Composition

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##### Stephanie C.

University of Central Florida

##### Allea C.

University of Maryland - University College

##### Jake R.

University of Toronto

### Video Transcript

Okay, so here we have 1000 grams of the current zirconia, and we're trying to find how much with this way if we replace their zirconium Adams with lithium. So same number of atoms in lake both cases. But how much would it wave? We just swapped out for Cicconi. Um, Adams for lithium atoms. Okay, so the given here is same number of atoms. Okay, So what is our equation for a number of Adams? So it's malls multiplied by average Gar Joe's constant and a So we're working with mass. You were looking for a math and were given the mass. So let's see, What do we know about moles? We know that moles is equivalent to mass over more mouse, and we'll just keep avocados constant right there. Waas. All we did was substitute out malls for a massive for more because, you know, it's a little right. So here we have our equation opus. So we know that the number Adams of zirconium equal that of lithium. So will confer massive lithium looks do m l for massive lithium and ml again capital M o for Mullah. Massive. With you, this is a constant is equivalent to the massive zirconium over molar mass of zirconium Times of God Rose Constant. Okay, so let's simplify this of it. So what, Henry, you know, cancel out, cancel out any on both sides and the numerator. Yeah, and we condense. So for the massive lithium by multiplying by massive molar mass of lithium on both sides, so then that will give us just formula multiplied by more massive lithium over Moeller mass of zirconium. All right, so then, to find these more masses, we can just look at the atomic mass of each of these elements on the periodic table. So we have 1000 grams well supplied by mm seven grams per mole. Just approximating the small amounts is you can use, you know, the full proper ones and tried by the molar mass of zirconium. Just 91. There was formal. All right, so if we do that, the quick mouth, we'll find out that this is equivalent to 76 0.8 Gramps. Yeah. Looks like we took the mass Zucconi and multiplied it by a Moeller mass factor ratio. The elements of both own mints ratio and on her new a mass

McMaster University

#### Topics

Composition

##### Stephanie C.

University of Central Florida

##### Allea C.

University of Maryland - University College

##### Jake R.

University of Toronto