Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord! Numerade Educator ### Problem 36 Easy Difficulty # What is the empirical formula and empirical formula mass for each of the following compounds?$\begin{array}{ll}{\text { (a) } \mathrm{C}_{2} \mathrm{H}_{4}} & {\text { (b) } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}} & {\text { (c) } \mathrm{N}_{2} \mathrm{O}_{5}} \\ {\text { (d) } \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}} & {\text { (e) } \mathrm{Te}_{4} \mathrm{I}_{16}}\end{array}$### Answer ## a)$\mathrm{CH}_{2} ; \mathrm{Mr}\left(\mathrm{CH}_{2}\right)=14,01 \mathrm{g} / \mathrm{mol}$b)$\mathrm{CH}_{3} \mathrm{O} ; \mathrm{Mr}\left(\mathrm{CH}_{3} \mathrm{O}\right)=31,01 \mathrm{g} / \mathrm{mol}$c)$\mathrm{N}_{2} \mathrm{O}_{5} ; \mathrm{Mr}\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)=108,02 \mathrm{g} / \mathrm{mol}$d)$\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ; \mathrm{Mr}\left(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right)=601,93 \mathrm{g} / \mathrm{mol}$e)$\mathrm{TeL}_{4} ; \mathrm{Mr}\left(\mathrm{TeL}_{4}\right)=635,2 \mathrm{g} / \mathrm{mol}\$

#### Topics

Chemical reactions and Stoichiometry

### Discussion

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##### Allea C.

University of Maryland - University College

##### Morgan S.

University of Kentucky

### Video Transcript

Okay, so we had to wear the empirical foreigner off various compounds. So we have a situation for your stuff. Aces waste for for empirical foreigner. We're going to show the simplest way show victory and mint so you can see that. See, to wish for we can further, um, simplify it. So we have c h to We're probably ate the common factors too. Okay. So for sea to exist Oh, to really also seemed to fight. And we should be able to find a C 1/8. We and oh, what? Yeah, but we don't show the one, but I show here, just let you know that. Okay? And then we found the end 205 we can have in and all 2.5. Yeah, most of time. When you're looking waiting and coefficient, we're showing as a whole number. But we don't have half of the atom so far. This case 2 to 5. You cannot further simplify it. So and to all fight. Also, it's ah, the empirical formula. Okay. And we're from the barium force fate by reinforce way where the eighth we were about to force Wait. And also we have eight oxygen. So we have faith. We Pete's you 08 socials to expand that to have looked if we can't identify it. Um, unfortunately, we cannot simplify it. So the empirical former just be a three p 208 where we don't have a common factor for free to an eight. Okay, that's been obvious. We have t d for ice sakes. So we can for the offensive if I buy the wide by four to come a factor. So we're TV one eye for

University of Toronto

#### Topics

Chemical reactions and Stoichiometry

##### Allea C.

University of Maryland - University College

##### Morgan S.

University of Kentucky