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Chapter 3: Problem 66

An oxide of tungsten (symbol W) is a bright yellow solid. If \(5.34 \mathrm{~g}\) of the compound contains \(4.23 \mathrm{~g}\) of tungsten, what is its empirical formula?

Short Answer

Expert verified
The empirical formula is \( \text{WO}_3 \).

Step by step solution

01

- Find Mass of Oxygen

To find the mass of oxygen in the compound, subtract the mass of tungsten from the total mass of the compound:\[\text{Mass of oxygen} = 5.34\, \text{g} - 4.23\, \text{g} = 1.11\, \text{g}\]
02

- Calculate Moles of Tungsten

Find the number of moles of tungsten. The molar mass of tungsten (W) is approximately 183.84 g/mol.\[\text{Moles of } W = \frac{4.23\, \text{g}}{183.84\, \text{g/mol}} \approx 0.023\, \text{mol}\]
03

- Calculate Moles of Oxygen

Find the number of moles of oxygen. The molar mass of oxygen (O) is approximately 16.00 g/mol.\[\text{Moles of } O = \frac{1.11\, \text{g}}{16.00\, \text{g/mol}} \approx 0.0694\, \text{mol}\]
04

- Determine the Simplest Whole Number Ratio

Find the simplest whole number ratio of moles of W to O by dividing each by the smaller number of moles.\[\frac{0.023}{0.023} = 1\quad \text{and}\quad \frac{0.0694}{0.023} \approx 3\]Thus, the mole ratio is 1:3.
05

- Write the Empirical Formula

The empirical formula is derived from the simplest whole number ratio of moles of elements. Since the ratio is 1:3, the empirical formula is \( \text{WO}_3 \).

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Key Concepts

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

Mass of Oxygen
When determining the empirical formula of a compound, one key step is to figure out how much oxygen is present in the mixture. This is done by calculating the mass of oxygen. To achieve this, subtract the mass of tungsten from the total mass of the compound. In this case, we have a total mass of 5.34 grams in which tungsten accounts for 4.23 grams.
This subtraction gives us:
  • Mass of oxygen = 5.34 g - 4.23 g = 1.11 g.
This step is fundamental because it provides the necessary data to calculate molar quantities, making it easier to compare different elements within the compound.
Moles of Tungsten
Understanding how to calculate the number of moles of an element is crucial. Moles provide a linkage between the mass of a substance and the number of entities, like atoms or molecules, it contains. To find the moles of tungsten (W), we use its molar mass. Tungsten's molar mass is about 183.84 g/mol.
So for tungsten, if you have 4.23 grams, you calculate the moles as:
  • Moles of W = \( \frac{4.23\, \text{g}}{183.84\, \text{g/mol}} \approx 0.023\, \text{mol} \).
Remember, this figure represents the amount of tungsten atoms present.
Whole Number Ratio
The concept of a whole number ratio is at the heart of deriving the empirical formula. Once the moles of each element in the compound are known, the next task is to express these in the simplest ratio of whole numbers.
Here, with tungsten at 0.023 moles and oxygen at 0.0694 moles, we find the simplest ratio by dividing both values by the smaller number of moles, i.e., 0.023:
  • For tungsten: \( \frac{0.023}{0.023} = 1 \).
  • For oxygen: \( \frac{0.0694}{0.023} \approx 3 \).
This resultant 1:3 ratio indicates that for every atom of tungsten, there are approximately three atoms of oxygen. Using these ratios is vital for writing the correct empirical formula.
Moles of Oxygen
Just as we calculated for tungsten, determining the moles of oxygen gives insight into the compound's stoichiometry. The moles of oxygen are derived using its mass and molar mass. With 1.11 grams of oxygen and a molar mass of approximately 16.00 g/mol, the calculation is:
  • Moles of O = \( \frac{1.11\, \text{g}}{16.00\, \text{g/mol}} \approx 0.0694\, \text{mol} \).
Moles offer a common ground for comparing quantities of each element, crucial for establishing the empirical formula in a straightforward, comprehensible manner.

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Most popular questions from this chapter

An exciting, and often loud, chemical demonstration involves the simple reaction of hydrogen gas and oxygen gas to produce water vapor: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ The reaction is carried out in soap bubbles or balloons that are filled with the reactant gases. We get the reaction to proceed by igniting the bubbles or balloons. The more \(\mathrm{H}_{2} \mathrm{O}\) that is formed during the reaction, the bigger the bang. Explain the following observations. a. A bubble containing just \(\mathrm{H}_{2}\) makes a quiet "fffft" sound when ignited. b. When a bubble containing equal amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) is ignited, a sizable bang results. c. When a bubble containing a ratio of 2 to 1 in the amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) is ignited, the loudest bang results. d. When a bubble containing just \(\mathrm{O}_{2}\) is ignited, virtually no sound is made.

Compounds of boron with hydrogen are called boranes. One of these boranes has the empirical formula \(\mathrm{BH}_{3}\) and a molecular mass of 28 amu. What is its molecular formula?

Aniline, a starting compound for urethane plastic foams, consists of \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). Combustion of such compounds yields \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{N}_{2}\) as products. If the combustion of \(9.71 \mathrm{mg}\) of aniline yields \(6.63 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) and \(1.46 \mathrm{mg} \mathrm{N}_{2}\), what is its empirical formula? The molecular mass of aniline is 93 amu. What is its molecular formula?

Calculate the percentage composition for each of the following compounds (three significant figures). a. \(\mathrm{CO}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) d. \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\)

Alloys, or metallic mixtures, of mercury with another metal are called amalgams. Sodium in sodium amalgam reacts with water. (Mercury does not.) $$ 2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) $$ If a \(15.23-\mathrm{g}\) sample of sodium amalgam evolves \(0.108 \mathrm{~g}\) of hydrogen, what is the percentage of sodium in the amalgam?
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