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Chemistry 15 Online
OpenStudy (anonymous):

calculate the number of C, N, O, and H atoms in 1.0 grams of C17H21NO4

OpenStudy (paxpolaris):

molecular mass of \( C_{17}H_{21} NO_4= 17\cdot 12+21\cdot1+1\cdot15+4\cdot16\)

OpenStudy (anonymous):

ok so molecular mass would be 304

OpenStudy (paxpolaris):

right. 17*12 / 304 comes from C 21/304 comes from H 15/304 comes from N 4*16/304 comes from O

OpenStudy (anonymous):

so carbon would have .67, Hydrogen would have .069, Nitroge would have 0.049 and Oxygen would have 0.211

OpenStudy (anonymous):

are those all the correct # of significant figures??

OpenStudy (anonymous):

thanks by the way!

OpenStudy (paxpolaris):

sorry atomic mass of N is 14 not 15 ... so it add to 303

OpenStudy (anonymous):

haha so I just do everything over 303...no sweat. its 3 significant figures correct??

OpenStudy (paxpolaris):

yes ... 303.353 if you want more precision

OpenStudy (paxpolaris):

btw ... why am i helping helping you make 1g of cocaine ... lol

OpenStudy (anonymous):

i swear its my chemistry homework...hahaha. so will my final answer be 3 sig figs? my teacher doesn't give ANY mercy with significant figures

OpenStudy (paxpolaris):

1.0g is just 2 sig figs .... so you are good

OpenStudy (paxpolaris):

@maggielucci : SORRY, this is the answer to the question you had asked earlier about % of C,H, N, O in the molecule (by mass)

OpenStudy (paxpolaris):

so C is 0.67 or 67% by mass and so on

OpenStudy (paxpolaris):

@maggielucci . for this one: Once you have the molar mass of cocaine as 303 grams per mole. So you know 1.0g has 1.0/303 moles. Since there are 17 C in every molecule, you have 1/303 *17 = 0.05604 moles of C then multiply this by Avagadro number to get the number of C atoms.

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