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Chapter 3: Problem 44

An oxide of nitrogen contains 30.45 mass \(\%\) N. (a) What is the empirical formula of the oxide? (b) If the molar mass is \(90 \pm 5 \mathrm{~g} / \mathrm{mol},\) what is the molecular formula?

Short Answer

Expert verified
The empirical formula is \(\text{NO}_2\). The molecular formula is \(\text{N}_2\text{O}_4\).

Step by step solution

01

Finding the Masses of Each Element

Assume 100 g of the compound. This means we have 30.45 g of nitrogen (N) and 69.55 g of oxygen (O), as the rest of the compound is oxygen.
02

Converting Mass to Moles

Convert the given masses to moles. Use the molar masses of nitrogen (14.01 g/mol) and oxygen (16.00 g/mol). \[ \text{Moles of } N = \frac{30.45 \text{ g}}{14.01 \text{ g/mol}} = 2.174 \text{ moles} \] \[ \text{Moles of } O = \frac{69.55 \text{ g}}{16.00 \text{ g/mol}} = 4.347 \text{ moles} \]
03

Finding the Simplest Ratio

Divide the moles by the smallest number of moles to find the simplest whole number ratio: \[ \text{Ratio of } N = \frac{2.174}{2.174} = 1 \] \[ \text{Ratio of } O = \frac{4.347}{2.174} = 2 \] So, the empirical formula is \(\text{NO}_2\).
04

Determining the Molecular Formula

First, find the molar mass of the empirical formula \( \text{NO}_2 \): \[ 14.01 \text{ g/mol (N)} + 2 \times 16.00 \text{ g/mol (O)} = 46.01 \text{ g/mol} \] Then, divide the given molar mass (90 g/mol) by the empirical formula mass (46.01 g/mol): \[ \frac{90 \text{ g/mol}}{46.01 \text{ g/mol}} \text{ approx } 2 \]
05

Finding the Final Molecular Formula

Multiply the empirical formula by this ratio: \[ \text{Molecular formula} = ( \text{NO}_2) \times 2 = \text{N}_2\text{O}_4 \]

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Key Concepts

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

Mass Percent
Mass percent is a way of expressing the concentration of an element in a compound. It tells us what percentage of the compound's total mass is made up of a particular element. For example, if we're given that an oxide of nitrogen contains 30.45 mass percent nitrogen, that means in 100 grams of this compound, there are 30.45 grams of nitrogen. The remaining mass (69.55 grams) is oxygen.

Calculating mass percent helps in determining the composition of compounds, which is crucial for figuring out their empirical formulas.
Molar Mass
The molar mass of an element or compound is the mass of one mole of that substance. For instance, the molar mass of nitrogen (N) is 14.01 grams per mole, and for oxygen (O), it is 16.00 grams per mole.

This value is essential in converting grams to moles, a fundamental step in deriving both empirical and molecular formulas. By knowing the molar masses, we can use the mass of each element given in a problem to find out how many moles of each we have.
Moles Conversion
To determine the empirical formula, we need to convert the mass of each element into moles. This involves dividing the mass of the element by its molar mass.

For example, in the given problem:
  • Moles of N = \(\frac{30.45 \text{ g}}{14.01 \text{ g/mol}} = 2.174 \text{ moles}\)
  • Moles of O = \(\frac{69.55 \text{ g}}{16.00 \text{ g/mol}} = 4.347 \text{ moles}\)
Converting mass to moles ensures that we are working with the number of atoms rather than their masses, which simplifies the comparison between different elements.
Simplest Ratio
Once we have the moles of each element, we can find the simplest ratio by dividing each mole value by the smallest number of moles from the values calculated.

In the exercise, the smallest mole value is 2.174. Thus:
  • Ratio of N = \(\frac{2.174}{2.174} = 1\)
  • Ratio of O = \(\frac{4.347}{2.174} = 2\)
This gives us the empirical formula \(\text{NO}_2\). The simplest ratio is vital because it shows the simplest whole-number ratio of elements in a compound.
Chemical Formula Calculation
To find the molecular formula, we first determine the molar mass of the empirical formula and then compare it to the given molar mass of the compound.

For \(\text{NO}_2\), the empirical formula mass is \(14.01 \text{ g/mol (N)} + 2 \times 16.00 \text{ g/mol (O)} = 46.01 \text{ g/mol}\). Given that the molar mass of the compound is about 90 g/mol, we can find the ratio by:
  • \(\frac{90 \text{ g/mol}}{46.01 \text{ g/mol}} \approx 2\)
This ratio means the molecular formula is twice the empirical formula, leading to \(\text{N}_2\text{O}_4\). This process is critical for determining the exact formula of a compound, especially when the empirical and molecular formulas differ.

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Most popular questions from this chapter

Iron reacts slowly with oxygen and water to form a compound commonly called rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 4 \mathrm{H}_{2} \mathrm{O}\right) .\) For \(45.2 \mathrm{~kg}\) of rust, calculate (a) moles of compound; (b) moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3} ;\) (c) grams of Fe.

High-temperature superconducting oxides hold great promise in the utility, transportation, and computer industries. (a) One superconductor is \(\mathrm{La}_{2-x} \mathrm{Sr}_{x} \mathrm{CuO}_{4} .\) Calculate the molar masses of this oxide when \(x=0, x=1,\) and \(x=0.163\). (b) Another common superconducting oxide is made by heating a mixture of barium carbonate, copper(II) oxide, and yttrium(III) oxide, followed by further heating in \(\mathrm{O}_{2}\) : \(4 \mathrm{BaCO}_{3}(s)+6 \mathrm{CuO}(s)+\mathrm{Y}_{2} \mathrm{O}_{3}(s) \longrightarrow\) $$ \begin{array}{c} 2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{6.5}(s)+4 \mathrm{CO}_{2}(g) \\\ 2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{6.5}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7}(s) \end{array} $$ When equal masses of the three reactants are heated, which reactant is limiting? (c) After the product in part (b) is removed, what is the mass \(\%\) of each reactant in the remaining solid mixture?

Ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuric acid, ethanol forms diethyl ether and water: $$ 2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(g) $$ In a side reaction, some ethanol forms ethylene and water: $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{2} \mathrm{CH}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) If \(50.0 \mathrm{~g}\) of ethanol yields \(35.9 \mathrm{~g}\) of diethyl ether, what is the percent yield of diethyl ether? (b) If \(45.0 \%\) of the ethanol that did not produce the ether reacts by the side reaction, what mass (g) of ethylene is produced?

Lead can be prepared from galena [lead(II) sulfide] by first roasting the galena in oxygen gas to form lead(II) oxide and sulfur dioxide. Heating the metal oxide with more galena forms the molten metal and more sulfur dioxide. (a) Write a balanced equation for each step. (b) Write an overall balanced equation for the process. (c) How many metric tons of sulfur dioxide form for every metric ton of lead obtained?

Is \(\mathrm{MgCl}_{2}\) an empirical or a molecular formula for magnesium chloride? Explain.
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