From the given molar mass and empirical formula of several compounds, find the molecular formula of each compound. a. C₄H₉, 114.22 g/mol

Question

Accepted Answer

All right. Hi, everyone. So this question says that when zirconium reacts with Bromy at 500 °C, it produces a metal Halide with a composition of 36.34% zirconium by weight. It is a molecular compound and not ionic as indicated by the hides melting and boiling points which are negative 20 °C and 200 °C respectively. Upon interaction with magnesium, this metal Halide gives rise to pure zirconium metal provided the balanced equation for this reaction, specifically the reaction of our Halide with magnesium metal. So first, we have to find what the hale actually looks like because zirconium is a transition metal, we cannot assume its oxidation state, meaning we cannot assume the ratio of zirconium to bromide or bromine in the hay light itself. So it's important that we are given the percent by weight of zirconium because this is going to help us find the simplest ratio of zirconium to bromium. Therefore, generating the molecular formula. This means that we have defined the empirical formula. So if we assume 100 g of this compound of the hide, then that means that there are 36.34 g of zirconium in the compound. So with respect to the mass of grooming, if we assume 100 g in total, then the mass of bromine is 100 subtracted by the mass of zirconium. And that results in a mass of S oops, sorry, 63 0.66 g of bro meat. But we cannot find a ratio based on the mass, right. We have to take this a step further and convert the mass of both zirconium and bromine to the moles of both zirconium and bromine. So this means that we have to divide each mass in grams by the molecular weight or excuse me, the molar mass of each element. So if we take zirconium, for example, that is 36.34 g of zirconium divided by the molar mass, which is 91.22 g per mole, right? This means that my units of grams cancel out yielding an answer of zero point 3984 moles of zirconium rounded to the same amount of significant figures which is four. So now we can do the same thing with Bromy. I will take my mass of bromine which is 63.66 g and divided by the molecular or the molar mass of bromine which is 79.90 g per mole. Once again, my units of grams cancel out and this results in an answer of 0.7967 moles of bromine routed to four significant figures. So our next step is to go ahead and find the mole ratio. And we will do this by dividing the moles of B roaming that we calculated previously by the moles of zirconia. So that is 0.7967 malls of prom divided by 0.3984 moles of zirconia. Now, in this case, my units of moles actually cancel out and my ratio simplifies to approximately 2/1. This means that the ratio of bromine to zirconium is 2 to 1, which means that the empirical formula and also the molecular formula is ZRBR two. So now the last part of the question is to find the balanced chemical reaction of our highlight here as a solid with magnesium metal. Now, as mentioned in the beginning of this video, when reading the text of the question, one of our products is going to be free zirconium metal. But as for the other product recall that this is going to be a single displacement reaction because here we have one compound. In this case, the hide reacting with an element that can replace are metal in this case, right? Because zirconium and magnesium are both metals, magnesium can displace zirconium from the bromide compound that results in pure zirconium metal as well as MG BR two. And because magnesium is a metal in group two A its charge is positive too, right. So the ratio of magnesium to bromine would also be 1 to 2. But recall that I cannot be certain that this is completed until I confirm that this reaction is actually balanced. Recall that a reaction is considered to be balanced when there is the same quantity of each element on both sides of the equation. So if I consider zirconium first, I have one atom of zirconium on the left as well as one on the right, I have two elements or atoms of bromine on the left and two on the right as well. Also I have one atom of magnesium on the left and one atom of magnesium on the right. This means that the chemical reaction equation is actually balanced as it's written. So here is our final answer. One mole of ZR BR two as a solid reacts with one mole of magnesium metal to generate one mole of zirconium metal and one mole of M GBR two. So with that being said, thank you so very much for watching. And I hope you found this helpful.

From the given molar mass and empirical formula of several compounds, find the molecular formula of each compound. a. C₄H₉, 114.22 g/mol

Verified Solution

Key Concepts

Empirical Formula

Molar Mass

Molecular Formula

From the given molar mass and empirical formula of several compounds, find the molecular formula of each compound. a. C4H9, 114.22 g/mol

Verified Solution

Key Concepts

Empirical Formula

Molar Mass

Molecular Formula

From the given molar mass and empirical formula of several compounds, find the molecular formula of each compound. a. C₄H₉, 114.22 g/mol