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Problem 90 Hard Difficulty

Calculate the number of atoms of each element present in each of the following samples.
a. 4.21 g of water
b. 6.81 g of carbon dioxide
c. 0.000221 g of benzene, $\mathrm{C}_{6} \mathrm{H}_{6}$
d. 2.26 moles of $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$

Answer

Hence limit will be 1

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Theodore D.

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Video Transcript

Okay, here we have water and a particular mass of the four point E. One grams of water. We want to know the number of Adams of each element are present. Okay, So first not derive off formula to find in this, so to find a normative, my number of elements are Adams. We need to find number of molecules and the equation for a number of molecules is malls multiplied by Have a Gar Joe's number and a you know, right. But we have mass. So let's find another way. We can represent moles. Three new about morals is equivalent to mass Ramallah, Mass. And we can find more months while adding up the atomic mass of each element present in the compound. So now that we have a formula, we could just insert it and substituted for Morse. I think mass over more Must I want to play that by and a Okay, So this is our formula. Oh, this. So here we have a mass we can divide by molar mass just for water. 18.1 damps per mole multiplied by of a garters number. Bush is six point 022 times 10 toothy exponents. 23 e here means time sent the exponents. And this is Nina odds of Permal. Okay, so if we carry through this formula we at 1.4 times ton to the exponents. Times tend to the exponents. 23 molecules of water present here. Knowledge als, iss kiss you. It's a bit tricky. E represents times 10 to the exponents. Any exponents over here and explain That's what e means quickly him. This number is wells in the exponents. Okay, so now that we have a number of molecules, we want to find number of atoms and eat of each element. Okay, So if you have one molecule of water, then we have two atoms of hydrogen and one atom oath. Oxygen, so but is simple enough. And we're gonna take that mm logic to apply it for large numbers. So if we have so many molecules of water, let's say 10. When we have 20 you have twice the number of hydrogen atoms versus water molecules, and we have always the same number of water molecules as we do have atoms of oxygen. All right, so that's the larger Find what we're about to do. So them now to find to convert this into a mal Adams of hydrogen. We just say this number multiplied by two. 1.4 times two is 2.8. Let's keep that exponents times 10 to the 23 Adams. Oh, hydrogen. And then for oxygen. It's the same amount of molecules for everyone. Molecule, we have the exact same one Oxygen, Adam. So yeah, one point time teat of 23 Adams. Oh, that next up, if a couple. So now we have carbon dioxide and 6.81 grams of it. So we take mass divided by more mouse. But I have summed up to being 44 grams per mole when you multiply that by the six point 022 times 10 to the 23. I've got a constant or any goods quite long. So this is equivalent to Mina. 0.3 times tend to the 22 um, molecules. Excuse me. Molecules is what that says. So no carbon and oxygen. Hominy Adams, do you have of each? So you just look a the ratios again for everyone. Um carbon Adam, we have for every one carbon atom, we have curry one carbon dioxide molecule. You have one carbon atom cysts, 1 to 1 for carbon. So that's always gonna be this thing. And then for oxygen, it's a 2 to 1 ratio. So we just doubled this number. Now, this is time since exponents of 20 three items. Oh, and now we have benzene C six h looks Okay, so we take this mass Tried it by Mullah Maas Years 78.11 grams per mole toe. Find malls. Multiply that by have a guard Joe's constant the six point all to to take him to the 23 and units of per moments. And this gives us one points six from sent to 18 molecules of Benty molecule is what that says. Oh, son. Here we have six carbon atoms, six kilograms per free compound of venting. So you're saying this number Multiply it by, um six. And here we we then no, the carbon and hydrogen our, um you know, equivalent in the number of items, it's 66 each. Just six multiplied by laughed. This number which will give us 9.7 1.714 Who'll freak People keep some more. 66 times e Did you the 18 Adams? Next, we have sugar C 12 h 22 0 11 to be suit girls to be other sugars. Frutos other Any dye sack right? If you've got. If you know that what that is, what any is is just sugar. Call it you also. You'll have to know that yet I think. But here we have a little bit of a catch. We're not working in mass. We're working with malls. Let's just quickly remind ourselves of what we're supposed to do. So we will just use this top for 11. Take most wanted by by Agriculture is constant and I will give us number of molecules. So it's more shit for it. Take this number most way by N A. And that will give us 1.36 times 10 to 24 molecules. Oh, so let me just quickly fix this and molecules of sure modules way he have Harbin hydrogen and oxygen. No, sure, OK, purple, hydrogen oxygen, simplest organic compound. So again we take this number and we multiplied by 12 defined number of carbon atoms which will give us 1.63 times into 25 carbon alums. This one over here for hydrogen number Ha Jin Adams, who takes the number of molecules of sugar and multiply by 22 which gives us 2.99 times tend to the Experiment 25. And for oxygen, we take a member of modules and multiply by. 11 gives this 1.49 times 10 to the 25 Adams, so yeah, quite large numbers.

McMaster University
Top Chemistry 101 Educators
TD
Theodore D.

Carleton College

Lizabeth T.

Numerade Educator

Morgan S.

University of Kentucky

Jake R.

University of Toronto