Arsenic(III) sulfide sublimes readily, even below its melting point of 320 °C. The molecules of the vapor phase are found to effuse through a tiny hole at 0.52 times the rate of effusion of Xe atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

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Welcome back everyone under the same conditions of temperature and pressure molecules of an acid in the gas phase, a fused through a small hole at 0.702 times the rate of effusion of argon gas, calculate the molecular formula of the acid if it only has one hydrogen atom and one other atom, we want to recall that we can utilize Graham's Law of effusion, in which we want to recall the following formula. And sorry, this should have to or no, that's spelled correctly. So recall the following formula in which we take the rate of gas one divided by the rate of effusion of gas to. And this quotient is inversely proportional to the square root. And sorry, we have the square root here of the falling quotient where we take the molar mass of our second gas gas two divided by the molar mass of gas one. So, in terms of our prompt, we would consider the rate of effusion of gas one equal to our rate of our unknown acid in gaseous form. And our rate of gas too, will consider as equal to our rate of a fusion of argon Based on what the prompt tells us. They were given that the rate of a fusion of gas one Is equal to .702, multiplied or times the rate of a fusion of argon. And so therefore, we would say it's .702 times the rate of gas too. And in our formula for graham's law of effusion. Our rate of gas one is divided by our rate of gas too. And so substituting our understanding of the rate of gas one as 10. times the rate of gas to, We would plug in for rate of gas one divided by our rate of gas too. We would substitute our numerator as again, 0.702 times the rate of gas to I'll say of argon instead of saying gas too, Divided by our rate of gas. And sorry, this should say two in the denominator. So our rate of argon, This is really just simplified. 2.702 divided by one, which is going to result in a value of just .702. As the value of our rate of effusion of gas, one divided by gas to so the value of this quotient simplifies 2.702. And so we can go back to our Graham's law of effusion and plug this in so that we have .70 to equal to the square root and sorry, equal to the square root of our molar mass of argon divided by our molar mass of our unknown. So plugging in our molar mass of argon, we would find from our periodic table That we have .702 on the left equal to the square root of our moller massive organ being 39.95 g per mole. And this is divided by the molar mass of our unknown gas, which we need to solve for. And so our first step is to get rid of that square root term, we can square both sides of our equation To a power of two. And so what we'll have is .702 squared is equal to .493, which simplifies the right hand side where our square root cancels out once we square it. And then we'll just have 39.95 grams per mole divided by the molar mass of our unknown gas. So simplifying this, recall that when we have diagonals in algebra we can just switch places. And so we have now our molar mass of our unknown gas equal to 39.95 g per mole divided by 0.4 93. And this quotient simplifies to a value of 81. g per mole as our molar mass of our unknown gas in the prompt. And so we are told that our unknown gas from the prompt, we're told that it has one hydrogen atom and one other atom as part of its molecular formula. So we'll note that the unknown contains one More of Hydrogen. And so we will say therefore, we would take 81.3 g per mole as our molar mass of our unknown gas subtracted from the molar mass of hydrogen. And so in this case we would subtract from our molar mass of hydrogen on our product table, which is 1.1 g per mole. And so this difference results in a value equal to 80.02 g per mole. And on our periodic table We have our Adam Bro mean, which has a molar mass of 79. g per mole. And that's pretty close to what we calculated as our molar mass of our unknown gas. And so we would say that therefore our molecular formula of our unknown gas is most likely going to just be hBr. And this formula highlighted in yellow, will be our final answer to complete this example corresponding to choice C as the correct choice. So I hope that this made sense. And let us know if you have any questions.

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Key Concepts

Effusion and Graham's Law

Molecular Formula

Molar Mass Calculation