A particular form of cinnabar (HgS) adopts the zinc blende
struc...

Question

A particular form of cinnabar (HgS) adopts the zinc blende 
structure. The length of the unit cell edge is 5.852 Å. 
(c) Which of the two substances has 
the higher density? How do you account for the difference 
in densities?

Accepted Answer

Hello everyone. So in this video we're trying to see which of the two substances has a greater density given that this structure has to formula units of the compound per unit cell. So first let's go ahead and calculate for the volume of C. D. S. So the dimensions are given to us in the problem here. Go ahead and highlight this for us. So we have 4. Angstrom Times 4.136 Angstrom Times 6.713. Putting these values into my calculator, I get the value of 114.84. And units being Angstrom cute. Now for the volume of my C. D. S. E. Again using the values given to us in the problem we'll have 4.271 Angstrom Times 4.271 Angstrom Times 6.969 Angstrom Again, putting this into my calculator, I get the value of 1-7.12 Angstrom cute. Let's go ahead and recognize that the molar mass of my C. D. S is equal to 144. g per mole. And my molar mass of my c. d. s. e. is equal to 191.36 g per mole. So there's no need to convert volume. We need mass per volume units to compare. So starting off then with my density of C. D. S will do some dimensional analysis. So we have two molecules of a compound per every one unit cell. Everyone unit cell. According to our calculations has 1 1 4. Angstrom cubed. Then we're gonna go ahead and use of ricardo's number. So for every one mole of our compound We'll have 6.022 times 10-23rd power molecules. And finally using our molar mass We'll have 144.46 g of our compound. For every one mole. You can see here that the units will cancel out. And finally give us the value of 4.178 times 10 to the negative 24 g per angstrom cute thus scrolling down to give us a little bit more of space. We need to calculate next for the density of C. D. S. E. So again we have our two molecules of a compound for every one unit cell, then we have our one unit cell to contain 1 to 7.12 extreme cubed again using avocados number, we have one more of our compound. And on the bottom we'll just have the avocados number. Now finally using our molar mass on top we'll have our mass. So 191.36 g for every one mole. Now again putting this into my calculator, I'll get the value of 4.999 times 10 to the negative 24 units being grams per angstrom cute. Now if you're comparing those two densities then we can see that this value right here is definitely a large larger. So we can say that this molecule at c d. S is going to still be denser even with a larger unit cell, because S is much heavier and can compensate for the larger volume of the unit cell. So this is going to be my final answer for this question. Thank you all so much for watching.

A particular form of cinnabar (HgS) adopts the zinc blende structure. The length of the unit cell edge is 5.852 Å. (c) Which of the two substances has the higher density? How do you account for the difference in densities?

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