# Question Video: Determining the Empirical Formula of an Unknown Compound of Tungsten and Chlorine Using Indirect Titration

A compound containing only tungsten and chlorine is reduced with hydrogen to produce hydrogen chloride gas and 0.2232 g of tungsten metal. Complete reaction of the hydrogen chloride in water requires 46.2 mL of a 0.1051 M sodium hydroxide solution. Determine the empirical formula of the unknown compound.

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### Video Transcript

A compound containing only tungsten and chlorine is reduced with hydrogen peroxide to produce hydrogen chloride gas and 0.2232 grams of tungsten metal. Complete reaction of the hydrogen chloride in water requires 46.2 milliliters of a 0.1051 molar sodium hydride solution. Determine the empirical formula of the unknown compound.

The empirical formula of a compound tells us the relative number of atoms for each element in a molecule. And, it’s always expressed in the simplest whole-number ratios. Our unknown compound contains only tungsten and chlorine, so our empirical formula will be something of the form W 𝑥 Cl 𝑦, where 𝑥 and 𝑦 are the moles of tungsten and the moles of chlorine, respectively, of course expressed in the simplest whole-number ratios.

So to solve this problem, we’re going to find both the moles of tungsten and the moles of chlorine that were present in the sample of our unknown compound. The problem describes a series of reactions that were done to determine the amount of tungsten and the amount of chlorine that are in the unknown compound.

In the first step, our unknown compound was reacted with hydrogen peroxide and formed hydrogen chloride gas and tungsten metal. This reaction produced 0.2232 grams of tungsten, which we can use to determine the amount of tungsten in moles that was present in the unknown compound. We can find the amount of tungsten in moles by dividing the mass of tungsten by the moler mass of tungsten. The mass of tungsten is 0.2232 grams and the molar mass of tungsten, which we can find on the periodic table, is 183.84 grams per mole. This gives us 0.0012141 moles of tungsten.

Now that we’ve found the amount of tungsten that was present in the unknown compound, let’s find the amount of chlorine. In the second reaction described in the problem, the hydrogen chloride, which was produced in the first reaction, is reacted with sodium hydride. This reaction would form NaCl and H₂. The problem gives us the molarity of the sodium hydride solution and the volume that was required to completely react with the hydrogen chloride. So, let’s use this information to find the amount of chlorine that was in the original unknown compound.

We used a 0.1051 molar solution of sodium hydride, which means that there’s 0.1051 moles of sodium hydride per liter. So, we can find the amount of sodium hydride in moles by simply multiplying by the volume, which is 46.2 milliliters. But, we need to convert this from milliliters to liters. We can do this conversion by dividing by 1000 because there’s 1000 milliliters in a liter. This gives us 0.00485562 moles of sodium hydride.

Since this amount of sodium hydride was needed to completely react or totally use up the hydrogen chloride, we can use the stoichiometry of the chemical equation to figure out the amount of hydrogen chloride that we started with. According to our chemical equation, for every one mole of hydrogen chloride we have, we need one mole of sodium hydride to react with it. This means that the number of moles of sodium hydride that we calculated will be equivalent to the number of moles of HCl. But ultimately, we want to know the amount of chlorine. Well, for every one molecule of hydrogen chloride, there’s one chlorine atom. So, the amount of moles of hydrogen chloride will be equivalent to the moles of chlorine.

Now that we know both the amount of tungsten and the amount of chlorine in moles, we have everything we need to find the empirical formula of our unknown compound. The empirical formula requires that we express these amount of moles in the simplest whole-number ratio. Right now, the numbers we have certainly aren’t in simple whole numbers. We can divide these numbers into each other to find a simpler ratio. This would give us 0.25004 moles of tungsten to moles of chlorine. This is approximately 0.25, which is equivalent to one over four. So, there’s one mole of tungsten for every four moles of chlorine. So, the empirical formula for our unknown compound is WCl₄.