{ atoms} \]
04

Calculate moles of manganese atoms

Mn2O3 has 2 manganese atoms per molecule, and MnO2 has 1 manganese atom per molecule. We have:\[ 0.250 \text{ mol of Mn}_2\text{O}_3 \times 2 = 0.500 \text{ mol Mn} \]\[ 0.230 \text{ mol of MnO}_2 \times 1 = 0.230 \text{ mol Mn} \]Total moles of manganese: \[ 0.500 \text{ mol} + 0.230 \text{ mol} = 0.730 \text{ mol Mn} \]
05

Convert moles of manganese to grams

Find the mass of manganese:\[ 0.730 \text{ mol} \times 54.94 \text{ g/mol} = 40.08 \text{ g} \]

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Key Concepts

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

molar mass calculation
Calculating the molar mass of a compound is essential in chemistry. It involves adding up the atomic masses of all the atoms in a molecule. Atomic masses can be found on the periodic table.
For example, to find the molar mass of \(\text{MnO}_2\) (Manganese Dioxide), you need the atomic masses of Manganese (Mn) and two Oxygens (O).
  • Manganese (Mn) = 54.94 g/mol
  • Oxygen (O) = 16.00 g/mol

Add them together: 54.94 + 2*16.00 = 86.94 g/mol.
So, the molar mass of \(\text{MnO}_2\) is 86.94 g/mol.
conversion of moles to atoms
One mole of any substance contains Avogadro's number of entities, which is \(\text{6.022} \times \text{10}^{23}\) atoms/molecules.
For instance, converting moles of oxygen atoms to actual atoms:
If you have 1.210 moles of oxygen, you multiply by Avogadro's number:
\(1.210 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 7.29 \times 10^{23} \text{ atoms} \)
This means there are approximately 7.29 x 10^23 oxygen atoms.
stoichiometry
Stoichiometry deals with the quantitative relationships in a chemical reaction. It helps us predict the amounts of products and reactants involved.
Consider \(\text{Mn}_2\text{O}_3\) and \(\text{MnO}_2\):
  • \(\text{Mn}_2\text{O}_3\) has 3 oxygens per molecule and 2 manganese atoms
  • \(\text{MnO}_2\) has 2 oxygens per molecule and 1 manganese atom

If you mix \(\text{0.250 mol Mn}_2\text{O}_3\) with \(\text{0.230 mol MnO}_2\): for oxygen:
  • \(0.250 \text{ mol} \text{Mn}_2\text{O}_3 \times 3 = 0.750 \text{ mol O} \)
  • \(0.230 \text{ mol} \text{MnO}_2 \times 2 = 0.460 \text{ mol O} \)
Total = 1.210 mol O
molecular composition
Understanding the molecular composition of a compound means knowing the types and numbers of atoms in it.
Take \(\text{Mn}_2\text{O}_3\) and \(\text{MnO}_2\):
  • \(\text{Mn}_2\text{O}_3\) has 2 Manganese (Mn) and 3 Oxygen (O) atoms.
  • \(\text{MnO}_2\) has 1 Manganese (Mn) and 2 Oxygen (O) atoms.

By knowing this, you can determine the total atoms of each element in mixtures.
  • Total Manganese atoms in mixture: Moles from \(\text{Mn}_2\text{O}_3\) + Moles from \(\text{MnO}_2\) = 0.500 mol + 0.230 mol = 0.730 mol
mass to mole conversion
Converting mass to moles is done using the molar mass.
If you have 20.0 grams of \(\text{MnO}_2\) and need to convert to moles:
First, know the molar mass of \(\text{MnO}_2\) (86.94 g/mol).
  • \(\text{Moles of MnO}_2 = \frac{20.0 \text{ g}}{86.94 \text{ g/mol}} = 0.230 \text{ mol} \)

This is crucial in stoichiometry for finding the amounts involved in reactions.

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Most popular questions from this chapter

Calculate each of the following in \(2.00 \mathrm{~mol}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4}:\) a. moles of \(\mathrm{H}\) b. moles of \(\mathrm{O}\) c. atoms of \(\mathrm{P}\) d. atoms of \(\mathrm{O}\)

Sorbic acid, an inhibitor of mold in cheese, has a mass percent composition of \(64.27 \% \mathrm{C}, 7.19 \% \mathrm{H}\), and \(28.54 \% \mathrm{O}\). If sorbic acid has an experimental molar mass of \(112 \mathrm{~g}\), what is its molecular formula? \((7.4,7.5)\)

Calculate the molar mass for each of the following: a. \(\mathrm{FeSO}_{4}\), iron supplement b. \(\mathrm{Al}_{2} \mathrm{O}_{3}\), absorbent and abrasive c. \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\), saccharin, an artificial sweetener

Calculate the empirical formula for each of the following compounds: (7.4) a. \(2.20 \mathrm{~g}\) of \(\mathrm{S}\) and \(7.81 \mathrm{~g}\) of \(\mathrm{F}\) b. \(6.35 \mathrm{~g}\) of \(\mathrm{Ag}, 0.825 \mathrm{~g}\) of \(\mathrm{N}\), and \(2.83 \mathrm{~g}\) of \(\mathrm{O}\) c. \(43.6 \% \mathrm{P}\) and \(56.4 \% \mathrm{O}\) d. \(22.1 \% \mathrm{Al}, 25.4 \% \mathrm{P}\), and \(52.5 \% \mathrm{O}\)

Write the empirical formula for each of the following substances: a. \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{3}\), pyrogallol, a developer in photography b. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), galactose, a carbohydrate c. \(\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}\), terephthalic acid, used in the manufacture of plastic bottles d. \(\mathrm{C}_{6} \mathrm{Cl}_{6}\), hexachlorobenzene, a fungicide e. \(\mathrm{C}_{24} \mathrm{H}_{16} \mathrm{O}_{12}\), laccaic acid, a crimson dye
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