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Chapter 2: Problem 11

A 1.0 -g sample of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 \(\mathrm{g}\) of oxygen. (a) What is the ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?\) (b) If a sample of a different compound decomposes into 0.429 \(\mathrm{g}\) of carbon and 0.571 \(\mathrm{g}\) of oxygen, what is its ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?(\mathbf{c})\) According to Dalton's atomic theory, what is the empirical formula of the second compound?

Short Answer

Expert verified
The mass ratio of O to C for the first compound is \( \frac{0.727}{0.273} \approx 2.66 \) and for the second compound is \( \frac{0.571}{0.429} \approx 1.33 \). The empirical formula of the second compound can be found using the mole ratio, resulting in the empirical formula as \( C_{1}O_{1} \) or simply CO.

Step by step solution

01

Calculation of mass ratio for the first compound

To calculate the mass ratio of O to C for the first compound, we have the following masses: Mass of Carbon (C) = 0.273 g Mass of Oxygen (O) = 0.727 g Now we can find the ratio of the mass of O to C: \( Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} \) \( Ratio_{O:C} = \frac{0.727}{0.273} \)
02

Calculation of mass ratio for the second compound

For the second compound, we have the following masses: Mass of Carbon (C) = 0.429 g Mass of Oxygen (O) = 0.571 g Now we can find the ratio of the mass of O to C: \( Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} \) \( Ratio_{O:C} = \frac{0.571}{0.429} \)
03

Calculating the empirical formula of the second compound

To find the empirical formula for the second compound, we will look at the ratio of O to C and compare it with the atomic masses of Oxygen and Carbon. The atomic mass of Oxygen (O) is 16 g/mol, and Carbon (C) is 12 g/mol. Now, we can calculate the mole ratio of O to C: \( Mole \ Ratio_{O:C} = \frac{\frac{0.571}{16}}{\frac{0.429}{12}} \) Simplify the ratio by finding the simplest whole number ratio: Now that we have the mole ratio in simplest whole number ratio, we can write the empirical formula for the second compound. Let the empirical formula is CXOY, where X and Y are the simplest whole-number quantities of C and O, respectively. Now we have the empirical formula for the second compound.

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Most popular questions from this chapter

Give the chemical names of each of the following familiar compounds:\((\mathbf{a})$$\mathrm{NaCl}(\) table salt \()\) \((\mathbf{b})$$\mathrm{NaHCO}_{3}\) (baking soda),\((\mathbf{c}) \mathrm{NaOCl}\) (in many branches)\((\mathbf{d})\) \(\mathrm{NaOH}(\) caustic soda \()\) \((\mathbf{e})\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\) smelling salts \()$$(\mathbf{f})$$\mathrm{CaSO}_{4}\) (plaster of Paris).

The most common charge associated with scandium in its compounds is \(3+.\) Indicate the chemical formulas you would expect for compounds formed between scandium and (a) iodine, ( b) sulfur, (c) nitrogen.

Name the following ionic compounds: (a) \(\mathrm{Li}_{2} \mathrm{O},(\mathbf{b})\) FeCl_ \((\mathbf{c}) \mathrm{NaClO},(\mathbf{d}) \quad \mathrm{CaSO}_{3}\)(e) \(\mathrm{Cu}(\mathrm{OH})_{2}, \quad(\mathbf{f}) \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}$$(\mathbf{g}) \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2},(\mathbf{h}) \mathrm{Cr}_{2}\left(\mathrm{CO}_{3}\right)_{3}\)(i) \(\mathrm{K}_{2} \mathrm{CrO}_{4},(\mathbf{j})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Identify the element represented by each of the following symbols and give the number of protons and neutrons in each: \((\mathbf{a})_{33}^{74} X,(\mathbf{b}) \frac{1}{33} X$$(\mathbf{c})_{63}^{152} \mathrm{X},(\mathbf{d}) \frac{209}{83} \mathrm{X}\).

Give the names and charges of the cation and anion in each of the following compounds: (a) \(\mathrm{CaO},(\mathbf{b}) \mathrm{Na}_{2} \mathrm{SO}_{4},(\mathbf{c}) \mathrm{KClO}_{4}\) (\mathbf{d} ) ~ \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{e}) \operatorname{Cr}(\mathrm{OH})_{3}\)
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